This is the second of three volumes that, together, give an exposition of the mathematics of grades 9 12 that is simulta

*617*
*141*
*10MB*

*English*
*Pages 375
[417]*
*Year 2020*

- Author / Uploaded
- Hung-Hsi Wu

*Table of contents : CoverTitle pageContents of the Companion Volumes and Structure of the ChaptersPrefaceTo the InstructorTo the Pre-Service TeacherPrerequisitesSome ConventionsChapter 1. Linear Functions 1.1. Definition of a function and its graph 1.2. Why functions? 1.3. Linear functions of one variable 1.4. Linear inequalities and their graphs 1.5. Linear programming 1.6. Optimization: The general case 1.7. Appendix: Mathematical inductionChapter 2. Quadratic Functions and Equations 2.1. Quadratic functions 2.2. A theorem on the graphs of quadratic functions 2.3. Graphs of equations of degree 2 2.4. The concept of an asymptoteChapter 3. Polynomial and Rational Functions 3.1. Some basic facts about polynomials 3.2. Descartes’ rule of signs 3.3. Rational functionsChapter 4. Exponential and Logarithmic Functions 4.1. An interpolation problem 4.2. Rational exponents 4.3. Exponential functions 4.4. LogarithmsChapter 5. Polynomial Forms and Complex Numbers 5.1. Polynomial forms 5.2. Complex numbers 5.3. Fundamental theorem of algebra 5.4. Binomial theoremChapter 6. Basic Theorems of Plane Geometry 6.1. Review 6.2. SSS and first consequences 6.3. Pedagogical comments 6.4. Proof of FTS 6.5. The angle sum of a triangle 6.6. Characterization of isometries 6.7. Some basic properties of a triangle 6.8. Basic properties of the circle 6.9. Power of a point with respect to a circle 6.10. Two interesting theorems about the circleChapter 7. Ruler and Compass Constructions 7.1. The basic constructions 7.2. The regular pentagon 7.3. A short history of the construction problemsChapter 8. Axiomatic Systems 8.1. The concept of an axiomatic system 8.2. The role of axioms in school geometry 8.3. Hilbert’s axioms 8.4. Hyperbolic geometryAppendix: Facts from [Wu2020a]Glossary of SymbolsBibliographyIndexBack Cover*

Algebra and Geometry Hung-Hsi Wu

Algebra and Geometry

Algebra and Geometry Hung-Hsi Wu

2010 Mathematics Subject Classiﬁcation. Primary 97-01, 97-00, 97D99, 97-02, 00-01, 00-02.

For additional information and updates on this book, visit www.ams.org/bookpages/mbk-132

Library of Congress Cataloging-in-Publication Data Names: Wu, Hongxi, 1940- author. Title: Algebra and geometry / Hung-Hsi Wu. Description: Providence, Rhode Island : American Mathematical Society, [2020] | Includes bibliographical references and index. Identiﬁers: LCCN 2020008693 | ISBN 9781470456764 (paperback) | ISBN 9781470460051 (ebook) Subjects: LCSH: Algebra. | Geometry. | AMS: Mathematics education – Instructional exposition (textbooks, tutorial papers, etc.). | Mathematics education – General reference works (handbooks, dictionaries, bibliographies, etc.). | Mathematics education – Education and instruction in mathematics – None of the above, but in this section. | Mathematics education – Research exposition (monographs, survey articles). | General – Instructional exposition (textbooks, tutorial papers, etc.). | General – Research exposition (monographs, survey articles). Classiﬁcation: LCC QA154.3 .W84 2020 | DDC 512–dc23 LC record available at https://lccn.loc.gov/2020008693

Copying and reprinting. Individual readers of this publication, and nonproﬁt libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected] c 2020 by the author. All rights reserved. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

25 24 23 22 21 20

To NiuNiu with the hope that she will ﬁnd it useful when she grows up

Contents Contents of the Companion Volumes and Structure of the Chapters

ix

Preface

xi

To the Instructor

xix

To the Pre-Service Teacher

xxxiii

Prerequisites

xxxvii

Some Conventions

xxxix

Chapter 1. Linear Functions 1.1. Deﬁnition of a function and its graph 1.2. Why functions? 1.3. Linear functions of one variable 1.4. Linear inequalities and their graphs 1.5. Linear programming 1.6. Optimization: The general case 1.7. Appendix: Mathematical induction

1 1 16 22 30 42 54 57

Chapter 2. Quadratic Functions and Equations 2.1. Quadratic functions 2.2. A theorem on the graphs of quadratic functions 2.3. Graphs of equations of degree 2 2.4. The concept of an asymptote

63 63 87 96 116

Chapter 3. Polynomial and Rational Functions 3.1. Some basic facts about polynomials 3.2. Descartes’ rule of signs 3.3. Rational functions

121 121 128 131

Chapter 4. Exponential and Logarithmic Functions 4.1. An interpolation problem 4.2. Rational exponents 4.3. Exponential functions 4.4. Logarithms

135 136 142 153 162

Chapter 5. Polynomial Forms and Complex Numbers 5.1. Polynomial forms 5.2. Complex numbers

175 175 189

vii

viii

CONTENTS

5.3. Fundamental theorem of algebra 5.4. Binomial theorem

196 203

Chapter 6. Basic Theorems of Plane Geometry 6.1. Review 6.2. SSS and ﬁrst consequences 6.3. Pedagogical comments 6.4. Proof of FTS 6.5. The angle sum of a triangle 6.6. Characterization of isometries 6.7. Some basic properties of a triangle 6.8. Basic properties of the circle 6.9. Power of a point with respect to a circle 6.10. Two interesting theorems about the circle

213 214 220 229 230 238 244 254 269 297 300

Chapter 7. Ruler and Compass Constructions 7.1. The basic constructions 7.2. The regular pentagon 7.3. A short history of the construction problems

305 306 318 325

Chapter 8. Axiomatic Systems 8.1. The concept of an axiomatic system 8.2. The role of axioms in school geometry 8.3. Hilbert’s axioms 8.4. Hyperbolic geometry

331 334 336 339 345

Appendix: Facts from [Wu2020a]

351

Glossary of Symbols

361

Bibliography

363

Index

367

Contents of the Companion Volumes and Structure of the Chapters Rational Numbers to Linear Equations [Wu2020a] Chapter Chapter Chapter Chapter Chapter Chapter

1: 2: 3: 4: 5: 6:

Fractions Rational Numbers The Euclidean Algorithm Basic Isometries and Congruence Dilation and Similarity Symbolic Notation and Linear Equations

Pre-Calculus, Calculus, and Beyond [Wu2020c] Chapter Chapter Chapter Chapter Chapter Chapter Chapter

1: 2: 3: 4: 5: 6: 7:

Trigonometry The Concept of Limit The Decimal Expansion of a Number Length and Area 3-Dimensional Geometry and Volume Derivatives and Integrals Exponents and Logarithms, Revisited

ix

x

CONTENTS OF THE COMPANION VOLUMES AND STRUCTURE OF THE CHAPTERS

Structure of the chapters in this volume and its two companion volumes (RLE= Rational Numbers to Linear Equations A&G = Algebra and Geometry PCC = Pre-Calculus, Calculus, and Beyond)

RLE-Chapter 1

RLE-Chapter 4 hhhh hhh

RLE-Chapter 2 ` ``` ``` RLE-Chapter 5

A&G-Chapter 6

``` `

``

RLE-Chapter 3

RLE-Chapter 6

A&G-Chapter 1 A&G-Chapter 2

A&G-Chaptera3 aa

c

c aa c

aa c

c A&G-Chapter 5 A&G-Chapter 4

! c !!

c ! c

!!

c !!

A&G-Chapter 8 A&G-Chapter 7 PCC-Chapter 1 ! !! !! PCC-Chapter 2 ! !! !! ! ! !! PCC-Chapter 3 PCC-Chapter 4 PCC-Chapter 6

PCC-Chapter 5

PCC-Chapter 7

Preface Explain it! The most important thing is, that you are able to explain it! You will have exams, there you have to explain it. Eventually, you pass them, you get your diploma and you think, that’s it!—No, the whole life is an exam,. . . . So learn to explain it! A quote attributed to Rudolf M¨ ossbauer (Nobel Prize in Physics, 1961) [Wiki-M¨ ossbauer] This volume and its two companion volumes, [Wu2020a] and [Wu2020c], are written for high school teachers and mathematics education researchers. Formally, they assume only a knowledge of whole numbers. Informally, they also assume a level of mathematical maturity that welcomes reasoning, i.e., proofs. Teachers have to be at ease with such reasoning in their teaching before they can convince their students to do likewise, and mathematics educators must base their research on correct content knowledge that takes reasoning for granted. It is unfortunate that, despite a plethora of publications on high school mathematics, none has given a systematic, grade-level appropriate exposition of all the mathematical topics (i.e., not including probability and statistics) in the curriculum of grades 9–12 that also embraces mathematical integrity. The main purpose of these volumes is to make an initial attempt to ﬁll this void.1 Because this is the second volume of a three-volume set, there are copious references throughout to the ﬁrst volume, [Wu2020a]. In an eﬀort to make this volume as self-contained as possible, we have collected the relevant assumptions, deﬁnitions, and theorems from [Wu2020a] in an appendix (page 351ﬀ.) These three volumes, together with [Wu2011a], [Wu2016a], and [Wu2016b], give a complete exposition of the mathematics of K–12 that respects the normal progression of topics from grade to grade. Given that there is no lack of publications on school mathematics, from school textbooks to education research monographs on proofs, one may question why these six volumes (totaling some 2,500 pages) needed to be written. A simple answer is the presence of what we call Textbook School Mathematics (TSM) in the school curriculum. TSM is the irreparably ﬂawed mathematics underlying the overwhelming majority of standard school mathematics textbooks and professional development materials for teachers in roughly the ﬁve decades after 1970; it is characterized by a lack of clarity, persistent replacement of reasoning by rote memorization, and overall incoherence. This is the body of knowledge passed from teachers to students, so the cycle is repeated when some of these 1 While these three volumes touch on probability only lightly in Section 5.4 of this volume, we should point out that there is a long section (Section 1.10) on ﬁnite probability in [Wu2016a].

xi

xii

PREFACE

same students become teachers. The fact that TSM is unlearnable is a main cause of the current crisis in school mathematics education. It is hoped that these six volumes will help break this vicious cycle by oﬀering, in suﬃcient detail, a usable replacement for TSM. They demonstrate, in a systematic and grade-appropriate manner, how the mathematics of the school curriculum can be developed coherently and purposefully by the use of precise deﬁnitions and reasoning. They open a curtain to reveal how school mathematics can be transformed into learnable mathematics in the school classroom. In fact, parts of these six volumes2 have served as a blueprint for the writing of CCSSM, the Common Core State Standards for Mathematics ([CCSSM]). We refer to the Preface of [Wu2020a] as well as pp. xx–xxv below for a more in-depth discussion of the issues involved. This volume covers high school algebra and high school geometry. In so doing, it has to confront two major instructional problems in the school mathematics curriculum. A current battle cry in school mathematics education is “Algebra for All” (compare [NMAP]), but algebra is also known as a notorious gatekeeper course that has kept too many students away from mathematics and other STEM subjects. In fact, the quadratic formula has become the poster child of arcane information in the curriculum—it is claimed—that only serves the purpose of thwarting the rising ambitions of many deserving students.3 A second problem is the high school geometry course; its reputation for incomprehensibility and unlearnability is too well known for further comments here (see, e.g., Part II of the book review [Wu2004]). Let us take up these issues one at a time. There are at least seven hurdles in students’ learning path of school algebra: (i) the proper use of symbols, (ii) the concept of the slope of a line and its relation to graphs of linear equations and functions, (iii) the concept of a function, (iv) quadratic functions and equations and the quadratic formula, (v) laws of exponents and exponential functions, (vi) complex numbers and the fundamental theorem of algebra, and (vii) the binomial theorem. Items (i) and (ii) have already been discussed at length in Chapter 6 of the ﬁrst volume, [Wu2020a]. We will make some passing comments on the remaining ﬁve issues. Let us start with (iii), functions. It is well known that school students ﬁnd the concept of a function to be elusive. On the one hand, functions are the passport to higher mathematics and, without a doubt, this concept is of a higher order of abstraction than any concept students have encountered previously. Therefore getting to know about functions does require real eﬀort on the part of teachers and students. On the other hand, students’ learning diﬃculties with functions are also partly due to TSM’s chronic neglect of precise deﬁnitions4 and even of the purpose behind studying functions, with the inevitable consequence that, in all things

2 Broadly speaking, CCSSM made substantial use of the expositions of fractions, rational numbers, parts of algebra, and the geometry of middle school and high school in [Wu2010b] and [Wu2010c], which were the drafts for [Wu2016a] and [Wu2016b]. 3 In May of 2017, Jonathan Rochelle, the then-director of Google’s education apps group, was quoted ([Singer]) as saying about his children that “I cannot answer for them what they are going to do with the quadratic equation. I don’t know why they are learning it. . . . And I don’t know why they can’t ask Google for the answer if the answer is right there.” 4 For example, the concept of inverse function does not seem to have ever been correctly deﬁned in TSM.

PREFACE

xiii

related to functions, reasoning gets shortchanged and often ignored entirely in favor of the memorization of rules. Such neglect then spawns the not uncommon practice in many school classrooms—when presented with a function f (x)—to switch over immediately to the equation y = f (x) and graph the equation instead. The idea seems to be to take the easy way out by looking at the graphs of equations instead of considering functions directly. More striking still is TSM’s obliviousness to the purpose of teaching the concept of a function. TSM fails to inform students that, whereas numbers are the alphabet of the language of arithmetic (and therefore of elementary school mathematics), functions are the alphabet of the language of modern mathematics and science as a whole. The fact that even a rough description of any natural or social phenomenon requires the use of functions to make sense of it has not been clearly and forcefully articulated in TSM school textbooks or professional development materials. We will plead the case for the critical need to study functions in Section 1.2, pp. 16ﬀ. Moreover, we will intentionally emphasize the functional aspect of the many skills in the discussion of quadratic, polynomial, and especially the exponential functions. For example, we begin the discussion of the laws of exponents by ﬁrst explaining the phenomenon of the interpolation of a given function (see Section 4.1) before turning to these laws and correctly identifying them as properties of the exponential functions rather than as “number facts” as TSM would have students believe (now we are addressing (v)). This clariﬁes the seemingly abrupt shift from √ √ the radical notation ( α, 3 α, etc.) to the exponential notation (α1/2 , α1/3 , etc.). See Section 4.4 on pp. 142ﬀ. The motivation for deﬁning exponential functions requires an unﬂinching look at the domain of deﬁnition of a function (see Section 4.1 on page 136), something TSM seems generically unwilling to confront. Similarly, we emphasize in the discussion of quadratic functions that for a given quadratic function—whose domain of deﬁnition is the x-axis—there is a point x0 on the x-axis so that if the function is increasing (respectively, decreasing) on the left side of x0 , then it is decreasing (respectively, increasing) on the right side of x0 . Thus quadratic functions exhibit a diﬀerent kind of behavior from the only other collection of functions students are familiar with up to this point, namely, the linear functions. For polynomial functions, the discussion of the behavior of p(x) as x approaches +∞ or −∞ again focuses attention on the function itself rather than on the equation y = p(x) (see Chapter 3). We can add a few more comments on (iv), quadratic equations and functions. First of all, we clarify the fact that the study of quadratic equations is only a small part of the study of quadratic functions; namely, the former is about how to locate the zeros of quadratic functions. Moreover, TSM makes this topic more diﬃcult than it needs to be, partly by presenting the technique of completing the square as a rote skill for one purpose only: getting the quadratic formula. Consequently, the quadratic formula ends up also being a rote skill and, likewise, the formula for the vertex of the graph of a quadratic function. In Section 2.1, however, we show that completing the square is the major idea that (a) leads to the proof of the quadratic formula and the formula of the vertex of the graph (see page 75), (b) proves that the graph of f (x) = ax2 +bx+c is congruent to the graph of fa (x) = ax2 , (c) exhibits the commonality between the study of linear and quadratic functions, namely, the fact that both revolve around the shape of the graphs of the representative functions ax and ax2 (see page 73), and (d) can be approached from a diﬀerent perspective that

xiv

PREFACE

becomes generalizable to polynomials of all degrees (see page 80). By presenting a comprehensive picture that makes sense of the diverse formulas and results, we hope to ease the learning of the subject of quadratic functions and equations. We should not fail to make mention of a quite spectacular application of the quadratic formula that testiﬁes once again to the remarkable coherence of mathematics—the construction of the regular pentagon by ruler and compass. See Section 7.2. Finally we come to (vi) and (vii): complex numbers, the fundamental theorem of algebra, and the binomial theorem. Since we have already made a line—a geometric object—into the algebraic object R, it is natural to also try to make the plane into an algebraic object, and this is the complex numbers C. Because the fundamental theorem of algebra leads to the factoring of polynomials into products of linear polynomials, we have no choice but to consider polynomials with complex coeﬃcients. This then prompts us to introduce the concept of formal algebra into school mathematics; this is another (entirely necessary) step toward abstraction in school mathematics. In technical language, we have to consider not just the polynomial ring over R, R[X], but also the polynomial ring C[X] over C, where X is just an abstract symbol (see Section 5.2). Whereas up to this point we have only computed with numbers, we now begin to compute with an abstract symbol X and consider the factoring of these polynomial forms. This paves the way for more advanced considerations where X will be not just a number but also square matrices or other algebraic objects. This formal setting is also the correct one for a discussion of the ubiquitous binomial theorem (X + Y )n for two symbols X and Y (see Section 5.4).5 Again, we hope that by providing a proper setting for the exposition of these topics, we facilitate and expedite their learning by showcasing the coherence and purposefulness of mathematics. The second half of this volume is a straightforward course on plane geometry. The troubling nature of TSM’s ﬂawed presentation of the foundations of geometry has been discussed in the Overview in Chapter 4 and Section 4.5 of the earlier companion volume, [Wu2020a], and our approach on how to overcome this diﬃculty has been presented in Chapters 4 and 5 of the same volume. The substance of Chapters 6–8 in this volume is a presentation of the geometry of the triangle and the circle, constructions by ruler and compass, and the nature of an axiomatic system. Before discussing these topics, however, we must ﬁrst bring closure to two pieces of unﬁnished business that were started in [Wu2020a], namely, a proof of the fundamental theorem of similarity (FTS) and a proof of the equivalence of the concepts of congruence and isometry in the plane. These proofs are not simple and are in any case not part of the standard school geometry curriculum. Nevertheless, we take pains in these volumes to oﬀer complete, self-contained proofs of these theorems because they are of fundamental importance for the mathematical education of teachers and educators (the ﬁnal step of the proof of FTS is given in Section 2.6 of [Wu2020c]). Teachers and educators need a clear understanding of the delicate issues involved to be able to responsibly discharge their professional duties. At the same time, we explicitly caution against bringing some of the theorems and their proofs into the school classroom because of the inherent time constraints of the

5 See

[Wiki-q-analog] for a general introduction to this kind of abstraction.

PREFACE

xv

school year and because of the level of sophistication that is part and parcel of these proofs. Only time can tell how much of these proofs will become an integral part of the school geometry curriculum. Now the geometry of the triangle and the circle in TSM presents diﬃculties that are no less problematic than the foundational issues when judged by the criterion of mathematical coherence. Faced with a jumble of seemingly disconnected theorems, one has to wonder what they are all about. Because the high school geometry curriculum is in such a state of ﬂux at this point after years of devastation by TSM, we have tried to be extra careful in the selection of the theorems about the triangle in Section 6.7. Overall, we want to give some indication that the geometry of the triangle, on the whole, is coherent when viewed as a study of the remarkable points of concurrence in a triangle (the circumcenter, the incenter, the orthocenter, the centroid, etc.). A good illustration of this coherence is the Euler line, which exhibits the collinearity of the orthocenter, the centroid, and the circumcenter. Thus these points of concurrence turn out not to be random after all, and a deeper study will further reveal the intricate interconnections among these and other points (the center of the nine-point circle, the Lemoine points, the Brocard points, etc.; see [Altshiller-Court] and [Johnson]). For this reason, although the Euler line is not traditionally part of the school mathematics curriculum, we will strenuously argue that it should be because of the coherence it brings to the geometry of the triangle. The basic theorem in the geometry of the circle is of course the remarkable fact that an arc on a circle subtends equal angles at any point on the opposite arc. Standard theorems about tangents, secants, chords, and cyclic quadrilaterals then round out the presentation in Section 6.8. This section also makes the eﬀort to prove two theorems that are usually taken for granted, namely, that a circle is symmetric with respect to any of its diameters and that a disk is convex. These proofs are not entirely trivial. The geometry of the triangle and that of the circle then come together in a most satisfactory fashion in the nine-point circle (Section 6.10). It is remarkable that the Euler line also appears naturally in that discussion—coherence again. We hope that at least the visual appeal of this theorem will not be lost on readers. This is the right place to bring up a special feature about the theorems in plane geometry. There is now a common consensus that students should be regularly exposed to multiple solutions of a given problem. Since TSM has hardly any valid reasoning at all, one can forget about looking for multiple solutions in TSM because getting one correct solution is already a rare event. However, in these three volumes on high school mathematics, it is common to ﬁnd many problems solved in more than one way. In the earlier volume ([Wu2020a]), for example, there are two proofs of equation (1.9) in Section 1.2, diﬀerent ways of doing the addition and subtraction of fractions and diﬀerent ways of doing the subtraction of mixed numbers in Section 1.3, two proofs for each of the basic formulas for complex fractions in Section 1.6, two ways to solve the percent problems in Examples 1 and 2 of Section 1.7 (see Exercise 5 in Exercises 1.7), etc., etc. Skipping ahead a few chapters, we can point to the two diﬀerent ways to get the equation of a line when geometric conditions are prescribed: see Examples 2 and 3 in Section 6.5. In the present volume, we can point to two diﬀerent proofs of Theorem 3.1 on page 125 and page 200, and two diﬀerent proofs of Theorem 3.3 on page 126 and page 184. Perhaps a more dramatic

xvi

PREFACE

example is the two diﬀerent presentations of the theory of quadratic functions, one using the traditional technique of completing the square (Section 2.1 on pp. 63ﬀ. in this volume) and the other using diﬀerential calculus (see the appendix in Section 6.4 of [Wu2020c]). But the special feature about the theorems in plane geometry that we alluded to above is the fact that they seem to routinely lend themselves to multiple proofs. Consider the two chapters on geometry in [Wu2020a]: there, we oﬀer two proofs of Lemma 4.3 in Section 4.1, two proofs of Theorem G1 in Section 4.4, two proofs of Theorem G15 in Section 5.1, two proofs of Lemma 5.1, and two proofs of Theorem G18 in Section 5.2. In the present volume, we have two proofs of Theorem G25 (the ﬁrst proof is in Section 5.3 of [Wu2020a] and the second proof is given in Exercise 1 on page 227) three proofs of Theorem G26(a) (for the third proof, see Exercise 7 on page 244), ﬁve proofs of Theorem G29 (see page 226), two proofs of Theorem G32 (see page 238), two proofs of the corollary on page 246 of Theorem G33 (see Exercise 3 on page 253), three proofs of Theorem G34 (all three proofs are given on page 246ﬀ.), four proofs of Theorem G40 (see page 255), etc. This aspect of plane geometry is well worth the attention of teachers and educators. Chapter 7 begins with a fairly standard discussion of the basic construction problems (with ruler and compass). What is not so standard is the construction of the regular pentagon; there we see how both algebra (the quadratic formula) and geometry (similar triangles) are naturally put in service of the solution of this construction problem. Equally nonstandard is a brief historical and intuitive account of the four classical construction problems and their eventual complete solutions. While such information may be considered optional for students, it should be part of the basic repertoire of every high school mathematics teacher and mathematics educator. Having just given a reasonable outline of plane Euclidean geometry, we can ﬁnally take a backward glance to discuss the basic concept of an axiomatic system and, more speciﬁcally, a correct and usable axiomatic system for the development of plane geometry (that of David Hilbert, [Hilbert]). This is the content of Chapter 8 on pp. 331ﬀ. It is our intention to provide a balanced view of the purpose of an axiomatic system in general, the complexity of the Hilbert system—the prototype of an adequate axiomatic system for plane geometry—in particular, and most importantly, why the usual axiomatic systems set up for plane geometry in TSM have no merit, either mathematically or pedagogically. In the course of this discussion, the singular importance of the parallel postulate—the postulate that led to one of mathematics’ internal revolutions—will naturally emerge. A major defect of TSM is its failure to impart to students an appreciation of this postulate. In order to reinforce this message, we conclude this volume by outlining the proof of a counterintuitive theorem (Theorem 8.2 on page 347) in a geometry where the parallel postulate does not hold, a geometry that has come to be known as hyperbolic geometry. We consider Chapter 8, or its equivalent, to be absolutely indispensable for the mathematical education of mathematics teachers and educators. Without this background knowledge, it is diﬃcult to imagine how anyone can discuss plane geometry in a way that makes mathematical sense.

PREFACE

xvii

Algebra and geometry are the core of the middle school and high school mathematics curriculum, so the content of this volume may be regarded as the bread and butter of teachers and educators involved with high school mathematics. I hope these teachers and educators will make the eﬀort to become familiar with the mathematics presented in this volume—better yet, with the mathematics in all three volumes! For a more extended discussion of the raison d’ˆetre for the content of these volumes, please see To the Instructor on pp. xix ﬀ. Acknowledgements The drafts of this volume and its companion volumes, [Wu2020a] and [Wu2020c], have been used since 2006 in the mathematics department at the University of California at Berkeley as textbooks for a three-semester sequence of courses, Math 151–153, that was created for pre-service high school teachers.6 The two people most responsible for making these courses a reality were the two chairs of the mathematics department in those early years: Calvin Moore and Ted Slaman. I am immensely indebted to them for their support. I should not fail to mention that, at one point, Ted volunteered to teach an extra course for me in order to free me up for the writing of an early draft of these volumes. Would that all of us had chairs like him! Mark Richards, then Dean of Physical Sciences, was also behind these courses from the beginning. His support not only meant a lot to me personally, but I suspect that it also had something to do with the survival of these courses in a research-oriented department. It is manifestly impossible to write three volumes of teaching materials without generous help from students and friends in the form of corrections and suggestions for improvement. I have been fortunate in this regard, and I want to thank them all for their critical contributions: Richard Askey,7 David Ebin, Emiliano G´omez, Larry Francis, Ole Hald, Sunil Koswatta, Bob LeBoeuf, Gowri Meda, Clinton Rempel, Ken Ribet, Shari Lind Scott, Angelo Segalla, and Kelli Talaska. Dick Askey’s name will be mentioned in several places in these volumes, but I have beneﬁtted from his judgment much more than what those explicit citations would seem to indicate. I especially appreciate the fact that he shared my belief early on in the corrosive eﬀect of TSM on school mathematics education. David Ebin and Angelo Segalla taught from these volumes at SUNY Stony Brook and CSU Long Beach, respectively, and I am grateful to them for their invaluable input from the trenches. I must also thank Emiliano G´omez, who has taught these courses more times than anybody else with the exception of Ole Hald. Some of his deceptively simple comments have led to much soul-searching and extensive corrections. Bob LeBoeuf put up with my last-minute requests for help, and he shows what real dedication to a cause is all about. Section 1.9 of the third volume ([Wu2020c]) on the importance of sine and cosine could not have been written without special help from Professors Thomas Kailath and Julius O. Smith III of Stanford University, as well as from my longtime collaborator Robert Greene of UCLA. I am grateful to them for their uncommon courtesy. 6 Since the fall of 2018, this three-semester sequence has been pared down to a two-semester sequence. A partial study of the eﬀects of these courses on pre-service teachers can be found in [Newton-Poon]. 7 Sadly, Dick passed away on October 9, 2019.

xviii

PREFACE

Last but not least, I have to single out two names for my special expression of gratitude. Larry Francis has been my editor for many years, and he has pored over every single draft of these manuscripts with the same meticulous care from the ﬁrst word to the last. I want to take this opportunity to thank him for the invaluable help he has consistently provided me. Ole Hald took it upon himself to teach the whole Math 151–153 sequence—without a break—several times to help me improve these volumes. That he did, in more ways than I can count. His numerous corrections and suggestions, big and small, all throughout the last nine years have led to many dramatic improvements. My indebtedness to him is too great to be expressed in words. Hung-Hsi Wu Berkeley, California January 17, 2020

To the Instructor These three volumes (the other two being [Wu2020a] and [Wu2020c]) have been written expressly for high school mathematics teachers and mathematics educators.1 Their goal is to revisit the high school mathematics curriculum, together with relevant topics from middle school, to help teachers better understand the mathematics they are or will be teaching and to help educators establish a sound mathematical platform on which to base their research. In terms of mathematical sophistication, these three volumes are designed for use in upper division courses for math majors in college. Since their content consists of topics in the upper end of school mathematics (including one-variable calculus), these volumes are in the unenviable position of straddling two disciplines: mathematics and education. Such being the case, these volumes will inevitably inspire misconceptions on both sides. We must therefore address their possible misuse in the hands of both mathematicians and educators. To this end, let us brieﬂy review the state of school mathematics education as of 2020. The phenomenon of TSM For roughly the last ﬁve decades, the nation has had a de facto national school mathematics curriculum, one that has been deﬁned by the standard school mathematics textbooks. The mathematics encoded in these textbooks is extremely ﬂawed.2 We call the body of knowledge encoded in these textbooks TSM (Textbook School Mathematics; see page xi). We will presently give a superﬁcial survey of some of these ﬂaws,3 but what matters to us here is the fact that institutions of higher learning appear to be oblivious to the rampant mathematical mis-education of students in K–12 and have done very little to address the insidious presence of TSM in the mathematics taught to K–12 students over the last 50 years. As a result, mathematics teachers are forced to carry out their teaching duties with all the misconceptions they acquired from TSM intact, and educators likewise continue to base their research on what they learned from TSM. So TSM lives on unchallenged. These three volumes are the conclusion of a six-volume series4 whose goal is to correct the universities’ curricular oversight in the mathematical education of 1 We use the term “mathematics educators” to refer to university faculty in schools of education. 2 These statements about curriculum and textbooks do not take into account how much the quality of school textbooks and teachers’ content knowledge may have evolved recently with the advent of CCSSM (Common Core State Standards for Mathematics) ([CCSSM]) in 2010. 3 Detailed criticisms and explicit corrections of these ﬂaws are scattered throughout these volumes. 4 The earlier volumes in the series are [Wu2011a], [Wu2016a], and [Wu2016b].

xix

xx

TO THE INSTRUCTOR

teachers and educators by providing the needed mathematical knowledge to break the vicious cycle of TSM. For this reason, these volumes pay special attention to mathematical integrity5 and transparency, so that every concept is precisely deﬁned and every assertion is completely explained,6 and so that the exposition here is as close as possible to what is taught in a high school classroom. TSM has appeared in diﬀerent guises; after all, the NCTM reform (see [NCTM1989]) was largely ushered in around 1989. But beneath the surface its essential substance has stayed remarkably constant (compare [Wu2014]). TSM is characterized by a lack of clear deﬁnitions, faulty or nonexistent reasoning, pervasive imprecision, general incoherence, and a consistent failure to make the case about why each standard topic in the school curriculum is worthy of study. Let us go through each of these issues in some detail. (1) Deﬁnitions. In TSM, correct deﬁnitions of even the most basic concepts are usually not available. Here is a partial list: fraction, multiplication of fractions, division of fractions, one fraction being bigger or smaller than another, ﬁnite decimal, inﬁnite decimal, mixed number, ratio, percent, rate, constant rate, negative number, the four arithmetic operations on rational numbers, congruence, similarity, length of a curve, area of a planar region, volume of a solid, expression, equation, graph of a function, graph of an inequality, half-plane, polygon, interior angle of a polygon, regular polygon, slope of a line, parabola, inverse function, etc. Consequently, students are forced to work with concepts whose mathematical meaning is at best only partially revealed to them. Consider, for example, the concept of division. TSM oﬀers no precise deﬁnition of division for whole numbers, fractions, rational numbers, real numbers, or complex numbers. If it did, the division concept would become much more learnable because it is in fact the same for all these number systems (thus we also witness the incoherence of TSM). The lack of a deﬁnition for division leads inevitably to the impossibility of reasoning about the division of fractions, which then leads to “ours is not to reason why, just invert-and-multiply”. We have here a prime example of the convergence of the lack of deﬁnitions, the lack of reasoning, and the lack of coherence. The reason we need precise deﬁnitions is that they create a level playing ﬁeld for all learners, in the sense that each person—including the teacher—has all the needed information about a given concept from the very beginning and this information is the same for everyone. This eliminates any need to spend time looking for “tricks”, “insider knowledge”, or hidden agendas. The level playing ﬁeld makes every concept accessible to all learners, and this fact is what the discussion of equity in school mathematics education seems to have overlooked thus far. To put this statement in context, think of TSM’s deﬁnition of a fraction as a piece of pizza: even elementary students can immediately see that there is more to a fraction than just being a piece of pizza. For example, “ 58 miles of dirt road” has nothing to do with pieces of a pizza. The credibility gap between what students are made to learn and what they subconsciously recognize to be false disrupts the learning process, often fatally. 5 We

will provide a deﬁnition of this term on page xxv. other words, every theorem is completely proved. Of course there are a few theorems that cannot be proved in context, such as the fundamental theorem of algebra. 6 In

TO THE INSTRUCTOR

xxi

In mathematics, there can be no valid reasoning without precise deﬁnitions. Consider, for example, TSM’s proof of (−2)(−3) = 2 × 3. Such a proof requires that we know what −2 is, what −3 is, what properties these negative integers are assumed to possess, and what it means to multiply (−2) by (−3) so that we can use them to justify this claim. Since TSM does not oﬀer any information of this kind, it argues instead as follows: 3 · (−3), being 3 copies of −3, is equal to −9, and likewise, 2 · (−3) = −6, 1 · (−3) = −3, and of course 0 · (−3) = 0. Now look at the pattern formed by these consecutive products: 3 · (−3) = −9,

2 · (−3) = −6,

1 · (−3) = −3,

0 · (−3) = 0.

Clearly when the ﬁrst factor decreases by 1, the product increases by 3. Now, when the 0 in the product 0 · (−3) decreases by 1 (so that 0 becomes −1), the product (−1)(−3) ceases to make sense. Nevertheless, TSM urges students to believe that the pattern must persist no matter what so that this product will once more increase by 3 and therefore (−1)(−3) = 3. By the same token, when the −1 in (−1)(−3) decreases by 1 again (so that −1 becomes −2), the product must again increase by 3 for the same reason and (−2)(−3) = 6 = 2 × 3, as desired. This is what TSM considers to be “reasoning”. TSM goes further. Using a similar argument for (−2)(−3) = 2 × 3, one can show that (−a)(−b) = ab for all whole numbers a and b. Now, TSM asks students to take another big leap of faith: if (−a)(−b) = ab is true for whole numbers a and b, then it must also be true when a and b are arbitrary numbers. This is how TSM “proves” that negative times negative is positive. Slighting deﬁnitions in TSM can also take a diﬀerent form: the graph of a linear inequality ax + by ≤ c is claimed to be a half-plane of the line ax + by = c, and the “proof” usually consists of checking a few examples. Thus the points (0, 0), (−2, 0), and (1, −1) lie below the line deﬁned by x + 3y = 2 and they all satisfy x + 3y ≤ 2, so it is believable that the “lower half-plane” of the line x + 3y = 2 is the graph of x + 3y ≤ 2. Further experimentation with other points below the line deﬁned by x + 3y = 2 adds to this conviction. Again, no reasoning is involved and, more importantly, neither “graph of an inequality” nor “half-plane” is deﬁned in such a discussion because these terms sound so familiar that TSM apparently believes no deﬁnition is necessary. At other times, reasoning is simply suppressed, such as when the coordinates of the vertex of the graph of ax2 + bx + c are peremptorily declared to be −b 4ac − b2 , . 2a 4a End of discussion. Our emphasis on the importance of deﬁnitions in school mathematics compels us to address a misconception about the role of deﬁnitions in school mathematics education. To many teachers and educators, the word “deﬁnition” connotes something tedious and nonessential that students must memorize for standardized tests. It may also conjure an image of cut-and-dried, top-down instruction that begins with a rigid and unmotivated deﬁnition and continues with the deﬁnition’s formal and equally unmotivated appearance in a chain of logical arguments. Understandably, most educators ﬁnd this scenario unappetizing. Their response is that, at least in school mathematics, the deﬁnition of a concept should emerge at the end—but

xxii

TO THE INSTRUCTOR

not at the beginning—of an extended intuitive discussion of the hows and whys of the concept.7 In addition, the so-called conceptual understanding of the concept is believed to lie in the intuitive discussion but not in the formal deﬁnition itself, the latter being nothing more than an afterthought. These two opposite conceptions of deﬁnition ignore the possibility of a middle ground: one can state the precise deﬁnition of a concept at the beginning of a lesson to set the tone of the subsequent mathematical discussion and exploration, which is to show students that this is all they will ever need to know about the concept as far as doing mathematics is concerned. Such transparency—demanded by the mathematical culture of the past century (cf. [Quinn])—is what is most sorely missing in TSM, which consistently leaves students in doubt about what a fraction is or might be, what a negative number is, what congruence means, etc. In this middle ground, a deﬁnition can be explored and explained in intuitive terms in the ensuing discussion on the one hand and on the other, put to use in proofs—in its precise formulation—to show how and why the deﬁnition is absolutely indispensable to any kind of reasoning concerning the concept. With the consistent use of precise deﬁnitions, the line between what is correct and what is intuitive but maybe incorrect (such as the TSM-proof of negative times negative is positive) becomes clearly drawn. It is the frequent blurring of this line in TSM that contributes massively to the general misapprehension in mathematics education about what a proof is (part of this misapprehension is described in, e.g., [NCTM2009], [Ellis-Bieda-Knuth], and [Arbaugh et al.]). These three volumes (this volume, [Wu2020a], and [Wu2020c]) will always take a position in the aforementioned middle ground. Consider the deﬁnition of a fraction, for example: it is one of a special collection of points on the number line (see Section 1.1 of [Wu2020a]). This is the only meaning of a fraction that is needed to drive the fairly intricate mathematical development of fractions, and, for this reason, the deﬁnition of a fraction as a certain point on the number line is the one that will be unapologetically used all through these three volumes. To help teachers and students feel comfortable with this deﬁnition, we give an extensive intuitive discussion of why such a deﬁnition for a fraction is necessary at the beginning of Section 1.1 in [Wu2020a] before giving the formal deﬁnition. This intuitive discussion, naturally, opens the door to whatever pedagogical strategy a teacher wants to invest in it. Unlike in TSM, however, this deﬁnition is not given to be forgotten. On the contrary, all subsequent discussions about fractions will refer to this precise deﬁnition (but not to the intuitive discussion that preceded it) and, of course, all the proofs about fractions will also depend on this formal deﬁnition because mathematics demands no less. Students need to learn what a proof is and how it works; the exposition here tries to meet this need by (gently) laying bare the fact that reasoning in proofs requires precise deﬁnitions. As a second example, we give the deﬁnition of the slope of a line only after an extensive intuitive discussion in Section 6.4 of [Wu2020a] about what slope is supposed to measure and how we may hope to measure it. Again, the emphasis is on the fact that this deﬁnition of slope is not the conclusion, but the beginning of a long logical development that 7 Proponents of this approach to deﬁnitions often seem to forget that, after the emergence of a precise deﬁnition, students are still owed a systematic exposition of mathematics using the deﬁnition so that they can learn about how the deﬁnition ﬁts into the overall logical structure of mathematics.

TO THE INSTRUCTOR

xxiii

occupies the second half of Chapter 6 in [Wu2020a], reappears in trigonometry (relation with the tangent function), calculus (deﬁnition of the derivative), and beyond. (2) Reasoning. Reasoning is the lifeblood of mathematics, and the main reason for learning mathematics is to learn how to reason. In the context of school mathematics, reasoning is important to students because it is the tool that empowers them to explore on their own and verify for themselves what is true and what is false without having to take other people’s words on faith. Reasoning gives them conﬁdence and independence. But when students have to accustom themselves to performing one unexplained rote skill after another, year after year, their ability to reason will naturally atrophy. Many students ﬁnd it more expedient to stop asking why and simply take any order that comes their way sight unseen just to get by.8 One can only speculate on the cumulative eﬀect this kind of mathematics “learning” has on those students who go on to become teachers and mathematics educators. (3) Precision. The purpose of precision is to eliminate errors and minimize misconceptions, but in TSM students learn at every turn that they should not believe exactly what they are told but must learn to be creative in interpreting it. For example, TSM preaches the virtue of using the theorem on equivalent fractions to simplify fractions and does not hesitate to simplify a rational expression in x as follows: x2 + 3 (x − 1)(x2 + 3) = . x(x − 1) x This looks familiar because “canceling the same number from top and bottom” is exactly what the theorem on equivalent fractions is supposed to do. Unfortunately, this theorem only guarantees a ca = bc b when a, b, and c are whole numbers (b and c understood to be nonzero). In the 2 previous rational expression, however, none of (x−1), √ (x +3), and x is necessarily a whole number because x could be, for example, 5. Therefore, according to TSM, students in algebra should look back at equivalent fractions and realize that the theorem on equivalent fractions—in spite of what it says—can actually be applied to “fractions” whose “numerators” and “denominators” are not whole numbers. Thus TSM encourages students to believe that “nothing needs to be taken precisely and one must be ﬂexible in interpreting what one learns”. This extrapolation-happy mindset is the opposite of what it takes to learn a precise subject like mathematics or any of the exact sciences. For example, we cannot allow students to believe that the domain of deﬁnition of log x is [0, ∞) since [0, ∞) is more or less the same as (0, ∞). Indeed, the presence or absence of the single point “0” is the diﬀerence between true and false. Another example of how a lack of precision leads to misconceptions is the statement that “β 0 = 1”, where β is a nonzero number. Because TSM does not use precise language, it does not—or cannot—draw a sharp distinction between a heuristic argument, a deﬁnition, and a proof. Consequently, it has misled numerous 8 There

is consistent anecdotal evidence from teachers in the trenches that such is the case.

xxiv

TO THE INSTRUCTOR

students and teachers into believing that the heuristic argument for deﬁning β 0 to be 1 is in fact a “proof” that β 0 = 1. The same misconception persists for negative exponents (e.g., β −n = 1/β n ). The lack of precision is so pervasive in TSM that there is no end to such examples. (4) Coherence. Another reason why TSM is less than learnable is its incoherence. Skills in TSM are framed as part of a long laundry list, and the lack of deﬁnitions for concepts ensures that skills and their underlying concepts remain forever disconnected. Mathematics, on the other hand, unfolds from a few central ideas, and concepts and skills are developed along the way to meet the needs that emerge in the process of unfolding. An acceptable exposition of mathematics therefore tells a coherent story that makes mathematics memorable. For example, consider the fact that TSM makes the four standard algorithms for whole numbers four separate rote-learning skills. Thus TSM hides from students the overriding theme that the Hindu-Arabic numeral system is universally adopted because it makes possible a simple, algorithmic procedure for computations; namely, if we can carry out an operation (+, −, ×, or ÷) for single-digit numbers, then we can carry out this operation for all whole numbers no matter how many digits they have (see Chapter 3 of [Wu2011a]). The standard algorithms are the vehicles that bridge operations with single-digit numbers and operations on all whole numbers. Moreover, the standard algorithms can be simply explained by a straightforward application of the associative, commutative, and distributive laws. From this perspective, a teacher can explain to students, convincingly, why the multiplication table is very much worth learning; this would ease one of the main pedagogical bottlenecks in elementary school. Moreover, a teacher can also make sense of the associative, commutative, and distributive laws to elementary students and help them see that these are vital tools for doing mathematics rather than dinosaurs in an outdated school curriculum. If these facts had been widely known during the 1900s, the senseless debate on whether the standard algorithms should be taught might not have arisen and the Math Wars might not have taken place at all. TSM also treats whole numbers, fractions, (ﬁnite) decimals, and rational numbers as four diﬀerent kinds of numbers. The reality is that, ﬁrst of all, decimals are a special class of fractions (see Section 1.1 of [Wu2020a]), whole numbers are part of fractions, and fractions are part of rational numbers. Moreover, the four arithmetic operations (+, −, ×, and ÷) in each of these number systems do not essentially change from system to system. There is a smooth conceptual transition at each step of the passage from whole numbers to fractions and from fractions to decimals and rational numbers; see Parts 2 and 3 of [Wu2011a] or Sections 2.2, 2.4, and 2.5 of [Wu2020a]. This coherence facilitates learning: instead of having to learn about four diﬀerent kinds of numbers, students basically only need to learn about one number system (the rational numbers). Yet another example is the conceptual unity between linear functions and quadratic functions: in each case, the leading term—ax for linear functions and ax2 for quadratic functions—determines the shape of the graph of the function completely, and the studies of the two kinds of functions become similar as each revolves around the shape of the graph (see Section 2.1 on pp. 63ﬀ.). Mathematical coherence gives us many such storylines, and a few more will be detailed below.

TO THE INSTRUCTOR

xxv

(5) Purposefulness. In addition to the preceding four shortcomings—a lack of clear deﬁnitions, faulty or nonexistent reasoning, pervasive imprecision, and general incoherence—TSM has a ﬁfth fatal ﬂaw: it lacks purposefulness. Purposefulness is what gives mathematics its vitality and focus: the fact is that a mathematical investigation, at any level, is always carried out with a speciﬁc goal in mind. When a mathematics textbook reﬂects this goal-oriented character of mathematics, it propels the mathematical narrative forward and facilitates its learning by making students aware of where the discussion is headed, and why. Too often, TSM lurches from topic to topic with no apparent purpose, leading students to wonder why they should bother to tag along. One example is the introduction of the absolute value of a number. Many teachers and students are mystiﬁed by being saddled with such a “frivolous” skill: “just kill the negative sign”, as one teacher put it. Yet TSM never tries to demystify his concept. (For an explanation of the need to introduce absolute value, see, e.g., the Pedagogical Comments in Section 2.6 of [Wu2020a]). Another is the seemingly inexplicable √ replacement √ of the square root and cube root symbols of a positive number b, i.e., b and 3 b, by rational exponents, b1/2 and b1/3 , respectively (see, e.g., Section 4.2 on pp. 142ﬀ.). Because TSM teaches the laws of exponents as merely “number facts”, it is inevitable that it would fail to point out the purpose of this change of notation, which is to shift focus from the operation of taking roots to the properties of the exponential function bx for a ﬁxed positive b. A ﬁnal example is the way TSM teaches estimation completely by rote, without ever telling students why and when estimation is important and therefore worth learning. Indeed, we often have to make estimates, either because precision is unattainable or unnecessary, or because we purposely use estimation as a tool to help achieve precision (see [Wu2011a, Section 10.3]). To summarize, if we want students to be taught mathematics that is learnable, then we must discard TSM and replace it with the kind of mathematics that possesses these ﬁve qualities: Every concept has a clear deﬁnition. Every statement is precise. Every assertion is supported by reasoning. Its development is coherent. Its development is purposeful. We call these the Fundamental Principles of Mathematics (also see Section 2.1 in [Wu2018]). We say a mathematical exposition has mathematical integrity if it embodies these fundamental principles. As we have just seen, we ﬁnd in TSM a consistent pattern of violating all ﬁve fundamental principles. We believe that the dominance of TSM in school mathematics in the past ﬁve decades is a principal cause of the ongoing crisis in school mathematics education. One consequence of the dominance of TSM is that most students come out of K–12 knowing only TSM but not mathematics that respects these fundamental principles. To them, learning mathematics is not about learning how to reason or distinguish true from false but about memorizing facts and tricks to get correct answers. Faced with this crisis, what should be the responsibility of institutions of higher learning? Should it be to create courses for future teachers and educators to help them systematically replace their knowledge of TSM with mathematics that is consistent with the ﬁve fundamental principles? Or should it be, rather, to leave TSM alone but make it more palatable by helping teachers infuse their classrooms

xxvi

TO THE INSTRUCTOR

with activities that suggest visions of reasoning, problem solving, and sense making? As of this writing, an overwhelming majority of the institutions of higher learning are choosing the latter alternative. At this point, we return to the earlier question about some of the ways both university mathematicians and educators might misunderstand and misuse these three volumes. Potential misuse by mathematicians First, consider the case of mathematicians. They are likely to scoﬀ at what they perceive to be the triviality of the content in these volumes: no groups, no homomorphisms, no compact sets, no holomorphic functions, and no Gaussian curvature. They may therefore be tempted to elevate the level of the presentation, for example, by introducing the concept of a ﬁeld and show that, when two fraction symbols m/n and k/ (with whole numbers m, n, k, , and n = 0, = 0) satisfying m = nk are identiﬁed, and when + and × are deﬁned by the usual formulas, the fraction symbols form a ﬁeld. In this elegant manner, they can eﬃciently cover all the standard facts in the arithmetic of fractions in the school curriculum.9 This is certainly a better way than deﬁning fractions as points on the number line to teach teachers and educators about fractions, is it not? Likewise, mathematicians may ﬁnd ﬁnite geometry to be a more exciting introduction to axiomatic systems than any proposed improvements on the high school geometry course in TSM. The list goes on. Consequently, pre-service teachers and educators may end up learning from mathematicians some interesting mathematics, but not mathematics that would help them overcome the handicap of knowing only TSM. Mathematicians may also engage in another popular approach to the professional development of teachers and educators: teaching the solution of hard problems. Because mathematicians tend to take their own mastery of fundamental skills and concepts for granted, many do not realize that it is nearly impossible for teachers who have been immersed in thirteen years or more of TSM to acquire, on their own, a mastery of a mathematically correct version of the basic skills and concepts. Mathematicians are therefore likely to consider their major goal in the professional development of teachers and educators to be teaching them how to solve hard problems. Surely, so the belief goes, if teachers can handle the “hard stuﬀ”, they will be able to handle the “easy stuﬀ” in K–12. Since this belief is entirely in line with one of the current slogans in school mathematics education about the critical importance of problem solving, many teachers may be all too eager to teach their students the extracurricular skills of solving challenging problems in addition to teaching them TSM day in and day out. In any case, the relatively unglamorous content of these three volumes (this volume, [Wu2020a], and [Wu2020c])—designed to replace TSM—will get shunted aside into supplementary reading assignments. At the risk of belaboring the point, the focus of these three volumes is on showing how to replace teachers’ and educators’ knowledge of TSM in grades 9–12 with mathematics that respects the fundamental principles of mathematics. Therefore, reformulating the mathematics of grades 9–12 from an advanced mathematical standpoint to obtain a more elegant presentation is not the point. Introducing 9 This is my paraphrase of a mathematician’s account of his professional development institute around year 2000.

TO THE INSTRUCTOR

xxvii

novel elementary topics (such as Pick’s theorem or the 4-point aﬃne plane) into the mathematics education of teachers and educators is also not the point. Rather, the point in year 2020 is to do the essential spadework of revisiting the standard 9–12 curriculum—topic by topic, along the lines laid out in these three volumes— showing teachers and educators how the TSM in each case can be supplanted by mathematics that makes sense to them and to their students. For example, since most pre-service teachers and educators have not been exposed to the use of precise deﬁnitions in mathematics, they are unlikely to know that deﬁnitions are supposed to be used, exactly as written, no more and no less, in logical arguments. One of the most formidable tasks confronting mathematicians is, in fact, how to change educators’ and teachers’ perception of the role of deﬁnitions in reasoning. As illustration, consider how TSM handles slope. There are two ways, but we will mention only one of them.10 TSM pretends that, by deﬁning the slope of a line L using the diﬀerence quotient with respect to two pre-chosen points P and Q on L,11 such a diﬀerence quotient is a property of the line itself (rather than a property of the two points P and Q). In addition, TSM pretends that it can use “reasoning” based on this defective deﬁnition to derive the equation of a line when (for example) its slope and a given point on it are prescribed. Here is the inherent danger of thirteen years of continuous exposure to this kind of pseudoreasoning: teachers cease to recognize that (a) such a deﬁnition of slope is defective and (b) such a defective deﬁnition of slope cannot possibly support the purported derivation (= proof) of the equation of a line. It therefore comes to pass that— as a result of the ﬂaws in our education system—many teachers and educators end up being confused about even the meaning of the simplest kind of reasoning: “A implies B”. They need—and deserve—all the help we can give so that they can ﬁnally experience genuine mathematics, i.e., mathematics that is based on the fundamental principles of mathematics. Of course, the ultimate goal is for teachers to use this new knowledge to teach their own students so that those students can achieve a true understanding of what “A implies B” means and what real reasoning is all about. With this in mind, we introduce in Section 6.4 of [Wu2020a] the concept of slope by discussing what slope is supposed to measure (an example of purposefulness) and how to measure it, which then leads to the formulation of a precise deﬁnition. With the availability of the AA-criterion for triangle similarity (Theorem G22 in Section 5.3 of [Wu2020a]), we then show how this deﬁnition leads to the formula for the slope of a line as the diﬀerence quotient of the coordinates of any two points on the line (the “rise-over-run”). Having this critical ﬂexibility to compute the slope—plus an earlier elucidation of what an equation is (see Section 6.2 in [Wu2020a])—we easily obtain the equation of a line passing through a given point with a given slope, with correct reasoning this time around (see Section 6.5 in [Wu2020a]). Of course the same kind of reasoning can be applied to similar problems when other reasonable geometric data are prescribed for the line. By guiding teachers and educators systematically through the correction of TSM errors on a case-by-case basis, we believe they will gain a new and deeper 10 A second way is to deﬁne a line to be the graph of a linear equation y = mx + b and then deﬁne the slope of this line to be m. This is the deﬁnition of a line in advanced mathematics, but it is so profoundly inappropriate for use in K–12 that we will just ignore it. 11 This is the “rise-over-run”.

xxviii

TO THE INSTRUCTOR

understanding of school mathematics. Ultimately, we hope that if institutions of higher learning and the education establishment can persevere in committing themselves to this painstaking work, the students of these teachers and educators will be spared the ravages of TSM. If there is an easier way to undo thirteen years and more of mis-education in mathematics, we are not aware of it. A main emphasis in using these three volumes should therefore be on providing patient guidance to teachers and educators to help them overcome the many handicaps inﬂicted on them by TSM. In this light, we can say with conﬁdence that, for now, the best way for mathematicians to help educate teachers and educators is to ﬁrm up their mathematical foundations. Let us repair the damage TSM has done to their mathematics content knowledge by helping them to acquire a knowledge of school mathematics that is consistent with the fundamental principles of mathematics. Potential misuse by educators Next, we address the issue of how educators may misuse these three volumes. Educators may very well frown on the volumes’ insistence on precise deﬁnitions and precise reasoning and their unremitting emphasis on proofs while, apparently, neglecting problem solving, conceptual understanding, and sense making. To them, good professional development concentrates on all of these issues plus contextual learning, student thinking, and communication with students. Because these three volumes never explicitly mention problem solving, conceptual understanding, or sense making per se (or, for that matter, contextual learning or student thinking), their content may be dismissed by educators as merely skills-oriented or technical knowledge for its own sake and, as such, get relegated to reading assignments outside of class. They may believe that precious class time can be put to better use by calling on students to share their solutions to diﬃcult problems or by holding small group discussions about problem-solving strategies. We believe this attitude is also misguided because the critical missing piece in the contemporary mathematical education of teachers and educators is an exposure to a systematic exposition of the standard topics of the school curriculum that respects the fundamental principles of mathematics. Teachers’ lack of access to such a mathematical exposition is what lies at the heart of much of the current education crisis. Let us explain. Consider problem solving. At the moment, the goal of getting all students to be proﬁcient in solving problems is being pursued with missionary zeal, but what seems to be missing in this single-minded pursuit is the recognition that the body of knowledge we call mathematics consists of nothing more than a sequence of problems posed, and then solved, by making logical deductions on the basis of precise deﬁnitions, clearly stated hypotheses, and known results.12 This is after all the whole point of the classic two-volume work, [P´ olya-Szeg¨ o], which introduces students to mathematical research through the solutions to a long list of problems. For example, the Pythagorean theorem and its many proofs are nothing more than 12 It is in this light that the previous remark about the purposefulness of mathematics can be better understood: before solving a problem, one should know why the problem was posed in the ﬁrst place. Note that, for beginners (i.e., school students), the overwhelming emphasis has to be on solving problems rather than the more elusive issue of posing problems.

TO THE INSTRUCTOR

xxix

solutions to the problem posed by people from diverse cultures long ago: “Is there any relationship among the three sides of a right triangle?” There is no essential diﬀerence between problem solving and theorem proving in mathematics. Each time we solve a problem, we in eﬀect prove a theorem (trivial as that theorem may sometimes be). The main point of this observation is that if we want students to be proﬁcient in problem solving, then we must give them plenty of examples of gradeappropriate proofs all through (at least) grades 4–12 and engage them regularly in grade-appropriate theorem-proving activities. If we can get students to see, day in and day out, that problem solving is a way of life in mathematics and if we also routinely get them involved in problem solving (i.e., theorem proving), students will learn problem solving naturally through such a long-term immersion. In the process, they will get to experience that, to solve problems, they need to have precise deﬁnitions and precise hypotheses as a starting point, know the direction they are headed towards before they make a move (sense making), and be able to make deductions from precise deﬁnitions and known facts. Deﬁnitions, sense making, and reasoning will therefore come together naturally for students if they learn mathematics that is consistent with the ﬁve fundamental principles. We make the eﬀort to put problem solving in the context of the fundamental principles of mathematics because there is a danger in pursuing problem solving per se in the midst of the TSM-induced corruption of school mathematics. In a generic situation, teachers teach TSM and only pay lip service to “problem solving”, while in the best case scenario, teachers keep TSM intact while teaching students how to solve problems on a separate, parallel track outside of TSM. Lest we forget, TSM considers “out of a hundred” to be a correct deﬁnition of percent, expands the product of two linear polynomials by “FOILing”, and assumes that in any problem about rate, one can automatically assume that the rate is constant (“Lynnette can wash 95 cars in 5 days. How many cars can Lynnette wash in 11 days?”), etc. In this environment, it is futile to talk about (correct) problem solving. Until we can rid school classrooms of TSM, the most we can hope for is having teachers teach, on the one hand, deﬁnition-free concepts with a bag of tricks-sans-reasoning to get correct answers and, on the other hand, reasoning skills for solving a separate collection of problems for special occasions. In other words, two parallel universes will co-exist in school mathematics classrooms. So long as TSM continues to reign in school classrooms, most students will only be comfortable doing one-step problems and any problem-solving ability they possess will only be something that is artiﬁcially grafted onto the TSM they know. If we want to avert this kind of bipolar mathematics education in schools, we must begin by providing teachers with a better mathematical education. Then we can hope that teachers will teach mathematics consistent with the fundamental principles of mathematics13 so that students’ problem-solving abilities can evolve naturally from the mathematics they learn. It is partly for this reason that the six volumes under discussion14 choose to present the mathematics of K–12 with explanations (= proofs) for all the skills. In particular, these three volumes on the 13 And, of course, to also get school textbooks that are unsullied by TSM. However, it seems likely as of 2020 that major publishers will hold onto TSM until there are suﬃciently large numbers of knowledgeable teachers who demand better textbooks. See the end of [Wu2015]. 14 These three volumes, together with [Wu2011a], [Wu2016a], and [Wu2016b].

xxx

TO THE INSTRUCTOR

mathematics of grades 9–12 provide proofs for every theorem. (At the same time, they also caution against certain proofs that are simply too long or too tedious to be presented in a high school classroom.) The hope is that when teachers and educators get to experience ﬁrsthand that every part of school mathematics is suﬀused with reasoning, they will not fail to teach reasoning to their own students as a matter of routine. Only then will it make sense to consider problem solving to be an integral part of school mathematics. The importance of correct content knowledge In general, the idea is that if we give teachers and educators an exposition of mathematics that makes sense and has built-in conceptual understanding and reasoning, then we can hope to create classrooms with an intellectual climate that enables students to absorb these qualities as if by osmosis. Perhaps an analogy can further clarify this issue: if we want to teach writing, it would be more eﬀective to let students read good writing and learn from it directly rather than to let them read bad writing and simultaneously attend special sessions on the ﬁne points of eﬀective written communication. If we want school mathematics to be suﬀused with reasoning, conceptual understanding, and sense making, then we must recognize that these are not qualities that can stand apart from mathematical details. Rather, they are ﬁrmly anchored to hard-and-fast mathematical facts. Take proofs (= reasoning), for example. If we only talk about proofs in the context of TSM, then our conception of what a proof is will be extremely ﬂawed because there are essentially no correct proofs in TSM. For starters, since TSM has no precise deﬁnitions, there can be no hope of ﬁnding a completely correct proof in TSM. Therefore, when teaching from these three volumes,15 it is imperative to ﬁrst concentrate on getting across to teachers and educators the details of the mathematical reformulation of the school curriculum. Speciﬁcally, we stress the importance of oﬀering educators a valid alternative to TSM for their future research. Only then can we hope to witness a reconceptualization—in mathematics education—of reasoning, conceptual understanding, problem solving, etc., on the basis of a solid mathematical foundation. Reasoning, conceptual understanding, and sense making are qualities intrinsic to school mathematics that respects the fundamental principles of mathematics. We see in these three volumes a continuous narrative from topic to topic and from chapter to chapter to guide the reader through this long journey. The sense making will be self-evident to the reader. Moreover, when every assertion is backed up by an explanation (= proof), reasoning will rise to the surface for all to see. In their presentation of the natural unfolding of mathematical ideas, these volumes also routinely point out connections between deﬁnitions, concepts, theorems, and proofs. Some connections may not be immediately apparent. For example, in Section 6.1 of [Wu2020a], we explicitly point out the connection between Mersenne primes and the summation of ﬁnite geometric series. Other connections span several grades: there is a striking similarity between the proofs of the area formula for rectangles whose sides are fractions (Theorem 1.7 in Section 1.4 of [Wu2020a]), the ASA congruence criterion (Theorem G9 in Section 4.6 of [Wu2020a]), the SSS congruence criterion (Theorem G28 on pp. 223ﬀ.), the fundamental theorem 15 As

well as from the other three volumes, [Wu2011a], [Wu2016a], and [Wu2016b]).

TO THE INSTRUCTOR

xxxi

of similarity (Theorem G10 on pp. 230ﬀ.), and the theorem about the equality of angles on a circle subtending the same arc (Theorem G52 on pp. 277ﬀ.). All these proofs are achieved by breaking up a complicated argument into two or more clearcut steps, each involving simpler arguments. In other words, they demonstrate how to reduce the complex to the simple so that prospective teachers and educators can learn from such instructive examples about the ﬁne art of problem solving. The foregoing unrelenting emphasis on mathematical content should not lead readers to believe that these three volumes deal with mathematics at the expense of pedagogy. To the extent that these volumes are designed to promote better teaching in the schools, they do not sidestep pedagogical issues. Extensive pedagogical comments are oﬀered whenever they are called for, and they are clearly displayed as such; see, for example, pp. 28, 32, 36, 80, 173, 229ﬀ., 242, 279ﬀ., 292, etc., in the present volume. Nevertheless, our most urgent task—the fundamental task—in the mathematical education of teachers and educators as of 2020 has to be the reconstruction of their mathematical knowledge base. This is not about judiciously tinkering with what teachers and educators already know or tweaking their existing knowledge here and there. Rather, it is about the hard work of replacing their knowledge of TSM with mathematics that is consistent with the fundamental principles of mathematics from the ground up. The primary goal of these three volumes is to give a detailed exposition of school mathematics in grades 9–12 to help educators and teachers achieve this reconstruction.

To the Pre-Service Teacher In one sense, these three volumes are just textbooks, and you may feel you have gone through too many textbooks in your life to need any fresh advice. Nevertheless, we are going to suggest that you approach these volumes with a diﬀerent mindset than what you may have used with other textbooks, because you will soon be using the knowledge you gain from these volumes to teach your students. Reading other textbooks, you would likely congratulate yourself if you could achieve mastery over 90% of the material. That would normally guarantee an A. More is at stake with these volumes, however, because they directly address what you will need to know in order to write your lessons. Ask yourself whether a mathematics teacher whose lessons are correct only 90% of the time should be considered a good teacher. To be blunt, such a teacher would be a near disaster. So your mission in reading these volumes should be to achieve nothing short of total mastery. You are expected to know this material 100%. To the extent that the content of these three volumes is just K–12 mathematics, this is an achievable goal. This is the standard you have to set for yourself. Having said that, we also note explicitly that many Mathematical Asides are sprinkled all through the text, sometimes in the form of footnotes. These are comments—usually from an advanced mathematical perspective—that try to shed light on the mathematics under discussion. The above reference to “total mastery” does not include these comments. You should approach these volumes diﬀerently in yet another respect. Students’ typical attitude towards a math course is that if they can do all the homework problems, then most of their work is done. Think back on your calculus courses or any of the math courses when you were in school, and you will understand how true this is. But since these volumes are designed speciﬁcally for teachers, your emphasis cannot be limited to merely doing the homework assignments because your job will be more than just helping students to do homework problems. When you stand in front of a class, what you will be talking about, most of the time, will not be the exercises at the end of each section but the concepts and skills in the exposition proper.1 For example, very likely you will soon have to convince a class on geometry why the Pythagorean theorem is correct. There are two proofs of this theorem in these volumes, one in Chapter 5 of [Wu2020a] and the other in Chapter 4 of [Wu2020c]. Yet on neither occasion is it possible to assign a problem that asks for a proof of this theorem. There are problems that can assess whether you

1 I will be realistic and acknowledge that there are teachers who use class time only to drill students on how to get the right answers to exercises, often without reasoning. But one of the missions of these three volumes is to steer you away from that kind of teaching. See To the Instructor on page xix.

xxxiii

xxxiv

TO THE PRE-SERVICE TEACHER

know enough about the Pythagorean theorem to apply it, but how do you assess whether you know how to prove the theorem when the proofs have already been given in the text? It is therefore entirely up to you to achieve mastery of everything in the text itself. One way to check is to pick a theorem at random and ask yourself: Can I prove it without looking at the book? Can I explain its signiﬁcance? Can I convince someone else why it is worth knowing? Can I give an intuitive summary of the proof? These are questions that you will have to answer as a teacher. To the extent possible, these volumes try to provide information that will help you answer questions of this kind. I may add that the most taxing part of writing these volumes was in fact to do it in a way that would allow you, as much as possible, to adapt them for use in a school classroom with minimal changes. (Compare, for example, To the Instructor on pp. xix ﬀ.) There is another special feature of these volumes that I would like to bring to your attention: these volumes are essentially school textbooks written for teachers, and as such, you should read them with the eyes of a school student. When you read Chapter 1 of [Wu2020a] on fractions, for instance, picture yourself in a sixth-grade classroom and therefore, no matter how much abstract algebra you may know or how well you can explain the construction of the quotient ﬁeld of an integral domain, you have to be able to give explanations in the language of sixth-grade mathematics (i.e., to sixth graders). Similarly, when you come to Chapter 6 of [Wu2020a], you are developing algebra from the beginning, so even the use of symbols will be an issue (it is in fact the key issue; see Section 6.1 of [Wu2020a]). Therefore, be very deliberate and explicit when you introduce a symbol, at least for a while. The major conclusions in these volumes, as in all mathematics books, are summarized into theorems. Depending on the author’s (and other mathematicians’) whims, theorems are sometimes called propositions, lemmas, or corollaries as a way of indicating which theorems are deemed more important than others. Roughly speaking, a proposition is not regarded to be as important as a theorem, a lemma is conceptually less important than a proposition, and a corollary is supposed to follow immediately from the theorem or proposition to which it is attached. (Incidentally, a formula or an algorithm is just a theorem.) This idiosyncratic classiﬁcation of theorems started with Euclid around 300 BC, and it is too late to do anything about it now. The main concepts of mathematics are codiﬁed into deﬁnitions. Deﬁnitions are set in boldface in these volumes when they appear for the ﬁrst time; a few truly basic ones are even individually displayed in a separate paragraph, but most of the deﬁnitions are embedded in the text itself, so you should watch out for them. The statements of the theorems, and especially their proofs, depend on the deﬁnitions, and proofs are the guts of mathematics. Please note that when I said above that I expect you to know everything in these volumes, I was using the word “know” in the way mathematicians normally use the word. They do not use it to mean simply “know the statement by heart”. Rather, to know a theorem, for instance, means know the statement by heart, know its proof, know why it is worth knowing, know what its potential implications are, and ﬁnally, know how to apply it in new situations. If you know anything short of this, how can you expect to be able to answer your students’ questions? At the very least, you should know by heart all the theorems and deﬁnitions as well as the main ideas of each proof because, if you do not, it will be futile to talk about the

TO THE PRE-SERVICE TEACHER

xxxv

other aspects of knowing. Therefore, a preliminary suggestion to help you master the content of these volumes is for you to copy out the statements of every deﬁnition, theorem, proposition, lemma, and corollary, along with page references so that they can be examined in detail when necessary, and also to form the habit of summarizing the main idea(s) of each proof. These are good study habits. When it is your turn to teach your students, be sure to pass along these suggestions to them. You should also be aware that reading a mathematics book is not the same as reading a gossip magazine. You can probably ﬂip through one of the latter in an hour or less. But in these volumes, there will be many passages that require slow reading and re-reading, perhaps many times. I cannot single out those passages for you because they will be diﬀerent for diﬀerent people. We do not all learn the same way. What you can take for granted, however, is that mathematics books make for exceedingly slow reading. (Nothing good comes easy.) Therefore if you get stuck, time and time again, on a sentence or two in these volumes, take heart, because this is the norm in mathematics learning.

Prerequisites In terms of the mathematical development of this volume, only a knowledge of whole numbers, 0, 1, 2, 3, . . . , is assumed. Thus along with place value, you are assumed to know the four arithmetic operations, their standard algorithms, and the concept of division-with-remainder and how it is related to the long division algorithm.1 Division-with-remainder assigns to each pair of whole numbers b (the dividend) and d (the divisor), where d = 0, another pair of whole numbers q (the quotient) and r (the remainder), so that b = qd + r

where 0 ≤ r < d.

Some subtle points about the concept of division among whole numbers will be brieﬂy recalled at the beginning of Section 1.5 of [Wu2020a]. A detailed exposition of the concept of “division” among whole numbers is given in Chapter 7 of [Wu2011a]. Note that 0 is included among the whole numbers. A knowledge of negative numbers, particularly integers, is not assumed. Negative numbers will be developed ab initio in Chapter 2 of [Wu2020a].

Because every assertion in these three volumes (this volume, together with [Wu2020a] and [Wu2020c]) will be proved, students should be comfortable with mathematical reasoning. It is hoped that as they progress through the volumes, all students will become increasingly at ease with proofs. In terms of the undergraduate curriculum, readers of this volume—as a rule of thumb—should have already taken the usual two years of college calculus or their equivalents.

1 Unfortunately, a correct exposition of this topic is diﬃcult to come by. Try Chapter 7 of [Wu2011a].

xxxvii

Some Conventions • Each chapter is divided into sections. Titles of the sections are given at the beginning of each section as well as in the table of contents. Each section (with few exceptions) is divided into subsections; a list of the subsections in each section—together with a summary of the section in italics—is given at the beginning of each section. • When a new concept is ﬁrst deﬁned, it appears in boldface but is not often accorded a separate paragraph of its own. For example: A function f : D → T is said to be injective (or one-toone) if for two distinct elements d and d of D, we always have f (d) = f (d ) (p. 5). You will have to look for many deﬁnitions in the text proper. (However, not all boldfaced words or phrases signify new concepts to be deﬁned, because boldface fonts are sometimes used for emphasis.) • When a new notation is ﬁrst introduced, it also appears in boldface. For example: If the function is denoted by f , then f : D → T is the correct notation to capture this information (p. 2). • Equations are labeled with (decimal) numbers inside parentheses, and the ﬁrst digit of the label indicates the chapter in which the equation can be found. For example, the “(1.17)” in the sentence “Thus (1.17) implies that . . . ” means the 17th labeled equation in Chapter 1. • Exercises are located at the end of each section. • Bibliographic citations are labeled with the name of the author(s) inside square brackets, e.g., [Ginsburg]. The bibliography begins on page 363. • In the index, if a term is deﬁned on a certain page, that page will be in italics. For example, the item linear programming, 57, 59, 73, 76 means that the term “linear programming” appears in a signiﬁcant way on all four pages, but the deﬁnition of the term is on page 59.

xxxix

CHAPTER 1

Linear Functions The main goals of this chapter are to introduce the concept of a function and to study the most elementary functions of one or two variables in school mathematics, namely, the linear functions. The present discussion of linear functions of two variables will be quite limited, and it will be given in the context of the graphs of linear inequalities in the plane and linear programming. Beyond supplying the most rudimentary information about functions, this chapter aims to make a contribution of a diﬀerent kind. At the moment, although the term “function” and the symbol of a function are formally introduced in middle and high school, the concept of a function is not seriously taught, probably because the idea of a function is known to be notoriously elusive to students. For example, if a function f (x) = 3x2 + 1 is given, the fact that this f is an assignment to each x a number 3x2 + 1 is pretty much ignored; instead, students are told to look at the equation y = 3x2 + 1 and its graph and to memorize that the latter satisﬁes the vertical line rule. Consequently, many students have no conception of what a “function” is and they only know about “equations” in two variables. Because TSM1 rarely addresses the question of why the function concept deserves to be learned, there is little incentive to teach this diﬃcult concept (compare the discussion of purposefulness in To the Instructor on pp. xix ﬀ.). It therefore comes to pass that since “equations” are mostly adequate for standardized tests, equations have generally replaced functions in TSM classrooms. Unfortunately, because functions are part of the basic vocabulary of higher mathematics and science, this failing of TSM in eﬀect prevents students from accessing STEM subjects in college. If we want to correct this disastrous trend, we have to explain why functions are important and actually demonstrate their importance in our mathematical discussions. With this in mind, we will devote Section 1.2 to a serious discussion of the raison d’ˆetre of functions. Furthermore, the ﬁrst four chapters of this volume should give ample evidence that functions will be indispensable in mathematics from this point onward. All of the third volume, [Wu2020c], is in fact about functions. 1.1. Deﬁnition of a function and its graph This section introduces the concept of a function. After giving some examples of functions, we take up the basic concepts of injectivity, surjectivity, equal functions, composition of functions, and graphs of functions. We call special attention to the importance of getting beginners to learn to graph real-valued functions of one variable by hand. We also comment on the subtlety of the common but misleading saying in TSM about “representing a function by its graph”. 1 See

page xi for a deﬁnition of TSM. 1

2

1. LINEAR FUNCTIONS

Basic deﬁnitions and examples (p. 2) Graphs of functions (p. 6) Some remarks on the subtleties of graphing functions (p. 10) Basic deﬁnitions and examples We ﬁrst recall some basic notation. We already have denoted the whole numbers, the integers, and the rational numbers by N, Z, and Q,, respectively. In Section 6.3 of [Wu2020a], we added to the list by using R to denote the number line, and we refer to the elements of R as the real numbers. Further recall that, with a pair of coordinate axes assumed to be ﬁxed, the plane is denoted by R2 . A function from a set D to a set T is a rule (i.e., a precise prescription)2 that assigns to each element of D exactly one element of T . Sometimes we also say a function is a rule that associates with each element of D an element of T . If the function is denoted by f , then f : D → T is the correct notation to capture this information. We then also say f is deﬁned on D and takes value in T . The set D is called the domain (of deﬁnition) of the function f , but the name for T is more problematic: the words range, image, codomain, and target have all been used for this purpose in the literature. The lack of universal agreement means that nothing should be taken for granted in this situation, and you are advised to check carefully in each case. In this volume, we resolve this confusion by not calling T anything other than the symbol T . We will denote by f (D) the set of all elements in T that are equal to f (x) for some element x in the domain D, and we will refer to f (D) as the image of D under f (as in Section 4.3 of [Wu2020a]). When D and T are understood, the function is often denoted generically by f or f (x) or, on occasion, by x → f (x). If f assigns to an element a of the set D the element b of T , then we write f (a) = b or, sometimes, a → b when the function f is understood. It is also common to say that f maps a to b. For example, if s is the function s : R → R which assigns to each number its square, then s can be succinctly given as s(x) = x2 for every number x. We note for emphasis that, since s(x) is always ≥ 0 for any x, we could equally well write s : R → {all numbers ≥ 0} so that D = R and T = {all numbers ≥ 0}. However, the function c : R → R deﬁned by c(x) = x3 for each number x cannot be described as c : R → {all numbers ≥ 0} because, since c(−2) = −8, c cannot assign to the number −2 any number from the set of {all numbers ≥ 0}. For a function f : D → T , if there is an element r0 in T so that f (d) = r0 for every element d of D, then we say f is a constant function. The concept of equality is being scrutinized in the education research literature3 as it seems to be a source of mystery as far as school mathematics education is concerned. For this reason if for no other, we should be explicit about the deﬁnition 2 In case you cringe upon setting your eyes on such an “imprecise” deﬁnition of a function, just hold on and read two more pages before you decide to give up. 3 As of 2020.

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

3

of the equality of two functions. Two functions f : D → T and g : D → T are said to be equal as functions, and we write f = g if they have the same domain of deﬁnition (D = D in this case), take values in the same set (T = T in this case), and for each x ∈ D, f (x) = g(x). (Recall that the symbol “∈” means belongs to or belonging to.) Observe that, by deﬁnition, one checks whether or not f = g holds for two given functions f and g by ﬁxing an x in their common domain D and checking whether the two elements f (x) and g(x) in T are equal and then doing the same for every element x ∈ D. The point we wish to make is that one checks this equality one ﬁxed element at a time; nothing ever “varies”. Thus students need not worry about the popular but erroneous notion that one must deal with a “variable” as “a quantity that changes or varies” when discussing functions. It is always a matter of one ﬁxed element at a time. This vital message should be made clear to all students from year 2020 forward. Before proceeding with the discussion of examples of functions, a few comments about the teaching of the concept of a function in the school classroom will be relevant. The formal deﬁnition of a function f : D → T is that it is a set of ordered pairs (d, r), where d ∈ D and r ∈ T , with the additional property that if (d, r) and (d, r ) are both elements in this set, then r = r . So formally, a function is not a “rule” but a collection of elements {(d, r)} where d ∈ D and r ∈ T . The formal statement that “(d, r) belongs to f ” then corresponds to our less formal statement above that “f (d) = r”. In addition, the last condition—that if (d, r) and (d, r ) are both elements in this set, then r = r —then corresponds to the less formal statement above that “f assigns to d exactly one element of T ”; i.e., if f (d) = r and also f (d) = r , then necessarily r = r .4 The reason for this requirement of the uniqueness of r in f (d) = r in the deﬁnition of a function becomes entirely understandable when we put it in a real-world context. For example, suppose D and T are the real numbers R and the function f (t) is the distance of a spaceship from earth t hours after launching. Imagine how the engineers in Kennedy Space Center would react if at a certain time t0 the information they get about f (t0 )— instead of being a single unambiguous number—is two numbers: 9 × 107 (miles) and 2 × 108 (miles). So which of the two numbers is the distance of the spaceship from earth!? From a strictly mathematical point of view, the preceding formal deﬁnition as a set of ordered pairs is the only correct deﬁnition of a function because it is entirely unambiguous and does not involve any “rule” or “assignment”. Hence, students should be exposed to this formal deﬁnition at least once, if only for the sake of their mathematical education. However, this deﬁnition is so turgid that few school students would ﬁnd it satisfying or even palatable.5 Our recommendation is therefore that this deﬁnition should be soft-pedaled, because schools are not the place to strive for maximum generality or impose an unnecessary formalism at the risk of turning oﬀ the majority of the students.6 Along these lines, there is a more general concept called a relation on D × T , which is just a set of ordered pairs (d, r) with d ∈ D and r ∈ T but has no further restrictions. For school mathematics, 4 In

terms of calculus, this is the precise way of stating the vertical line rule. See page 7. is somewhat analogous to the deﬁnition of a rational number as an equivalence class of ordered pairs of integers, where the equivalence relation is basically the cross-multiplication algorithm. It is entirely correct, but one would not want to use it in a ﬁfth-grade classroom. 6 Keep the New Math in mind. 5 This

4

1. LINEAR FUNCTIONS

the concept of a relation is not of fundamental importance. It may be mentioned for the purpose of providing contrast, but one should certainly not go into the details, as many algebra textbooks do by rhapsodizing about the “domain and range of a relation”. This kind of nonessential formalism at the school level should be minimized. Most functions in school mathematics are deﬁned on R and take values also in R, and they are also given by speciﬁc symbolic expressions, such as the function s(x) = x2 above. In general, a function f : I → R, where I is a subset of R (thus I could be R itself), is called a real-valued function of one variable. An important class of real-valued functions of one variable is the collection of polynomial functions; i.e., f : R → R so that f (x) = p(x) for all x ∈ R, where p(x) is a polynomial in x (see page 354 for the deﬁnition of a polynomial). A common abuse of language is to conﬂate a “polynomial function” with a “polynomial”. In case the degree of the polynomial p(x) is 2, this f is called either a quadratic function or a quadratic polynomial, and if the degree of p(x) is 3, f is called either a cubic function or a cubic polynomial. There is some value, however, in recognizing from the outset that some functions cannot be described in symbols. Consider, for example, the function G : D → T whose domain of deﬁnition D consists of a deck of cards and with T consisting of four elements which are the possible suits of a card, so that G : {a deck of cards} → {club, diamond, heart, spade} and G assigns to each card its suit. Then what G does to each card would be diﬃcult to describe in symbols. One way is to tabulate all possible assignments of G. The beginning and the end of the tabulation would then look something like this: Ace of spades spade Ace of clubs club Ace of hearts heart Ace of diamonds diamond Two of spades spade Two of clubs club .. .. . . King of hearts heart King of diamonds diamond The deﬁnition of this function G, strictly speaking, can only be described by the complete tabulation above. A little thought would reveal that such a tabulation could equally well be given by a listing of 52 ordered pairs: (Ace of spades, spade), (Ace of clubs, club), (Ace of hearts, heart), . . . , (King of hearts, heart), (King of diamonds, diamond). From this example, one gets the essential idea why the formal deﬁnition of a function is in terms of a set of ordered pairs. In the case of the function G above, 52 ordered pairs suﬃce to pin it down completely. In the general case of a function such as f : R → R, however, it would take an (uncountably) inﬁnite number of ordered pairs to describe f completely if we insist on a tabulation of all possible ordered pairs. This is one reason why functions are almost never deﬁned by a table of values but are deﬁned instead by symbolic methods. It is for this reason that the

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

5

glib phrase in TSM, to the eﬀect that “a function can be deﬁned by a table of values”, should be avoided no matter what TSM tells you. This point will resurface in the discussion below. The terminology of a “function” may be new to some readers, but in Section 4.2 of [Wu2020a], we already encountered some examples of functions: transformations of the plane are just functions from the plane R2 to R2 . In particular, all translations, rotations, and reﬂections are functions deﬁned on the plane and taking value in the plane. In Section 1.6 of [Wu2020c], we will describe all the basic isometries symbolically in terms of coordinates, but such formulas are, in general, not particularly useful except in the case of translations, rotations around the origin of a coordinate system, and reﬂections across the coordinate axes or the lines x ± y = 0; see Lemma 6.20 in Section 6.6 of [Wu2020a] (which is also stated on page 358 of this volume) as well as Exercises 8–10 of the same section. There is still another kind of function that you are already familiar with but which is nevertheless diﬃcult to describe in symbols. Let A = {the ﬁrst eight letters of the alphabet, a, b, c, d, e, f, g, h}, and let τ : A → A be a function that “rearranges” the eight letters of A; e.g., τ may be the following function: τ (a) = b, τ (b) = c, τ (c) = d, τ (d) = a, τ (e) = g,

τ (f ) = h,

τ (g) = f,

τ (h) = e.

It is clear that there is no simple symbolic description of such a τ . The τ above is an example of what is called a permutation of A. We proceed to formally deﬁne a permutation in general by using terminology that is an obvious extension of what we have already deﬁned for transformations of the plane (see Section 4.3 of [Wu2020a]). A function f : D → T is said to be injective (or oneto-one) if for two distinct elements d and d of D, we always have f (d) = f (d ). We also say f is surjective (or onto) if every element r of T is equal to f (d) for some element d in D. A function that is both injective and surjective is said to be bijective. With this terminology understood, we then deﬁne—for the above set A consisting of eight letters—a permutation of A to be any bijective function A → A. This terminology of a permutation is in fact also used for any ﬁnite set, in particular for the set Nn consisting of the integers from 1 to n (n is a positive integer). A permutation of Nn is usually called a permutation on n letters. We can also deﬁne the composition of functions as in Section 4.3 of [Wu2020a]. Given functions f : D → T and g : T → S, we deﬁne the composite function g ◦ f to be the function g ◦ f : D → S so that for every d of D, (g ◦ f )(d) = g(f (d)). Schematically, we have D d

f

- T -f (d)

g

- S - g(f (d))

Notice that because f (d) is an element of T , which is the domain of g, it makes sense to talk about g(f (d)). We leave as an exercise (Exercise 6 on page 15) the proof of the analog of Theorem 4.2 in Section 4.2 of [Wu2020a], to the eﬀect that a function f : D → T is bijective if and only if there is a function g : T → D so that f ◦ g = idT and g ◦ f = idD , where idD and idT are the identity functions on D and T , respectively; i.e., by deﬁnition, idD (x) = x for every x ∈ D and idT (t) = t

6

1. LINEAR FUNCTIONS

for every t ∈ T . In this case, we say g and f are each other’s inverse functions. We will have much more to say about inverse functions on pp. 162 ﬀ. Mathematical Aside: The study of permutations is a basic part of algebra (speciﬁcally, ﬁnite groups), ﬁnite probability, and combinatorial mathematics, among many other ﬁelds. Other than real-valued functions of one variable, the functions one encounters in schools most often are those deﬁned on a subset of the plane and taking value in R. We call these real-valued functions of two variables, where the “two” refers to the number of coordinates in the coordinate plane. Thus one is tempted to concentrate on real-valued functions of one or two variables in school mathematics, but there are good reasons why we should not overdo it. Consider, for example, the purchase of a textbook that costs $65. If we buy n books for a positive integer n, then the cost is $(65n). Therefore we have a cost function C(n), the total cost of n books, which is clearly deﬁned on all the whole numbers N, so that C : N → R and C(n) = 65n. In this case, it would not do to pretend (as TSM generally does) that the domain of deﬁnition of C is the x-axis because C(3.24) has no meaning (what could “3.24 textbooks” mean, for example?). Here then is a function that arises from “real-world” considerations whose domain of deﬁnition is nevertheless N and not R. Of course the function F : R → R so that F (x) = 65x makes sense for any real number x, and furthermore, F (n) = C(n) for any whole number n. If one wishes, one may choose to use the function F in place of the function C for ease of discussion, but if this is done, the distinction between F and C should still be clearly drawn. Graphs of functions We next take up the concepts of graphs of functions and graphs of equations, as there is apparently some confusion about the relationship between these two in school mathematics. Let f : D → T be a function as usual, where D and T are arbitrary sets. We deﬁne the direct product D × T to be the set of all ordered pairs (a, b), where a ∈ D and b ∈ T . If D = T = R, then we recognize that R × R is nothing but the coordinate plane R2 . In general, the graph of a function f : D → T is deﬁned to be the subset in D × T consisting of all the ordered pairs of the form (a, f (a)), where a is any element of the domain D. (It should not escape your attention that the graph of a function coincides with the formal deﬁnition of a function given on page 3.) Again, if we have a real-valued function of one variable f : I → R, where I ⊂ R (recall: “⊂” means “is contained in”), then the graph of f lies in R2 , and it consists of the collection of all ordered pairs of numbers (x, f (x)) where x ∈ I. We have consistently stressed the importance of deﬁnitions in these volumes, and one reason is that the precise deﬁnition of something as basic as the graph of a function is often not even found in TSM. A vivid illustration of TSM’s neglect is given by Dan Meyer in his blog [Meyer]. He wrote: I left high school adept at graphing functions. I could complete the square and change forms easily. I knew how to identify the asymptotes, holes, and limiting behavior of those thorny rational expressions. But it wasn’t until I had graduated university

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

7

math and was several years into teaching that I really, really understood that the graph is a picture of all the points that make the function true. This was diﬃcult for me because graphs don’t often look like a bunch of points. They look like a line.7 This quote, together with readers’ responses in [Meyer], shows clearly how TSM has ruined school mathematics education by denying students the most basic information such as the fact that the graph of a function f is, by deﬁnition, the totality of all the points {(x, f (x))}, where x belongs to the domain of f . Incidentally, the deﬁnition of a function implies that if G is the graph of a function f : I → R where I ⊂ R, then G satisﬁes the vertical line rule, to the eﬀect that no vertical line can intersect G at more than one point. Indeed, suppose a vertical line x = c intersects G at two points, (c, y) and (c, y ). Since (c, y) lies in G, we have by the deﬁnition of G that y = f (c). Likewise, (c, y ) being in G implies that y = f (c). However, f can assign to c only one element, by the deﬁnition of a function, so y = y and therefore (c, y) and (c, y ) coincide, as desired. There is little or no mystery to the concept of the graph of a function, at least not when the function is a real-valued function of one variable. Just follow the precise deﬁnition and plot as many points as you like in the coordinate plane to get a feel for what the graph looks like. The importance of actually plotting points on the graph by hand cannot be overstated, and this is especially true in the age of aﬀordable graphing calculators. The cumulative experience of many in the mathematics community is that the learning that comes from the act of doing something with one’s ﬁngers is, in a mysterious way, transmitted to the brain. This kind of learning does not seem to be replaceable by technology. Be sure to insist that your students get used to plotting points on the graph. For example, take the square function s : R → R, s(x) = x2 for all numbers x. The graph G of s consists of all the points of the form (x, x2 ), where x is arbitrary. Since (−x)2 = x2 , we see that G includes both (x, x2 ) and (−x, x2 ), no matter what x may be. This means the graph of s has bilateral symmetry (see page 352 for the deﬁnition) with respect to the y-axis. The point (0, 0) is an obvious point on the graph. We can put in values of x = ±1, ±2, ±3 to get the points (±1, 1),

(±2, 4),

(±3, 9).

Let us also throw in (±0.5, 0.25),

(±1.5, 2.25), (±2.5, 6.25),

(±3.5, 12.25)

for good measure, and we get a sequence of points on G; see the ﬁgure on the left below. Note that in order to make the graph manageable, we have shrunk the scale of the y-axis by a factor of 4. It is not diﬃcult to extrapolate from these points to envision the graph itself, which is the curve on the right. This curve is an example of a parabola. See Sections 2.2 and 2.3 below for a deﬁnition as well as a fuller discussion of parabolas. 7 Most

likely, he meant to say, “They look like a curve.”

8

1. LINEAR FUNCTIONS

r

r

16

r r

−4

r

−2

6

r

r r r2 r r O

r

r

r

r

−4

r

10

r

4

r

14

r

2

r

16

r

10

r

r

r

14

r

6

r

−2

r r r2 r r O

r

r

r

2

4

There are some functions that are naturally associated with the function s (or in fact with any given function) that will be important in Section 2.1 below. We refer to these loosely as the cognate functions of s and discuss three of them brieﬂy. The simplest is the function s1 deﬁned by s1 (x) = s(x)+q, where q is a constant. For deﬁniteness, let q = 2 so that s1 (x) = x2 + 2. The graph G1 of s1 is the set of all points of the form (x, x2 + 2), where x is any number. Recall that the graph G of s is the set of all points of the form (x, x2 ) (where x is any number). Therefore, if TV is the translation along the vector from the origin O to the point (0, 2)—so that TV (x, y) = (x, y + 2) for any numbers x and y (Lemma 6.20 of [Wu2020a]; see page 358 of this volume)—then G1 = TV (G);

(1.1)

i.e., G1 the image of G by a vertical translation. This is shown in the left ﬁgure below. G1

10

10

G

G 6

G2

6

2

2

6 6 −2

O

−2

2

O

2

4

Next, consider the function s2 deﬁned by s2 (x) = (x − p)2 . Again, for deﬁniteness, let us ﬁx p to be 3, so that s2 (x) = (x − 3)2 . Now the graph G2 of s2 is the set of all points of the form (x, (x − 3)2 ), where x is any number. If we plot the graph of G2 (see the right ﬁgure above), it suggests that G2 can be obtained from G by a horizontal translation. To achieve greater clarity, we can look at the points of G2 in the following way. Let t = x − 3; then G2 is the set of all points of the form (t + 3, t2 ), where t is any number, or equivalently, G2 is the set of all points of the form (x + 3, x2 ), where x is any number. Since the graph G of s is the set of all points of the form (x, x2 ), it is then clear that if we deﬁne TH to be the horizontal translation along the vector from (0, 0) to (3, 0), then again by Lemma 6.20 of [Wu2020a], TH maps a point (x, y) to the point (x + 3, y) (for any numbers x and y). Now we see as before that (1.2)

G2 = TH (G).

Caution. While s2 = (x − 3) is obtained from s(x) = x2 by subtracting 3 from x, the horizontal translation TH in (1.2), TH (x, y) = (x + 3, y), is one that adds 3 to the x-coordinate. 2

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

9

Another function associated with the function s(x) = x2 is the function s3 (x) = cx for some nonzero constant c. Let G3 be the graph of s3 . Then the relationship between the graph G of s(x) and G3 is less straightforward and can only be described on a point-by-point basis. For each point (x, s(x)) on the graph G of s(x), we will describe the location of the point (x, cs(x)) on G3 . If s(x) = 0, then of course (x, s(x)) = (x, cs(x)). Now suppose s(x) = 0. First suppose c > 1, then (x, cs(x)) lies above (c, s(x)), as in the left ﬁgure below. If 0 < c < 1, however, then (x, cs(x)) lies below (x, s(x)) but is still above the x-axis, as in the middle ﬁgure below. If c < 0, then (x, cs(x)) lies below the x-axis, and its distance to the x-axis is |c| times the distance of (x, s(x)) to the x-axis, as in the right ﬁgure below. 2

r(x, s(x))

r(x, s(x))

q(x, cs(x))

x

O

q(x, cs(x)) r(x, s(x)) x

O

q(x, cs(x)) x

O

Here is an example of the graphs of s(x) and cs(x) for c = 14 ; see the left picture below. Observe that each vertical thickened segment is 14 of the length of the thin vertical segment containing it. x2

4

x2 − 1

3

3

2

r 2

1 1 2 4x

1

−2

r

−1

x r

1 (x2 4

− 1)

O 1

2

r −2

−1

O

1

x

2

−1

For a given function f (x), we can likewise consider the graph of cf (x) for a nonzero constant c. The complexity of the relationship between the graphs of f (x) and cf (x) can be readily seen from the right picture above, where f (x) = x2 − 1 and c = 14 . The graph of f (x) is now seen to be sometimes below that of 14 f (x) and sometimes above it. In general, let f : R → R be a function and b a number. As with s(x), let G denote the graph of f and G1 denote the graph of the function f1 deﬁned by f1 (x) = f (x) + q. Then G1 is similarly obtained from G by a vertical translation. Precisely, we have (1.3)

G1 = TV (G),

10

1. LINEAR FUNCTIONS

where TV is the vertical translation along the vector from the origin O to (0, q); i.e., TV (x, y) = (x, y + q) for all (x, y). In a similar manner, if G2 denotes the graph of the function f2 deﬁned by f2 (x) = f (x − p), then G2 is obtained from G by a horizontal translation. Precisely, G2 = TH (G),

(1.4)

where TH is the horizontal translation along the vector from the origin O to (p, 0); i.e., TH (x, y) = (x + p, y) for all (x, y). As we have just noted, the relationship between the graphs of f (x) and cf (x) (for a nonzero constant c) cannot be described by a single basic isometry as in (1.3) and (1.4). The information encoded in (1.3) and (1.4) about the relationships between the graph of a function f (x) and the graphs of the cognate functions f (x) + q and f (x − p) for ﬁxed constants q and p is standard fare in school mathematics. However, such a discussion does not carry much substance when it is presented as a collection of rote skills without a persuasive mathematical context. Pedagogically, it becomes more meaningful to students when it is presented with a mathematical purpose, such as in Lemma 2.2 on page 68 and Theorem 2.6 on page 73 in the study of quadratic functions. Compare the discussion of purposefulness in To the Instructor on pp. xix ﬀ. Some remarks on the subtleties of graphing functions With the availability of scientiﬁc calculators, teachers should not hesitate to ask students to graph quite sophisticated functions, e.g., a function such as x −→

x4 − 7x + 5 . x2 + 2

Let us illustrate with a simpler function such as G : R → R given by G(x) = x3 − 3x + 6, which is a cubic polynomial, or more simply, a cubic. A key issue in the graphing of a function deﬁned on all of R is that, R being of inﬁnite length, it is impossible to graph the whole function. We have to be selective and try to graph it only on a “well-chosen” (ﬁnite) closed interval [a, b] (i.e., all the points x satisfying a ≤ x ≤ b).8 Naturally, which interval to select will be critical, and we will address this issue presently. Since we have no idea of what to expect, we begin by looking at the graph of G(x) = x3 − 3x + 6 on the closed interval [−4, 4]. One reason for this choice is that G(−4) = −46 while G(4) = 58, so at least intuitively, the values of G on [−4, 4] seem to comprise a good range from a negative number to a positive number. Let us get the value of G at some obvious numbers, e.g., G(0), G(±1), G(±2), G(±3), G(±4), obtaining the following points on the graph of G: (−4, −46), (−3, −12), (−2, 4), (−1, 8), (1, 4),

(2, 8),

(3, 24),

(0, 6),

(4, 58).

8 Up to this point, we have mostly referred to [a, b] as a segment, but gradually we will have to transition to the more standard terminology of closed intervals.

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

11

Because the jumps in the values between G(−4) and G(−3), G(−3) and G(−2), G(2) and G(3), and ﬁnally G(3) and G(4) are so great, we will also plot the following additional points on the graph of G to ﬁll in the picture: (−3.5, −26.375), (−2.5, −2.125),

(2.5, 14.125), (3.5, 38.375).

By compressing the y-axis, we can exhibit these points as follows: r

60

r

40

−4

r

r

−3r −2

−1

r

r r

r

20

O

r

r

1

2

r

r 3

4

−20 −40 −60

The graph seems to cross the x-axis between −3 and −2. Suppose it crosses the x-axis at (x0 , 0); then 0 = G(x0 ) by the deﬁnition of the graph of G. This means x30 − 3x0 + 6 = 0. Such an x0 is called a root or a zero of the cubic polynomial equation x3 − 3x + 6 = 0. In general, a number c is a zero of the polynomial P (x) or a root of the polynomial equation P (x) = 0 if P (c) = 0. Sometimes we also refer to such a c as a root of the polynomial P (x). The roots of a polynomial equation are of great interest in mathematics (see Section 2.1 below as well as Sections 3.1–3.2 and Section 5.3 on the roots of polynomial equations, quadratic or otherwise). For this reason, we may try to get a better estimate of this x0 . We have G(−2.5) = −2.125, G(−2.4) = −0.624, G(−2.3) = 0.733, so it is intuitively clear that x0 is between −2.4 and −2.3. By experimenting with G(−2.31), etc., we can get even better estimates of this root if we so wish. Now we face the critical question: does the graph of G(x) = x3 − 3x + 6 over the closed interval [−4, 4], in some sense, “capture all the important information” about G? The answer is aﬃrmative, but the reason is far from simple. We can explain this intuitively by pointing to the fact that the graph of G on [−4, 4] has a “peak” above −1 (roughly) and has a “valley” above 1 (roughly). It is a known fact, which can be proved using calculus (but see the discussion in Section 3.1 of Chapter 3), that the graph of a cubic polynomial can have at most one “peak” and one “valley”, and if the “peak” is to the left of the “valley”, then the graph of the function will go down as one moves to the left of the “peak”, and go up as one moves to the right of the “valley”. Since all the “peaks” and “valleys” of this particular cubic G are already displayed in its graph over [−4, 4], we are assured that the graph will continue to go down as we go to the left of −4 on the x-axis, and continue to go up as we go to the right of +4 on the x-axis. In particular, nothing of interest happens outside the interval [−4, 4]. Therefore the theory suggests that, in a sense, the graph of G over the ﬁnite interval [−4, 4] does “capture all the important information” about G.

12

1. LINEAR FUNCTIONS

To put the theory to the test, we now graph the same function G(x) = x3 − 3x + 6 over the larger interval [−10, 10], as shown. One can see that, indeed, it has nothing essential to add to the graph of G over [−4, 4]. (The reason the part of the graph on the right over the interval [−4, 4] seems to look diﬀerent from the preceding graph of G over the same interval [−4, 4] is that the former is extremely compressed in the vertical direction.) We now come to the main point in this preliminary discussion of graphing functions. In TSM, it is usually blandly stated that “you can use a graph to represent a function”. Here, a “function” is understood to refer a real-valued function of one variable. What “represent” means is of course never deﬁned, but the implicit message is unmistakable that “graphing” such a function is something so routine that it is not worth discussing: just draw any graph and you will get a “representation” of the given function. The truth is that it is a nontrivial matter to “graph” a function that can truly “represent it” if the latter is taken to mean “exhibit all the salient characteristics of the function”. Consider the cubic f (x) = x3 − 100x, for example. Suppose we also graph f over the same closed interval [−4, 4]. What we get is the left picture below, which strongly suggests f will continue to rise as we move left beyond −4, and it will continue to fall as we move right beyond +4.

However, the behavior of the function f (x) = x3 − 100x over R is badly represented by its behavior over [−4, 4] because, by looking at the graph of f over a larger interval such as [−12, 12], we already get a completely diﬀerent picture; see the right picture above. It can be shown, by a reasoning similar to the preceding case of G, that the latter graph does give a faithful representation of the graph of f on all of R. Speciﬁcally, the graph will continue to fall as we move left beyond −12 on the x-axis, and it will continue to rise as we move right beyond +12 on the x-axis. The moral is that if we want the graph of a function to “represent” it well, there are serious mathematical issues to address. So far, we have only looked at cubics, which are relatively simple objects compared with polynomial functions of degree > 3, rational functions (see Section 3.3), and other functions soon to be introduced (exponential, logarithmic, trigonometric, etc.). One can easily imagine that the issue of whether the graph of a general

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

13

function over a chosen ﬁnite closed interval is “representative” of the function over the number line R will be even more complicated. A substantial mathematical knowledge about functions would be required—and most of this knowledge requires calculus—to be able to choose a ﬁnite interval [a, b] over which the graph of a given function f manages to display all the key features of f deﬁned on R. The importance of getting to know the graph of a function is beyond dispute (see this and the next three chapters, especially page 63 about graphs of quadratic functions), but we cannot carry on this TSM tradition of misleading students with such a glib statement without a proper explanation about the mathematical details that are involved. Please be aware of this fact when you teach functions. Next, we introduce a diﬀerent way of looking at the graph of a real-valued function of one variable. We have just seen that the graph of a function f : R → R is the subset of the plane R2 consisting of all the ordered pairs (x, f (x)) where x ∈ R. Recall that the graph of an equation in two variables, h(x, y) = 0 where h(x, y) is some expression in the numbers x and y, consists of all the points (x, y) in the plane R2 that satisfy h(x, y) = 0 (see page 353). That said, it is time to point out that, with f : R → R given, the graph of the equation y − f (x) = 0 in two variables consists of all the points (x, f (x)) and therefore coincides with the graph of the function f .9 Therefore we have: The graph of the function f : R → R is equal to the graph of the equation y − f (x) = 0 deﬁned on R2 , where x, y ∈ R. Because this is the ﬁrst instance where we relate the graph of a function to the graph of an equation, let us point out the key feature of this relationship. The graph of the function f : R → R is not a subset of R, but a subset of the plane R2 . On the other hand, the graph of the equation y − f (x) = 0 deﬁned on the plane R2 is a subset of the plane R2 itself. The fact that both graphs are subsets of the plane R2 is the reason we can talk about them being equal. For illustration, consider the function of one variable h : R → R deﬁned by h(x) = 2x + 1for any real number x. The graph of the function h is then all the points in the plane R2 of the form (x, 2x + 1) where x is an arbitrary real number. On the other hand, consider the equation in two variables, y = 2x + 1, which is of course the same as the set of all the points (x, 2x + 1) where x is an arbitrary number. Thus the graphs of the function h and the equation y = h(x) are the same subset of R2 . (Although we know from Section 6.5 of [Wu2020a] that the set of all the points (x, 2x + 1) is a line, this fact plays no role in the present consideration.) We have just shown that the graph of a real-valued function of one variable is always the graph of a certain equation in two variables. The converse is not true, however. There are many equations in two variables whose graphs are not graphs of any function of one variable. See Exercise 8 on page 15 for some simple examples. The discussion for functions of one or two variables which are deﬁned only on subsets of R or R2 is of course entirely analogous. 9 As noted on page 14 below, we are in the process of expanding the deﬁnition of an expression to include numbers obtainable by the usual operations on not only a collection of numbers but also values of given functions at these numbers. In this sense, y − f (x) = 0 is an equation since it is an equality between the expression “y − f (x)” and the expression “0”.

14

1. LINEAR FUNCTIONS

Among functions of one or two variables, the linear ones are the simplest, and also the most basic. A real-valued function f : R → R is said to be a linear function of one variable if it is of the form f (x) = ax + b for all numbers x, where a, b are constants. In the terminology of Section 6.1 in [Wu2020a], a linear function is a linear polynomial of one variable (or more correctly, “the expression that deﬁnes a linear function is a linear polynomial of one variable”, but it is hard to ﬁnd anyone who talks like that). The graph of such a linear function f in R2 is the graph of the equation y = ax + b, i.e., the graph of the equation ax − y = −b, which we have already seen is a line. Thus the graphs of real-valued linear functions of one variable are lines in the plane. This is why these functions are called “linear” functions. We deﬁne a linear function of two variables : R2 → R to be a function of the form (x, y) = ax + by + d, where a, b, d are arbitrary constants. Unlike the deﬁnition of a linear equation in two variables,10 there is no restriction in this deﬁnition on a and b: they could both be 0, in which case is the constant function which assigns to every ordered pair (x, y) the number d. We have already seen a very natural example of a real-valued function of two variables that emerges from a linear equation of two variables ax + by = c, namely, the linear function of two variables : R2 → R deﬁned by (x, y) = ax + by − c. As noted, the graph of the equation (x, y) = 0 for a linear function in two variables is a line. For the purpose of a later discussion (see page 33), we will very brieﬂy take up the case of the graph of a function of two variables. In case g : R2 → R is a real-valued function of two variables, the deﬁnition on page 6 implies that the graph of g is the set of all ordered pairs ((x, y), g(x, y)) in R2 × R, where x, y ∈ R. Note that the direct product R2 × R may be considered to be the triple direct product R × R × R consisting of all ordered triples (x, y, z), where x, y, and z are any numbers. The latter is usually denoted by R3 . Therefore the graph of g(x, y) will be considered as a subset of R3 consisting of all the ordered triples (x, y, g(x, y)), where x, y ∈ R. You may recall from multi-variable calculus that the graph of a “nice” function g(x, y) is a surface in R3 . Furthermore, the graph of the function g(x, y) is also identical to the graph of the equation z − g(x, y) = 0 deﬁned on R3 , for exactly the same reason as the one-variable case above. Just as the graph of a linear function of one variable is a line in the plane, we also know from multi-variable calculus that if the function g(x, y) is linear, then its graph is a plane in R3 . We would like to make a general comment about linear functions of one or two variables, which will be the subjects of discussion in Sections 1.3–1.5. We spend a lot of time in mathematics discussing linear functions of one kind or another, but in the real world, it is rare that a function is ever exactly linear. The importance of linear functions therefore lies less in their ability to precisely model the real world, but more in the fact that they are suﬃciently simple so that we can get to know a lot about them and can use them to approximate more complicated functions. The study of linear functions is, for this reason, truly basic. Before leaving the general topic of functions, we tie up a loose end from Section 6.1 of [Wu2020a] by expanding the deﬁnition of a number expression, or more simply an expression, in a given collection of numbers x, y, . . . , w: it is a number 10 Recall that, by the deﬁnition of a linear equation of two variables, at least one of a or b is nonzero.

1.1. DEFINITION OF A FUNCTION AND ITS GRAPH

15

obtained from these x, y, . . . , w, from a collection of speciﬁc real numbers and from the values of a given collection of functions f , . . . , h at these x, y, . . . , w using a speciﬁc combination of arithmetic operations (i.e., +, −, ×, ÷). For example, if we know what the exponential function 2x means (see Theorem 4.1 on page 141), then 13 · 25x − 2−x + 8 would be considered an expression in the number x.

Exercises 1.1. (1) Let f (x) = x. Compare the graph of f (x) with that of (a) G(x) = f (x)+3, (b) g(x) = f (x + 3), (c) H(x) = f (x) − 2, (d) h(x) = f (x − 2), (e) K(x) = 3f (x), (f) F (x) = 13 f (x) + 4, (g) k(x) = −2f (x), (h) (x) = −2f (x) + 1. (2) Let f (x) = x2 . Repeat Exercise 1. (3) Let f (x) = x3 . Repeat Exercise 1. (4) If f (x − 4) = 13 (x + 7)3 − 1, what is f (x)?11 (5) Let f : D → T and g : T → S be bijective functions between sets D, T , and S. Prove that the composition g ◦ f : D → S is also bijective. (6) Prove that a function f : D → T is bijective if and only if there is a function g : T → D so that f ◦ g = IT and g ◦ f = ID , where ID and IT are the identity functions on D and T , respectively. (7) Recall that a subset of the plane S is said to be symmetric with respect to a line L if the reﬂection Λ across L has the property that Λ(S) = S. (a) Prove that the graph of s(x) = ax2 is symmetric with respect to the y-axis, where a is a constant. (b) Is the graph of g(x) = 3x2 −1 symmetric with respect to a line? (c) Does the graph of g in part (b) have a highest point? Lowest point? (d) Is the graph of f (x) = 3(x − 4)2 − 1 symmetric with respect to a line? (e) Does the graph of f in part (d) have a highest point? Lowest point? (8) (a) Prove that the graph of the equation x2 = 1 in R2 is not the graph of a function f : R → R. (b) Prove that the graph of the equation x2 + y 2 − 25 = 0 is not the graph of a real-valued function of one variable. (9) Now let f and g be real-valued functions deﬁned on R. We say f ≤ g if f (x) ≤ g(x) for every x ∈ R, and similarly for f < g. For each of the following assertions, give a proof if it is true and give a counterexample if it is false. (a) If f < 0 and g < 0, then f ◦ g > 0.12 (b) If f > 1, then g ◦ f > 1. (c) If f > 1, then f ◦ g > 1. (d) Suppose f is injective. If f ◦ f = f , then f is the identity function on R; i.e., f (x) = x for all numbers x. (e) Suppose f is surjective. If f ◦ f = f , then f is the identity function on R; i.e., f (x) = x for all numbers x. (f) Give an example of a function f deﬁned on R so that f ◦f = f but f is neither constant nor the identity function on R.13

11 Due

to Gowri Meda. to Judy Roitman. 13 Parts (d)–(f) are due to N. Ackerman. 12 Due

16

1. LINEAR FUNCTIONS

1.2. Why functions? This section gives an explanation, suitable for middle or high school, of why functions are critically important to mathematics and science. At some point, the explanation will make a reference to the concepts of diﬀerentiation and integration from calculus, but a precise knowledge of what they are is not necessary for an overall understanding of this explanation. The free-falling stone (p. 16) Temperature as a function (p. 20) The free-falling stone Before proceeding further with our study of functions, we should ﬁrst ask why functions? After all, function is a conceptually complex concept, so one should not impose it on students without a compelling reason. Before students encounter functions, their exposure to mathematics has been limited to numbers and some elementary geometry. To them, other than the recognition of shapes, numbers are all suﬃcient. To convince them that, henceforth, functions—and not numbers—will take center stage will be a bit of a hard sell. Functions are to mathematics as words are to writing,14 but an understanding of this statement will require a familiarity with the mathematics beyond middle school. Any explanation of why we must take up the study of functions therefore necessarily involves exposing students to future needs. All we can do here is give such an exposition in intuitive language and hope that it makes enough sense so that you can use it with your own students. Suppose we try to describe the various processes and phenomena that occur in the human and the natural worlds. These are evolving processes that change over time so that numbers arise only when we freeze a given moment in time and take measurements. For example, consider the movement of a car between time t1 and time t2 . In between these two points in time, the car’s motion is likely to be very erratic: fast and smooth on freeways but slow and jerky on surface streets because of city traﬃc and traﬃc lights. Armed with only numbers, basically the only sensible way we can describe the car’s motion in the time interval from t1 to t2 is to give its average speed (see page 352); i.e., freeze the car at time t1 and record its position, freeze the car again at time t2 and record its position, and from these two measurements, we get the total distance the car has traveled from time t1 to time t2 . Then we report the car’s average speed from t1 to t2 as the ratio the distance traveled from t1 to t2 . t2 − t1 Note that the information conveyed by this one number (the average speed from t1 to t2 ) is extremely meager. For example, if the average speed over a time interval of 1 hour is 30 mph, it could mean that the car actually traveled at a constant speed of 30 mph throughout, or it could also mean that the car traveled 25 miles in the ﬁrst 40 minutes and got stuck in heavy traﬃc and went only 5 miles in the last 20 minutes, or it could mean that the car only went 10 miles in the ﬁrst 30 minutes but made it up on the freeway by covering 20 miles in the last 30 minutes. Or it could mean a host of other possibilities. One thing is certain, however: if we 14 If we want to push this analogy one step further, we can say that numbers are to mathematics as the alphabet is to writing.

1.2. WHY FUNCTIONS?

17

only have numbers at our disposal to describe what is happening around us, then a massive loss of information would be the inevitable result. If we may illustrate this with an analogy, using only numbers to describe what is happening around us is akin to using snapshots to describe an evolving natural event. Snapshots are better than no visual record at all, but ultimately they are unsatisfactory. What we need is a video to show what is really happening from moment to moment. In this analogy, numbers are snapshots, but functions are the videos of the mathematical world. To describe the phenomena of change in both the human and natural worlds, we need functions. We can be a bit more detailed about the need to go beyond the mathematics of numbers in grades K–6 by looking at the example of a freely falling object. Suppose a stone is dropped from a point A that is 160 ft above the ground and a video is made of the motion.

A

160 ft

The video shows that it takes about 3.16 seconds for the stone to hit the ground, but students coming to algebra for the ﬁrst time would really want to know more about the motion of the free fall. For example, they may very well assume that the stone drops at a constant speed (because this is all they have been taught), but could this be true? A tabulation of the distances of the stone from the point of release, point A, at t = 1, 2, and 3 seconds dispels this notion (see the left table): t 0 1 2 3

dist. from A after t sec 0 16 64 144

t 0.5 1.5 2.5

dist. from A after t sec 4 36 100

Thus the stone drops 16 ft in the ﬁrst second, 48 (= 64 − 16) ft in the second, and 80 (= 144 − 64) ft in the third. This is deﬁnitely not constant-speed motion! Suppose students want to know even more about the motion, so they further mine the data in the video to get the height of the stone at shorter time intervals, say every half-second. The heights are tabulated in the table on the right above, and it gives a more detailed proﬁle of how the stone drops to the ground. If you are the teacher, you may wish to prod your students into realizing that what they really want is not a table of a ﬁnite number of the distances the stone has fallen at various moments of the free fall, but the distance the stone has fallen at any given moment, i.e., the distance the stone is below point A t seconds after it

18

1. LINEAR FUNCTIONS

was dropped for any t ≥ 0. Thus, what they need to know is an explicit description of the following function: f : {all numbers ≥ 0} → R so that f (0) = 0 and f (t) = the distance of the stone from point A t seconds after it is dropped. What may not be obvious is that, even if students are provided with the values of f (t) on demand for every t, it will still not be enough to give them an understanding of the phenomenon of free fall, any more than knowing the meanings of individual English words will enable someone to write intelligibly in English. In a vague sense, what they need is a “holistic” understanding of this free-falling phenomenon. To this end, they have to get hold of the function that describes the free fall and try to achieve a “holistic” understanding of this function. The ability to attain such an understanding was the great discovery of Newton (1643–1727) and Leibniz (1646–1716) in the seventeenth century.15 What they found was that, with the availability of the new tools of diﬀerentiation and integration, one could directly compute with functions as one could with numbers in elementary school. Whereas unknown numbers in elementary school can be determined by solving an equation, it is the case that unknown functions can also be made to satisfy a new type of equation involving diﬀerentiation and integration. The solution of such an equation then determines the function, which in turn yields information that would be inaccessible to mere number-crunching.16 We will return at the end of this subsection to give more substance to these vague ideas about the freely falling stone. Consider the fundamental concept of “speed”. Up to this point, the only way to express how fast an object moves—when it is not moving at constant speed—is to appeal to the concept of average speed over a certain time interval, which is clumsy and not always informative. For example, we observed above that the distances fallen in the ﬁrst, second, and third seconds are 16 ft, 48 ft, and 80 ft, respectively. Thus the average speeds in the ﬁrst three successive 1-second time intervals are 16 ft/sec., 48 ft/sec, and 80 ft/sec. What we can gather from these numbers is that they get bigger, but they have nothing to say about the average speeds in the ﬁrst three successive 0.5-second time intervals or the average speeds in the ﬁrst three successive 0.1-second time intervals. Does each such collection of numbers continue to get bigger? We see clearly that such a discussion will always be dependent on the length of the time interval that happens to be used, but this is not germane to the question that lies at the heart of the matter: what is “speed”? By making use of the above function f (t) that describes the distance the stone has traveled from time 0 to time t, however, we can now deﬁne the speed of the free fall at time t0 to be f (t0 ), the value of the “derivative” function f at time t0 . We will refer to f as the speed function of the freely falling stone. It will be seen presently that, in fact, f (t) = 32t. This then immediately implies that if 0 ≤ t1 < t2 , then 15 We will have more to say about both Newton and Leibniz in Chapters 5 and 6 of [Wu2020c]. 16 To be more accurate in the historical context, the evolution of the concept of a function cannot be separated from that of calculus. Until the middle of the nineteenth century, the whole subject of calculus was on shaky mathematical ground so that, for example, the precise meaning of a function was not clear during the same period. See volume 2 of [Kline] (especially pp. 403–406, 949–954) and pp. 723–726 of [Katz].

1.2. WHY FUNCTIONS?

19

f (t1 ) < f (t2 ); i.e., the speed of the free fall increases with time. This is the critical ﬁrst step toward a “holistic” understanding of the speed of the free fall. Incidentally, we can now be more precise about dropping the stone from point A at time t = 0. Inasmuch as we are not “throwing” the stone in any way, the speed of the stone at time t = 0 is assumed to be 0. In other words, part of the hypothesis is that f (0) = 0. More can be said, however. Like “speed”, acceleration is part of the everyday language, but until the advent of calculus, there was no precise deﬁnition of the concept; therefore it could not be incorporated into science or mathematics. All this changed when the concept of a function—as well as the operation of diﬀerentiation—became available. If we continue with the example of the freely falling stone, the acceleration function of the freely falling stone is the derivative function of the speed function, thus f . The acceleration of the stone at time t0 is then f (t0 ). The ability to deﬁne acceleration precisely became a cornerstone of Newtonian physics. To continue with the freely falling stone, we will use it to expand on the idea that one could compute with functions to obtain precise information about the changing phenomena in the natural world almost as eﬀectively as one could compute with numbers to solve problems in arithmetic. First, consider the following problem in arithmetic: A train runs at constant speed. If it travels 70 miles in 48 minutes, how long does it take to travel 85 miles at the same speed? If it takes t minutes for this train to travel 85 miles, then we have to ﬁnd out what this unknown number t is. The given hypothesis of constant speed means that the average speed in any time interval is equal to a ﬁxed constant v (see Theorem 1.10 on page 79 of [Wu2020a]). Thus the average speed of the train over the given 48-minute interval is equal to the average speed over the t minutes it takes to travel 85 miles, because both are equal to v mi/min. Thus we have the equation 85 − 0 70 − 0 = . 48 − 0 t−0 We ﬁnd the number t by solving equation (1.5). Now look at the freely falling stone. It is no longer an issue of ﬁnding a unknown number about the free fall. What is unknown is the function f (t) itself. In other words, starting with f (0) = 0, f (0) = 0, we want to know the whole phenomenon of the free fall, including the number t0 so that f (t0 ) = 160 (i.e., t0 is the time the stone hits the ground) and the value of f (t0 ) (i.e., the speed at which the stone hits the ground). What do we know about this function f ? All we know is that, by Newton’s second law, the acceleration of any freely falling object is a constant g, which is approximately 9.8 m/sec2 , or 32 ft/sec2 (rounded to the nearest decimeter/sec2 and ft/sec2 , respectively).17 Thus f (t) satisﬁes the following (diﬀerential) equation for any t so that f (t) ≤ 160: (1.5)

(1.6)

f (t) = g,

with f (0) = 0 ft and f (0) = 0 ft/sec.

Just as the solution of the train problem requires the solution of the equation (1.5) about numbers, the freely falling stone problem now requires the solution of the 17 Near

the surface of the earth and not counting air resistance.

20

1. LINEAR FUNCTIONS

equation (1.6) about functions. The solution of (1.6) turns out to be easy: √ 1 (1.7) f (t) = gt2 = 16t2 ft, for 0 ≤ t ≤ 10. 2 √ √ 18 Since f ( 10) = 160, the stone hits the ground after 10 second, i.e., approximately 3.16 seconds. From a conceptual standpoint, using the diﬀerential equation (1.6) to determine the function f (t) is a bold step. At the risk of belaboring the point, whereas the unknown in equation (1.5) is a single number, the unknown in equation (1.6) is a function. The boldness comes from the fact that this function contains complicated information about the evolving motion of the freely falling stone, so we are now conﬁdently asserting that the evolving motion of the stone, considered as a single entity, can be determined by an equation such as (1.6). The ability to solve this equation then allows for the decoding of these “moment-to-moment changes”. This ability marks the beginning of modern science and, of course, modern mathematics as well. At this point, we are in a position to say more about what we meant earlier by the need to achieve a “holistic” understanding of the phenomenon of free fall. Now that we know that the distance an object falls in t seconds is described by f (t) = 16t2 ft, we also know, for example, how long it would take a freely falling stone to hit the ground if it is released from 255 ft or, for that matter, 817 ft. In fact, we also know the speed of the free fall at a given moment of t0 seconds; namely, f (t0 ) = 32t0 ft/sec. In addition, we now understand that the free-falling data, far from just a collection of numbers, is completely described by a single function that is the solution of an equation: f (t) = the gravitational constant g. Clearly all this information would not have emerged if we had stubbornly clung to using numerical data—and numerical data alone—to understand the phenomenon of free fall. From this simple discussion, we get a glimpse of the reason that the concept of a function is truly fundamental in science and mathematics. Temperature as a function We have space for another example of a function, and this one is about the temperature of the city of Berkeley on a certain day. We are so used to hearing things like “It’s 67◦ (Fahrenheit) in Berkeley” that we no longer notice how much information is left out of such a statement. For example, is the temperature taken in the early dawn or in the late afternoon? In Berkeley, this could mean a 25◦ diﬀerence. And where is the temperature measured: at the top of the hill (about 1,300 feet high), downtown, or by the Bay? The diﬀerence there could be another 15◦ . If it is taken in Evans Hall,19 the diﬀerence between the temperature on the balcony of the tenth ﬂoor and that on the ground ﬂoor would be at least 5◦ on most days. We therefore see that the very concept of the temperature of Berkeley really makes no sense until we can be more precise. Let us limit ourselves to the temperature in a ﬁxed day and measure time in terms of t hours from midnight; then 0 ≤ t ≤ 24. To specify the geographic location, we need three more numbers √ that 10 = 3.16 when rounded to the nearest hundredth. This is therefore consistent with the above video data that the stone hits the ground about 3.16 seconds after being dropped. 19 The building that houses the mathematics department of UC Berkeley and most of its classes. 18 Note

1.2. WHY FUNCTIONS?

21

which may be thought of as the x-, y-, z-coordinates of an idealized 3-dimensional coordinate system with the origin being a point that we call the city center. Berkeley being a small city, 3 miles from the city center in any direction on land would include everything, and there is no building in Berkeley that is more than half a mile above sea level. Therefore, a scientiﬁcally usable description of the temperature of Berkeley would make use of a function T that depends on 4 numbers (t, x, y, z), so that T (t, x, y, z) gives the temperature at t hours after midnight on that day, at the location corresponding to the point (x, y) from city center and z miles above sea level at (x, y). In other words, if S is the region in 4-space consisting of all ordered quadruples of numbers (t, x, y, z) so that 0 ≤ t ≤ 24 (hours), |x|, |y| ≤ 3 (miles),20 and 0 ≤ z ≤ 12 (miles), then the function T :S→R can serve to legitimately describe the temperature of Berkeley on that particular day. It is a function, not a single number. Incidentally, such a function T is said to be a function of four variables, because four numbers t, x, y, and z are needed to specify its domain of deﬁnition. It may not have escaped your attention that the word “variable” is being used here merely as an afterthought. There is no need to explain what it means, and this is as it should be. (Compare the discussion of the term “variable” given at the beginning of Section 6.1 in [Wu2020a].) An additional comment is that people often wonder why mathematicians need to investigate n-dimensions for large n; aren’t three dimensions suﬃcient? This example shows that, in order to describe something as mundane as “the temperature of Berkeley”, we already need to use four numbers (t, x, y, z); such an ordered quadruple is a point in 4-dimensional Euclidean space. We are by no means suggesting that, in everyday conversation, we must use a function of four variables to describe the temperature of a city from now on. However, what we must accept is the fact that numbers are grossly inadequate for the scientiﬁc purpose of describing the world around us. Functions must take over eventually. For example, if we stay with the second example about temperature, one can easily see that, to describe an oncoming hurricane, we need to know, either explicitly or approximately, the functions that describe the air pressure and the temperature (much like the above function T (t, x, y, z) which describes air temperature) in the region surrounding the hurricane. To be able to read out the values of these functions at time t > t0 (where t0 represents the time at present) is what is normally called weather forecasting. These functions are obtained as solutions of equations similar to, but much more complicated than, equation (1.6) on page 19. When global weather forecasting is sought, not in terms of weeks but years or even decades, it belongs to the science of climatology. Without functions, there would be no climatology, and without the science of climatology, we would not have known about the perils of human-caused climate change. Think about that for a moment! So the point is that if we are stuck with numbers only but know nothing about the mathematics of functions, climatology would not be a scientiﬁc discipline and nor would physics, chemistry, biology, etc. Functions are truly basic indeed. 20 Notice the natural use of absolute value to describe the physical extent of the city. It means of course that −3 ≤ x ≤ 3 and −3 ≤ y ≤ 3. See the discussion of absolute value in Section 2.6 of [Wu2020a].

22

1. LINEAR FUNCTIONS

Exercises 1.2. (1) Does the “population of a city” make sense? If not, show how to make sense of it at least approximately. Do the same for a person’s “weight” (you may assume that this person stays at sea level). (2) Suppose you have a cup of freshly brewed coﬀee. Does it make sense to talk about “its temperature”? How can you make sense of “temperature” in this case?

1.3. Linear functions of one variable This section illustrates the function concept by taking up the simplest example, that of a linear function. It revisits and clariﬁes the concepts of constant rate— a topic in Section 1.7 of [Wu2020a]—and makes sense of the concept of scale drawing, which is a staple of middle school mathematics. It also explains, again from the perspective of linear functions, why the bogus concept of “proportional reasoning” must be handled with extreme care in the school classroom. Linear function and constant rate (p. 22) Scale-drawing (p. 27) Linear function and constant rate A linear function of one variable is a function f deﬁned on R which can be brought to the form f (x) = ax + b for all x ∈ R by the use of the commutative, associative, and distributive laws on numbers, where a and b are ﬁxed numbers. If a = 0, then f is the constant function equal to b. In general, b is called the constant term of f . If b = 0, then f (x) = ax and such an f is called a linear function without constant term. Linear functions without constant term are very important for middle school mathematics as well as for beginning algebra; every time one deals with proportions (i.e., equality of two ratios) or constant rates, there is always a linear function without constant term that lurks in the background. A main purpose of this section is to make explicit the role played by such functions in school mathematics. The usual failure of TSM to do so has made most problems related to rate unlearnable.21 We will also give a new interpretation of slope. Consider a moving object which starts at some point at time t = 0. Let f (t) be the total distance (measured, let us say, in meters) traveled by the object from the starting point after t seconds. We now formally recognize that this f is a function, and we will refer to it as the distance function of the motion or of the object. Thus we are given that f (0) = 0. We deﬁned in Section 1.7 of [Wu2020a] the concept of constant speed; let us recall this deﬁnition in terms of the function concept. A motion is of constant speed if there is a ﬁxed constant v so that the distance function satisﬁes f (t) = vt for all t ≥ 0. We then say the speed is v meters per second (m/sec). It was proved in Section 1.7 of [Wu2020a] that an object moves at a constant speed of v if and only if its average speed (see page 352 in this volume) over any time interval is v. It suits our purpose here to brieﬂy recall this reasoning. 21 See

page xi of the preface for a deﬁnition of TSM.

1.3. LINEAR FUNCTIONS OF ONE VARIABLE

23

Let t1 and t2 be two positive numbers, with t1 < t2 . Recall the standard notation (see page 10) of denoting all the numbers t satisfying t1 ≤ t ≤ t2 by [t1 , t2 ], called the closed interval from t1 to t2 . Then the average speed in the time interval [t1 , t2 ] of the object is, by deﬁnition, the number total distance traveled between time t1 and time t2 . t2 − t1 By the deﬁnition of f (t), the total distance traveled between t1 and t2 is f (t2 )−f (t1 ) meters. It follows that f (t2 ) − f (t1 ) m/sec. average speed in [t1 , t2 ] = t2 − t1 The right-hand side is seen to be the slope of a line, in the following way. On the graph of f , let the two points (t1 , f (t1 )) and (t2 , f (t2 )) be denoted by P1 and P2 , respectively. Let L1,2 be the line joining P1 and P2 .

f

L1,2

P2 = (t 2, f (t 2)).

P1 = (t 1, f (t 1)). t1

O

t2

Then the above equality becomes (1.8)

average speed in [t1 , t2 ] = slope of L1,2 .

This is the new interpretation of slope that we are after. Of course the foregoing discussion applies to any other kind of rate problem, be it water ﬂow, lawn mowing, or house painting. For example, let g(t) be the number of gallons of water coming out of a faucet after t minutes. Then the average rate of water ﬂow in a time interval [t1 , t2 ] is exactly the slope of the line joining the points (t1 , g(t1 )) and (t2 , g(t2 )) on the graph of g. Note that an average rate of change has to refer to a ﬁxed time interval before it makes any sense. The phrase “average rate of change” is common in TSM but it has no meaning by itself, and it is only after a time interval has been speciﬁed that we can begin to talk about the average rate of change in the given time interval. We now revisit the simple proof that a motion is of constant speed v if and only if the average speed of the motion over any time interval is equal to v (see Section 1.7 of [Wu2020a]). Suppose the motion is of constant speed v, so the total distance traveled by the object in the time interval [0, t] is f (t) = vt meters for any t ≥ 0. Therefore in the time interval [t1 , t2 ], the average speed is v because average speed =

f (t2 ) − f (t1 ) vt2 − vt1 v(t2 − t1 ) = = = v m/sec. t2 − t1 t2 − t1 t2 − t1

24

1. LINEAR FUNCTIONS

It is the proof of the converse that is revealing. Suppose the average speed of a motion over any time interval is always v m/sec and we want to prove that the distance function f (t) is the linear function without constant term f (t) = vt. In place of the algebraic proof given in Section 1.7 of [Wu2020a], we now give a geometric proof of this fact. Let F be the graph of f and let be the line with slope v passing through the origin. Now both F and have the origin (0, 0) in common. We pause to note that the equation of is y = vt in the vt-plane so that is the set of all points of the form (t, vt) for all t ∈ R. We are going to prove that F ⊂ (recall: “⊂” means “is contained in”). To this end, take an arbitrary point (t, f (t)) on F, t > 0, and we have to show that (t, f (t)) ∈ . Let Lt be the line joining (0, 0) to (t, f (t)). By (1.8), the average speed of the motion in [0, t] is equal to the slope of Lt . By hypothesis, the average speed is v, so the slope of Lt is also v. Thus Lt and are two lines passing through the same point (namely, (0, 0)) and have the same slope (namely, v). By Theorem 6.9 in Section 6.4 of [Wu2020a],22 Lt = and, in particular, (t, f (t)) lies on . Thus F ⊂ , proving the claim. Consequently, for any t > 0, (t, f (t)) is equal to (t, vt), so that, because the second coordinates of these points must be equal, we have f (t) = vt for all t ≥ 0, as desired. We note as usual that entirely analogous considerations apply to other rate problems, such as water ﬂow, lawn mowing, house painting, etc. For example, suppose a faucet is turned on at time t = 0 and suppose f (t) is the number of gallons of water coming out of a faucet after t minutes (t ≥ 0); then we say the rate of water ﬂow is constant if there is a positive constant r so that f (t) = rt. We also deﬁne the average rate of water ﬂow in time interval [t1 , t2 ] to be amount of water out of the faucet between time t1 and time t2 . t2 − t1 Then one proves as before that f (t) = rt for some positive constant r if and only if the average rate of water ﬂow in any time interval is r gallons per minute. We now give some examples to illustrate how linear functions provide the underpinning for all kinds of mathematical considerations related to proportions and constant rate problems. The discussions underscore the common pitfalls in the presentations of these topics in TSM which hide—apparently with the “good intention” of making mathematics easier to learn—the necessary assumption that the rate is constant, or equivalently, that the function in question is linear without constant term. We hope all teachers will make a concerted eﬀort to correct these entrenched errors in their teaching. Example 1. Ellen walks 285 meters in 9 minutes. How far does she walk in 2 minutes if her speed is constant? We ﬁrst transcribe the given information into symbolic language. Let the distance that Ellen walks in t minutes from her starting point be f (t) meters, with f (0) = 0 understood. Since her speed is constant, we see that f (t) = vt for all t ≥ 0, where v is her speed. We are given that f (9) = 285 and we want to know what f (2) is. This means we know v · 9 = 285, and we want to ﬁnd out the value 22 See

page 359 in this volume for the statement.

1.3. LINEAR FUNCTIONS OF ONE VARIABLE

25

of v · 2. The answer is straightforward: from v · 9 = 285, we get v = 285/9, so 1 meters. 3 Of course our interest in this problem lies not so much in getting the answer as in seeing how the concept of a linear function can shed light on its standard solutions. To this end, let us ﬁrst consider the common practice in TSM of “setting up a proportion” to do this problem. Students are told that if x is the number of meters Ellen walks in 2 minutes, then “the ratio of 285 to x is proportional to the ratio of 9 to 2”, and therefore v · 2 = (285/9) · 2 = 63

(1.9)

9 285 = . x 2

Solving for x by cross-multiplying, we get x = 63 31 as before. All appears to be well, until we ask why “the ratio of 285 to x is proportional to the ratio of 9 to 2”. Teachers and students alike are usually at a loss about how to explain this reasoning beyond saying that they “are used to it”, or, less frequently, that this is part of their “conceptual understanding of proportional reasoning”. Clearly, if we want to make school mathematics learnable, we cannot instill the skill of “setting up a proportion” by rote—which TSM wants us to do—but must ﬁnd ways to explain to students that the proportion is the logical conclusion of a deductive process using the assumption of constant speed. Let us do this now. Let f (t) be the distance (in meters) that Ellen walks in t minutes. Then the assumption of constant speed leads to f (t) = vt as before and therefore, for any t, is always the same, namely, v. We want the value of f (2). Let the quotient f (t) t f (2) 285 x us denote f (2) by x. Then from f (9) 9 = 2 (= v), we get 9 = 2 which, by the cross-multiplication algorithm (see p. 356), is equivalent to 285 9 = . x 2 But this is exactly equation (1.9). We ﬁnally get an explanation for “setting up a proportion” to solve this problem. TSM cannot provide such an explanation because TSM has no correct deﬁnition of constant speed. To cover up this failing, TSM has made up the bogus concept of proportional reasoning. We have used the concept of a linear function to explain a correct method of solution, but one can bypass any mention of a function, for example, by sacriﬁcing a bit of precision and simply explain constancy of speed to mean that there is a ﬁxed number v so that, for any t > 0, the distance traveled in a time duration of t minutes = v. t This deﬁnition of constant speed in terms of division is of course correct because it is a reformulation of the multiplicative statement f (t) = vt (see Section 1.5 of [Wu2020a]). As we have just seen, this reformulation has the advantage of leading directly to the use of “proportions” (equality of two ratios) in doing the problem (or any constant speed problem) and is therefore entirely adequate for doing such problems without invoking functions. The trouble with mathematics education as of 2020 is that students are generally not given any deﬁnition of “constant speed”

26

1. LINEAR FUNCTIONS

and, consequently, cannot be given any explanation as to why it is legitimate to “reason proportionally” by setting up the requisite proportion. Let us now inspect another common approach in TSM to this kind of problem, both in terms of the formulation of the problem as well as in the eventual solution. First of all, such a problem is often posed in the following form: Ellen walks 285 meters in 9 minutes. How far does she walk in 2 minutes? There is no mention of constant speed, which is equivalent to not assuming that the function f (t) that describes how many meters Ellen walks in t minutes is a linear function. In any case, the way to do a problem such as this, according to TSM, is to ﬁrst ﬁnd the “unit rate”, or in this case, the “unit speed”, i.e., the distance Ellen walks in unit time, which in this case is 1 minute. Since the given data implies that 23 and the Ellen walks 285 meters in 9 minutes, she walks 285 9 meters in a minute, 2 “unit rate” is therefore 285 = 31 m/min. It follows that, in 2 minutes, she walks 9 3 285 1 2 × 9 = 63 3 meters, as before. This solution seems attractive until we inquire how we can be sure that Ellen would walk the same distance in every 1-minute interval. Consider, for example, the following possibility for the way Ellen walks in the ﬁrst 4 minutes: time

distance she walks

ﬁrst 30 seconds

31 23 meters

30 seconds to 1 minute

at rest

1 minute to 1-minute-30-seconds

at rest

1-minute-30-seconds to 2 minutes

31 23 meters

2 minutes to 2-minutes-30-seconds

31 23 meters

2-minutes-30-seconds to 3 minutes

at rest

3 minutes to 3-minutes-30-seconds

31 23 meters,

3-minutes-30-seconds to 4 minutes

at rest

Assume that Ellen repeats the same walking pattern every 4 minutes. One should ﬁrst verify that in this motion Ellen always walks 31 23 meters in the ﬁrst minute, second minute, third minute, etc., so that she still walks 285 meters in 9 minutes. However, during the 1-minute interval from 30 seconds after she starts walking to 1:30 minutes, she walks 0 meters. Moreover, the distance she walks in the 2-minute interval from 1:30 minutes to 3:30 minutes is 2 2 2 31 + 31 + 0 + 31 = 95 meters. 3 3 3 It is not 2 × 31 23 = 63 13 meters. 23 This pedagogical strategy is clearly designed to trigger students’ conditioned reﬂex acquired =2 in elementary school, to the eﬀect that, e.g., if 5 books cost 10 dollars, then 1 book costs 10 5 dollars.

1.3. LINEAR FUNCTIONS OF ONE VARIABLE

27

The point should be clear. Unless we deﬁne what constant speed means and make an explicit assumption of constant speed, this kind of problem cannot be solved by considerations of “unit speed”. It cannot be overstated that, to make school mathematics learnable,24 precise deﬁnitions and explicit assumptions are absolutely essential. Scale-drawing We now come to a concept, scale-drawing, that is never adequately explained in TSM, and yet students’ failure to solve problems related to scale-drawing is often blamed on students’ putative lack of “conceptual understanding”. The following is a prototypical scale-drawing problem and is taken verbatim from a standard text. It exempliﬁes TSM like nothing else. Example 2. On a certain map, the scale indicates that 5 centimeters represents an actual distance of 9 km. Suppose the distance between two cities on this map measures 2 centimeters. Explain how you would ﬁnd out the actual distance between these two cities. What makes such problems confusing to students is that in TSM, there is no explanation of what it means to have maps with a certain scale.25 Without such an explanation, students are given no chance to learn about scale drawings because this is a subtle concept for at least two reasons. The ﬁrst reason for this subtlety is that, from the problem itself, one gets the false impression that a map somehow manages to be a 2-dimensional representation of a 3-dimensional object, namely, the part of the earth containing the two cities, together with streets, buildings, trees, mountains, etc. Such is not the case. One must ﬁrst imagine that there is a true-to-size 2-dimensional representation of this part of the earth; call this 2-dimensional representation N . Intuitively, one can think of N as what one gets when all the trees, buildings, mountains, etc., in this curved part of the earth have been squashed into a ﬂat surface. A map is then understood to be a contracted 2-dimensional representation of this imagined 2-dimensional object N , and we will next describe this contracted representation in greater detail. In any case, a map is emphatically not a representation of the original 3-dimensional region. A second reason is that the meaning of the “scale” of the map requires a detailed explanation. For this purpose one has to make use of, in one way or another, the concept of similarity (see the deﬁnition on page 355; a detailed discussion is given in Section 5.3 of [Wu2020a]). Let this N be considered as a region in the usual coordinate plane. Then a map of N is a region M in the plane similar to N . In greater detail, this means there is a similarity ϕ of the plane so that ϕ(M) = N , and so that if the scale factor of ϕ is r, then for any two points A and B in M, |A B | = r · |AB| 24 By “learnable”, we mean that most students understand the mathematics enough so that they can use it for reasoning and so that their future teachers can use it for teaching them more sophisticated concepts. 25 The lack of such an explanation is duly reﬂected, for example, in discussions of this topic in case books, e.g., Case 3 on page 19 of [Merseth].

28

1. LINEAR FUNCTIONS

where A = ϕ(A) and B = ϕ(B) are the corresponding points in N . (We recall that |AB| (respectively, |A B |) denotes the length of the segment AB (respectively, A B ) joining A and B (respectively, A and B ).) Because the given data of Example 2 calls for the conversion of centimeters in M to kilometers in N , we will make explicit the unit of distance in the plane by letting it be centimeters (cm) and rewrite the preceding equality as |A B | cm = r · |AB| cm. Now 1 km is equal to 105 cm, so if we let μ denote the number 105 , then 1 cm = km. Thus, what we have is r (1.10) |A B | cm = · |AB| km, for any A, B ∈ M. μ

1 μ

The scale s of the map is, by deﬁnition, the number μr km/cm; s is sometimes expressed in words: “1 cm to s km”. At this point, the introduction of a linear function without constant term will clarify the meaning of (1.10). Let x be the length in cm between two points on the map, and let h be the function so that h(x) is the distance in km between their two image points in N under ϕ. Then what (1.10) says is that h(x) = sx km,

where s =

r . μ

In terms of the function h, we may rephrase Example 2 as follows: we are given h(5) = 9 and we want the value of h(2). Now h(5) = 9 implies s · 5 = 9, so s = 95 and 9 h(x) = x km. 5 Therefore, h(2) = 95 × 2 = 3 35 km. In other words, if two cities are 2 cm apart on the map, the actual distance between them is 3 35 km. All students should be given the information in equation (1.10), but TSM does not see ﬁt to do that. Hence the nonlearning. Pedagogical Comments. It seems that there is a tradition in TSM to withhold from students the meaning of scale in order to see if they possess “a conceptual understanding of proportional reasoning”. We would like to gently remind readers that such an act of withholding information is defective mathematics education, because mathematics is not about students’ ability to guess the meanings of concepts. Precise deﬁnitions must be given explicitly as a matter of routine to facilitate learning (see page xx). In TSM, a great deal of importance is attached to problems based on so-called “proportional reasoning”. The message we want to convey with the two preceding examples and their accompanying discussions is that, unless we get out of the habit of formulating problems with hidden assumptions or without deﬁning explicitly the meaning of terms such as “scale of a map”, we are shortchanging students who want to learn about rates or scale-drawing. To make scale-drawing and related concepts learnable, we must make explicit the assumption that certain functions are linear functions without constant term and we must also make the eﬀort to explain to students the basic facts about such linear functions. Lest we forget, mathematics is precise (see the discussion of the Fundamental

1.3. LINEAR FUNCTIONS OF ONE VARIABLE

29

Principles of Mathematics on pp. xx–xxv) Let us make sure that we give students the information they need to make learning a reality in school math classrooms. End of Pedagogical Comments. Exercises 1.3. Each of Exercises 1–7 is based on actual TSM texts. Do them in three steps: (i) make explicit the assumption that makes a solution possible, (ii) transcribe this assumption into symbolic form using the concept of a function, and (iii) solve the problem. (1) A group of 9 people are going camping for two days and need to carry their own water. They read in a guide book that 12.5 liters are needed for a party of 5 persons for 1 day. How much water should they carry? (2) If the local sales tax is 7.5 cents per dollar, how much sales tax must you pay for a $12.3 item? (3) Cheese costs $4.25 per pound. How much does 7.75 pounds (approx.) of cheese cost? (4) On a certain map, the scale indicates that a length of 3 centimeters represents an actual distance of 10 miles. What actual distance does 7.3 centimeters on the map represent? (5) Bill’s grandfather enjoys knitting. He can knit a scarf 32 inches long in 10 hours, and he always knits for 2 hours each day. a. How many inches can he knit in 1 hour? b. How many days will it take Grandpa to knit a scarf 32 inches long? c. How many inches long will the scarf be at the end of 2 days? Explain how you ﬁgured it out. d. How many hours will it take Grandpa to knit a scarf 25 inches long? Explain your reasoning. (6) If a person weighing 98 pounds on earth weighs only 86 pounds on Venus, how much would a person weigh on Venus if she weighs 132 pounds on earth? (Some facts in Newtonian physics are needed to make sense of this exercise.) (7) Suppose 25 cows consume 400 pounds of hay in 6 days. How many cows can you feed in a week with 260 pounds of hay? (8) Discuss the inherent danger of giving students problems such as Exercises 1–7 above. (9) (i) Find a linear function f so that, for two distinct numbers a and b, f (a) = A and f (b) = B, where A and B are pre-assigned numbers. (Hint: The following is the standard way to do problems of this kind. Clearly if k is a number, the function f (x) = k(x − a) + A is linear and satisﬁes f (a) = A. Now adjust k to make this f satisfy f (b) = B.) (ii) Is this function f unique? (iii) Can you do part (i) by making use of Lemma 6.13 in [Wu2020a] (quoted on page 357 of this volume)? (10) (i) Let f : R → R be a linear function. Prove that there is a constant m so that for all numbers x and d, we have f (x + d) − f (x) = md. (ii) Let g : R → R be a function. Suppose that for some constant m, we have g(x + 1) − g(x) = m for any number x. Is g a linear function?

30

1. LINEAR FUNCTIONS

1.4. Linear inequalities and their graphs The goal of this section is to make sense of the rote procedure taught in TSM26 about “graphing linear inequalities in two variables”. We will make use of the deﬁnition of a half-plane from geometry (assumption (L4), recalled on page 215) and we will deﬁne the graph of a linear inequality on page 30. Using these deﬁnitions, we will be able to prove the theorem that the graph of a linear inequality is a halfplane (Theorem 1.1). The solutions to Examples 3–4 starting on page 40 reveal the reasoning that underlies the rote procedures taught in TSM about the graphing of linear inequalities, which we hope will make the procedures learnable. Most of the work in this section will be devoted to integrating the geometric concept of a half-plane into the present algebraic setting.27 Graphs of linear inequalities (p. 30) Upper and lower half-planes of nonvertical lines (p. 37) How to graph linear inequalities (p. 40) Graphs of linear inequalities In Section 6.5 of [Wu2020a], we discussed at some length the graph of a linear equation in two variables, ax + by = c, including the proof that it is a line L. It is only natural that we also take a look at the associated linear inequalities of the equation ax + by = c, namely, ax + by > c (or ax + by ≥ c) and ax + by < c (or ax+by ≤ c). We have to explicitly recall at this juncture our convention concerning such equations ax + by = c in two variables, namely, the fact that at least one of the coeﬃcients a and b is nonzero. The same convention will continue to hold for linear inequalities in two variables in general: at least one of a and b is understood to be nonzero in ax + by > c, for example. Since there is no conceptual diﬀerence between the inequalities ax + by > c or ax + by < c, we will frequently limit the discussion to just one of them, for example, ax + by > c, with the understanding that whatever we ﬁnd out about ax + by > c will have a counterpart in ax + by < c. The graph of a linear inequality ax + by > c is by deﬁnition the set of all the points (x0 , y0 ) in R2 that satisfy this inequality; i.e., ax0 + by0 > c. Similarly, the graph of ax + by ≥ c is by deﬁnition the set of all the points (x0 , y0 ) so that ax0 + by0 ≥ c. It follows from the deﬁnitions that the graph of ax + by ≥ c is the union of the graph of ax + by > c and the line L deﬁned by the associated equality ax + by = c. As mentioned above, the deﬁnition of the graph of an inequality is almost never precisely given in TSM, with the inevitable result that the teaching and learning about graphs of inequalities have to be done by rote. We must change this part of the school culture. The importance of having a deﬁnition of the graph of an inequality is that it allows us to prove the following basic fact in school mathematics—Theorem 1.1— about such a graph. This theorem is intuitively obvious, but its proof is long and 26 See

page xi for a deﬁnition of TSM. is another example of coherence mentioned in the Fundamental Principles of Mathematics on page xxiv. 27 This

1.4. LINEAR INEQUALITIES AND THEIR GRAPHS

31

requires some preparation because it relates a geometric concept (that of a halfplane of a line) to the algebra of linear inequalities. Such theorems are usually not simple to prove, and the following proof of Theorem 1.1 is more for the beneﬁt of teachers and educators than for students. (But see the Pedagogical Comments on page 37 for a more nuanced commentary on this proof.) Note that Theorem 1.1 cannot be proved in TSM because, if there is no deﬁnition for the graph of an inequality, there can be no proof of any theorem (i.e., no reasoning) about these graphs. Theorem 1.1. The half-planes of the line deﬁned by the linear equation ax + by = c are the graphs of the linear inequalities ax + by > c and ax + by < c. As a reminder from Section 4.1 in [Wu2020a] (but see also page 215 below), a half-plane of a line L does not include L.28 The relevance of the concept of a function to Theorem 1.1 will be made explicit in Theorem 1.4 on page 40. To prepare for the proof of the theorem (which begins on page 36), we ﬁrst recall the basic assumption about the half-planes of a line from Section 4.1 of [Wu2020a] (it is also summarized on page 215 of this volume). Let L denote the graph of ax + by = c. Then the half-planes H+ and H− of L are the two nonempty subsets of the plane that satisfy the following three properties: (i) The plane is the disjoint union29 of H+ , L, and H− ; i.e., the three sets are disjoint and their union is the whole plane. (ii) Both half-planes are convex; i.e., if two points A and B belong to the same half-plane, then the segment AB lies in the same half-plane (and therefore does not intersect L). (iii) If two points A and B in the plane belong to diﬀerent halfplanes, then the segment AB intersects L. Our task is to identify these H+ and H− , abstractly deﬁned in (i)–(iii), with the graphs of the linear inequalities ax + by > c and ax + by < c. The general idea of the proof of Theorem 1.1 is very simple and is given in the following abstract characterization of half-planes. Lemma 1.2. Let a line L be given, with its associated half-planes H+ and H− satisfying (i)–(iii) above. Suppose E and F are nonempty subsets of the plane so that: (a) The plane is the disjoint union of E, L, and F. (b) Both E and F are convex. Then either E = H+ and F = H− , or E = H− and F = H+ . The proof of Lemma 1.2 is analogous to the proof of Lemma 6.16 in Section 6.5 of [Wu2020a] (see page 358 of this volume for the statement of Lemma 6.16). From the vantage point of Lemma 1.2, the path to proving Theorem 1.1 is clear. Indeed, let E be the graph of ax + by > c and let F be the graph of ax + by < c. If these graphs can be shown to satisfy conditions (a) and (b) of Lemma 1.2, then Theorem 1.1 would follow from the lemma. 28 Mathematical Aside: In terms of standard mathematical terminology, a “half-plane” in these volumes refers to an open half-plane rather than a closed one. 29 This terminology is used in Section 4.1 of [Wu2020a].

32

1. LINEAR FUNCTIONS

Pedagogical Comments. The following proof of Lemma 1.2 is purely formal and nonintuitive. Although it is phrased in geometric language, it is in fact diﬃcult to draw pictures to illustrate the abstract geometric arguments (the reason will be obvious once we get into the details). Insofar as mathematics is a technical subject, this kind of reasoning cannot be avoided. Although it is unlikely that you will ever present such a proof in its entirety in a high school classroom, we nevertheless suggest that you make an eﬀort to learn it for your own beneﬁt because this kind of reasoning is a basic part of mathematics. End of Pedagogical Comments. Proof of Lemma 1.2. The overall strategy is to show that if E contains a point of H+ , then E = H+ and F = H− . We are only given that the plane can be expressed as a disjoint union of nonempty convex sets in two diﬀerent ways (see (i) below Theorem 1.1 and (a) in Lemma 1.2, respectively): (1.11)

the plane = H+ ∪ L ∪ H−

where “∪” stands for “union”, and (1.12)

the plane = E ∪ L ∪ F.

So we will have to do the best we can with (1.11) and (1.12). Let P be a point of E. Since P is not in L, then P lies in H+ or H− according to the right side of (1.11). There is no loss of generality in assuming that P lies in H+ (e.g., switch the notation of H+ and H− if necessary). Thus E contains a point P of H+ . Our main observation is this: ()

If E contains a point of H+ , then E ⊂ H+ .

For the proof of (), remember that we have a point P in both E and H+ . Let Q be another point of E and we have to prove that Q also belongs to H+ . In view of (1.11), we only need to show that Q does not belong to L or H− . If Q belongs to L, then Q ∈ E ∩ L where “∩” stands for “intersection”, so that E and L will not be disjoint. This contradicts assumption (a) of the lemma. If, on the other hand, Q belongs to H− , then since P lies in H+ , property (iii) of half-planes listed on page 31 implies that P Q contains a point Z of L. Now E is convex (see assumption (b) of the lemma), and both P and Q lie in E. Thus P Q lies in E and therefore Z also lies in E. Thus E and L again will not be disjoint because they have Z in common. Contradiction again. Thus we have shown that Q cannot lie in either L or H− and, therefore, must lie in H+ after all. It follows that E ⊂ H+ , thereby proving (). Clearly, the same argument will also prove the following: ()

If F contains a point of H+ , then F ⊂ H+ .

Now back to the proof of the lemma. Still assuming E contains a point P of H+ , we will improve on () by proving E = H+ . This means we have to show H+ ⊂ E. So let Z be a point of H+ , and we will prove Z is also a point of E. Suppose not, i.e., suppose Z ∈ E, and we will deduce a contradiction. According to (1.12), Z not lying in E means Z lies in L or F. If Z lies in L, then L and H+ cannot be disjoint because they have Z in common. This contradicts property (i) of half-planes on page 31. Thus the only possibility is that Z lies in F. Since Z ∈ H+ , this means

1.4. LINEAR INEQUALITIES AND THEIR GRAPHS

33

F contains a point of H+ . But then () implies that F ⊂ H+ . We now have the situation that both E and F are contained in H+ . By (1.12), the plane = E ∪ L ∪ F ⊂ H+ ∪ L = the plane where the inequality at the end is because of (1.11) and the fact that H− is nonempty. Thus we have the absurd statement that the plane is a subset of, but not equal to, itself. Consequently, the point Z of H+ lies in E after all and E = H+ , as desired. Next, we prove that E = H+ implies F = H− . By (1.12), we have a disjoint union: the plane = E ∪ L ∪ F. + If E = H , then we have a disjoint union: the plane = H+ ∪ L ∪ F. Thus F is exactly the collection of all the points in the plane not in H+ or L. However, we also have the disjoint union in (1.11); i.e., the plane = H+ ∪ L ∪ H− . This says that H− too is the collection of all the points in the plane not in H+ or L. Therefore F = H− , as desired. We have therefore proved that if E contains a point of H+ , then E = H+ and F = H− . Now if E contains no point of H+ , then (1.11) implies that E must contain a point of H− . Then the same argument (with H− replacing H+ ) will now show that E = H− and F = H+ (see Exercise 8 on page 42). The proof of Lemma 1.2 is complete. We now head toward the proof of Theorem 1.1. To this end, it will be convenient to look at the graphs of linear inequalities in terms of functions of two variables. Thus, with the linear equation ax + by = c as given, we consider the function : R2 → R deﬁned by (x, y) = ax + by.30 Observe that in this case, at least one of a and b is nonzero because we are given that ax + by = c deﬁnes the line L. Then the line L is exactly all the points (x, y) in the plane satisfying (x, y) = c. It is standard practice in mathematics to denote the set of all the points (x, y) at which (x, y) = c by the symbol { = c}. Thus L = { = c}. Any such { = c}, for a ﬁxed c, is called a level set of .31 The set of all the points (x, y) at which (x, y) > c is denoted, in like manner, by { > c}. Similarly, we deﬁne { ≥ c}, { < c}, and { ≤ c}. For example, {the graph of ax + by = c} = L = { = c}, {the graph of ax + by > c} = { > c}, {the graph of ax + by ≥ c} = { ≥ c} = { > c} ∪ L, {the plane} = { < c} ∪ L ∪ { > c}.

30 Strictly speaking, is a function of the point (x, y) and we should therefore write ((x, y)), but it is customary to abbreviate the symbol to (x, y). In addition, we could equally well use, instead, the linear function (x, y) = ax + by − c in this discussion, in which case we would be discussing the sets { = 0}, { > 0}, etc. 31 You may recall this terminology from multi-variable calculus. See the next section for a more detailed discussion.

34

1. LINEAR FUNCTIONS

In terms of this new notation, Theorem 1.1 can be restated as follows: The half-planes of the line deﬁned by the linear equation ax + by = c are the sets { > c} and { < c}, where is the linear function deﬁned on the plane by (x, y) = ax + by. To prove the theorem in this form, we will make use of Lemma 1.2. There is no question that the plane is a disjoint union of L, { > c}, and { < c}. Therefore the lemma implies that it suﬃces to prove that { > c} and { < c} are both convex. The key that unlocks the secrets of linear functions of two variables is the following lemma. Lemma 1.3. Let U and V be two points in the plane and let Q be a point in the segment U V distinct from the endpoints U and V . Also let : R2 → R be a linear function on the plane. Then one and only one of the following holds:

(U )

< (Q)

< (V ),

(U )

> (Q)

> (V ),

(U )

= (Q)

= (V ).

r V s Q r U

The intuitive content of this lemma is easy to bring out. Let U and V lie on a coordinate axis; let us say the x-axis. Let (x, y) = ax+by +d for some constants a, b, and d. Now on the x-axis, the linear function becomes a linear function of one variable: (x, 0) = ax + d. Then the lemma is obviously true in this special case, as the following three pictures show: the graph of (x, 0) is a line and therefore, unless the line is horizontal, the value of (x, 0) at a point Q in the segment U V is trapped between the values of (x, 0) at the end-points U and V . Y

Y

Y

? d

PP d PP PP PP? PP P

O

O

graph of (x, 0)

U

Q

a>0

V

graph of (x, 0) graph of (x, 0)

U

Q

a 0, then x < y ⇐⇒ xz < yz. (E) For any x, y, z, if z < 0, then x < y ⇐⇒ xz > yz. Note that these assertions were only shown to be correct for rational numbers x, y, z in Section 2.6 of [Wu2020a], but they are valid for all real numbers x, y, z, as will be proved in Section 2.1 of [Wu2020c]. For now, we simply invoke FASM (whose precise statement is given on page 351 of this volume) to justify using them for real numbers x, y, z. Proof of Lemma 1.3. Let us ﬁrst assume that the line LUV is not vertical; then LUV is the graph of an equation of the form y = mx + k for some constants m and k (see Lemma 6.14 on p. 357). Since U and V lie on LUV , we may let U = (u, mu + k) and V = (v, mv + k). Without loss of generality, we may let u < v. Let Q = (s, ms + k) be a point of the segment U V , where s is a number. If Q is distinct from U and V , Lemma 6.15 of [Wu2020a] (see page 358 in this volume) implies that the x-coordinate s of Q must satisfy u 0. Intuitively, equation (1.13) implies that (Q) increases as s increases from u to v. We now make it precise: if u < s, then according to assertion (D) on p. 35, we have (a + bm)u < (a + bm)s By assertion (B) on p. 35, (a + bm)u + (bk + d) < (a + bm)s + (bk + d). This is equivalent to (U ) < (Q). Similarly, if s < v, then (Q) < (V ). Therefore (U ) < (Q) < (V ) in this case. Suppose, instead, a+bm < 0. Intuitively, equation (1.13) implies (Q) decreases as s increases from u to v. We now make this reasoning precise. If u < s, then according to assertion (E) on p. 35, (a + bm)u > (a + bm)s. Then as before, (a + bm)u + (bk + d) > (a + bm)s + (bk + d) so that we have (U ) > (Q). Similarly, s < v leads to (Q) > (V ). Therefore (U ) > (Q) > (V ) if a + bm < 0.

36

1. LINEAR FUNCTIONS

Finally, if a + bm = 0, then the number (Q) stays the same and equals bk + d no matter what s is. In this case, (U ) = (Q) = (V ). Hence the lemma is proved in case LUV is not vertical. Suppose LUV is vertical; then the equation of LUV is x = k for some constant k. In this case, we may let U = (k, u ) and V = (k, v ), with u < v . For a point Q in U V , Q must have coordinates Q = (k, s ) where u ≤ s ≤ v . Now (Q) = ak + bs + d = bs + (ak + d). If b > 0 and u < q , then assertion (D) on p. 35 implies bu < bs so that (U ) = bu + (ak + d) < bs + (ak + d) = (Q). Similarly, if b > 0 and s < v , then assertion (D) implies bs < bv so that (Q) < (V ). Thus if b > 0, then (U ) < (Q) < (V ), as before. If, however, b < 0, then exactly the same argument—but with assertion (E) replacing assertion (D)—shows that (U ) > (Q) > (V ). Finally, if b = 0, then (Q) = ak + d for every point Q in U V so that (U ) = (Q) = (V ). The proof of Lemma 1.3 is complete. Pedagogical Comments. The importance of Lemma 1.3 for understanding linear functions in the plane cannot be overstated. Furthermore, one virtue of the proof is that it puts to use the inequalities about rational numbers from Section 2.6 of [Wu2020a] (recalled on page 356 of this volume), so students get to see in a very small way the coherence of mathematics. For these reasons, some form of this proof should be given in the school classroom. We suggest that, at the very least, this lemma be proved in class when LUV is vertical. Another suggestion is to at least verify the lemma for a special case; e.g., let U = (−2, 3) and V = (1, 4), and let (x, y) = 2x − y + 4. End of Pedagogical Comments. We are now ready to give the Proof of Theorem 1.1. We are given a line L deﬁned by the linear equation ax + by = c and the linear function deﬁned on the plane by (x, y) = ax + by. In view of Lemma 1.2 on p. 31, it suﬃces to prove that the sets { > c} and { < c} are nonempty and that (i) the plane is the disjoint union of L, { > c}, and { < c} and (ii) { > c} and { < c} are both convex. The fact that { > c} and { < c} are nonempty is shown as follows. If a = 0, then it is easy to verify that c+1 c−1 , 0 ∈ { > c} while , 0 ∈ { < c}. a a If a = 0, then b = 0 and a similar argument shows that both { > c} and { < c} are nonempty. To prove (i), we observe that the three sets L, { > c}, and { < c} are clearly disjoint. To show that their union is the whole plane, let P be any point in the plane and we have to show that P lies in one of them. This follows from the trichotomy law (see page 359): (P ) is either > c, = c, or < c. This proves (i). For (ii), let us prove that { > c} is convex. Let U and V be in { > c} and let Q be a point lying in the segment U V . By Lemma 1.3 on p. 34, either (Q) > (U ) or (Q) > V or (Q) = (U ) = (V ). In any case, since U , V are points in { > c},

1.4. LINEAR INEQUALITIES AND THEIR GRAPHS

37

it will always be the case that (U ) > c and (V ) > c. Consequently, (Q) > c in all cases and Q lies in { > c}. This proves the convexity of { > c}. The convexity of { < c} is proved by an entirely analogous argument. The proof of Theorem 1.1 is complete. Pedagogical Comments. Each teacher will have to decide how to present, in a meaningful way, the proof of Theorem 1.1 in a high school classroom. We can only make a generic suggestion. First, state Lemma 1.2 and Lemma 1.3 without proof, but make some eﬀort to indicate what they mean and intuitively why they are true (more on this later). Then it would be reasonable to give the complete proof of Theorem 1.1 on the basis of Lemma 1.2 and Lemma 1.3. If there is time left, how much you want to say about these two lemmas will depend on the mathematical sophistication of your students. In general, a complete proof of Lemma 1.2 is out of the question, but the proof of () on page 32 is suﬃciently elementary and short enough to be worth a try in a high school classroom. We have already made some suggestions on how to present the proof of Lemma 1.3 on page 36. End of Pedagogical Comments. Upper and lower half-planes of nonvertical lines Theorem 1.1 succeeds in relating the abstract concept of a half-plane of a line L (see assumption (L4) on page 215 below) to what may be called the deﬁning function of L; i.e., if the equation of L is ax + by = c, then its deﬁning function is the function of two variables given by (x, y) = ax + by. For the applications in the next section, the relationship between L and described in Theorem 1.1 suﬃces. For the purpose of an intuitive understanding of a half-plane, however, a more concrete representation of the half-planes would be desirable. The remainder of this section describes this more concrete representation. First, we introduce a new concept. Given a nonvertical line L, then a point (x0 , y) is said to lie above L if the vertical line passing through (x0 , y) intersects L at the point (x0 , y0 ) and y > y0 . See the left picture below. r(x0 , y)

" " " " " O

" " " " r " " (x0 , y0 ) "

L

" " " " r " " (x0 , y0 ) " O "" " " r(x , y) "

L

0

Similarly, we say (x0 , y) lies below L if the vertical line passing through (x0 , y) intersects L at the point (x0 , y0 ) and y < y0 ; see the right picture above. For a nonvertical line L, the set of all the points above L (respectively, below L) is called the upper half-plane of L (respectively, the lower half-plane of L). This nomenclature strongly suggests that an “upper half-plane of L” (respectively, a “lower half-plane of L”) is indeed a half-plane of L. We are going to prove that such is the case. Precisely, we will show that the upper and lower half-planes of a

38

1. LINEAR FUNCTIONS

nonvertical line L are directly related to the two sets { > c} and { < c}, where is the deﬁning function of L = { = c} (see Theorem 1.4 on page 40). Thanks to Theorem 1.1, the latter are the half-planes of L. We can give a proof of Theorem 1.4 using Lemma 1.2 on page 31, but it turns out that there is a more direct proof. This will become clear once we look at the following two examples. Example 1. If (x, y) = 3x + 2y, describe the set { > 6}. By deﬁnition, the level set { = 6} is the line L deﬁned by 3x + 2y = 6 and the set { > 6} is the set of all (x, y) so that 3x + 2y > 6. A point in { > 6} clearly cannot lie on the line L and therefore must lie above L or below L. We now show that { > 6} = the upper half-plane of L. We ﬁrst show that {the upper half-plane of L} ⊂ { > 6}. So let P = (x0 , y) be a point in the upper half-plane of L and we will prove that P lies in { > 6}. Let the vertical line passing through P intersect L at a point (x0 , y0 ) for some number y0 . By the assumption that P is above L, we have y > y0 . Therefore 2y > 2y0 , and 3x0 + 2y > 3x0 + 2y0 . But this is equivalent to (x0 , y) > (x0 , y0 ). Since (x0 , y0 ) ∈ L, we have (x0 , y0 ) = 6. Thus (x0 , y) > 6, i.e., P ∈ { > 6}. J LJ

O

J J

rP = (x , y) 0 J

Jr(x0 , y0 ) J J J J J J

Next, we show that { > 6} ⊂ {the upper half-plane of L}. The proof is similar. Let P = (x0 , y1 ) ∈ { > 6} and let the vertical line passing through P intersect L at a point Q = (x0 , y0 ) for some number y0 . Then 3x0 + 2y1 > 6 and 3x0 + 2y0 = 6. Thus 3x0 + 2y1 > 3x0 + 2y0 , which implies 2y1 > 2y0 , which in turn implies y1 > y0 . This means P is above L and is therefore in the upper half-plane of L. To summarize, we have proved that (1.14)

{3x + 2y > 6} = the upper half-plane of {3x + 2y = 6}.

A similar argument gives (1.15)

{3x + 2y < 6} = the lower half-plane of {3x + 2y = 6}.

Remarks. We have just shown that basically the same reasoning proves both assertions (1.14) and (1.15). It is instructive to also be aware that these two assertions are in fact equivalent to each other; i.e., knowing one is equivalent to knowing the other. Let us ﬁrst show that (1.14) implies (1.15). So assume (1.14). To prove (1.15), we ﬁrst show that if a point U lies in {3x + 2y < 6}, then U lies in the lower half-plane of L (= {3x + 2y = 6}). Suppose not; then U is either on L or above L. The former clearly cannot happen; if the latter, then by (1.14), U ∈ {3x + 2y > 6}, a contradiction. Therefore the U lies in the lower half-plane of L. This proves one

1.4. LINEAR INEQUALITIES AND THEIR GRAPHS

39

half of (1.15). Next, we have to show that if a point U lies in the lower half-plane of L, then U ∈ {3x + 2y < 6}. If this is false, then U ∈ L or U ∈ {3x + 2y > 6}. The former clearly cannot be true. If the latter holds, then (1.14) implies that U lies in the upper half-plane of {3x + 2y = 6}, a contradiction. Therefore U lying in the lower half-plane of L implies U ∈ {3x + 2y < 6}. This proves the second half of (1.15). Thus if (1.14) is true, then (1.15) is also true. The fact that (1.15) implies (1.14) can be proved in like manner and will be left as an exercise (Exercise 9 on page 42). Example 2. With (x, y) = 3x − 2y, describe the set { < 6}. By deﬁnition, { < 6} denotes the set of all points (x, y), so that (x, y) < 6, i.e., the set of all (x, y) so that 3x − 2y < 6. Let P ∈ { < 6} and let P = (x0 , y). Then by deﬁnition, 3x0 − 2y < 6. Let L denote the line { = 6} and let the vertical line passing through P intersect L at a point (x0 , y0 ) for some number y0 . Then by the deﬁnition of L, we have 6 = 3x0 − 2y0 . Thus 3x0 − 2y < 6 implies 3x0 − 2y < 3x0 − 2y0 . Therefore, −2y < −2y0 . Multiplying both sides by − 12 and making use of assertion (E) on p. 35, we get y > y0 . Thus we have proved that P ∈ { < 6} implies P lies above L; i.e., { < 6} ⊂ {the upper half-plane of L}.

r

P = (x0 , y)

O

L

r

(x0 , y0 )

Conversely, we will prove that {the upper half-plane of L} ⊂ { < 6}. Thus let P = (x0 , y) lying above L, and let the vertical line passing through P intersects L at a point (x0 , y0 ) as before. Thus we are assuming y > y0 . Since (x0 , y0 ) lies on L, 3x0 − 2y0 = 6. But y > y0 and assertion (E) on p. 35 imply that −2y < −2y0 . Thus, 3x0 − 2y < 3x0 − 2y0 = 6. In other words, (x0 , y) < 6, so that P = (x0 , y) ∈ { < 6}. Together, we have proved that (1.16)

{3x − 2y < 6} = the upper half-plane of {3x − 2y = 6}.

The same reasoning also shows that (1.17)

{3x − 2y > 6} = the lower half-plane of {3x − 2y = 6}.

The key observation to be made in these two examples is that, given a linear function (x, y) = ax+by, the correlation of the two half-planes { > c} and { < c} with the upper and lower half-planes of the line { = c} depends on whether b > 0 or

40

1. LINEAR FUNCTIONS

b < 0. This can be seen by comparing (1.14) with (1.17) and comparing (1.15) with (1.16). Also observe that the reasoning in Examples 1 and 2 is perfectly general and can be used to prove the following theorem. Theorem 1.4. Given a nonvertical line L deﬁned by the equation ax + by = c, let (x, y) = ax+by. Then the graph of the inequality { < c} (respectively, { > c}) is either the upper half-plane of L or the lower half-plane of L. More precisely: (i) If b > 0, then { > c} = the upper half-plane of L, { < c} = the lower half-plane of L. (ii) If b < 0, then { < c} = the upper half-plane of L, { > c} = the lower half-plane of L. In an exercise (Exercise 11 on page 42), you will be asked to write out the details of the proof. How to graph linear inequalities Note that Theorem 1.4 is very precise, but the precision also entails an abundance of details that make it almost impossible to remember which half-plane corresponds to which inequality. Fortunately there is no need to memorize anything, as the following two examples show. Bear in mind, however, that the reasoning in these two examples is made possible by Theorem 1.4, which in turn is based on Theorem 1.1. Example 3. What is the graph of 3x − 5y < −7? By Theorem 1.4, it is either the points above the line L given by 3x−5y = −7 or the points below it. Rather than going through the precise description of Theorem 1.4 to decide which is which, there is a shortcut. With the graph of L pictured as L " r" 7 " " (0, 5 ) " " " " " r " (0, 0) since (0, 75 ) is on L, (0, 0) is below L. But (0, 0) is obviously not in the graph of 3x − 5y < −7 as 0 > −7; the points below L cannot be the graph of 3x − 5y < −7. Therefore by Theorem 1.4, the graph of 3x − 5y < −7 is the set of points above L; i.e., it is the upper half-plane of L. The preceding procedure for getting the graph of 3x − 5y < −7 is one that is taught by rote in TSM, without any deﬁnition of what a “half-plane” is, and therefore without any reasoning to support the prescribed procedure. Our explanation— let it be noted—is rooted in an understanding of Theorem 1.4. Example 4. What is the graph of 5x + 4y < 0?

1.4. LINEAR INEQUALITIES AND THEIR GRAPHS

41

The line L given by 5x + 4y = 0 already goes through the origin (0, 0), so we cannot make use of the origin as we did above. L

\

\

\

r(0, 1) \

\

\O \\

However, we can choose a point on the y-axis above the origin (which is on L), such as (0, 1), and ask for the value of the function (x, y) = 5x + 4y at (0, 1). Since (0, 1) = 4 > 0, the point (0, 1) is not in the half-plane { < 0}; i.e., the upper half-plane of L is not { < 0}. By Theorem 1.4, { < 0} is the lower half-plane of L. It remains to note that when the line is vertical, Theorem 1.4 gives no information. However, it is easy to see that, in that case, since the equation of a vertical line is x = c, then the function becomes (x, y) = x, and the sets { > c} and { < c} are just {x > c} and {x < c}), respectively. Then it is easy to see that the latter are the set of all points to the right and to the left of the vertical line, respectively. These are usually called the right half-plane and left half-plane of the vertical line, respectively. To summarize then, we have proved the following theorem, which gives an explicit description of the half-planes of a line. Theorem 1.5. If the line is vertical, its half-planes are the left and right halfplanes of the vertical line; if the line is nonvertical, then its half-planes are the upper and lower half-planes of the line. Exercises 1.4. (1) Graph each of the following inequalities and explain clearly why your graph is correct: (a) 52 x + 34 y ≤ 2. (b) − 52 x + 34 y ≤ 2. (c) 52 x + 34 y ≤ −2. (2) Describe the points (x, y) that satisfy the following three inequalities simultaneously: 52 x + 34 y ≤ 2, 4x − y > −8, and x − 2y < 3. (3) Let L be the line y = 2x − 5, and let H+ = {y > 2x − 5}. Also let A be the set of all the points above L. Without making use of Theorem 1.1 or Theorem 1.4, prove directly that H+ = A. (4) Let be the function in the plane deﬁned by (x, y) = 3x − 4y. (a) Sketch the two level sets { = 11} and { = 2}, and clearly label their x-intercepts. (b) Prove that these level sets are parallel lines. (5) Does the half-line y = −2x + 4, x ≤ 0, lie inside the region deﬁned by the inequalities y ≥ x, y ≥ − 12 x + 3? (6) Let P = (56, 5) and Q = (2, −7 76 ) be two points on the line Λ whose equation is 27 x − 65 y = 10. Let U = (s, t) be a point on the segment P Q. (a) If (x, y) = 8x + 2y, ﬁnd an expression of (U ) in terms of s.

42

1. LINEAR FUNCTIONS

(b) Compute the explicit values of (P ), (Q), and (U ) in terms of s, and directly verify that Lemma 1.3 is valid in this special case. (c) Find an expression of (U ) in terms of t when U is a point in P Q. (7) Assume a linear function (x, y) = ax + by + d with a < 0 and b = 0. Suppose we have two lines both with slope − ab as shown: y (p, q) r r r x (s, t) r (p , q ) r

(8) (9) (10)

(11) (12)

(13)

(a) Which is bigger: (s, t) or (p, q)? (b) Which is bigger: (p, q) or (p , q )? Complete the proof of Lemma 1.2 on page 31 by showing that if E contains a point of H− , then E = H− and F = H+ . Use (1.15) on page 38 to prove (1.14). (a) Let be the function on the plane deﬁned by (x, y) = 5x + 3y. Graph { > 4}, and explain as if to a ninth grader why your graph is correct. (b) Let be the function on the plane deﬁned by (x, y) = 5x − 3y. Graph { > 4}, and explain as if to a ninth grader why your graph is correct. (a) Give a detailed proof of part (i) of Theorem 1.4. (b) Give a detailed proof of part (ii) of Theorem 1.4. (a) Let be the function on the plane deﬁned by (x, y) = 12 x − 43 y. Let Q be a point on the line Λ deﬁned by 3x+y = −1. If Q = (s, t), express (Q) in terms of s. As s increases (and therefore Q moves from left to right in the plane), how does (Q) behave? (b) Repeat part (a) if now Λ is deﬁned by 6x − 2y = 1. (c) Repeat part (a) if Λ is deﬁned by −3x + 8y = 2. Let a, b be constants, b = 0, and let (x, y) = ax + by be a function on the plane. Let L be the level set { = c} for some constant c. (a) Prove that any line parallel to L is also a level set of . (b) Prove that the level sets of are precisely all the lines with slope − ab . (c) Let L1 , L2 be two lines parallel to L, and let P1 , P2 be points on L1 , L2 , respectively. Suppose L1 meets the x-axis at (−3.5, 0) and L2 meets the y-axis at (0, 25 ). What are the values of (P1 ) and (P2 )?

1.5. Linear programming Linear programming is the study of eﬃcient methods for ﬁnding where a linear function achieves its maximum or minimum in a region deﬁned by a system of linear inequalities. Such problems arise naturally in industry and military operations. The aim of this section is to give a hint of the core idea of linear programming by working out two highly idealized examples in two variables. In the process, we will get to see some of the main issues that stand in the way of a solution. More important for

1.5. LINEAR PROGRAMMING

43

our purpose is the fact that solutions to these examples showcase the mathematical relevance of the information garnered in the preceding section about the graphs of linear inequalities. Optimization and a manufacturing problem (p. 43) A general strategy towards a solution (p. 46) An advanced manufacturing problem (p. 49) Optimization and a manufacturing problem Recall the following familiar problem in multi-variable calculus. Let S be a region32 in the plane that includes its boundary points, and let f (x, y) be a function deﬁned in some neighborhood of S. Then we want to locate the maxima (respectively, minima) of f (x, y) in S, i.e., the points P ∈ S so that for all Q which are in S, f (P ) ≥ f (Q) (respectively, the points P ∈ S so that for all Q which are in S, f (P ) ≤ f (Q)). Also recall the standard terminology: in this case, we say f achieves a maximum on S at P (respectively, achieves a minimum on S at P ). The standard way in calculus to do this is to compute the partial derivatives of f and solve for the points (x, y) at which the partial derivatives vanish (i.e., equal zero). Then one applies the second derivative test to separate the possible maxima from the possible minima and then compares the values of f on the boundary of S to obtain the ﬁnal answer. This method, especially the last step, can be messy. What we are going to describe in this section is the core idea of an eﬃcient method—we note explicitly that the emphasis is on the eﬃciency of the method— for getting both the maxima and minima when f is a linear function and when S is the intersection of a ﬁnite number of closed half-planes (deﬁned below). We will use an oversimpliﬁed example (the Manufacturing Problem below) to illustrate this method. Please keep in mind that many complexities of real life will be intentionally glossed over in order to get across the basic idea. Let L be a line and H+ one of its half-planes. Recall from Section 4.1 of [Wu2020a] that the region which is the union of H+ and L is called the closed half-plane of L corresponding to H+ . Thus the diﬀerence between H+ and its closed half-plane is that the latter includes the boundary—which is L—but the former does not. In term of Theorem 1.4, we can express the diﬀerence as follows. Let (x, y) = ax + by, and let L = { = c}. Let us say H+ = { > c} for some constant c. Then the closed half-plane of L corresponding to H+ is simply { ≥ c}. The kind of region S that arises naturally in industrial applications will be the intersections of a ﬁnite number of closed half-planes. Consider a linear function of two variables f (x, y) = ax + by + d, where a, b, and d are constants (see page 14). The study of algorithms that lead to the optimization (i.e., the location of the maxima or minima) of a linear function in a region that is the intersection of a ﬁnite number of closed half-planes is generally

32 As in Chapter 4 of [Wu2020a], we are using the concept of a region informally. Think of a region as the inside of a polygon, with or without the boundary; a ﬁnite number of points or a curve will not be considered to be a region. If we want to include the boundary as part of the region, it will be explicitly mentioned as in the present situation. We avoid a precise deﬁnition of a region because of its level of sophistication and also because, ultimately, we only seriously consider regions that are the intersections of closed half-planes. See Theorem 1.8 on page 54.

44

1. LINEAR FUNCTIONS

referred to as linear programming in two variables. Here is an example that illustrates, in a stylized way, why linear programming is of interest. The corresponding study of a linear function in n variables is conceptually no diﬀerent, but the mathematical prerequisites for such a discussion are of course beyond the level of this volume. Manufacturing Problem. A plant has the capacity to produce 225 pieces of merchandise per day. It produces only two kinds of merchandise, A or B. The cost of manufacturing merchandise A is $48 per item, and each brings in a net proﬁt of $35. The cost of manufacturing merchandise B is $144 per item, and each brings in a net proﬁt of $70. The plant has an operating budget of $18, 000 per day. How many of each kind of merchandise should the plant produce each day in order to maximize the proﬁt? (Assume that every item will be sold.) To get a sense of what this kind of problem is all about, let us try to solve the problem the naive way. Since each item of merchandise B brings in a $70 proﬁt compared with only $35 for each piece of merchandise A, why not go all out to manufacture merchandise B only? If so, we would be committing the daily operating budget of $18,000 completely to producing merchandise B. Since each item of merchandise B costs $144 to manufacture, the plant can only produce 125 items ( 18,000 144 = 125) of merchandise B each day. The total proﬁt each day will be 125 × 70 = 8,750 dollars. There is an intuitive reason why $8,750 cannot be the maximum proﬁt: since the plant has the capability to produce not 125 but 225 pieces of merchandise per day, limiting the daily production to 125 items would mean wasting some of the available resources. Wasting available resources is probably not the way to maximize proﬁt. Indeed, once this idea surfaces, we can make a quick calculation to see that $8,750 is not the maximum possible proﬁt, as follows. We can increase the total number of items produced daily by producing only 124 items of merchandise B and using the freed-up $144 to produce 3 items of merchandise A (144 = 3 × 48). To make sure that the new manufacturing plan meets the requirements set forth in the problem, we check: the total daily production is 124 + 3 = 127 items, and 127 < 225, while the total daily manufacturing cost is (124 × 144) + (3 × 48) = 18,000 dollars, which is within the daily operating budget. Next, we check the new proﬁt: it is (124 × 70) + (3 × 35) = 8,785 dollars. Needless to say, 8,785 > 8,750, and we see conclusively that manufacturing only merchandise B does not maximize the proﬁt. Next, we try the other extreme. Observe that merchandize A is the better option for proﬁt-making because its proﬁt is 35 48 of its manufacturing cost, compared 70 with merchandise B whose proﬁt is only 144 of its manufacturing cost, and 72 1 35 70 < = < . 144 144 2 48 So why not manufacture merchandise A only? Suppose we do that, then we manufacture 225 items of merchandise A per day, and the total daily manufacturing cost would come to $10,800 dollars (225 × 48 = 10,800), and this amount is way below

1.5. LINEAR PROGRAMMING

45

the available $18,000. Again, we are under-utilizing the given resources. In this case, it is even easier to see that the resulting proﬁt cannot be a maximum because the total proﬁt that 225 items of merchandise A can bring comes to only 225 × 35 = 7,875 dollars. But we already saw that manufacturing merchandise B alone would bring in a daily proﬁt of $8,750. So we know for sure that $7,875 cannot be the maximum proﬁt. We are now intuitively convinced that, in order to maximize the proﬁt, we must manufacture some items of each of merchandise A and B, so the determination of the optimal number for each kind of merchandise is our next concern. Let us approach this problem systematically by ﬁrst transcribing the verbal information into symbolic form. Suppose each day x items of merchandise A and y items of merchandise B are manufactured. The daily net proﬁt is then given by the proﬁt function f (x, y), with f (x, y) = 35x + 70y. We are not trying to ﬁnd any two positive integers x and y to make f (x, y) as large as possible; clearly there is no sense in trying to do that because f (x, y) gets arbitrarily large when x and y increase without bound. Rather, we are trying to ﬁnd such a pair that satisﬁes some restrictions on both x and y. First, obviously, 0≤x

and

0 ≤ y.

Furthermore, in view of the limit on the maximum daily production capacity of the plant, we have x + y ≤ 225. Finally, there is a restriction on the manufacturing cost because the daily operating budget is $18,000. Thus, 48x + 144y ≤ 18,000, which simpliﬁes to the following by 1 : multiplying both sides by 48 x + 3y ≤ 375. (See assertion (D) on page 35.) Thus we are only interested in all the pairs of integers (x, y) that satisfy all of the following restrictions: ⎧ 0 ≤ x, ⎪ ⎪ ⎨ 0 ≤ y, (1.18) x + y ≤ 225, ⎪ ⎪ ⎩ x + 3y ≤ 375. We refer to (1.18) as a linear system of inequalities, or more simply as a system of inequalities. Suppose we look at (x, y) not as pairs of integers but as pairs of real numbers. Then the set of all (x, y) satisfying each inequality, i.e., the graph of that inequality, is a closed half-plane (Theorem 1.1 of Section 1.4). The set of all pairs of real numbers (x, y) satisfying all the inequalities in (1.18), to be denoted by S, is therefore the intersection of the four closed half-planes deﬁned by the four inequalities. In general, when we have a system of linear inequalities, we will refer to the intersection of the graphs of the linear inequalities as the graph of the system of inequalities. In the context of the optimization problem at hand, this graph S of the system (1.18) is usually called the feasibility region of the problem. In these terms, what we want to know is, among all the pairs of integers (x, y) in the feasibility region, which pair makes the value of the proﬁt function f (x, y) = 35x + 70y the largest? S is represented below by the dotted region; it

46

1. LINEAR FUNCTIONS

is a quadrilateral region (in the sense of a polygonal region on p. 354), being the intersection of the four closed half-planes deﬁned by (1.18). [email protected] @ @ @ @ x + y = 225 @ @ PP @ P PP 125 @ q q PP PP @ q q q q P q [email protected] PP q q q q q q @ q PPP PP @ PPx + 3y = 375 q q q qS q q q q @ PP PP q q q q q q q q @ q PP @ PP P O [email protected] 375 S looks convex, and indeed it is convex because it is the intersection of closed half-planes and closed half-planes are convex (see page 352 for the deﬁnition of convexity). Remark. To show that closed half-planes are convex, let be a line with its two half-planes H+ and H− . Let us show that the closed half-plane ∪ H+ is convex. Since H+ is convex, it suﬃces to show that the segment P Q joining a point P of and a point Q of H+ lies in ∪ H+ . Suppose not, then P Q does not lie in ∪ H+ ; therefore it must contain a point Z of H− . Then the segment ZQ intersects at a point Y . By converting LP Q into a number line (see (L3) on page 214), we have the following situation: Pq

Zq

Yq

Qq

This means the line LP Q intersects at two distinct points P and Y and therefore the two lines coincide. Contradiction. A general strategy towards a solution For the rest of this section, we will be making frequent references to points (x, y) whose coordinates x and y are integers; for simplicity, we will follow the standard mathematical practice of calling these points whose coordinates are integers lattice points. The number of lattice points that satisfy inequalities (1.18) is ﬁnite in number: this is because the ﬁrst two inequalities guarantee that the coordinates of these lattice points are nonnegative integers, and the third inequality, to the eﬀect that x + y ≤ 225, then ensures that the coordinates of these lattice points are among the ﬁnite number of nonnegative integers x and y satisfying 0 ≤ x, y ≤ 225. Therefore, let us say that there are N lattice points, {(xi , yi )} (1 ≤ i ≤ N ), in the feasibility region S. All we have to do now is to look at the N numbers {f (xi , yi )} (1 ≤ i ≤ N ), which are the values of the proﬁt function f (x, y) at these N lattice

1.5. LINEAR PROGRAMMING

47

points, and pick out the largest, let us say, f (xk , yk ), where k is an integer. Then, manufacturing xk items of merchandise A and yk items of merchandise B will yield the largest proﬁt. Problem solved. End of discussion. But is it? At this point, we recall the initial remark that this example is an oversimpliﬁed version of an industrial problem. If the maximum daily production capacity is a small number like 225, then the preceding number N would stay relatively small and this strategy of trial-and-error may be good enough. But if the capacity is not 225 but a six-digit number, if there are 50 rather than 2 diﬀerent kinds of merchandise to be produced,33 and if the daily operating budget is millions of dollars, then it would be impractical to rely on the strategy of trial-and-error to make this kind of industrial decision, for then it would mean getting hold of all such allowable ordered rows of 50 integers (instead of a few hundred ordered pairs of integers) and ﬁnding out the values of the proﬁt function on all these allowable ordered rows of 50 integers in order to see which is the largest. The resulting number of computations—even for computers—is generally too large for comfort and a shorter, more cost-eﬀective method becomes desirable, if not downright mandatory. With this in mind, we will describe a new strategy to solve optimization problems like the Manufacturing Problem on page 44 without resorting to the primitive method of trial-and-error. A more sophisticated discussion of the theory behind this strategy in the general case will be left to the next section. For the Manufacturing Problem about the two kinds of merchandise, A and B, we are supposed to locate a lattice point (x0 , y0 ) that lies in S so that f (x0 , y0 ) ≥ f (m, n) for all other lattice points (m, n) in S. It may not be obvious in the context of school mathematics, but ﬁnding integer solutions to any kind of problems in general is actually very awkward and usually quite subtle. We therefore break up this kind of optimization problem into two subproblems. Subproblem 1. Given a proﬁt function f (x, y) deﬁned on a feasibility region S (cf. (1.18) on p. 45), we look for the point (x0 , y0 ) ∈ S at which f (x, y) achieves a maximum (respectively, minimum); i.e., f (x0 , y0 ) ≥ f (x, y) for all other points (x, y) ∈ S (respectively, f (x0 , y0 ) ≤ f (x, y) for all other points (x, y) ∈ S). If this (x0 , y0 ) is a lattice point, then the optimization problem is already solved. If (x0 , y0 ) is not a lattice point, then we have more work to do. This is because, if (x0 , y0 ) is not a lattice point, let us say x0 = 81.4 and y0 = 92.7, then the solution will make no sense in terms of the given problem since one cannot manufacture 81.4 items of merchandise A and 92.7 items of merchandise B. In that event, the solution of the optimization problem will depend on solving a second problem: Subproblem 2. Devise an algorithm to pass from the maximum (respectively, minimum) point (x0 , y0 ) in Subproblem 1 to a lattice point (m0 , n0 ) in S so that f (m0 , n0 ) is the maximum (respectively, minimum) among all the values f (m, n) of the function f at the lattice points (m, n) in S. Subproblem 2 may turn out to be a formidable problem in real-world situations, but we will cross that bridge when we come face to face with it (see the Advanced Problem in the next subsection). For the beneﬁt of our general understanding of linear functions of two variables, we will concentrate on solving Subproblem 1. 33 Thereby

giving rise to a linear programming problem in 50 variables.

48

1. LINEAR FUNCTIONS

To this end, let us revisit S, the graph of inequalities in (1.18). Our goal is to prove the following lemma. Lemma 1.6. The proﬁt function f (x, y) of the Manufacturing Problem (on p. 44) achieves its maximum in the feasibility region S at one of the vertices of S. Notice that the lemma does not preclude the possibility that f (x, y) may also achieve its maximum at another point of S that is not a vertex. All it says is that if P is not a vertex of S, then f (P ) cannot exceed all the values of f (x, y) at the vertices of S. Assuming this lemma for the moment, we will solve Subproblem 1 of the Manufacturing Problem. So we label the four vertices of the quadrilateral S by A, O, B, and C, as shown, and will compute f (A), f (O), f (B), and f (C). The coordinates of A, O, and B are obvious, but the coordinates of C, being (150, 75), come from solving the following linear system (see Theorem 6.21 on page 359): x + y = 225, x + 3y = 375. @ @ @ x + y = 225 @ @ @ @ PPs @ P P A = (0, 125) q P q PP @ PP @ q q q q qPP s C = (150, 75) @ PP q q q q q @ q PPP PP s s q q q W s @ PP q q q q q P Q @ PP x + 3y = 375 PP q q q q q q q @ q PP s @s PP P O = (0, 0) B [email protected](225, 0) Now that we know the coordinates of these four vertices, we conclude from f (x, y) = 35x + 70y that f (A) = 8,750,

f (O) = 0,

f (B) = 7,875,

f (C) = 10,500.

Therefore, among the four vertices of the convex quadrilateral S, the proﬁt function f (x, y) attains its maximum at the vertex C = (150, 75), where f (C) = 10,500. According to Lemma 1.6, this vertex C in fact enjoys the property that f (x, y) ≤ f (C) for every point in S. In particular, if (m, n) is a lattice point in S, then f (m, n) ≤ f (C). Since C happens to be a lattice point of S, we have solved the Manufacturing Problem. Indeed, the preceding conclusion becomes the statement that if we manufacture 150 items of merchandise A and 75 items of merchandise B, then the proﬁt so obtained—$10,500—is the maximum possible under the given conditions of the problem. Let us give the Proof of Lemma 1.6. Take a point Q on the boundary of S so that Q is not a vertex of the boundary quadrilateral. We claim that f (Q) ≤ f (V ) for at least one vertex V of S. We may assume that Q lies on the side BC of the

1.5. LINEAR PROGRAMMING

49

quadrilateral. By Lemma 1.3 on page 34, f (Q) ≤ either f (B) or f (C). Let us say f (Q) ≤ f (C). Since C is a vertex, the claim is proved with V = C. Next, suppose W is not a boundary point of S. Again, we claim that there is a vertex V of S so that f (V ) ≥ f (W ). Let the horizontal line passing through W intersect side AO of the quadrilateral at P and the side BC at Q, as in the preceding picture. Then W is a point in the segment P Q. By Lemma 1.3 on page 34 again, f (W ) ≤ either f (P ) or f (Q). Let us say f (Q) is the bigger of the two; then (1.19)

f (W ) ≤ f (Q).

Now it could happen that Q is already a vertex of S (for example, one can imagine that W were a bit higher in the preceding picture so that Q would coincide with the vertex C). In that case, our claim would be proved with V = Q. So suppose Q is a boundary point but not a vertex and Q lies on side BC of the boundary of S. Then Lemma 1.3 again shows that f (Q) ≤ the value of f (x, y) at B or C. Let us say f (Q) ≤ f (C). Together with (1.19), we get f (W ) ≤ f (C). Then our claim is proved with V = C. What we have proved is that if we take any point P ∈ S which is not a vertex of the quadrilateral AOBC, then f (P ) ≤ the value of f (x, y) at one of the vertices A, O, B, and C. It follows that if among the four numbers f (A), f (O), f (B), and f (C) we say f (C) is the largest, then clearly f (C) ≥ f (P ) for every point P in S. In other words, f (x, y) achieves its maximum in S at the vertex C. Thus Lemma 1.6 is completely proved. To recapitulate: For the Manufacturing Problem, we found that its proﬁt function f (x, y) must achieve a maximum at a vertex of the feasibility region S. Since among the vertices O, A, B, and C, the value of f (x, y) is largest at C and C is (by luck) already a lattice point, our Manufacturing Problem is solved. An advanced manufacturing problem The reasoning that proves that the linear function f always achieves a maximum at a vertex of S will be seen to remain valid in other similar situations; the detailed justiﬁcation is given in the next section. However, as we mentioned earlier, the good fortune that the vertex at which f (x, y) achieves a maximum happens to be a lattice point is unlikely to prevail often. In that case, we will have to face up to Subproblem 2 on page 47 and take an additional step to get a lattice point (m0 , n0 ) in S so that f (m0 , n0 ) is nevertheless ≥ any f (m, n), where (m, n) is a lattice point in S. This process can be messy, and a complete discussion is beyond the scope of school mathematics. In the following (again oversimpliﬁed) example, which is a variation on the Manufacturing Problem, we give a glimpse of how it can be done. Advanced Problem. A plant has the capacity to produce 70 pieces of merchandise per day. It produces only two kinds of merchandise, A and B. The manufacturing cost of merchandise A is $600 per item, and each brings in a proﬁt of $300. The manufacturing cost of merchandise B is $1,500 per item, and each brings in a proﬁt of $450. The plant has an operating budget of $75,000 per day. How many of eachkind of merchandise should the plant produce each day in order to maximize the proﬁt? (Assume that every item will be sold.)

50

1. LINEAR FUNCTIONS

Suppose each day x items of merchandise A and y items of merchandise B are manufactured. The daily proﬁt is then f (x, y) = 300x + 450y. As before, the feasibility region S (the dotted region below) is deﬁned by 0 ≤ x, 0 ≤ y, x+y

≤ 70,

2x + 5y

≤ 250.

The last inequality, 2x + 5y ≤ 250, is the simpliﬁed version of the inequality 600x + 1,500y ≤ 75,000, which describes the restriction on the daily operating budget. @q 70 Q Q @ x + y = 70 Q @ Q @ Q a ara Q @ A = (0, 50) qaaQQ @ aaQ @ aasC = (33 1 , 36 2 ) q Q q q @Q 3 3 aa @Q aa q @QQaa q q q {f = 26500} @ Q aaa S Q aa q q q @ q q Q aa2x + 5y = 250 Q @ aa Q q q q q q @ q aa @r aa a O = (0, 0) B [email protected] (70, 0) 125

L

As one can see, S is again a quadrilateral, and we have labeled its four vertices by A, O, B, C as before. The coordinates of the point of intersection C of the lines x + y = 70 and 2x + 5y = 250 are obtained by solving the linear system (see Theorem 6.21 on page 359): x + y = 70, 2x + 5y = 250. The solution is seen to be (33 31 , 36 23 ). By Lemma 1.6 (or more precisely, the counterpart of this lemma for the Advanced Problem), the maximum of the proﬁt function f (x, y) = 300x + 450y in the feasibility region S is achieved at a vertex. Since f (O) = 0,

f (A) = 22,500,

f (B) = 21,000,

f (C) = 26,500,

we see that the maximum of f (x, y) is achieved at C, as expected. If we draw the level set L of f (x, y) passing through C—which is the line L = {f = 26,500} in the preceding picture—we see that S lies completely in the half-plane {f ≤ 26,500} (see the comments on pp. 40ﬀ. about how to identify half-planes such as {f ≤ 26,500}). Now, the maximum proﬁt will be 26,500 dollars if one could produce 36 23 items of merchandise A and 33 13 items of merchandise B each day. Of course, it makes no sense whatsoever to speak of “36 23 items” of merchandize or “33 13 items” of merchandise. Therefore, we must now ﬁnd a lattice point (m0 , n0 ) in the feasibility region S so that the value of f on (m0 , n0 ) is largest among all the lattice points in the feasibility region (see Subproblem 2 on page 47 at this point). As mentioned above, this is not easy to do in general, even in the plane. In a real-world situation

1.5. LINEAR PROGRAMMING

51

where the number of variables would not be 2 (the dimension of the plane) but very large and where the number of inequalities would also be very large, the passage from the theoretical maximum point (such as the point C = (36 23 , 33 13 ) in this problem) to a lattice point at which the maximum would be attained by the proﬁt function among all lattice points in the feasibility region would be even more troublesome and would require an elaborate discussion. For the Advanced Problem, however, a simpleminded strategy turns out to be suﬃcient to get it done, as follows (recall that f (C) = 26,500): Step 1. Locate a lattice point (x0 , y0 ) in S near C = (36 23 , 33 13 ). Let f (x0 , y0 ) = k. Observe that k ≤ f (C) because S ⊂ {f ≤ 26,500}. Step 2. Let T be the intersection of S and the closed half-plane {f ≥ k} (but observe that this half-plane contains C, because f (C) ≥ k). Among the lattice points in T , let f (x, y) achieve a maximum at (m0 , n0 ). Then f (m0 , n0 ) ≥ f (a, b) for all lattice points (a, b) in S. Let us brieﬂy explain the reasoning behind this strategy. Since we are trying to locate a lattice point (m0 , n0 ) in S at which f (x, y) is as large as possible and since f (x, y) is already known to achieve its maximum in S at C, it is natural to look for (m0 , n0 ) “near C”. Note that the strategy does not specify precisely what is meant by “near C”, nor does it oﬀer an algorithm for ﬁnding a lattice point near C. As we said, this is only a simpleminded strategy. In any case, let (x0 , y0 ) be one such lattice point near C and let f (x0 , y0 ) = k. The lattice point (m0 , n0 ) that we are looking for in Step 2 must then satisfy f (m0 , n0 ) ≥ f (x0 , y0 ) = k, and we see that (m0 , n0 ) ∈ {f ≥ k}; since (m0 , n0 ) is also in S, we see that (m0 , n0 ) ∈ S ∩ {f ≥ k}. Thus (m0 , n0 ) lies in the region T of Step 2, which is the striped region in the picture below (we have changed the shape of S near C to make it easier to display the region T ). This explains why it suﬃces to look for this (m0 , n0 ) among the lattice points in T , as in Step 2.

O

``@ ``` @ ```` ``` @ r `` C (x0 , y0 ) @ @ T {f ≥ k} @ S @ @ @ {f = k} @@ @

Now back to the Advanced Problem. We know f (x, y) achieves its maximum in S at the vertex C = (33 13 , 36 23 ) and f (C) = 26,500. An obvious candidate for (x0 , y0 ) is (34, 36) (be sure to check that, indeed, (34, 36) lies in S; in fact, it lies on the boundary of S), and k = f (34, 36) = 26,400. The region T is then the intersection of S and the closed half-plane {f ≥ 26,400}, which is the striped region in the picture below. This choice of (36, 34) turns out to be exceptionally felicitous because T traps no other lattice point in S except (36, 34) itself (all the lattice points near C are shown in the following picture and it is clear that there are no

52

1. LINEAR FUNCTIONS

dots in the region T ). Thus (34, 36) is our (m0 , n0 ), by default, and we conclude that by manufacturing 34 pieces of merchandise A and 36 pieces of merchandise B, the resulting proﬁt of $26,400 is the maximum possible under these restrictions. Q aa Q aa Q Q Qa Q r(32, 37) a r(33, 37) r(34, 37) Qaa Q Q a Q Q aaQ Q Q aa Q Q a -Q a {f = 26400} QsC = (33 31 , 36 23 ) Q Q Q T @Q @Q Q Q Q boundary of S Q @ Q Q @ Q S Q @ Q {f = 26500} Q Q [email protected] Q [email protected] Q Q Q @ Qs(34, 36) r(32, 36) r(33, 36) Q Q Q @Q Q @Q Q Q @ If we happened to have chosen (x0 , y0 ) to be (33, 36) instead of (34, 36), then f (33, 36) = 26,100 and the intersection {f ≥ 26,100} ∩ S would trap the following lattice points of S: (30, 38), (32, 37), (34, 36), (33, 36), (35, 35), and (36, 34). By following through Step 2, we see once again that f (x, y) is largest at (34, 36) and, consequently, among all lattice points in S, f (34, 36) is still the largest. Acivity. Verify the statement that (32, 37), (34, 36), (33, 36), (35, 35), and (36, 34) are all the lattice points in {f ≥ 26,100} ∩ S. Exercises 1.5. (1) Let S be the graph of the following system of inequalities: ⎧ 5 ≤ 2, ⎨ 2 x + 34 y −2x + 3y ≤ 12, ⎩ 1 ≤ 5. 3x − y Sketch S by explicitly labeling the coordinates of all the intersections of the lines and the x- and y-intercepts. (2) Let be the linear function on the plane deﬁned by (x, y) = 4x + 5y. Let L be the level set { = 10}, and let H be the set of all the points P so that the line λ passing through P and parallel to L has an x-intercept t < 10 4 . (a) Fixing a point P = (x0 , y0 ) in H. What is the equation of such a line λ in terms of x0 and y0 , and what is the x-intercept of this λ? (b) Prove that H = { < 10}. (3) (a) Let L be the graph of ax + by = c, where a > 0 and b = 0. Let H be the set of all the points P so that the line passing through P and parallel to L intersects the x-axis at a point > ac . Then H = {ax + by > c}. (b) Now suppose a < 0 in part (a), what should the description of H be and why?

1.5. LINEAR PROGRAMMING

53

(4) (a) Let (x, y) = 12 x − 3y be a function on the plane. Write down the equation of the level set of that passes through (−1, −7). (b) Let (x, y) = ax + by, where a, b are constants. Write down the equation of the level set of that passes through a given point (p, q). (5) Let S be the graph of the inequalities −x − 1 ≤ y and y ≤ −x + 1, and let (x, y) = 2x+2y +1. (i) What are the level sets of (x, y)? (ii) Does (x, y) achieve a maximum in S, and if so, where? (iii) Does (x, y) achieve a minimum in S, and if so, where? (6) Let S be the graph of the inequalities −x − 1 ≤ y and y ≤ −x + 1, and let 0 (x, y) = x − 2y + 1. (i) Show directly, without using Lemma 1.3 on page 34, that 0 (x, y) does not achieve a maximum in S. (ii) Show directly, without using Lemma 1.3 on page 34, that 0 (x, y) does not achieve a minimum in S. (7) Let S be the graph of the following inequalities: ⎧ 1 4 ⎪ ⎪ ⎨ 2 x + 5 y ≥ 4, −2x ⎪ ⎪ ⎩ 1 3x

+

5y

≤ 10,

−

1 4y

≤ 2.

Where would the linear function (x, y) = 2x + 13 y − 4 attain its maximum in S? Its minimum? What are the maximum and minimum values of the function? Use a scientiﬁc calculator. (8) Let S be the region in the plane deﬁned by ⎧ −6 ≤ y ≤ 11, ⎪ ⎪ ⎨ y ≥ −2x + 4, y ≤ −2x + 16, ⎪ ⎪ ⎩ y ≥ x − 9. Among the lattice points (see p. 46) in S, where does the function (x, y) = 3x − y achieve its maximum and minimum? (9) A farmer sells 100 bushels of Crop A for $450 and the same amount of Crop B for $650. (Business transactions for these crops are done only in terms of 100 bushels.) The available resources and restrictions are as follows: (a) He has 150 acres of land. It takes 1 acre to produce 100 bushels of Crop A but 3 acres to produce the same amount of Crop B. (b) He has a capital of $9,500. It takes $125 to produce 100 bushels of Crop A and $95 to produce the same amount of Crop B. Let x and y be, respectively, the number of 100 bushels of Crop A and Crop B the farmer produces. (a) Let (x, y) be the revenue (i.e., total dollar intake) the farmer gets for selling 100x bushels of Crop A and 100y bushels of Crop B. Write down an explicit expression of (x, y). (b) Describe precisely, using inequalities, the region S over which you want to locate the maximum of . Sketch S, and also sketch the level set { = 48,750}. (c) Without solving for the maximum of in S, explain as simply as possible, and without quoting theorems, where will achieve its maximum, and why.

54

1. LINEAR FUNCTIONS

(10) The nutritional values of a basic unit of two food items are tabulated below: vitamin C (i.u.) iron (mg) protein (mg) A 50 4.1 13 B 200 1.2 15 A mountain climber wants to bring enough of both items for her trip so that she would get at least 1,500 i.u. of vitamin C, 50 mgs of iron, and 300 mgs of protein. Suppose each unit of item A costs $4.20 and each unit of item B costs $2.80. How many units of each should she buy to minimize the total cost and still meet her nutritional requirements? Use a scientiﬁc calculator. 1.6. Optimization: The general case In this section, we will investigate the general case of linear programming in two variables to provide a more solid conceptual foundation for the consideration of the preceding section. The main theorem in this discussion is Theorem 1.8 whose proof depends on two very technical assertions—Lemma 1.7 below and assertion ( ) on p. 55. We suggest that the proofs of both technical facts be omitted from a class presentation and be relegated to a reading assignment instead. For the statement of the main theorem of this section, Theorem 1.8, we need a deﬁnition and a lemma. A nonempty region that is the intersection of a ﬁnite number of closed half-planes will be called a ﬁnite intersection of closed halfplanes for short. Now a ﬁnite intersection of closed half-planes can be unbounded; e.g., the intersection of the closed upper half-plane of the x-axis and the closed right half-plane of the y-axis is the region which is the union of the ﬁrst quadrant together with the nonnegative x-axis and the nonnegative y-axis. However, we have the following lemma. Lemma 1.7. If a ﬁnite intersection of closed half-planes is bounded, then it is a convex polygonal region.34 More precisely, if a bounded region S is the intersection of a closed half-plane of each of the lines L1 , L2 , . . . , Ln (where n is a positive integer ≥ 3), then S is a convex polygonal region so that each side of its boundary polygon is a segment on a line Lj for some j, 1 ≤ j ≤ n. The proof of this lemma is achieved by the use of mathematical induction on n (see the appendix to this chapter, which is the next section). Mathematical induction is a skill worth learning, but the proof of the lemma is intricate and not particularly instructive. For this reason, the complete proof will be posted instead on the author’s homepage, https://math.berkeley.edu/~wu/. It follows from the lemma that if we have a ﬁnite intersection of closed halfplanes that is bounded, then we can freely speak about its vertices and its sides. With this in mind, the sought-after theorem states: Theorem 1.8. If a bounded region S is a ﬁnite intersection of closed halfplanes, then every linear function achieves a maximum and a minimum at a vertex of S. 34 For a precise deﬁnition of the concept of polygon as well as the concept of convexity , see pp. 352 and 354.

1.6. OPTIMIZATION: THE GENERAL CASE

55

First we give a few comments before the proof. As in Lemma 1.6, the conclusion of the theorem should be understood strictly for what it says: it does not say that a maximum or a minimum of a linear function is achieved only at a vertex of the polygon, but merely that one can always ﬁnd a vertex of the polygon at which the linear function achieves a maximum or minimum. For example, let S be the unit square in the coordinate plane: 1

O

(1, 1)

1

The linear function (x, y) = x achieves a maximum at any point of the right vertical side of the square. For example, (1, 0) = (1, 12 ) = (1, 1) = 1. Thus achieves a maximum at (1, 12 ), which is not a vertex, but since also achieves a maximum at the vertices (1, 0) and (1, 1), there is no contradiction to Theorem 1.8. The assumption that S is bounded is crucial. Without it, the theorem would be false because the linear function may not achieve its maximum or minimum in S at all. For example, let S0 be the closed right half-plane of the y-axis; i.e., S0 = all the points (x, y) so that x ≥ 0. Then the linear function (x, y) = y does not achieve a maximum or a minimum in S0 . The situation does not improve even if S is “cut down” to a smaller region; e.g., we may let S1 be the intersection of the closed upper half-plane of the x-axis and the closed right half-plane of the y-axis, as at the beginning of this section. Even then, the linear function (x, y) = x − y has neither maximum nor minimum in S1 . Finally, we should put one part of Theorem 1.8 in perspective: the fact that a linear function must have a maximum or minimum in S is a very special case of a general theorem, and we will state this theorem without deﬁning all the relevant technical terms: a continuous function on a compact set always achieves a maximum and a minimum on that set. (A linear function is easily shown to be continuous, and a bounded ﬁnite intersection of closed half-planes is compact.) A proof of this general theorem can be found in any textbook on real analysis, e.g., [Rosenlicht, Chapter 4, §4]; the analog of Theorem 1.8 for a continuous function deﬁned on a closed bounded interval of R will be proved in Section 6.2 of [Wu2020c]). What makes Theorem 1.8 special is the precise speciﬁcation that the maximum or the minimum will be found at a vertex of S. Proof of Theorem 1.8. The proof is quite similar to the proof of Lemma 1.6 on page 48. We will prove the case of maximum and leave the case of minimum to Exercise 1 on page 57. By Lemma 1.7, S is a convex polygonal region. Let a linear function of two variables (x, y) be given. We claim that if a point W of S is not a vertex, then there is a vertex V of S so that (V ) ≥ (W ). First, let W be a point in S not lying on the boundary of S. We will show: () The horizontal line L passing through W intersects the boundary of S at two points P and Q so that the segment P Q contains W .

56

1. LINEAR FUNCTIONS

Granting () for the moment, we will ﬁnish the proof of the theorem. By Lemma 1.3 on page 34, (W ) ≤ (P ) or (Q). Let us say (W ) ≤ (P ). Suppose P lies on the side A1 A2 of S. By Lemma 1.3 again, (P ) ≤ (A1 ) or (A2 ). Thus (P ) is no bigger than the value of (x, y) at one of the vertices of S, namely, A1 or A2 . Let us say (P ) ≤ (A2 ). Then (W ) ≤ (A2 ). Our claim is therefore proved in this case with V = A2 . If, on the other hand, W is a point on the boundary of S, let us say W lies on the side A3 A4 , then Lemma 1.3 once more shows that (W ) ≤ (A3 ) or (A4 ). For deﬁniteness, let (W ) ≤ (A3 ). Then the claim is proved in this case too with V = A3 . In summary, we have proved that if we take any point W in S that is not a vertex, then there is a vertex V so that f (V ) ≥ f (W ). Therefore if, let us say, (A5 ) is the largest number among (A1 ), . . . , (An ), then for any W in S, we have (W ) ≤ (A5 ). So (x, y) achieves its maximum at the vertex A5 . The proof of Theorem 1.8 is complete if we assume (). We now confront the issue of proving assertion (). Looking at the preceding picture, you would wonder why we need to prove something this obvious. For the record, let it be known that ( ) is valid for what is known as a compact region S, and it would take something subtle like part (ii) of Theorem 4.13 from [Wu2020a] (recalled on page 358 of this volume) to prove ( ) in this generality (keep in mind that we did not provide a proof for Theorem 4.13 either because it would require the kind of reasoning that is suitable only for upper division college courses). In a high school classroom, you should just assume it and move on, the same way we are going to simply assume the intermediate value theorem and move on when we come to it on page 122. That said, we should point out that there is actually an elementary proof of ( ) that avoids any subtle reasoning; this proof exploits the fact that the S in ( ) is a ﬁnite intersection of closed half-planes. Because this proof is tedious and not particularly instructive, it too will be posted instead on the author’s homepage, https://math.berkeley.edu/~wu/. In summary, the main point of Theorem 1.8 is that in order to ﬁnd the points at which a linear function achieves a maximum or minimum in a convex polygon which is the graph of a system of inequalities, we locate the (ﬁnite number of) vertices of the region by solving (a ﬁnite number of) systems of linear equations associated with the inequalities. Then we examine the values of the given function at these vertices (ﬁnite in number): the vertices at which the function is biggest are the points we seek. This is the essence of the simplex method in linear programming. The great utility of this method stems from the fact that even when the number of inequalities deﬁning the feasibility region as well as the number of

1.7. APPENDIX: MATHEMATICAL INDUCTION

57

variables is very large, the present ability to solve systems of linear equations very eﬃciently by computers leads quickly to a determination of all the vertices and hence a quick solution of the optimization problem. Exercises 1.6. (1) Write out a self-contained proof of Theorem 1.8 on page 54 for the case of a minimum. (2) Let S be the graph of the following inequalities: ⎧ ⎨ 3x + 2y ≤ 6, 2x − y ≥ −2, ⎩ x − 5y ≤ 5. Let (x, y) = x + 3y. Without quoting Theorem 1.8, explain directly why achieves its minimum in S at a corner of S. Determine which corner it is and what the value of is at that corner. (3) Let S be the region deﬁned by the inequalities 3x − 3y ≤ 5, 2x + y ≤ −4. Does the linear function (x, y) = x + 32 y achieve a maximum in S? Does it achieve a minimum in S? Explain. (4) Let a region S be an intersection of closed half-planes so that its boundary does not contain a whole line. If a linear function (x, y) = ax + by + d achieves its maximum or minimum in S, prove that it already does so at a “vertex” of S (i.e., at the point of intersection of two noncollinear segments at the boundary). (5) Prove that if a linear function of two variables achieves either a maximum or a minimum in a general region S at some (x0 , y0 ) ∈ S, then (x0 , y0 ) lies on the boundary of S 1.7. Appendix: Mathematical induction In abstract terms, mathematical induction refers to a general method that conﬁrms that certain subsets of the positive integers, to be denoted by Z+ , must be Z+ itself. In practical terms, it is a method for writing down a proof—in a simple manner—that some statement about a positive integer n is true for all n ∈ Z+ . Both sound too vague, but everything will become clear as soon as we work through an example: see Example 1 immediately following. We also call attention to the fact that there is a slight variant of the standard form of mathematical induction, and it is explained on page 182. Example 1. For all positive integers n, the following equality is valid: n(n + 1) . 2 What this says is that if Pn denotes the equality (1.20), then P1 , P2 , P3 , . . . are all valid. Equivalently, this can be rephrased as claiming that if A is the subset of positive integers consisting of all the n so that Pn is valid, then A = Z+ . (1.20)

(1 + 2 + 3 + · · · + n) =

58

1. LINEAR FUNCTIONS

In principle, we can easily check the validity of any Pn when a speciﬁc n is given. For example, if n = 6, then 1 + 2 + 3 + 4 + 5 + 6 = 21 and clearly 6(6 + 1) = 21. 2 So P6 is valid. If n = 25, then more patience would be needed to check that both sides are equal to 325, but there is no doubt that it can be done. But to show that Pn is valid once and for all for any n ∈ Z+ clearly calls for something more than patience. This is where the principle of mathematical induction enters the picture. It says that to show A = Z+ , it suﬃces to show that 1 ∈ A. For any positive integer k, if k ∈ A, then (k + 1) ∈ A. In more down-to-earth language, what we are saying is that Pn will be valid for all n ∈ Z+ if we can prove the following two assertions: (MI-1) P1 is valid. (MI-2) For any positive integer k, if Pk is valid, then Pk+1 is also valid. This is the method ﬁrst precisely formulated by Blaise Pascal in the process of creating the theory of probability with Fermat.35 The reason behind the validity of this principle is very simple. (MI-1) guarantees that P1 is true. But with k = 1, (MI-2) says P2 is now true. Applying (MI-2) with k = 2, we see that P3 is true. Applying (MI-2) again but with k = 3, we see that P4 is true. Applying (MI-2) yet again with k = 4, we get the truth of P5 , etc. After repeating the same reasoning n − 1 steps, we get the truth of Pn for any n ∈ Z+ , which is exactly what we want. (You can think of this as a kind of domino eﬀect.) Remark. We start with P1 , because this is the most common situation. However, it does not have to be P1 . Later, we will see in the proof of the Lemma 6.1 on page 232 an example of a proof by induction that starts with P2 . In general, one could start with P for any positive integer and show the validity of P , P+1 , P+2 , . . . by proving the following two assertions: (MI-1) P is valid. (MI-2) For any positive integer k ≥ , if Pk is valid, then Pk+1 is also valid. It is time to go back to proving equation (1.20). First, we prove P1 is valid, but this is trivial as it merely says that 1 = 1. The second step is to prove that Pk =⇒ Pk+1 ; i.e., if Pk is valid, then so is Pk+1 . Now what Pk says is this: (1.21)

(1 + 2 + 3 + · · · + k) =

k(k + 1) . 2

35 Pascal (1623–1662) was a French mathematician and physicist and was a contemporary of Fermat and Descartes. We will encounter him again in Section 5.4 below. He was a prodigy: he proved the Pascal theorem on hexagons, a cornerstone of projective geometry, at age sixteen. He and Fermat were the cofounders of the theory of probability. Pascal’s Principle is basic in ﬂuid mechanics. But for all that, he is actually more famous in France as a writer and a theologian than as a mathematician or physicist.

1.7. APPENDIX: MATHEMATICAL INDUCTION

59

We have to prove that Pk+1 is valid. What Pk+1 says is (k + 1)(k + 2) . 2 We simply compute the left side to see if it is equal to the right side: (1 + 2 + 3 + · · · + k + (k + 1)) =

(1 + 2 + · · · + k + (k + 1)) =

(1 + 2 + · · · + k) + (k + 1)

=

k(k + 1) + (k + 1) 2

=

k(k + 1) + 2(k + 1) 2

(using (1.21)

(k + 1)(k + 2) (distributive law). 2 So (MI-2) is also veriﬁed. Therefore by the principle of mathematical induction, we can be conﬁdent that Pn is valid for all n. We pause to point out that equation (1.20) can be proved directly without using mathematical induction, as follows. With n ﬁxed, write S = 1 + 2 + · · · + n. Now S can also be written as a “decreasing sum”; i.e., S = n + (n − 1) + · · · + 2 + 1. Therefore,

2S = S + S = 1 + 2 + · · · + (n − 1) + n + n + (n − 1) + · · · + 2 + 1 . =

Now, Theorem 1 in the appendix of Chapter 1 in [Wu2020a] (recalled on page 358 of this volume) allows us to add the terms of the two sums “vertically” and we get 2S = (1 + n) + (2 + (n − 1)) + · · · + ((n − 1) + 2) + (n + 1). Therefore, 2S = (n + 1) + (n + 1) + · · · + (n + 1) = n(n + 1).

n

Thus S = 12 n(n + 1), which is equation (1.20). The point of the preceding observation is that it is not necessary to use mathematical induction to prove every assertion about whole numbers, but it never hurts to have another weapon in one’s arsenal. The reader may have already observed that one can use the principle of mathematical induction to prove something only when one knows ahead of time what is to be proved. Therefore, it is an eﬀective tool for a streamlined exposition of a special class of proofs when we have a correct guess to begin with (compare Exercise 5 on page 61). In that event, what mathematical induction can accomplish is to write down the proof of the validity of an inﬁnite sequence of assertions Pn without having to resort to sprinkling the phrase “and so on” all over the place. Instead, it reduces such proofs to the veriﬁcation of just two simple steps, (MI-1) and (MI-2). As a practical matter, it is usually helpful to write down the proofs of the ﬁrst few cases, P1 , P2 , P3 , P4 , etc., to discover the general structure of the induction argument before carrying out (MI-2). Example 2. by 36. 36 This

36

For all positive integers n, the integer 7n − 6n − 1 is divisible

is an exercise in [Ross], page 4, but see also Exercise 8 on page 61 of this volume.

60

1. LINEAR FUNCTIONS

Let Pn be the assertion “7n − 6n − 1 is divisible by 36”. To show that Pn is true for all positive integers n, we verify (MI-1) and (MI-2). Since 71 − 6 · 1 − 1 is 0, it is certainly divisible by 36. However, 0 is so special that one ought to be suspicious about such a beginning, so it would be advisable, psychologically, to try P2 . Then we get 72 − 6(2) − 1 = 36, and of course 36 is divisible by 36. This takes care of (MI-1). Next suppose Pk is true, and we must show Pk+1 is true. Now Pk says 7k − 6k − 1 is divisible by 36, and we have to check that 7k+1 − 6(k + 1) − 1 is also divisible by 36. Clearly if we could express 7k+1 − 6(k + 1) − 1 in terms of 7k − 6k − 1, we would be in good shape. Once this is realized, it becomes quite obvious what must be done: 7k+1 − 6(k + 1) − 1 = 7 · 7k − 6k − 7 = (7 · 7k − 7 · 6k − 7) + 36k = 7(7k − 6k − 1) + 36k. We know (7k − 6k − 1) is divisible by 36 because Pk is true, and 36k is of course divisible by 36. Thus 7(7k − 6k − 1) + 36k is divisible by 36, and therefore so is 7k+1 − 6(k + 1) − 1. This veriﬁes (MI-2), and the proof is complete. Incidentally, this example may be diﬃcult to prove without using mathematical induction. Exercises 1.7. (1) Out of a container containing marbles of diﬀerent colors, Tom randomly puts a few in a bag and tells everybody that his bag consists of marbles all of the same color. When challenged, he proves his claim as follows. Let Pn be the statement that a set of n marbles must be all of the same color. Tom uses mathematical induction to prove Pn is true for all whole numbers n. First, P1 is obviously true. Next, suppose Pk is true. He then proves that Pk+1 is true in the following way. In a set of k + 1 marbles, he names the marbles m1 , m2 , . . . , mk+1 . The set consisting of the k marbles {m1 , m2 , . . . , mk } must be marbles of the same color, by the truth of Pk . Likewise, the set consisting of the k marbles {m2 , m3 , . . . , mk , mk+1 } must also be marbles all of the same color, by the truth of Pk again. Since these two sets of marbles have the marbles {m2 , m3 , . . . , mk } in common, the two colors must be the same, and thus the original set of k + 1 marbles are of one and the same color. Do you believe his argument? Why or why not? (2) Prove that for all integers n ≥ 4, 3n + 1 ≤ 2n . (3) Prove that for any positive integer n, 13 + 23 + 33 + · · · + (n − 1)3 + n3 = (1 + 2 + 3 + · · · + n)2 . (4) (a) Is the following sum of ﬁve consecutive integers, 4+5+6+7+8, divisible by 5? (b) Is the following sum of six consecutive integers, 4+5+6+7+8+9 divisible by 6? (c) Formulate a conjecture about the divisibility of the sum of n consecutive integers by n and prove it.37 37 Due

to R. A. Askey.

1.7. APPENDIX: MATHEMATICAL INDUCTION

61

(5) Consider the sum of the ﬁrst n odd positive integers 1+3+5+· · ·+(2n−1). (a) Find the sum for n = 2, 3, 4, 5, and guess a general formula. (b) Prove the general formula by mathematical induction. (6) Use mathematical induction to prove the formula for the geometric series, i.e., 1 − xn+1 1 + x + x2 + · · · + xn = 1−x for any number x = 1 and for any positive integer n. (7) Let a, b be ﬁxed numbers. Find the following sum for any positive integer n: a + (a + b) + (a + 2b) + (a + 3b) + · · · + (a + (n − 1)b) + (a + nb). (8) Prove that if a is a positive integer ≥ 2, then (a − 1)2 divides the integer an − (a − 1)n − 1.38 (If a = 7, this exercise reduces to Example 2 on page 59.)

38 Due

to O. Hald.

CHAPTER 2

Quadratic Functions and Equations After linear equations of two variables and linear functions of one variable, the next topics we tackle are quadratic functions of one variable and equations of degree 2 in two variables. The study of quadratic functions of one variable has a venerable history; it can be traced to some of the earliest written records of human civilization. For example, the quadratic formula was essentially known to the Babylonians about thirty-seven centuries ago.1 Our exposition will underscore the importance of the technique of completing the square and the utility of the basic isometries for the understanding of quadratic functions of one variable. At the end of the chapter, we will initiate a discussion of the graphs of equation of degree 2 in two variables, i.e., the conic sections, concentrating on the case of equations “without a mixed term”. This topic, especially the concept of a parabola, has been much abused in TSM2 . The conclusion of this discussion will be found in Section 1.7 of [Wu2020c].

2.1. Quadratic functions The study of quadratic equations and functions of one variable is a mainstay of introductory algebra. Obviously, this study is only one step removed from the study of the most elementary topics in school algebra: linear equations and functions. Since the study of the latter is greatly simpliﬁed by the fact that their graphs are the simplest of geometric objects—lines—it is natural to investigate whether the graphs of quadratic functions might not be almost as simple. Indeed they are, in spite of the fact that this simplicity has remained hidden in TSM. Precisely, these graphs are all similar (in the precise sense given on page 355) to the graph of the familiar square function, f (x) = x2 (see Theorem 2.11 on page 88). Furthermore, the graph of any quadratic function f (x) = ax2 + bx + c is the translation of the graph of fa (x) = ax2 (with the same leading coeﬃcient a; see page 73). It is therefore not surprising that, by making the graphs the focus of the study of quadratic equations and functions, we restore simplicity and unity to this study. In particular, this geometric viewpoint sheds light on the quadratic formula, the technique of completing the square (see Lemma 2.2 on page 68 and Theorem 2.6 on page 73), and the behavior of roots in a general quadratic equation.3

1 See,

for example, Chapter 1 of [Katz]. page xi for the deﬁnition of TSM. 3 It may be remarked that, although at the heart of the study of quadratic functions is the technique of completing the square, it is possible to approach it from the vantage point of diﬀerential calculus without making use of this technique. See the appendix of Section 6.4 in [Wu2020c]. 2 See

63

64

2. QUADRATIC FUNCTIONS AND EQUATIONS

Square roots and basic properties (p. 64) A special class of quadratic functions and their graphs (p. 65) General quadratic functions and completing the square (p. 72) Quadratic equations of one variable (p. 78) Completing the square, revisited (p. 80) Signiﬁcance of the zeros of a quadratic function (p. 81) Square roots and basic properties We begin with a preliminary discussion of the concept of a square root, which is more subtle than usually realized. Let s be a number. A number r is called a square root of s if r 2 = s. To make sense of the concept of a square root, we begin by trying to make sense of the equality r 2 = s. Notice that, ﬁrst of all, if s < 0, then s cannot have a square root because (2.1)

r 2 > 0 for any number r = 0.

To see this, ﬁrst let r be rational. If r > 0, we multiply both sides of r > 0 by the positive number r and inequality (D) on page 35 shows that r · r > r · 0, so that r 2 > 0. If r < 0, we multiply both sides of r < 0 by the negative number r and inequality (E) on page 35 shows r · r > r · 0 again, so that r 2 > 0 and (2.1) is proved when r is rational. By FASM (page 351), the preceding reasoning also holds for an irrational r and, therefore, (2.1) holds in general. Next we consider r = 0. Obviously 0 is a square root of 0. We claim that 0 is the only square root of 0. Indeed, if r 2 = 0, then (2.1) shows that r cannot be nonzero. Thus r = 0, proving the claim. At this point, we know that we only need to consider an s > 0 and ask if it has a square root, i.e., if there is a real number r so that r 2 = s. The key fact in this context is: Lemma 2.1. Given a positive number √s, 2then there is one and only one positive √ number, to be denoted by s, so that ( √ s) = s. Moreover there are exactly two real numbers r so that r 2 = s, namely, ± s. Proof. The existence of a positive number whose square is equal to s is not trivial and, in fact, cannot be proved until after we have made explicit what we assume about the real numbers. (See Section 2.5 of [Wu2020c], but also Theorem 4.2 on page 143 of the present volume.) Because there is no fear of circular reasoning, we are free to assume for now that there is such a number and proceed to prove that there is only one; i.e., if r1 and r2 are both > 0 and (r1 )2 = (r2 )2 = s, then r1 = r2 . This can be proved very simply as follows. If (r1 )2 = (r2 )2 = s, then (r1 )2 − (r2 )2 = 0, so that (r1 − r2 )(r1 + r2 ) = 0. Since both r1 and r2 are positive, so is r1 + r2 . In particular, (r1 + r2 ) = 0. By Corollary 1 of Theorem 2.9 in [Wu2020a] (see page 358 of this volume), necessarily (r1 − r2 ) = 0; i.e., r1 = r2 . (Notice that the validity of this argument when r1 or r2 is irrational, or both, requires FASM again.) As stated in the lemma, we denote this positive √ number by s. To ﬁnish the proof of√Lemma 2.1,√we have to prove that if r is any have real so that r 2 = s,√then r = s or r = − s. Indeed, √ number √ because√we also √ 2 ( s) = s, therefore r 2 − ( s)2 = s − s = 0. Since r 2 − ( s)2 = (r − s)(r + s), we get √ √ (r − s)(r + s) = 0.

2.1. QUADRATIC FUNCTIONS

65

√ √ √ Therefore√a similar argument shows that r − s = 0 or r + s = 0. Thus r = s or r = − s, as claimed. The proof of Lemma 2.1 is complete. √ The mathematical terminology to express “only one” is “unique”. The number s is called the positive square root of s. Thus the positive √ square root of a positive number is unique. Notice that, by deﬁnition, if s > 0, s is always > 0. When there is no danger of confusion, the positive square root of s is sometimes referred to as just the square root of s. The signiﬁcance of the fact that √ every positive number s has a unique positive square root—which we denote by s—will emerge when we deal with complex numbers (see page 195). In general, there has been no eﬀort in TSM to bring out the importance of the concept of uniqueness; this should be done no later than high school. Because this concept is fundamental to mathematics, we have to make a concerted eﬀort to spread this message in school mathematics. √ Corollary. If a is a number, then a2 = |a|. √ Proof. If a = 0, then we already pointed out that 02 = 0 = |0|. Suppose a = 0. Then |a| > 0 and |a|2 = a2 . This says |a| is a positive square root of a2 . By the uniqueness part √ of the lemma, |a| has to be the unique positive square root of 2 a , or in symbols, a2 = |a|, as claimed. An application of Lemma 2.1 is the proof of the critical fact that √ √ √ ab = a b for all positive a, b. (2.2) Because we have√just seen that there is only one positive number whose square is equal to ab and ab is by deﬁnition that number, all√ we need to do is to show √ that √ √ √ the square of ( a b) is also equal to ab. Then a b must be equal to ab, by uniqueness. But this is so because4 √ √ √ √ √ √ √ √ √ √ ( a b)2 = ( a b)( a b) = ( a a)( b b) = ab, √ √ √ √ √ where the last step is by the deﬁnition of a and b. So ab = a b after all. In school mathematics, the identity (2.2) is often either left unexplained or relegated to veriﬁcation by calculator for a few special cases. We have to do much better than that. One will have a better appreciation of this seemingly mundane identity (2.2) after reading the discussion about complex numbers on page 195. A special class of quadratic functions and their graphs Recall from page 4 that a quadratic function of one variable is, by deﬁnition, a function f deﬁned by f (x) = ax2 + bx + c, where a, b, c are ﬁxed numbers and a = 0, or a function that can be brought to this form through the use of the associative, commutative, and distributive laws. By abuse of language, we say a quadratic function f (x) is a quadratic polynomial in x. The simplest quadratic function is the square function s : R → R deﬁned by s(x) = x2 for all numbers x. We have seen its graph in Section 1.1. The next 4 By

FASM again, the multiplication of real numbers is commutative.

66

2. QUADRATIC FUNCTIONS AND EQUATIONS

simplest quadratic function is fa (x) = ax2 , where it will always be understood that a = 0. We will see presently that these functions fa are no less simple than the square function s. It would seem that the lower-order terms bx + c in a general quadratic function f (x) = ax2 +bx+c confound our understanding of the latter, but we will show that such is not the case. As a preliminary ﬁrst step, this subsection focuses on a special class of quadratic functions of the form fa (x − p) + q for some numbers p and q. (It would be helpful to recall the discussion on pp. 9ﬀ. of the graphs of the cognate functions of a given function.) Observe that, accepting the notation of fa (x) = ax2 , we have s = f1 , and we shall henceforth use the notation f1 for s instead. Consider ﬁrst the case a > 0. Then its graph is the set of all ordered pairs (x, ax2 ), where x is any number. Let 0 < a < 1 < b. Because x2 > 0 for x = 0, we see that 0 < ax2 < x2 < bx2 for any x = 0 (see (D) on page 35). Thus 0 < fa (x) < f1 (x) < fb (x) for all x = 0. Therefore above a nonzero x on the x-axis, the point on the graph of fa is below that of f1 but the graph of fb would be above that of f1 . (All these graphs pass through the origin (0, 0), of course.) The following picture on the left gives the graphs of f1 , f2 , and f 12 for illustration. f2

10

f1

f 12

O −4

4

−6

2 −2

2 −2

6

−4

−2

O

2

4

−10

f−2 f −1

f− 12

For the case of a < 0, the graph of fa , being the set of all ordered pairs (x, ax2 ), is nothing but the reﬂection across the x-axis of the graph of f|a| , which is the set of all points (x, |a|x2 ). We would have, for instance, the graphs of function in the above picture on the right. We now make some observations about the graph of fa in general for any a = 0. For (c) below, we adopt the intuitive terminology that the graph of a function falls as we move to the right if, for any two points (a, b) and (c, d) on the graph satisfying a < c, we have b > d as shown in this picture:

b

.(a,b ) .(c,d )

d

O

a

c

2.1. QUADRATIC FUNCTIONS

67

Similarly, we say the graph rises if we move to the right, if, for any two points (a, b) and (c, d) on the graph satisfying a < c, we have instead b < d. Then the observations are the following: (a) The y-axis is a line of symmetry in the sense that the reﬂection with respect to the y-axis maps the graph of fa onto itself. (b) If a > 0, the lowest point of the graph, which is (0, 0), is on the line of symmetry. If a < 0, the highest point of the graph, also (0, 0), is likewise on the line of symmetry. (c) Let a > 0. Then on the negative x-axis, the graph of fa falls as we move to the right, and then the graph rises on the positive x-axis as we continue to move to the right. The situation in case of a < 0 is the opposite; it rises as we move to the right until (0, 0) and then falls as we continue to move to the right from (0, 0). The reason for (a) is that for each x, both points (x, ax2 ) and (−x, ax2 ) are on the graph and the reﬂection with respect to the y-axis maps (x, ax2 ) to (−x, ax2 ) and vice versa. The reason for (c) is that if a > 0, for instance, then x1 < x2 < 0 implies 0 < (−x2 ) < (−x1 ), so that 0 < (−x2 )2 < (−x1 )2 , which in turn implies 0 < a(−x2 )2 < a(−x1 )2 . Therefore 0 < ax22 < ax21 , which is the same as fa (x2 ) < fa (x1 ). Thus the point (x1 , f (x1 )) is higher than the point (x2 , f (x2 )). Similar reasoning holds for the other cases. The next simplest case is that of ga (x) = ax2 + q for some ﬁxed number q. For ease of discussion, let us assume for the time being that a > 0. The graph of ga consists of all ordered pairs (x, ax2 + q), which is clearly the image of the graph of fa under the translation (x, y) → (x, y + q) (see Lemma 6.20 on page 358). The following shows the three cases of q < 0, q = 0, and q > 0 in succession from left to right. Notice that there is no change in the line of symmetry and the fact that the lowest point of the graph is on the line of symmetry.

q

O

qq

q

O

qq q

O

At this point, it is not so clear what the “next simplest” case ought to be, but cumulative wisdom of the past (going all the way back to the Babylonians circa 1800 BC) saves the day. We learn that the decisive idea is to consider the quadratic function ha (x) = a(x − p)2 + q as the next step, where p, q are some ﬁxed numbers. Thus, ha (x) = fa (x − p) + q. (Recall at this point the discussion of the graphs of the cognate functions of a

68

2. QUADRATIC FUNCTIONS AND EQUATIONS

given function on pp. 8ﬀ.) Note that this a quadratic function because ha (x) = ax2 − (2ap)x + (ap2 + q). We will now make an eﬀort to understand this ha because, as we shall see presently, every quadratic function can be written in this form. Still with a > 0, we ﬁrst plot the graph of ha to get an intuitive idea of what it looks like. Recall that we are assuming a > 0. The following three graphs show the case of a positive p for q < 0, q = 0, and q > 0, as above.

O

.. . .. . .. . .. . .. .q ..p .q ..(p, q) .

O

.. . .. . .. . .. . .. .q ..p . .. .

O

.. . .. . .. . .. . ..q(p, q) .q ..p . .. .

From the pictures, it is intuitively clear that the graph of ha can be obtained from the graph of fa (recall fa (x) = ax2 ) ﬁrst by a translation along the x-axis, (x, y) → (x + p, y), followed by another translation along the y-axis, (x, y) → (x, y + q). In other words, the pictures suggest that the translation T deﬁned by T (x, y) = (x + p, y + q) maps the graph of fa to the graph of ha , where ha (x) = a(x − p)2 + q. The case of a < 0 can be handled in a similar manner. Thus, the graph of ha for a negative a is illustrated below, for q > 0, q = 0, and q < 0, in succession from left to right.

O

..(p, q) .q .. .q ..p . .. . .. . .. . .. .

O

.. . .. .q ..p . .. . .. . .. . .. .

O

.. . ..p .q ..(p, q) .q .. . .. . .. . .. .

Just as in the case of a > 0, it is intuitively clear also, when a < 0, that the translation T (x, y) = (x + p, y + q) maps the graph of fa onto that of ha . We now replace intuitive feelings with proofs. Lemma 2.2. For any a = 0, the graph of the quadratic function ha (x) = a(x − p)2 + q is congruent to the graph of fa (x) = ax2 under the translation T (x, y) = (x + p, y + q). Remark. Pay attention to the fact that the translation that maps the graph of fa to the graph of ha (x) = a(x − p)2 + q is T (x, y) = (x + p, y + q) but not T (x, y) = (x − p, y + q).

2.1. QUADRATIC FUNCTIONS

69

Proof. Let the graphs of fa and ha be Fa and Ha , respectively. Fa consists of all the ordered pairs (x, ax2 ) for all numbers x. Similarly, Ha consists of all ordered pairs (x, a(x − p)2 + q) for all numbers x. The picture below is for a positive a.

Y

Ha

Fa 2 1 r P = (x + p, ax + q) T (x, ax2 ) r q 1r V = (p, q) X O p

The claim is that T (Fa ) = Ha . This requires as usual that we prove T (Fa ) ⊂ Ha

and Ha ⊂ T (Fa ).

First, we prove T (Fa ) ⊂ Ha . Let a point P ∈ T (Fa ). This means P = T (x, ax2 ) for some number x. Thus P = (x + p, ax2 + q) = ((x + p), a((x + p) − p)2 + q). If we write u = x + p, then P = (u, a(u − p)2 + q), which is equal to (u, ha (u)). Thus P lies on the graph Ha of ha . We have proved T (Fa ) ⊂ Ha . Next we prove Ha ⊂ T (Fa ). Let P ∈ Ha . Then P = (x, a(x − p)2 + q) for some number x. By the deﬁnition of T , we have P = T (x − p, a(x − p)2 ). Writing v = x − p, we have P = T (v, av 2 ), and since (v, av 2 ) ∈ Fa , we conclude that P ∈ T (Fa ). The proof of T (Fa ) = Ha is complete. Lemma 2.2 plays a fundamental role in our quest to understand quadratic functions. By showing that the graph of ha is congruent to the graph of fa = ax2 under the translation T (x, y) = (x + p, y + q), we establish the fact that ha is as transparent as the simple function fa . For example, if a part of Fa (the graph of fa ) falls as we move to the right (see page 67), the same must be true of its image in Ha (the graph of ha ). Indeed suppose (u, v) and (c, d) are on Fa so that u < c and v > d. Let the translated images T (u, v) and T (c, d) in Ha be (m, n) and (s, t), respectively; i.e., T (u, v) = (m, n) and T (c, d) = (s, t). Then we have to show that (2.3)

m t.

This is because (m, n) = T (u, v) = (u + p, v + q) so that m = u + p and n = v + q. Similarly, (s, t) = T (c, d) = (c + p, d + q) so that s = c + p and t = d + q.

70

2. QUADRATIC FUNCTIONS AND EQUATIONS

Ha n v t

Fa

r(m, n) r(u, v) r(s, t) r(c, d)

d

O

u

c

m

s

Therefore m = u + p < c + p by (B) on page 35. Since s = c + p, we have m < s, which is the ﬁrst part of (2.3). Next, t = d + q < v + q = n, for exactly the same reasons, so the second part of (2.3) is also true. Our claim is proved. For similar reasons, if a part of Fa (the graph of fa ) rises as we move to the right (see page 67), the same is also true of its image in Ha (the graph of ha ). Obviously we know all there is to know about fa and, therefore, also about ha . It follows that as soon as we can show that every quadratic function is equal to some function of the form ha (x) = a(x − p)2 + q for a suitable p and q (Theorem 2.6 below), the theory of quadratic functions will be essentially complete. We proceed to translate the congruence-by-translation statement of Lemma 2.2 into function-theoretic terms. Lemma 2.3. Let ha (x) = a(x − p)2 + q for some constants a, p, and q, with a = 0, and let Ha be its graph. Then: (a) The vertical line x = p is a line of symmetry (or an axis of symmetry) of Ha in the sense that the reﬂection with respect to x = p maps Ha onto itself. (b) If a > 0, the lowest point of Ha is (p, q), which is on the line of symmetry. If a < 0, the highest point of Ha , which is also (p, q), is likewise on the line of symmetry. (c) Ha intersects the x-axis either twice, once, or not at all. In the ﬁrst case, the two points of intersection are symmetric with respect to the axis of symmetry in the sense that the reﬂection across the line of symmetry maps the points onto each other. In the second case, q = 0 and the point of intersection is (p, 0). Proof. For part (a), it is because the translation T deﬁned by T (x, y) = (x + p, y + q) clearly maps the y-axis to the vertical line x = p. For part (b), It suﬃces to note that T (0, 0) = (p, q). As to (c), ﬁrst let a > 0. Then by property (c) on page 67 of the graph fa and by the observation above (2.3), Ha falls (moving to the right) in the left half-plane of the vertical line x = p, and therefore if Ha intersects the x-axis to the left of x = p at all, it can only do so at most once. If it does intersect the x-axis once to the left of x = p, say at (p − t, 0) where t > 0, then it must also do so to the right of x = p at (p + t, 0) because the reﬂection of (p − t, 0) across x = p is exactly (p + t, 0). Thus we have two points of intersection between Ha and the x-axis. Suppose Ha does not intersect the x-axis to the left of the line of symmetry x = p; then Ha does not intersect the x-axis to the right of x = p either because of symmetry. In this case, either Ha does not intersect the x-axis altogether or Ha intersects the x-axis on the line of symmetry x = p. If

2.1. QUADRATIC FUNCTIONS

71

the latter, then the point of intersection between Ha and the x-axis is the same as the point of intersection between the x-axis and the line of symmetry x = p, i.e., (p, 0). By part (b), (p, 0) = (p, q) and therefore q = 0. The proof of Lemma 2.3 is complete. It is also worthwhile to give a direct proof of the fact that the vertical line x = p is a line of symmetry of the graph Ha , as follows. Any number x can be written as p + t for some number t; in fact, simply let t = x − p. Notice that ha (x) then becomes ha (p + t) = a((p + t) − p)2 + q = at2 + q, so that ha (p − t) = a(−t)2 + q = at2 + q = ha (p + t). Now since the point p − t on the x-axis is the reﬂection of p + t with respect to the line x = p, the reﬂection of the point (p − t, ha (p − t)) with respect to the line x = p is (p + t, ha (p + t)). Thus the reﬂection with respect to the vertical line x = p maps the graph Ha onto itself. We make two additional observations. Let a > 0. The point (p, q) being on the graph of ha means, by deﬁnition of the graph, that ha (p) = q. Now take any point (x, ha (x)) on the graph of ha . The fact that (p, q) is the lowest point of the graph of ha implies that ha (p) = q ≤ ha (x), for any x. We express this by saying that ha achieves a minimum at p, and the minimum value is q = ha (p). Again, it is valuable to be able to directly verify this fact. Thus, we have ha (x) = a(x − p)2 + q. Since a > 0 and the square of a number is never negative, we get a(x − p)2 ≥ 0. Thus ha (x) ≥ q, and of course ha (p) = 0 + q = q. We have therefore retrieved this information about ha achieving a minimum at p, with minimum value q. The lowest point of the graph of ha is called the vertex of the graph. Thus the vertex of the graph of ha lies on the line of symmetry of the graph. If on the other hand a < 0, then (p, q) is the highest point on the graph of ha . Then we get analogous conclusions about ha , but we will have to adjust the terminology: we say that ha achieves a maximum at p because ha (p) = q ≥ ha (x) for any x, and the maximum value is q = ha (p). We continue to refer to (p, q) as the vertex of the graph. An additional observation about the graph of ha is this. Suppose the graph of ha intersects the x-axis; we recall that this means they have at least a point in common. Let Z be a point of intersection of the graph of ha with the x-axis. Since Z is on the x-axis, the coordinates of Z are necessarily of the form (z, 0). But Z is also on the graph of ha , so Z must also be of the form (z, ha (z)). Since the two points are identical, we have (z, 0) = (z, ha (z)) and therefore ha (z) = 0. We have just proved that the x-coordinate of any point where the graph of ha meets the x-axis is a number at which ha is equal to zero. The converse fact, that if c is a number at which ha (c) = 0 then the graph of ha intersects the x-axis at (c, 0), is obvious. A number c at which a function f is equal to zero (i.e., f (c) = 0) is called a zero of the function f (compare page 11). What we have just proved may now be rephrased as follows: The zeros of ha are exactly the x-coordinates of the points of intersection of the graph of ha with the x-axis. This reasoning is of course valid not just for ha but for any function. The last relationship between the zeros of a function and the points of intersection of its graph with the x-axis may seem trivial, but it is a mystery to many students in algebra because TSM usually teaches this fact by rote. They may be able to memorize this fact for exams, but they would have no clue as to why it is

72

2. QUADRATIC FUNCTIONS AND EQUATIONS

true. Naturally, if the deﬁnition of the graph of a function is never taken seriously, no explanation of this fact is possible. Once again, the importance of deﬁnitions in the teaching and learning of mathematics cannot be overemphasized. We may summarize the main points of the foregoing commentary on Lemma 2.3 in the following lemma. Lemma 2.4. Let ha (x) = a(x − p)2 + q for some constants a, p, and q, with a = 0. Then: (a) If a > 0, then ha achieves a minimum at p and ha (p) = q. Moreover, ha has either two zeros or one zero or no zero. In the ﬁrst case, the two zeros are of the form p − t and p + t for some t > 0. In the second case, the zero is p. (b) If a < 0, then ha achieves a maximum at p and ha (p) = q. Moreover, ha has either two zeros or one zero or no zero. In the ﬁrst case, the two zeros are of the form p − t and p + t for some t > 0. In the second case, the zero is p. We proceed to translate Lemma 2.4 into algebraic symbols. Suppose ha has a 2 zero. With ha (x) = a(x − p)2 + q, ha having a zero x0 means a(x0 − p) + q = 0,

≥ 0, then −q a would make sense and one can easily verify that the two numbers p ± −q a are both zeros of ha . We so that

−q a

= (x0 − p)2 ≥ 0. Conversely, if

−q a

have therefore shown that ha has at least one zero if and only if −q a ≥ 0. The two −q zeros collapse into one if and only if a = 0 and therefore if and only if q = 0. Observe that the zeros of ha , p ± −q , are symmetric with respect to the a line of symmetry x = p of the graph of ha , as predicted by Lemma 2.3(c). Putting all these pieces together, we have proved the following theorem. Theorem 2.5. Let ha be the quadratic function ha (x) = a(x − p)2 + q, where a, p, q are ﬁxed numbers and a = 0. Then: (a) If a > 0, the function ha achieves a minimum at p, with minimum value equal to q. (b) If a < 0, the function ha achieves a maximum at p, with maximum value equal to q. (c) For any a = 0, ha has a zero ⇐⇒ −(q/a) ≥ 0, and when this happens, the zeros are equal to −q . p± a The two zeros collapse into one zero ⇐⇒ q = 0.

General quadratic functions and completing the square This subsection brings closure to the idea, broached on page 66 of the last subsection, that the lower-order terms bx + c in a general quadratic function f (x) = ax2 + bx + c do not essentially complicate our understanding of the latter. We will prove the following theorem, Theorem 2.6, which says every quadratic function f (x) = ax2 + bx + c can be written in the form of fa (x − p) + q for some appropriate

2.1. QUADRATIC FUNCTIONS

73

constants p and q. Precisely: Theorem 2.6. Every quadratic function f (x) = ax2 + bx + c can be written in the form f (x) = a(x − p)2 + q, where p=

−b 2a

and

q=

4ac − b2 . 4a

We will refer to a(x − p)2 + q as the normal form of f (x) = ax2 + bx + c. We also refer to the rewriting of f in the form f (x) = a(x − p)2 + q as reduction to normal form. Note that in the school mathematics education literature, the normal form is usually called the vertex form. The preceding two theorems together tell us almost everything there is to know about quadratic functions. The full implications of this somewhat cryptic statement will be systematically explored in the remainder of this section, but we hasten to single out a striking feature about quadratic functions that follows from these theorems. First observe that for a linear function g(x) = ax + b, where a and b are constants, its graph is the translated image of the graph of the linear function ga (x) = ax. In greater detail, if G denotes the graph of g and Ga denotes that of ga , then G is the line passing through B = (0, b) with slope equal to a while Ga is the line passing through the origin O also with slope equal to a. Clearly the translation −−→ along the vector OB maps Ga to G. Now, something entirely analogous holds with respect to quadratic functions. Given a quadratic function f (x) = ax2 + bx + c, Theorem 2.6 says that it can be reduced to the normal form, f (x) = a(x − p)2 + q, for some numbers p and q (the exact values of p and q are of no concern to us at the moment). But Lemma 2.2 on page 68 tells us that the graph of fa (x) = ax2 is congruent to the graph of f (x) = a(x − p)2 + q, the congruence being given by the translation along the vector from the origin to the point (p, q). Together, this says the graph of any quadratic function f (x) = ax2 + bx + c is the translated image of the graph of fa (x) = ax2 , where the leading coeﬃcient a is the same number in both functions. We may paraphrase the preceding paragraph as follows: Up to a translation, the graph of a quadratic function f (x) = ax2 + bx + c is completely determined by its leading coeﬃcient a, in the same way that, up to a translation, the graph of a linear function g(x) = ax + b is completely determined by its leading coeﬃcient a. Let us take a backward glance at this point. Recall that our study of linear equations and functions begins in Section 6.4 of [Wu2020a] with a detailed examination of lines passing through the origin O—which as we now know are nothing but the graphs of the linear functions ga (x) = ax for all numbers a = 0—and it leads to the concept of the slope of a line. This examination pays oﬀ in the long run because understanding slope is foundational to any understanding of linear functions, and the graph of a general linear function g(x) = ax + b is nothing but a translation of the graph of ga . From this perspective, it is easy to understand why we make the eﬀort to discuss at length the graph of fa (x) = ax2 for an arbitrary nonzero a, because the graph of a general quadratic function f (x) = ax2 + bx + c will also turn out to be the translation of the graph of fa (Theorem 2.6 and Lemma

74

2. QUADRATIC FUNCTIONS AND EQUATIONS

2.2). Therefore the study of quadratic functions parallels, literally, the study of linear functions, and such coherence expedites student learning. It remains to ﬁnd out why Theorem 2.6 is correct. In a sense, once the statement of Theorem 2.6 is known, it is entirely obvious how to show that it is correct. All we have to do is to substitute the announced values of p and q into a(x−p)2 +q and, by a routine computation, verify that a(x − p)2 + q is in fact equal to ax2 + bx + c. For the purpose of understanding Theorem 2.6, however, this is quite the wrong thing to do, because the way one actually derives the formulas for p and q in Theorem 2.6 is both instructive and useful. As is well known, this is the technique of completing the square. Proof of Theorem 2.6. The central issue is this: if we are given an expression involving a number x, x2 + kx, what number A2 can we add to it so that x2 + kx + A2 becomes a perfect square in the sense that x2 + kx + A2 = (x + A)2 is valid? It is easy enough to guess what A has to be: expand the right side by the distributive law and compare the coeﬃcients of both sides: x2 + kx + A2 = x2 + 2Ax + A2 . We need k = 2A, and therefore the guess is to take A = k2 . It is easy to verify that this works: (x + A)2 =

2 2 k k x+ = x2 + kx + = x2 + kx + A2 . 2 2

We have therefore proved:

2 2 k k = x+ . For any given x + kx, we have x + kx + 2 2 2

2

This is known as the technique of completing the square. We now apply it to our problem, with k = ab : 2

ax + bx + c

b 2 = a x + x +c a 2 2 b b b 2 = a x + x+ −a +c a 2a 2a 2 −b 4ac − b2 = a x− + . 2a 4a

This proves Theorem 2.6. It is instructive to give a graphic representation of the information encoded in Theorems 2.5 and 2.6. Let f (x) = ax2 + bx + c be given. For deﬁniteness, we let a > 0, so that we can look for the point at which f achieves a minimum. (The case

2.1. QUADRATIC FUNCTIONS

75

of a negative a is of course entirely similar, except that, in that case, one will be looking for the point at which f achieves its maximum.) According to Theorem 2.5, the point at which f achieves its minimum is (p, q), where, according to Theorem 2.6, p=

−b 2a

and q =

4ac − b2 . 4a

The vertical line x = p is the axis of symmetry of the graph of f , as shown: x = −b/2a

−b

2a −

s

−b 4ac−b2 2a , 4a

b2 −4ac ,0 4a2

−b s

s

2a +

b2 −4ac ,0 4a2

s

−b 4ac−b2 2a , 4a

x = −b/2a

Furthermore, Theorem 2.5(c) says that if f has a zero, then it has two zeros which are symmetric with respect to the axis of symmetry x = p, namely, the points

−q (p − −q a , 0) and (p + a , 0), and they are shown in the right picture above. Of course, these two zeros collapse into one, (p, 0), if q = 0. It may be noted that included in these pictures are the formula for the vertex of the graph of f (x) = ax2 + bx + c and the quadratic formula for the roots of the equation ax2 + bx + c = 0 (this information will be codiﬁed algebraically in Theorem 2.7 below). These are the high points of any instruction on quadratic equations and functions in school algebra, but the diﬀerence is that, here, we have provided a step-by-step graphic explanation for why these formulas are correct. Finally, let us retrace the historical origin for the terminology of “completing the square”. Let x2 + kx be given. For the moment, think of x and k as positive numbers. Then x2 + kx may be thought of as x2 + k2 · x + k2 · x, which is in turn equal to the total area of the ﬁgure consisting of a square with side length x and two rectangles each with sides of length k2 and x, as shown:

x k 2

x2

x k 2

Therefore, if we “complete” this geometric ﬁgure by adding the small square in the upper left corner (with area ( k2 )2 ), then we get a big square of area (x + k2 )2 .

76

2. QUADRATIC FUNCTIONS AND EQUATIONS

x k 2

( k2 )2

x2

x k 2

We can transcribe the geometric illustration above into symbolic terms. Given x2 + kx, x > 0 and k > 0, then adding ( k2 )2 to it would give us (x + k2 )2 ; i.e., 2

k 2 k 2 x + kx + = x+ . 2 2 What is remarkable is that, in this symbolic form, this equality of numbers is in fact an identity in x (x ∈ R)—as a routine computation will easily verify—regardless of whether x is positive or not or whether k is positive or not. The simple geometric consideration therefore has the virtue of suggesting the correct algebraic formula in spite of the fact that the geometry ceases to make sense when x or k is negative. It was the Babylonians (Babylon, about sixty miles south of Baghdad in present day Iraq) who ﬁrst thought of this geometric idea of “completing the square” some four thousand years ago. They had neither the symbolic notation nor the concept of negative numbers to express the quadratic formula in full generality, but despite those handicaps, the merit of ﬁrst conceiving this fruitful idea is theirs and, clearly, they already had all the essential ingredients of this general formula. The technique of “completing the square” has yet another interpretation. See the subsection Completing the square, revisited on page 80. Acivity. (a) A function f deﬁned on R is given by f (x) = 2bx − x2 , where b is a positive constant. Without writing anything down, determine the point x0 at which the function f achieves a maximum, and also determine f (x0 ). (b) Without writing anything down, determine the maximum value of the function g(x) = 30x − x2 . We now summarize in algebraic formalism some consequences of Theorems 2.5 and 2.6. With f (x) = ax2 + bx + c, Theorem 2.6 tells us that we can also write it as f (x) = a(x − p)2 + q, where p and q are explicitly given as −b 4ac − b2 and q = . 2a 4a In the form of f (x) = a(x − p)2 + q, it is clear that if a > 0 (respectively, a < 0), f achieves a minimum (respectively, maximum) at p and f (p) = q. In terms of (2.4), we see that if a > 0 (respectively, a < 0), f achieves a minimum (respectively, maximum) at −b 2a and −b 4ac − b2 f . = 2a 4a Furthermore, Theorem 2.5 states that a(x − p)2 + q has zeros if and only if −q a ≥ 0, which means if and only if b2 − 4ac ≥ 0. 4a2

(2.4)

p=

2.1. QUADRATIC FUNCTIONS

77

Since 4a2 is positive, we see that the function f (x) = ax2 + bx + c has zeros if and only if b2 − 4ac ≥ 0. The number b2 − 4ac is called the discriminant of the quadratic function. Thus a quadratic function has zeros if and only if the discriminant is ≥ 0. Suppose the discriminant is ≥ 0. Then Theorem 2.5 implies that the zeros r1 and r2 of f (x) = a(x − p)2 + q are p ± −q/a. In terms of (2.4), the zeros of f can therefore be given as b2 − 4ac −b ± r1 , r2 = . 2a 4a2 This can be further simpliﬁed, as follows. If A, B are positive numbers, then it is a simple exercise to show that (see Exercise 1 on page 85) √ A A =√ . B B Using this, we get √ √ b2 − 4ac b2 − 4ac b2 − 4ac √ . = = 4a2 2|a| 4a2 √ In the last step, we used the fact that 4a2 = 2|a| (see the corollary on page 65). Consequently, √ b2 − 4ac −b ± . r1 , r2 = 2a 2|a| But ±|a| is just ±a if a ≥ 0, and ∓a if a < 0. We therefore have √ b2 − 4ac −b r1 , r2 = ± . 2a 2a In summary, we have proved the following theorem. Theorem 2.7. A quadratic function f (x) = ax2 + bx + c has a zero if and only if its discriminant ≥ 0. If the discriminant ≥ 0, then the zeros of f are given by the following quadratic formula: √ −b ± b2 − 4ac . r1 , r 2 = 2a Furthermore, if a > 0 (respectively, a < 0), f achieves a minimum (respectively, maximum) at −b 2a and −b 4ac − b2 . f = 2a 4a The quadratic formula is one of the most basic formulas in mathematics. All high school students should not only know how to derive it and know the intuitive idea behind the derivation but also memorize it like a poem.5 One also cannot fail to notice that the maximum or minimum value of f (x) = ax2 + bx + c is given in terms of its discriminant: it is −1 4a (discriminant). One consequence of this fact is that if the discriminant vanishes (i.e., = 0), then the maximum or minimum (depending on whether a < 0 or a > 0, respectively) of f is 0. Then the graph of f 5 The

“like a poem” part comes from [Lang, page 86].

78

2. QUADRATIC FUNCTIONS AND EQUATIONS

touches the x-axis at one point and therefore f has only one zero. This is consistent with the quadratic formula. It follows from Theorem 2.7 that a quadratic function has at most two zeros, which gives a purely algebraic proof of part (c) of Lemma 2.3 on page 70. Quadratic equations of one variable Let ax2 + bx + c be a quadratic polynomial, where a, b, and c are ﬁxed real numbers. The question of whether there is any real number x so that ax2 +bx+c = 0 is called a quadratic equation in the variable x. We can now look at the quadratic formula (Theorem 2.7) from the vantage point of a quadratic equation in one variable. Consider the quadratic function f (x) = ax2 + bx + c. In the terminology of Section 6.1 of [Wu2020a] (also see page 355 in this volume), solving the quadratic equation ax2 + bx + c = 0 is equivalent to ﬁnding the zeros of the quadratic function f . In greater detail, if we ﬁnd two zeros r1 and r2 for f (so that f (r1 ) = f (r2 ) = 0), then ar12 + br1 + c =

0,

ar22

0.

+ br2 + c =

It follows that r1 and r2 are solutions of the quadratic equation ax2 + bx + c = 0. The converse is equally obvious: any solution of the quadratic equation is a zero of the quadratic function f . Traditionally, r1 and r2 are also called the roots of this equation. We have in particular derived the quadratic formula for a quadratic equation which exhibits the roots in terms of the coeﬃcients of the quadratic polynomial by the use of the technique of completing the square. If the discriminant vanishes (i.e., is equal to 0), then r1 = r2 and the two roots coincide. In this case, the common value r1 = r2 is called a double root of the quadratic equation, or a double zero of the quadratic function. Traditionally, the study of quadratic equations precedes the study of quadratic functions in the school curriculum. This is the correct sequencing in the high school curriculum, because the concept of a function is diﬃcult for most students and it makes sense to ﬁrst expose them to the easier topic of quadratic equations. However, because you will be teachers and educators, we purposely reverse the order of presentation here to underscore the message that the study of the quadratic equation ax2 + bx + c = 0 is nothing but a small part of the study of the quadratic function f (x) = ax2 +bx+c. More precisely, seeking the solutions of ax2 +bx+c = 0 is equivalent to seeking the zeroes of the function f . Teachers and educators need this kind of mathematical perspective for their work. In particular, a teacher has to impress on students, in the course of teaching quadratic equations, that the technique of completing the square is much more than a skill to solve quadratic equations. Rather, it is the key to the understanding of quadratic functions in general. See Section 2.3 below for further discussions. Incidentally, when quadratic equations are viewed in this way, it sheds light on why an equation is always a question: which numbers are the zeros of the function? For the case of a quadratic equation ax2 + bx + c = 0, it is desirable (especially in the school classroom) to give a direct derivation of the quadratic formula without the intervention of the p’s and the q’s, as follows. Suppose ax2 + bx + c = 0 has a solution, which, at this point, we may boldly continue to denote by x without any

2.1. QUADRATIC FUNCTIONS

79

fear of confusion. Thus with x clearly understood to be a solution of the equation (i.e., x now denotes a ﬁxed number), we have b 2 2 ax + bx + c = a x + x + c. a By completing the square of the expression in the parentheses on the right, we get ⎧ 2 2 b b b ⎪ 2 2 ⎪ ⎪ ax + bx + c = a x + x + +c −a ⎪ ⎪ a 2a 2a ⎪ ⎪ ⎨ 2 b2 b (2.5) + c − = a x + ⎪ ⎪ 2a 4a ⎪ 2 ⎪ ⎪ ⎪ 4ac − b2 b ⎪ ⎩ + = a x+ . 2a 4a In summary, if x is a ﬁxed number, then 2 4ac − b2 b + (2.6) ax2 + bx + c = a x + . 2a 4a Thus x being a solution of ax2 + bx + c = 0 is equivalent to 2 2 b − 4ac b x+ = . 2a 4a2 Since the square of a number is always ≥ 0 and since 4a2 ≥ 0, we see from the right side of this equality that the discriminant b2 − 4ac ≥ 0. Thus if the quadratic equation has a root, its discriminant ≥ 0. Then by Lemma 2.1 on page 64, √ √ √ b2 − 4ac b2 − 4ac b2 − 4ac b =± , (2.7) x+ = ± =± √ 2a 2|a| 2a 4a2 so that (2.8)

x=

−b ± 2a

√ √ b2 − 4ac −b ± b2 − 4ac = . 2a 2a

We have therefore shown that if ax2 + bx + c = 0 has a solution x, then necessarily, the discriminant b2 − 4ac ≥ 0 and √ −b ± b2 − 4ac . (2.9) x= 2a What we have accomplished thus far is that if ax2 + bx + c = 0 has a solution, then it is one of the numbers on the right side of (2.9). It remains to show that both of these numbers are actually solutions of ax2 + bx + c = 0 (the reader may wish to consult the discussion in Section 6.2 of [Wu2020a] on the solution of linear equations). So let X be either of the numbers on the right side of (2.9). Then we need to check that X is in fact a solution of ax2 + bx + c = 0; i.e., we have to show that aX 2 + bX + c = 0. This can be veriﬁed by a direct computation, but we can simplify matters by making use of equation (2.6). Thus we will show that 2 4ac − b2 b + a X+ = 0. 2a 4a

80

2. QUADRATIC FUNCTIONS AND EQUATIONS

This is so because the left side of this equation is equal to √ 2 2 b2 − 4ac 4ac − b2 4ac − b2 b − 4ac + + a =a = 0, 2a 4a 4a2 4a as desired. Thus we have given a direct proof of the quadratic formula: Theorem 2.7 . If b2 − 4ac ≥ 0, then the equation ax2 + bx + c = 0 has the following two roots (which coincide if the discriminant b2 − 4ac is 0 ): √ −b ± b2 − 4ac . 2a Conversely, if ax2 + bx + c = 0 has a root, then b2 − 4ac ≥ 0. Pedagogical Comments. [This is a continuation of the discussion started in Section 6.2 of [Wu2020a] about what an equation is and what it means to solve an equation.] In TSM, an equation such as ax2 + bx + c = 0 is supposed to be “some kind of equality involving a variable x”, so the solution of ax2 + bx + c = 0 is accomplished by blindly computing with the variable x as in equations (2.5)–(2.8). We have to point out that such computations with a variable make no sense because equation (2.7) cannot be true when x is “a quantity that varies” while a, b, and c are ﬁxed constants. To us, the x in (2.5)–(2.8) is not a variable but a hypothetical solution of ax2 + bx + c = 0—and therefore a ﬁxed number—so that each of the (2.5)–(2.8) is an equality between numbers. Then when we do get a candidate for a solution as in (2.9), we have to perform a second step to actually verify that the number in (2.9) is a solution. A pedagogical suggestion on teaching the solution of quadratic equations is to make sure that students understand the correct proof that (2.9) is a solution before allowing them to save time by going through the usual TSM algorithm of (2.5)–(2.8) to get to the solution in (2.9). Then also make sure that they “check their answers” by substituting the value of x in (2.9) into the original equation ax2 + bx + c = 0 to verify the correctness of the solution. In retrospect, the second step of “checking their answers” has added signiﬁcance: it is actually the second half of the proof that the solution method in equations (2.5)–(2.8) indeed produces the correct solutions. End of Pedagogical Comments. Completing the square, revisited We now rephrase the preceding proof of Theorem 2.7 in a way that sheds a diﬀerent light on the technique of completing the square. We observe that a quadratic equation without a ﬁrst-degree terms (the term “bx” in ax2 + bx + c = 0)

is immediately solvable. Indeed, ax2 + c = 0 has solutions ± −c a . From this standpoint, the main thrust of equation (2.6) is to reduce the original equation in x b so that the latter no longer has a ﬁrst-degree term in t. Let us to one in t = x + 2a b make this explicit: if we let t be the number x + 2a , then the right side of equation (2.6) is a quadratic polynomial in t without a ﬁrst-degree term so that the original equation in x now becomes the following equation in t: 4ac − b2 2 = 0. at + 4a

2.1. QUADRATIC FUNCTIONS

81

b As noted, this can be solved for t and, from x = t − 2a , we get the solutions in terms of x. Therefore we may look at the proof of Theorem 2.7 as “ﬁnding the correct substitution t = x + γ for a suitable constant γ so that the original equation b ) with no ﬁrst-degree term”. in x becomes a quadratic equation in t (where t = + 2a With this in mind, we may reformulate the preceding proof of Theorem 2.7 as follows:

Second Proof of Theorem 2.7 . The quadratic equation in x, ax2 +bx+c = 0, is given, where b2 − 4ac ≥ 0. Let x be a solution and let t = x + γ for some number γ. Then the equality ax2 + bx + c = 0 can be rewritten as one for the number t: a(t − γ)2 + b(t − γ) + c = 0. After expanding and collecting terms in decreasing powers of t, we get (2.10)

at2 + (b − 2aγ)t + (aγ 2 − bγ + c) = 0.

b Note that (2.10) is valid for any γ so long as t = x + γ. Now we set γ = 2a ; then b−2aγ = 0 and (2.10) no longer has a ﬁrst-degree term with respect to t. Moreover, b . Thus (2.10) now becomes the constant term also simpliﬁes when γ = 2a 2 4ac − b (2.11) at2 + = 0. 4a

As noted earlier, equation (2.11) has solutions √ b2 − 4ac b4 − 4ac . = ± (2.12) t=± 2 4a 2a b In summary, if x is a solution of ax2 + bx + c = 0 and t = x + 2a , then t is given by (2.12). It follows that the solution x is given by the usual quadratic formula: √ −b ± b2 − 4ac . x= 2a The fact that both numbers are actually solutions of ax2 + bx + c = 0 is proved in the same way (by substitution) as before. The proof is complete.

The reason we point out the second proof is that it generalizes to equations of all degrees; namely, given a polynomial equation of degree n, n ≥ 2, one can show that there is always a substitution x = t − γ for a suitable constant γ so that the resulting polynomial equation in t has no term of degree n − 1. This fact is important in the derivation of the formulas for the roots of cubic and quartic equation. See Chapter 4 of [Rotman], or see [Osler] and the references given therein. Signiﬁcance of the zeros of a quadratic function Let us return to the quadratic function f (x) = ax2 + bx + c. If its discriminant b − 4ac is ≥ 0, then we have seen that f has two zeros given by √ −b ± b2 − 4ac r1 , r2 = , 2a it being understood that the two zeros collapse into one if the discriminant is 0. We now point out a remarkable property of these roots. 2

82

2. QUADRATIC FUNCTIONS AND EQUATIONS

Theorem 2.8. Suppose r1 , r2 are the zeros of a quadratic function f (x) = ax2 + bx + c (possibly r1 = r2 ). Then f (x) = a(x − r1 )(x − r2 )

for all x ∈ R.

This theorem is conceptually so important that, before giving its proof, we want to point out its many ramiﬁcations. The theorem implies that two quadratic functions with the same two zeros diﬀer from each other by a multiplicative constant; i.e., one is a constant multiple of the other. This is not true for general functions. For example, assuming we know what an exponential function is (see Section 4.3 on pp. 153ﬀ.), consider the functions G(x) = ex (x2 − x − 6) and H(x) = e2x (x2 − x − 6). Then G and H share the two zeros, 3 and −2, but H is deﬁnitely not a constant multiple of G because H(x) = ex G(x). However, we hasten to point out that if we allow complex roots, then two polynomials of degree n with the same n (complex) roots again diﬀer from each other by a multiplicative constant. See Theorem 5.6 on page 197. Given a quadratic function f (x) = ax2 + bx + c, we have by now given two additional, distinct representations of f in Theorems 2.6 and 2.8, and each reveals a diﬀerent aspect of the quadratic function f : the normal form in Theorems 2.6 reveals the line of symmetry of the graph of f as well as the maximum or minimum of f , whereas the representation in Theorem 2.8 exhibits the zeros of f . Conceptu3 ally, this is no diﬀerent from rewriting a fraction such as 23 4 either as 5 4 (revealing its position on the number line), as 5.75 (in the context of scientiﬁc measurements, for instance), or as 575% (if percentage is important). In mathematics, seeking different representations of the same concept is a way of life, because a comprehensive understanding of that concept usually depends on piecing together these diﬀerent representations. The common mistake about the teaching of “representations” in TSM is that, even before a concept has been clearly deﬁned, diﬀerent representations are already oﬀered on the grounds of emphasizing “conceptual understanding”. The most obvious transgression of this genre (among many) occurs in the teaching of fractions. In the absence of a clear deﬁnition of what a fraction is, diﬀerent “representations” of a fraction are already paraded in front of students, e.g., as a part-of-a-whole, as a division, and as a ratio. Such a perversion of mathematics must be strenuously avoided in the classroom and in education research. This is then a good opportunity to point out that (i) a sizable part of mathematics is devoted to uncovering diﬀerent representations of a mathematical concept such as a quadratic polynomial above, (ii) mathematics cannot talk about the “different representations” of a concept that has not been precisely deﬁned, and (iii) the validity of each representation must be explained (i.e., proved). Theorem 2.8 gives us the proper perspective to look at the preoccupation in TSM with factoring quadratic polynomials whose coeﬃcients are integers. Theorem 2.8 tells us that there is a much more eﬃcient method to obtain the factorization of any quadratic polynomial regardless of whether its coeﬃcients are integers or not;

2.1. QUADRATIC FUNCTIONS

83

namely, formula to ﬁnd its roots. For example, since the roots of √ use the quadratic √ √ x2 − 2x − 1 are 12 ( 2 ± 6), we get the following factorization by Theorem 2.8: √ √ √ 1 √ 1 √ x2 − 2x − 1 = x − 2+ 6 2− 6 . x− 2 2 It will follow from (5.18) on page 194 and Theorem 5.6 on page 197 that this method of factoring quadratic polynomials is valid even for all such polynomials with complex coeﬃcients. This is one reason why the overemphasis in TSM on the skill of factoring quadratic polynomials with simple integer coeﬃcients is unhealthy. Now we give the proof of Theorem 2.8. It will go more smoothly if we make use of the following beneﬁcial lemma ﬁrst discovered by F. Vi`ete (1540–1603), a French mathematician who was regarded as the ablest mathematician in Europe in his day.6 Lemma 2.9 (Vi`ete’s theorem). If r1 and r2 are the zeros of f (x) = ax2 +bx+c, then r1 + r2 = − ab and r1 r2 = ac . Proof of Lemma 2.9. This is a routine computation using the quadratic formula. Since √ −b ± b2 − 4ac , r1 , r2 = 2a the assertion about r1 + r2 is trivial. As to the assertion about their product, we make use of the identity (x − y)(x + y) = x2 − y 2 to get √ √ −b b2 − 4ac b2 − 4ac −b − + r1 r2 = 2a 2a 2a 2a =

b2 b2 − 4ac 4ac c − = 2 = . 4a2 4a2 4a a

This proves the lemma. Lemma 2.9 is useful not just for the proof of Theorem 2.8. It gives a nice relationship between the zeros of a quadratic function and its coeﬃcients. In particular, if we have a monic quadratic polynomial in the sense that its ﬁrst coeﬃcient is 1, i.e., x2 + bx + c, then its roots r1 and r2 satisfy r1 + r2 = −b and

r1 r2 = c.

Although we will not pursue these relationships any further in these volumes, you should be aware that they generalize to polynomial functions of all degrees (see e.g., page 133 of [Jacobson]). Proof of Theorem 2.8. We have no choice but to directly verify that a(x − r1 )(x − r2 ) = ax2 + bx + c. 6 Among the many accomplishments of Vi` ete was his advocacy for the use of symbols in mathematics. He made great strides, but the person who put the ﬁnishing touches on this development was Ren´ e Descartes (1596–1650). With a little exaggeration, it may be said that a modern reader would have no trouble reading the mathematics after Descartes but would ﬁnd the mathematical writings before him almost unreadable because of the strange notations or the absence of any of the notation we are familiar with today.

84

2. QUADRATIC FUNCTIONS AND EQUATIONS

We make use of Lemma 2.9 to get a(x − r1 )(x − r2 ) = a(x2 − (r1 + r2 )x + r1 r2 ) c b = a x2 + x + a a

(by Lemma 2.9)

= ax2 + bx + c. This proves Theorem 2.8. The preceding proof of Theorem 2.8 depends on the quadratic formula. For polynomials of higher degree, there will be no analog of the quadratic formula in general, but the analogous factorization phenomenon persists for polynomials of any degree. See Theorem 5.6 on page 197. We can anticipate Theorem 5.6 by giving another approach to Theorem 2.8 that does not depend on the quadratic formula. It will make use of the following analog of the division-with-remainder for whole numbers. Theorem 2.10. Given a quadratic polynomial ax2 + bx + c and a number r, there is a linear polynomial (ax + β) and a constant γ so that for all x, ax2 + bx + c = (ax + β)(x − r) + γ.

(2.13)

This is the division algorithm for quadratic polynomials. As in the case of whole numbers, the linear polynomial ax + β is called the quotient of the division algorithm of ax2 + bx + c by x − r, the polynomial (x − r) the divisor of the division algorithm, and γ the remainder. The only diﬀerence between this theorem and the division-with-remainder for whole numbers is that, whereas in the case of whole numbers we can characterize the remainder as a whole number less than the divisor, the comparison of the remainder γ with the divisor x − r has to be done via their respective degrees: deg γ < deg(x − r). Proof of Theorem 2.10. Expanding the right side of the asserted equality, we get (ax + β)(x − r) + γ = ax2 + (β − ar)x + (γ − rβ). In order to match the left side with the right side, we must have b = β − ar

and

c = γ − rβ.

Thus if we set β = b + ar

and γ = c + r(b + ar),

then ax + bx + c = (ax + β)(x − r) + γ and Theorem 2.10 is proved. 2

It is seen from equation (2.13) that by letting x = r, the remainder γ satisﬁes γ = ar 2 + br + c. Thus we see directly that γ = 0 ⇐⇒ r is a zero of f (x) = ax2 +bx+c. But Theorem 2.10 implies that γ = 0 ⇐⇒ ax2 + bx + c = (ax + β)(x − r). Therefore, a number r is a zero of f ⇐⇒ f (x) = (ax + β)(x − r). Writing r = −β a , we see that r is a zero of f ⇐⇒ f (x) = a(x − r )(x − r) for some r .

2.1. QUADRATIC FUNCTIONS

85

Obviously, the number r is also a zero of f . Therefore we have fulﬁlled our promise of giving a second proof of Theorem 2.8 expressing a quadratic function in terms of its two zeros. The ideas behind Theorem 2.10 will be generalized to polynomials of higher degrees in Theorem 5.1 on page 180.

Exercises 2.1. (1) (a) If q > 0, prove that √ p prove pq = √q .

1 q

=

√1 . q

(b) If p, q are positive numbers, then

(2) Prove that √ the distance between the two points (a, b) and (b, a) in the plane is 2 · |a − b|. (3) Reduce each of the following quadratic functions to normal form (see page 73). Do so directly in each case, without quoting any formulas, and be sure you get it right because the next two exercises depend on your answers here. (i) f (x) = 2x2 −8x+7. (ii) g(x) = −2x2 +6x−9. (iii) h(x) = − 23 x2 +x+1. (4) What are the zeros (if any) of each of the functions in Exercise 3? (Do not use the quadratic formula; instead, answer this question directly.) At which point does the function achieve a maximum or minimum, and what is the maximum or minimum value of the function? (5) Sketch the graph of each of the functions in Exercise 3; indicate the axis of symmetry, the vertex of the graph, and the points at which it crosses the x-axis (if it does at all). (6) Write each of the following quadratic polynomials as a product of linear polynomials: (a) 330x2 − 131x − 133, (b) 104x2 − 331x + 92, (c) 272x2 + 481x + 204, (d) 1407x2 + 787x − 52. (Use a 4-function calculator.) (7) Write each of the following quadratic polynomials as a product of linear 2 2 2 polynomials: (a) 3x2 − √2x − 19, (b) x + 29x − 1, (c) 13x − 182x + 637. the online (8) (a) Solve for x: x − 2 x − 3 = 0. (b) Around April of 2017, √ version of the New York Times had an ad that said: “What is x: x + 15+ √ x = 15.” Solve it. (9) Prove: (a) A function f (x) = ax2 + bx + c, where a, b, c are constants, is (identically) equal to 0 ⇐⇒ a = b = c = 0. (b) Two quadratic functions are equal ⇐⇒ their coeﬃcients are pairwise equal. (10) Given two numbers r1 and r2 , prove that here is one and only one monic quadratic polynomial (recall that this means its leading coeﬃcient is 1) whose zeros are r1 and r2 . (11) Explain to a ninth grader how to locate the zeros of the function f (x) = −3x2 + 5x + 6 and the point at which it achieves its maximum, as well as how to sketch the graph of this function. Do not assume that he knows anything about quadratic functions. (12) Given two numbers r1 and r2 , let additional numbers c1 , D1 , c2 , and D2 be given so that c1 , c2 , r1 , and r2 are all distinct. Then prove that there is a function F so that F (r1 ) = F (r2 ) = 0 and F (c1 ) = D1 , F (c2 ) = D2 . [This should give you some idea how special quadratic polynomials are vis-` a-vis Theorem 2.8; they are essentially completely determined by their zeros.]

86

2. QUADRATIC FUNCTIONS AND EQUATIONS

(13) Two workmen can paint a house together in 6 13 days. In how many days can each paint it alone if it takes one of them 2 days longer than the other? (Assume as usual that both paint at a constant rate and that when they work together they do not interfere with each other’s work.) (14) A train makes a run of 120 miles. A second train starts one hour later and, traveling at 6 mph faster, reaches the end of the same run 20 minutes after the ﬁrst train. Find the time of the run of each train. (Assume that both trains make the run at a constant speed.) (15) If the height of an object from the ground t seconds after it is thrown from a height of 55 feet is −4.9t2 + 7t + 55 feet, what is the highest point of the object above ground? (16) Prove, using quadratic functions, the fact that among rectangles with the same perimeter p, the one with the largest area is a square. Is there one with the smallest area? Explain. (Compare Exercise 16 in Exercises 2.6 of [Wu2020a], quoted on page 356 of this volume.) (17) Prove that among all rectangles with the same area A, the one with the smallest perimeter is a square. (18) Suppose that against one side of a (long) wall you want to build a rectangular enclosure of 300 square feet using a part of the wall as one of its sides. Materials for the fence being exorbitantly expensive, you want to make sure the total length of the three sides of the fence is the shortest possible. What is this length? (Hint: Exploit Exercise 16(a) in Exercises 2.6 of [Wu2020a], quoted on page 356 of this volume.) (19) (a) For what value(s) of k would the function f (x) = 4x2 + 7x − k have a double zero? (b) For what value(s) of would the function g(x) = 2x2 − 2x + 8 have a zero? (c) If we want to make sure that the function h(x) = tx2 + x − 8 has no zero, what range of values can the number t assume? (20) (a) In the quadratic function g(x) = 2x2 + x + 5u, u is a number. For what value of u would g have two zeros? One zero? No zero? (b) Solve for x in y 2 x2 + 3yx − 7 = 0 for any nonzero number y. (21) For each number t, consider the quadratic function f (x) = t2 x2 − 5tx + 6. (a) Deﬁne a function φ(t) for all positive t so that φ(t) = the bigger of the two zeros of f . Find a formula for φ. (b) Deﬁne a function λ(t) for all negative t so that λ(t) = the bigger of the two zeros of f . Find a formula for λ. (c) Let β : {nonzero numbers} → R be deﬁned by β(t) = the sum of the two roots of f . Find a formula for β. (22) Is there a quadratic function whose graph crosses the x-axis at (11, 0) and (20, 0) and reaches it highest point at (15, 3)? If so, write down the function. If not, explain why not. (23) (a) Write down a quadratic function whose graph crosses the x-axis at (5, 0) and (15, 0) and passes through the point (20, −5). (b) If a quadratic function g has a zero at x = 1 and achieves its maximum at x = −5, what is its other zero? (24) Let f (x) = ax2 + bx + c be a quadratic function with coeﬃcients a, b, and c. Let the zeroes of f be r1 and r2 . Prove that the discriminant of f is equal to a2 (r1 − r2 )2 . (This gives another way to understand why a quadratic equation with zero discriminant has double roots.)

2.2. A THEOREM ON THE GRAPHS OF QUADRATIC FUNCTIONS

87

2.2. A theorem on the graphs of quadratic functions Our goal in this section is to clear up some misleading statements perpetuated by TSM about graphs of quadratic functions. Since we already have a precise deﬁnition of similarity, we will ﬁrst prove that all graphs of quadratic functions are similar to each other. Then we give a deﬁnition of parabola and prove that all graphs of quadratic functions are parabolas. The general fact that all parabolas are similar to each other is simple enough to be left as an exercise (Exercise 4 on page 95). We conclude this section with some basic facts about parabolas. Similarity of graphs of quadratic functions (p. 87) Graphs of quadratic functions are parabolas (p. 89) The vertex and focal length of a parabola (p. 94)

Similarity of graphs of quadratic functions What can one say about the graphs of quadratic functions? In TSM, it is asserted that “the graph of a quadratic function is a parabola”. Such a statement is generally understood to mean one of the following:7 A parabola is the general shape of the graph of a quadratic function. The graphs of quadratic functions all curve in a similar way. Such a graph is called a parabola. These statements seem to make sense until you stop to think. For example, do you believe that the following graphs “curve in a similar way”?

−70

O

70

−70

O

70

Well, they have to if you believe that “graphs of quadratic functions all curve in a similar way”. This is because the left curve is the graph of f (x) = x2 + 10 while 1 x2 + 10, and both functions are quadratic. the right curve is the graph of g(x) = 360 Keeping this phenomenon in mind, you are now all too willing to concede that the

7 Both

of these statements are taken directly from existing textbooks.

88

2. QUADRATIC FUNCTIONS AND EQUATIONS

following graph “has the general shape of the graph of a quadratic function”:

But this time you are wrong, because it is the graph of h(x) = 14 x4 + x2 + 1, which is deﬁnitely not quadratic (it is not a parabola; see Exercise 8 on page 96). Things are clearly getting out of hand, and all because we don’t know what a “parabola” is. The moral of this discussion is that students should not be given such a vague, meaningless statement as “the graph of a quadratic function is a parabola”. Rather, they need a precise deﬁnition of what a parabola is and then a proof that the graph of a quadratic function is a parabola. This is how mathematics works. Precision matters in mathematics.8 Let us ﬁrst conﬁrm that, in a precise sense, the graphs of quadratic functions indeed “all curve in a similar way”. Theorem 2.11. The graphs of all quadratic functions are similar to each other. Once we have given the simple proof of this theorem, we will return to the serious business at hand: deﬁne what a parabola is, and show that the graphs of all quadratic functions are parabolas (Theorem 2.15 on page 92). We note that there is more than one way to present this body of facts; see, e.g., Section 10.4 of [Wu2016b]. For the proof of Theorem 2.11, let us agree to call the graph of f1 (x) = x2 the standard parabola.9 In symbols, this is F1 . Thus by deﬁnition, F1 is the set of all ordered pairs {(x, x2 )} for all real numbers x. Proof. Given a quadratic function g(x) = ax2 + bx + c, let G be its graph. We will prove that G ∼ F1 (recall: “∼” is the symbol for “is similar to”); i.e., there is a ﬁnite composition of congruences and dilations which maps G to F1 (see page 355 for the deﬁnition of similarity). Assuming this for the moment, we will complete the proof, as follows. Suppose G is the graph of another quadratic function; then the same reasoning shows that G ∼ F1 . Since similarity is a symmetric relation (Lemma 5.2(iii) of [Wu2020a]; see page 357 of this volume), we have F1 ∼ G . Now from G ∼ F1 and F1 ∼ G , we get G ∼ G because similarity is a transitive relation (Lemma 5.2(iv) of [Wu2020a]). This then proves the theorem. 8 Compare

page xxiii. Compare also the Fundamental Principles of Mathematics on page

xxv. 9 There should be no fear of confusion with this terminology because the graph of f (x) will 1 turn out to be a parabola, by the corollary on page 92.

2.2. A THEOREM ON THE GRAPHS OF QUADRATIC FUNCTIONS

89

We proceed to the proof of G ∼ F1 for the graph G of g(x) = ax2 + bx + c. Let fa be the function fa (x) = ax2 and let Fa be the graph of fa . By Lemma 2.2 and Theorem 2.6 of Section 2.1, page 68 and page 73, respectively, we see that a translation T maps G to Fa . Now, we give an observation: (∗) If a > 0, the dilation centered at (0, 0) with scale factor a maps Fa onto the standard parabola F1 . If a < 0, the reﬂection across the x-axis followed by the dilation centered at (0, 0) with scale factor (−a) maps Fa onto F1 . Let us assume (∗) for the time being and we will ﬁnish the proof of the theorem. If a > 0 and if D is the dilation (guaranteed by (∗)) that maps Fa onto F1 , then D ◦ T maps G onto the standard parabola F1 . Since D ◦ T is a similarity, we are done if a > 0. Now suppose a < 0. Then by (∗), there is a reﬂection Λ and a dilation D so that (D ◦ Λ)(Fa ) = F1 . Therefore the composition D ◦ Λ ◦ T maps G onto the standard parabola F1 . Since D ◦ Λ ◦ T is a ﬁnite composition of congruences and dilations, it is a similarity and we are done again. To ﬁnish the proof of the theorem, it remains to prove (∗). Consider ﬁrst the case of a > 0. Let D be the dilation with center at the origin O and scale factor a so that D(x, y) = (ax, ay) (see Exercise 2 in Exercises 6.6 on page 356). We have to prove D(Fa ) = F1 . As usual, this means we have to prove D(Fa ) ⊂ F1

and

F1 ⊂ D(Fa ).

To prove D(Fa ) ⊂ F1 , let P ∈ D(Fa ). Then P = D(x, ax2 ) for a number x, so that P = (ax, a2 x2 ) = (ax, (ax)2 ) ∈ F1 . Thus D(Fa ) ⊂ F1 . Conversely, let P ∈ F1 and we must show that P ∈ D(Fa ). Indeed, let P = (t, t2 ) for some number t. Then the point Q = ( at , a( at )2 ) is a point of Fa . Now, 2

t 2 t D(Q) = a , a = t, t2 = P. a a This implies that P is equal to D(Q), where Q is a point of Fa . So P ∈ D(Fa ), and we have proved that D(Fa ) = F1 . The proof of (∗) is complete when a > 0. Now suppose a < 0 in fa (x) = ax2 . Let Λ denote the reﬂection across the x-axis and let, as usual, F−a denote the graph of f−a = (−a)x2 . It is clear that Λ(Fa ) = F−a . Notice that (−a) is now positive (and F−a is above the x-axis), so that the ﬁrst part of (∗) now implies that the dilation D with center at (0, 0) and scale factor (−a) maps F−a onto F1 . Therefore the composition D ◦ Λ maps Fa onto the standard parabola F1 . The proof of (∗) is complete and, by an earlier remark, so is the proof of Theorem 2.11.

Graphs of quadratic functions are parabolas We are now in a position to deﬁne a parabola. Recall that a geometric ﬁgure, or more simply a ﬁgure, is just a subset of the plane. Deﬁnition. A parabola P is a geometric ﬁgure consisting of all the points equidistant from a ﬁxed point A and a ﬁxed line L not containing A.

90

2. QUADRATIC FUNCTIONS AND EQUATIONS

Recall that the distance of a point Q from a line L is the length of the segment QQ , where Q is the point of intersection of L and the perpendicular line from Q to L, as shown: rQ L @ @ @ @ @r Q @ @ @ The point A and the line L are called the focus and the directrix of the parabola P, respectively. We will refer to the given property of P in the deﬁnition as the focus-directrix property. The name parabola may have been coined by Apollonius (circa 250–175 BC), a Greek mathematician who was born in Perga (in present-day Turkey) but spent most of his life in Alexandria (in present-day Egypt)10 and was a contemporary of Archimedes (287–212 BC). More than any other mathematician, Apollonius put conic sections (see Section 2.3 below), including parabolas, on the mathematical map. He deﬁned a parabola in terms of the intersection of an oblique cone with a plane. The deﬁnition of a parabola in terms of the focus-directrix property is apparently due to Pappus of Alexandria (circa 290–350 AD). The following sequence of simple observations is designed to get you acquainted with parabolas. Lemma 2.12. The line passing through the focus and perpendicular to the directrix of a parabola is an axis of symmetry (see page 70) of the parabola. Proof. Let the focus and the directrix of the parabola P be A and L as usual, and let the line passing through A and perpendicular to L meet L at the point B. Let the reﬂection across LAB be Λ. We will prove Λ(P) = P. q Q qH Q HH HH q A L

M

B

M

Suppose Q is a point on P. We ﬁrst prove Λ(Q) ∈ P, so that Λ(P) ⊂ P. Let Q = Λ(Q). To show that Q ∈ P, it suﬃces to show Q that is equidistant from A and L. Let the line passing through Q and perpendicular to L intersect L at the point M . Then Q being on P implies that |AQ| = |QM |. Next, because L ⊥ LAB , Λ(L) = L by the deﬁnition of a reﬂection. Therefore Λ(M ) is another point M on 10 Turkey was part of the Greek empire until about the ﬁrst century BC. One may note that many of the most famous ancient Greek mathematicians and scientists did not live in what is present-day Greece: Euclid lived in Egypt, Archimedes in Sicily, Eudoxus in Turkey, Ptolemy in Egypt, and, as noted, Apollonius in Egypt.

2.2. A THEOREM ON THE GRAPHS OF QUADRATIC FUNCTIONS

91

L. However, since Λ preserves degrees of angles, we also have LQ M ⊥ L. Therefore the distance of Q from L is just |Q M |. Since Λ(Q) = Q and Λ(M ) = M and since Λ preserves distance, we have |AQ| = |AQ | and |QM | = |Q M |. So from |AQ| = |QM |, we get |AQ | = |Q M | and therefore Q is equidistant from A and L. This proves Q ∈ P, so that Λ(P) ⊂ P. It remains to prove that P ⊂ Λ(P); i.e., for every Q ∈ P, there is a point Q ∈ P so that Q = Λ(Q ). Let Λ(Q) = Q . We know Q is in P because Λ(P) ⊂ P. Moreover, Λ(Q) = Q implies Λ(Λ(Q)) = Λ(Q ), which is equivalent to Q = Λ(Q ) because Λ ◦ Λ is the identity map of the plane. Thus we have proved that P ⊂ Λ(P). The proof of Lemma 2.12 is complete. Lemma 2.13. A geometric ﬁgure congruent to a parabola is also a parabola. Proof. Suppose P is the parabola with focus A and directrix L, and suppose ϕ is a congruence so that P = ϕ(P). We will show that P is the parabola with focus A = ϕ(A) and directrix L = ϕ(L), as shown in the picture below.

L

P

P

q Q HH HH HA q B L

M

B M

qA A A A A Aq Q

Take an arbitrary point Q of P ; we will ﬁrst show that Q is equidistant from A and L . To this end, let Q be the point on P so that Q = ϕ(Q). If M is the point on L so that QM ⊥ L and if M = ϕ(M ), then also Q M ⊥ L because the congruence ϕ preserves degrees of angles. We also have |AQ| = |A Q | and |QM | = |Q M | because ϕ also preserves lengths (distances). Since |AQ| = |QM | (Q is equidistant from A and L), Q is also equidistant from A and L . Thus every point on P is equidistant from A and L . On the other hand, if a point Q is equidistant from A and L , let Q be the point so that ϕ(Q) = Q . Then by the same argument as above, Q is equidistant from A and L and therefore must lie in P. Consequently, Q lies in P , and P is the set of all the points equidistant from A and L . This proves the lemma.

Acivity. Go through the details of the reasoning that a point Q equidistant from A and L must lie in P . Next, we will determine explicitly the parabola with focus A = (0, ) ( = 0) and directrix L deﬁned by y = −.

92

2. QUADRATIC FUNCTIONS AND EQUATIONS

y

P = (x, y)

A = (0, )

x

O

L = {y = −} Since the origin O is now equidistant from A and L, the parabola P with this choice of focus and directrix will pass through the origin. The picture above is for the case of > 0. The picture for the case of < 0 is just a reﬂection of this picture across the x-axis. Thus take a point P = (x, y) on P. The distance of P from A is x2 + (y − )2 (see page 356), while the distance of P from L is |y − (−)| = |y + |. By the deﬁnition of P, we have x2 + (y − )2 = |y + |. Because |y + | > 0, P being on P is equivalent to x2 + (y − )2 = |y + |2 = (y + )2 (compare Exercise 17 in Exercises 2.6 of [Wu2020a], quoted on page 356 of this volume). Expanding the squares, we get, x2 + y 2 − 2y + 2 = y 2 + 2y + 2 , which can be written as x2 = 4y, or 1 2 x . 4 We have therefore proved that any point equidistant from both A and L lies on the 1 2 x . By essentially running this reasoning graph of the quadratic function h(x) = 4 backwards, we can show that the converse is also true; i.e., if a point is on the graph 1 2 x , then it has to be equidistant from (0, ) and the line y = −. of h(x) = 4

(2.14)

y=

Acivity. Supply the details of this reasoning. At this point, we have proved that P = (x, y) being on the parabola P is equivalent 1 2 x . Thus, we have to (x, y) being on the graph of the quadratic function h(x) = 4 proved the following lemma. Lemma 2.14. Let = 0. Then the graph of the quadratic function h(x) = is the parabola with the point (0, ) as focus and the line y = − as directrix. Let a = 0 be given. By letting =

1 4a ,

1 2 4 x

we obtain the following corollary.

Corollary. The graph Fa of the function fa (x) = ax2 is the parabola with focus 1 1 ) and directrix y = − 4a . (0, 4a We are now ready for the following general theorem. Theorem 2.15. The graphs of all quadratic functions are parabolas.

2.2. A THEOREM ON THE GRAPHS OF QUADRATIC FUNCTIONS

93

We note that, conversely, every parabola is congruent to the graph of a quadratic function. See Exercise 4 on page 95. It follows that all parabolas are similar to each other, by Theorem 2.11 (it is also possible to give a direct proof of this fact without going through graphs of quadratic functions). Proof. Let G be the graph of f (x) = ax2 + bx + c as usual. We know by Lemma 2.2 on page 68 that there is a translation T so that T (G) = Fa , where Fa is the graph of fa (x) = ax2 . Thus G is congruent to Fa . By the corollary to 1 1 Lemma 2.14, Fa is a parabola with focus (0, 4a ) and directrix deﬁned by y = − 4a . Therefore, by Lemma 2.13, G being congruent to a parabola is itself a parabola. The theorem is proved. Remark. The proof of the corollary to Lemma 2.14 is obtained by determining the parabola with a special focus and a special directrix. It is very instructive to give a second proof from the opposite direction by starting with the graph Fa of fa (x) = ax2 and to ask whether it is a parabola and, if so, what are its focus and directrix? We will consider the case of a > 0, as the case of a < 0 is entirely similar. To prove Fa is a parabola, a > 0, one way is to make an educated guess about where its focus and directrix ought to be, and then it would be easy to prove that Fa consists of points equidistant from them. The rest of this paragraph will be about how to make this educated guess. So imagine that Fa is a parabola. By Lemma 2.12 on page 90, we should look for its focus on the line of bilateral symmetry of Fa , which is obviously the y-axis. Let us say that the focus is A = (0, ) for some positive number . We want to know what is in terms of the number a of Fa . Since the origin O has to be equidistant from A and the directrix, the directrix has to be the line L deﬁned by y = −. Here is a picture for the case a > 0. y

Fa

A

P

O

B

{y = } x

L = {y = −} Q Let the horizontal line deﬁned by y = intersect Fa at a point P ; from P drop a perpendicular to L and let it intersect the x-axis at B and the line L at Q. Since we are working from the assumption that Fa is the parabola with focus A = (0, ) and directrix L, |P A| = |P Q|. Since P lies on the line y = and Q lies on the line y = −, we see that |P B| = |BQ|. But |P B| is just because |AO| = , so |P A| = |P Q| = 2|P B| = 2. Now the coordinates of P are (|AP |, |P B|); therefore P = (2, ). Since P lies on Fa and Fa is the graph of fa (x) = ax2 , we also have P = (2, a(2)2 ). Comparing the 1 . Therefore, y-coordinates of the two expressions of P , we get a(2)2 = , or = 4a 1 our guess is that if Fa is a parabola, its focus and directrix “ought to be” (0, 4a ) 1 and {y = − 4a }, respectively.

94

2. QUADRATIC FUNCTIONS AND EQUATIONS

Once we know this, we can easily prove that Fa is the parabola with focus 1 1 ) and the line y = − 4a as directrix, as in the proof of Lemma 2.14 on page (0, 4a 92. Theorem 2.15 ﬁnally enables us to make sense of the statement that “the graph of a quadratic function is a parabola”. Having a precise deﬁnition for the parabola also helps to clear up another common confusion. Is the graph of the equation x = y 2 − 5 (i.e., the equation x − y 2 + 5 = 0) a parabola? It should be, because it looks like the graph of f (x) = x2 −5 rotated 90 degrees clockwise around the origin and therefore, by Lemma 2.13, it should be a parabola. With the deﬁnition of a parabola available, we can convert the preceding intuitive argument into a valid proof of the fact that the graph of x = y 2 −5 is a parabola, as follows. Let be the counterclockwise rotation of 90 degrees around the origin O and let G, G be the graphs of x = y 2 − 5 and y = x2 − 5, respectively. Then maps the positive x-axis onto the positive y-axis and the negative y-axis onto the positive x-axis, so that (x, y) = (−y, x); it is now straightforward to check that (G) = G and G is congruent to the graph G of the quadratic function f (x) = x2 − 5. By Lemma 2.13 on page 91 and Theorem 2.15 on page 92, G is a parabola. We will discuss the graphs of second-degree equations more generally in Section 2.3 below. The vertex and focal length of a parabola For later discussions of parabolas, it will be convenient to have available the concept of the vertex of a parabola. Assume a parabola P with focus A = (h, k) and directrix L. Let the line containing A and perpendicular to L intersect P at M and intersect L at B. Then M is the midpoint of AB, by the deﬁnition of a parabola. P

L

Aq rX y XX M B

By deﬁnition, M is the vertex of the parabola P. Now recall that for the graph of a quadratic function we already had a concept of its vertex (see page 71). Because graphs of quadratic functions are now known to be parabolas (Theorem 2.14), let us make sure that for graphs of quadratic functions, these two concepts coincide. From the proof of Theorem 2.14, we know that every graph of a quadratic function is congruent to the graph Fa of the quadratic function fa (x) = ax2 . It therefore suﬃces to verify that, for Fa , the two concepts of “vertex”

2.2. A THEOREM ON THE GRAPHS OF QUADRATIC FUNCTIONS

95

coincide. Now the vertex of Fa , as the graph of a quadratic function, is obviously (0, 0), the origin. Next, consider Fa as a parabola. By the corollary on page 92, the 1 1 parabola Fa has focus (0, 4a ) and has the line deﬁned by y = − 4a as its directrix. 1 Since the perpendicular from the focus (0, 4a ) to the directrix is just the y-axis, 1 this perpendicular intersects the directrix at (0, − 4a ). The midpoint of the segment 1 1 from (0, 4a ) to (0, − 4a ) is then the origin (0, 0). Thus the vertex of the parabola is also the origin, and the two concepts of “vertex” coincide. The distance from the focus of a parabola to its vertex is called the focal length of the parabola. It follows that the focal length of a parabola is half the distance of the focus from the directrix. For the parabola with focus at (0, ) and with directrix deﬁned by y = −, its vertex is O = (0, 0) so that its focal length is ||. We conclude this section by explaining the signiﬁcance of the focus of a parabola in terms of everyday life. With the parabola y = ax2 as in the following picture, the theory of optics shows that any light ray coming from above and parallel to the y-axis is reﬂected by the parabola to hit the focal point A (this fact explains the common, nontechnical meaning of the word “focus”). This property is critical to the construction of high-powered telescopes in observatories: by putting the light signal receptor precisely at the focus of a paraboloid telescope (i.e., the surface of the telescope is obtained by rotating a parabola around its axis of symmetry) the receptor can capture all incoming light signals. By A the same token, the light rays emanating from a light source placed at the focus A will be reﬂected O by the parabolic mirror into upward-pointing parallel beams along the y-axis. This is the principle behind the design of the headlights of your cars. For a more recent application of this “focusing property” of parabolas, see the video https://youtu.be/lrRTCbXE0Jc. Exercises 2.2. (1) Find the focus and directrix of the graph of each of the following functions: (a) f (x) = 10x2 − 5. (b) g(x) = − 12 x2 + x. (c) h(x) = x2 + 5x − 36. (2) According to Theorem 2.11, the graphs of f (x) = 2x2 + 1 and g(x) = x2 + 4x − 5 are similar to each other. Write down explicitly a similarity (see page 355 for a deﬁnition) that maps the graph f to the graph of g. (3) As usual, let f1 (x) = x2 and f3 (x) = 3x2 , and let F1 and F3 be their respective graphs. (a) Take a point P = (x0 , 3x20 ) on F3 , and let K be the ray from the origin O to P . If K intersects F1 at Q, determine the coordinates of Q in terms of x0 . (b) How would you explain to an eighth grader that there is a dilation of the plane that maps F3 to F1 ? (4) Prove: (a) Two parabolas with equal focal lengths (see page 95) are congruent. (b) Every parabola is congruent to the graph of fa (x) = ax2 for some positive a. (c) Any two parabolas are similar to each other. (5) (a) Find c > 0 such that the parabola G with focus at (c, −c) and directrix y = 0 is congruent to the standard parabola F1 (use the preceding exercise if you like). (b) Write down explicitly the congruence ϕ such that ϕ(G) = F1 .

96

2. QUADRATIC FUNCTIONS AND EQUATIONS

(6) (a) Prove that if D is a dilation and P is a parabola, then the image D(P) is a parabola. (b) Prove that any geometric ﬁgure that is similar to a parabola is a parabola. (7) (a) Prove that if a geometric ﬁgure is congruent to a convex set, then it is itself convex. (b) Prove that the region “inside” the graph of the quadratic function fa (x) = ax2 , where a > 0, is convex. More precisely, prove that the region of all the points (x, y) satisfying y ≥ ax2 is convex. (c) Given a quadratic function f (x) = ax2 + bx + c, where a > 0, prove that the region of all the points (x, y) satisfying y > ax2 + bx + c is convex. (8) Prove that the graph of h(x) = 14 x4 + x2 + 1 is not a parabola. (Hint: Compare the reasoning in the remark on page 93, and note that it suﬃces to prove that the graph of g(x) = 14 x4 + x2 is not a parabola.) 2.3. Graphs of equations of degree 2 Among curves, we have so far studied lines,11 circles, and parabolas. Lines are diﬀerent from the other two, obviously, but it may not be obvious that circles and parabolas are closely related. Their geometric kinship comes from the fact that both can be obtained as the intersection of a plane with a circular cone (see, e.g., [Wolfram]). Their algebraic kinship is the subject of this section: both belong to a distinguished family of curves, which are graphs of what have come to be known as equations of degree 2 in two variables. These graphs have a venerable history, going back to the very beginnings of geometry. General equations of degree 2 (p. 96) Parabolas (p. 97) Circles (p. 102) Ellipses (p. 103) Hyperbolas (p. 109) General equations of degree 2 An equation of degree 2 in two variables is by deﬁnition an equation of the form (2.15)

Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0,

where A, B, . . . , F are constants and at least one of A, B, and C—the coeﬃcients of the second degree terms x2 , xy, and y 2 —is nonzero. Recall what such an equation means: it is a question that asks which points (x0 , y0 ) have the property that their coordinates x0 and y0 satisfy the equation in the sense that Ax20 + Bx0 y0 + Cy02 + Dx0 + Ey0 + F = 0. Such a point (x0 , y0 ) is said to be a solution of (2.15). For example, (0, −2) and (0, − 21 ) are solutions of the equation 3x2 + 2y 2 − 3x + 5y + 2 = 0. Recall that by the graph of an equation (2.15), we mean the set of all its solutions. Denote the graph of (2.15) by G. Note that the graph of an equation of degree 2 11 A straight line is straight, but it is a “curve” nonetheless, in the same way that a square is a rectangle.

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

97

in two variables can be a pair of lines; e.g., x2 − y 2 + x − y = 0 can be rewritten as (x − y)(x + y + 1) = 0, so that its graph is the union of the two lines deﬁned by x − y = 0 and x + y + 1 = 0. (Make sure you understand the last assertion.) The presence of the mixed term Bxy in the equation (2.15) complicates the discussion of its graph G. It turns out that if we rotate G appropriately, then the equation of the rotated graph will no longer have a nonzero mixed term. Such a rotation will be introduced in Section 1.7 of [Wu2020c]. Until then, we will ignore equations with a nonzero mixed term. The graphs of second degree equations have a long history, dating back to at least the Greeks in the fourth century BC, under the name of conic sections, or more simply conics. Conic sections fall into three major categories: ellipses (including circles), parabolas, and hyperbolas.12 These names were codiﬁed by Apollonius’ treatise Conics, which did for conic sections what Euclid’s Elements did for plane geometry. The reason for the name “conic sections” is that every pair of intersecting lines, every ellipse, every parabola, and every hyperbola can be realized as the intersection of a cone with a plane.13 We have more to say about conics at the end of this section. It should be pointed out that, without some basic facts from linear algebra on the diagonalization of (real) symmetric matrices by orthogonal transformations and the classiﬁcation of quadratic forms in two variables—topics found in every textbook on linear algebra—our study of the graphs of equations of degree 2 in two variables is bound to be somewhat haphazard even for the special case of nomixed-terms. What we show below is that the graphs of these equations include parabolas, ellipses, and hyperbolas, but we will not be able to prove the theorem that this list—together with pairs of lines—exhausts all the possibilities. What follows below is a uniform approach to parabolas, circles, ellipses, and hyperbolas. We will begin with the geometric deﬁnition of the curve in each case and derive what is called the normal form of the equation for the curve; it is of course an equation of the form of (2.15). Then we show, conversely, that the graph of an equation of the form of (2.15) with suitable restrictions on the coeﬃcients is such a curve. Parabolas We begin with the parabola, the curve that satisﬁes the focus-directrix property (see page 89). It is a fact that, for the equation of a parabola to avoid having a nonzero mixed term, the directrix of the parabola has to be either horizontal or vertical. Assuming this fact, let us start with a parabola whose directrix is horizontal and whose vertex (page 94) is the origin O = (0, 0). Let its focus be (0, ) and the directrix be the line deﬁned by y = −. Then its focal length (page 95) is || ( can be positive or negative). By Lemma 2.14 on page 92, this parabola is known to be the graph of the equation 1 (2.16) y = x2 . 4 12 Mathematical Aside: These are called the nondegenerate conics. A pair of lines, including parallel lines, is called a degenerate conic. By going to the projective plane, all conics— nondegenerate or degenerate—can be obtained as graphs of second degree equations. 13 See, for example, the Wolfram MathWorld article on conic sections, [Wolfram].

98

2. QUADRATIC FUNCTIONS AND EQUATIONS

Next, we look at the general case. Let the vertex of the parabola P be (h, k) and let its directrix be a horizontal line. If its focal length is || for some number , then the focus is (h, k + ) and its directrix is the horizontal line y = k − . The following picture shows the case where > 0: P (h, k + ) ((( ( r (x, y)

(h, k)

-r y =k−

(x, k − )

Now if (x, y) ∈ P, then by deﬁnition, (x, y) is equidistant from (h, k + ) and the directrix y = k − . Therefore we have (x − h)2 + (y − (k + )2 = |y − (k − )| (x, y) ∈ P ⇐⇒ ⇐⇒

(x − h)2 + (y − (k + ))2 = |y − (k − )|2 = (y − (k − ))2

⇐⇒

(x − h)2 = (y − (k − ))2 − (y − (k + ))2 .

In the second equivalence, we have made use of Exercise 17 in Section 2.6 of [Wu2020a] (which is quoted on page 356) and the fact that |y − (k − )| > 0 (because P is disjoint from the directrix). Now observe that the right side of the last equation is equal to 4(y − k). Therefore, (x, y) ∈ P ⇐⇒ (x − h)2 = 4(y − k). Thus a point (x, y) lies on the parabola P if and only if (2.17)

(y − k) =

1 (x − h)2 . 4

Equation (2.17) is called the normal form, or the standard form, of the equation of a parabola with a horizontal directrix. In mathematics, “normal form” generally refers to something that is “simple” but still captures the essence of a mathematical object. Equation (2.17) looks neat, without a doubt, but it does more, because one can immediately read oﬀ from (2.17) that the vertex of this parabola (see page 94) is (h, k) (from (y − k) and (x − h)2 ), that its focal length is || (from the denominator of the coeﬃcient on the right side), and therefore its focus is (h, k + ) and its directrix has equation y = k − . P (h, k + r ) r (h, k − )

(h, k) y =k−

Also observe that if the denominator of the coeﬃcient on the right is positive, then the focus is above the directrix (as shown), but if the denominator is negative, then

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

99

the focus is below the directrix. Therefore one glance at equation (2.17) is enough to allow us to construct the parabola. This is why (2.17) is called the normal form of the equation for the parabola. 1 (x − 2)2 , exhibit its focus Acivity. For the parabola deﬁned by (y + 1) = − 12 and directrix, and sketch the parabola.

Incidentally, if the vertex is (0, 0) (thus h = k = 0) and the focus is (0, ) and the directrix is the line y = −, then equation (2.17) reduces to equation (2.16). There is another way to approach equation (2.17). See Exercise 3 on page 115. It is time to recall that our main concern is with the graph of equation (2.15). We therefore hasten to point out that (2.17) is a special case of (2.15) because, expanding both sides of (2.17) and simplifying, we get (2.18)

x2 + (−2h)x + (−4)y + (h2 + 4k) = 0.

Equation (2.18) is the special case of equation (2.15) with A = 1, B = C = 0, D = −2h, E = −4, and F = h2 + 4k. We are now in a position to ask: given an equation of the form (2.15), how do we recognize that its graph is a parabola with a horizontal directrix? We begin by taking note of the fact that the coeﬃcients of x2 and y in equation (2.18) are nonzero ( = 0 since the focus does not lie on the directrix, by deﬁnition) and the fact that there is no y 2 term. This being true of all equations of parabolas with a horizontal directrix, we may therefore pose an equivalent question: is the graph of the following special case of equation (2.15) a parabola with a horizontal directrix: Ax2 + Dx + Ey + F = 0,

with AE = 0?

(Note that the inequality AE = 0 is nothing but a shorthand for “A = 0 and E = 0”.) We now show that the answer is aﬃrmative. Dividing both sides of the preceding equation by E, we get D A F A = 0. y= − x2 + − x+ − , with E E E E In the interest of notational brevity, we rewrite this as y = ax2 + bx + c,

(2.19) where now

D F A , b=− , c=− . E E E Since the graph of (2.19) is the graph of the quadratic function f (x) = ax2 + bx + c, by Theorem 2.15 on page 92, it is a parabola. We are done, or so it would seem. But we cannot let this discussion end by appealing to a fact about quadratic functions because we will also have to tackle the case of parabolas with a vertical directrix, in which case we will not have any theorems about graphs of quadratic functions to fall back on. For this reason, we look for a way to directly identify the graph of (2.19) as a parabola whose focus, directrix, and vertex can be read oﬀ from a, b, and c. We will achieve this goal “through the backdoor”, so to speak, as follows. a=−

100

2. QUADRATIC FUNCTIONS AND EQUATIONS

Let us start with an arbitrary point (h, k) as the vertex and an arbitrary horizontal line, y = k − , as the directrix of a parabola, where = 0. We have seen that the equation of the resulting parabola is (2.18), which is equivalent to 2 h 1 (−h) x+ +k . (2.20) y = x2 + 4 2 4 As stated, the coeﬃcients on the right side of (2.20) immediately yield the information about the fact that its graph is a parabola with vertex (h, k) and directrix y = k − . Now, with a, b, and c given as in (2.19), we will choose the h, k, and in (2.20) so that (2.21)

1 −4

=

a,

(2.22)

(−h) 2

=

b,

h2 + k = c. 4 This will imply that the equation of the parabola with (h, k) as vertex and y = k − as directrix—for these particular choices of h, k, and —coincides with (2.19). Therefore the graph of (2.19) is this parabola. To ﬁnd h, k, and so that the system of equations (2.21)–(2.23) is satisﬁed, we ﬁrst let −b 1 and h = . (2.24) = 4a 2a (2.23)

Indeed, such choices are dictated by comparing the coeﬃcients of x2 and x in (2.19) and (2.20). From (2.23) and making use of (2.24), we see that, necessarily, k =c−

(−b/2a)2 h2 =c− . 4 (1/a)

Therefore, we also let 4ac − b2 . 4a Equations (2.24) and (2.25) together furnish the solution to equations (2.21)–(2.23). In fact, it is exceedingly simple to directly verify that the values of h, k, and in (2.24) and (2.25) satisfy (2.21)–(2.23). Therefore, we have proved the following lemma.

(2.25)

k=

Lemma 2.16. The graph of y = ax2 + bx + c, where a = 0, is a parabola so that 2

b 4ac−b 1 , 4a + 4a ), its focus is (− 2a its directrix is the line deﬁned by y = b 4ac−b2 , 4a ). its vertex is (− 2a

4ac−b2 4a

−

1 4a ,

and

We can now summarize the foregoing discussion of parabolas in the context of the general equation of degree 2, equation (2.15), in the ﬁrst half of the following theorem.

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

101

Theorem 2.17. (i) A parabola with a horizontal directrix is the graph of an equation (y − k) =

1 (x − h)2 4

where (h, k) is the vertex of the parabola, its focus is (h, k + ), its directrix is the line deﬁned by y = k − , and the focus is above (respectively, below) the directrix if > 0 (respectively, < 0). Conversely, the graph of the second degree equation in two variables Ax2 + Dx + Ey + F = 0

with

AE = 0

is a parabola with a horizontal directrix. Its vertex, focus, and directrix are given by Lemma 2.16 once the equation is reduced to the form y = ax2 + bx + c. P r 2

b 4ac−b (− 2a , 4a )

2

b 4ac−b (− 2a , 4a +

1 4a )

- r 1 4a

y=

4ac−b2 4a

−

1 4a

(ii) A parabola with a vertical directrix is the graph of an equation (x − h) =

1 (y − k)2 4

where (h, k) is the vertex of the parabola, its focus is (h + , k), its directrix is the line deﬁned by x = k − , and the focus is to the right (respectively, to the left) of the directrix if > 0 (respectively, < 0). Conversely, the graph of the second degree equation in two variables Cy 2 + Dx + Ey + F = 0

with

CD = 0

is a parabola with a vertical directrix. Its vertex, focus, and directrix are given in Exercise 9 on page 115 once the equation is reduced to the form x = ay 2 + by + c. We will skip the proof of part (ii) because the proof of part (i) of Theorem 2.17 has been presented in such a way that the modiﬁcation it takes to change the latter to the former is minimal. We conclude this discussion of the parabola by relating Lemma 2.16 to what we have learned about the graph of a quadratic function f (x) = ax2 + bx + c. Indeed, b 4ac−b2 with f as given, Theorem 2.6 on page 73 shows that the point (− 2a , 4a ) is the vertex of the graph. Furthermore, by Lemma 2.2 on page 68 and by the corollary b 4ac−b2 1 , 4a + 4a ) and its to Lemma 2.14 on page 92, the focus of the graph is (− 2a 4ac−b2 1 directrix is y = 4a − 4a . These facts are consistent with Lemma 2.16.

102

2. QUADRATIC FUNCTIONS AND EQUATIONS

Circles Next, we consider circles. The circle of radius r with center Q = (a, b) is, by deﬁnition, the set of all the points P = (x, y) so that the distance of P from Q satisﬁes |P Q| = r. Since both |P Q| and r are positive numbers, |P Q| = r ⇐⇒ |P Q|2 = r 2 . Therefore, by the distance formula (see page 356), this circle is the set of all the points (x, y) so that (x − a)2 + (y − b)2 = r 2 .

(2.26)

This is called the normal form of the equation of the circle of radius r centered at (a, b). The fact that this is an equation of the form of (2.15) on page 96 can be seen by expanding the left side to get x2 + y 2 − 2ax − 2by + (a2 + b2 − r 2 ) = 0. This is then a special case of equation (2.15) when A = C(= 1), B = 0, D = −2a, E = −2b, and F = a2 + b2 − r 2 . The most important features of this equation are that the coeﬃcients of x2 and y 2 are equal (both equal 1 here, but if we multiply the equation through by A, then they would both be A) and that the radius r of the 2 2 2 circle can be√expressed in terms √ of D, E, and F , as follows. From F = a + b − r , 1 we get r = a2 + b2 − F = 2 D2 + E 2 − 4F . In particular, D2 + E 2 − 4F > 0.

(2.27)

We now consider the converse: given an equation of the form of (2.15), how do we recognize that its graph is a circle? The preceding discussion implies that it suﬃces to limit our considerations to the following special case: Ax2 + Ay 2 + Dx + Ey + F = 0 where A = 0 and (2.27) holds.

(2.28)

We claim that the graph of such an equation is a circle. To prove this claim, we may henceforth assume A > 0 without loss of generality because we can just multiply both sides of the equation by (−1) otherwise. Now we multiply the equation by A1 to get an equivalent equation: x2 + y 2 + dx + ey + f = 0,

(2.29)

where the constants are now D , A Because (2.27) holds, we now have

(2.30)

d=

e=

E , A

f=

F . A

d2 + e2 − 4f > 0.

(2.31)

By completing the square in equation (2.29), we get ⇐⇒ ⇐⇒ (2.32) ⇐⇒

x2 + y 2 + dx + ey + f = 0 (x2 + dx) + (y 2 + ey) = −f 2 e 2 d 2 e 2 d 2 2 + −f + y + ey + = x + dx + 2 2 2 2 2 e 2 1 2 d d + e2 − 4f . + y+ = x+ 2 2 4

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

103

The right side of (2.32) is positive, by assumption (2.31). Thus,comparing (2.32) with (2.26), we see that the graph of (2.29) is a circle of radius 12 d2 + e2 − 4f and centered at (− d2 , − 2e ). Taking (2.30) into account, we conclude that the graph of √ −E 1 2 2 equation (2.28) is a circle with center ( −D 2A , 2A ) and radius 2A D + E − 4AF . We have therefore proved the following characterization of the circle. Theorem 2.18. A circle with center (a, b) and radius r is the graph of (x − a)2 + (y − b)2 = r 2 . Conversely, the graph of a second degree equation Ax2 + Ay 2 + Dx + Ey + F = 0 −E where A > 0 and D2 + E 2 − 4AF > 0 is a circle with center ( −D 2A , 2A ) and radius √ 1 2 2 2A D + E − 4AF .

Ellipses The next natural curve to consider is an ellipse, which we proceed to deﬁne. Let F1 and F2 be two points in the coordinate plane so that |F1 F2 | = 2c ≥ 0 and let a be a number so that a > c. We deﬁne the ellipse with foci14 F1 , F2 and semimajor axis a to be the set of all points P so that |P F1 | + |P F2 | = 2a. The midpoint of the segment F1 F2 is called the center of the ellipse. Observe that F1 and F2 coincide if and only if c = 0. In this case, |P F1 | = |P F2 | and the preceding equation becomes |P F1 | = a; then the ellipse is just the circle of radius a with center F1 . We see that a circle is just an ellipse whose foci coincide. From now on, we will assume that the foci of ellipses are distinct. Equivalently, we assume c > 0. We are going to show that every ellipse is the graph of an equation of the form of (2.15). It is a fact that if the foci of an ellipse lie on a line that is neither horizontal nor vertical, then this equation will have a nonzero mixed term. In the following discussion, we will therefore look only at ellipses so that the line containing the foci is either horizontal or vertical. Let us ﬁrst look at the simplest case where F1 and F2 are two points on the x-axis symmetric with respect to the origin O. Then they have coordinates (−c, 0) and (c, 0), respectively, so that the center of the ellipse is the origin O. Let a > c as above. The condition that a point P satisﬁes |P F1 | + |P F2 | = 2a can be expressed equivalently as an equation in terms of the coordinates x and y of P = (x, y). We now derive this equation. The equality |P F1 | + |P F2 | = 2a is equivalent to (x + c)2 + y 2 + (x − c)2 + y 2 = 2a. 14 “Foci”

is the plural of “focus”.

104

2. QUADRATIC FUNCTIONS AND EQUATIONS

If we rewrite this as (2.33)

(x + c)2 + y 2 = 2a − (x − c)2 + y 2 ,

then the left side is positive and therefore so is the right side. Since for positive numbers A and B, A = B ⇐⇒ A2 = B 2 , the preceding equality is equivalent to the equality of the squares of both sides. Hence, we have (x + c)2 + y 2 = 4a2 − 4a

(x − c)2 + y 2 + (x − c)2 + y 2 .

Expanding (x + c)2 on the left and (x − c)2 on the right and making the obvious cancellations, we obtain (2.34)

a (x − c)2 + y 2 = a2 − cx.

Again, the left side is positive and therefore so is the right side. The same reasoning as before then shows that this is equivalent to the equality of the squares of both sides. Thus after simplifying, we arrive at (2.35)

(a2 − c2 )x2 + a2 y 2 − a2 (a2 − c2 ) = 0.

This is an equation of degree 2 in x and y where, in the notation of equation (2.15) on 96, A = (a2 − c2 ), B = 0, C = a2 , D = E = 0, and F = −a2 (a2 − c2 ). Equivalently, we can also write equation (2.35) as (2.36)

x2 y2 + 2 = 1. 2 a a − c2

In this form, equation (2.36) explicitly displays a wealth of information about the ellipse. To explain this, it will be convenient to have a change of notation. Let √ 2 b = a − c2 . Because c > 0, we see that a > b. Then what we have shown is that for two distinct points F1 and F2 on the x-axis symmetric with respect to the origin, a point P = (x, y) satisﬁes |P F1 | + |P F2 | = 2a if and only if (2.37)

x2 y2 + 2 = 1, 2 a b

where 0 < b < a.

This is the normal form of the equation of an ellipse whose center is the origin and whose foci lie on the x-axis. When the equation of an ellipse is written in the form of (2.37), it clearly exhibits the points of intersection of the ellipse with the x-axis, namely, (±a, 0), which are called the x-intercepts of the ellipse. The same equation also exhibits the points of intersection of the ellipse with the y-axis, namely, (0, ±b), which are called the √ y-intercepts of the ellipse. In addition, because c2 = a2 − b2 , we have c = a2 − b2 . Since 2c is the distance between the foci of the ellipse, c = |OF1 | = |OF2 |. By the Pythagorean theorem, the distance from F1 or F2 to (0, b) is therefore equal to a. Also observe that, because a > b, the segment between the x-intercepts is longer than the segment between the y-intercepts. This is why a is called the semimajor axis of the ellipse.

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

105

(0, b )

P a F1

F2

O

(a, 0)

For example, the graph of 2.5x2 + 4y 2 = 7, when rewritten as x2 7 2.5

+

y2 7 4

= 1,

7 . This shows that the denominators of the two terms on the left satisfy 0 < 74 < 2.5 7 2 therefore puts this equation in the context of equation (2.37). With a = 2.5 and b2 = 74 , we have 7 7 21 2 2 − = . c= a −b = 2.5 4 20 21 This then shows that the graph of 2.5x2 + 4y 2 = 7 is an ellipse with foci (± 20 , 0) 7 on the x-axis and with semimajor axis 2.5 . In general, we want to write down the equation of an ellipse whose two foci F1 and F2 lie on a horizontal line deﬁned by y = k for some constant k. As usual, we let F1 and F2 be of distance 2c apart (c > 0), and we let the semimajor axis of the ellipse be a, where a > c > 0. We may therefore write F1 = (h − c, k) and F2 = (h + c, k) for some ﬁxed number h. The center of the ellipse is then the point (h, k). The derivation of the equation for this ellipse does not diﬀer from that of (2.36), with the only diﬀerence that (0, 0) is now replaced by (h, k). See the picture below.

P = r(x, y) @ @ @ @r r r F1 = (h − c, k) F2 = (h + c, k) (h, k) Then for P = (x, y), |P F1 | + |P F2 | = 2a (a > c) is equivalent to (x − (h − c))2 + (y − k)2 + (x − (h + c))2 + (y − k)2 = 2a, or equivalently, ((x − h) + c)2 + (y − k)2 + ((x − h) − c)2 + (y − k)2 = 2a. As in the case when F1 = (−c, 0) and F2 = (c, 0), we are going to simplify this equation by getting rid of the square roots. Replacing (x − h) by X and (y − k) by Y , we may rewrite the equation as (X + c)2 + Y 2 = 2a − (X − c)2 + Y 2 .

106

2. QUADRATIC FUNCTIONS AND EQUATIONS

But this equation is formally identical with equation (2.33) on page 104, so the same calculations as before lead us to the analog of equation (2.34): a (X − c)2 + Y 2 = a2 − cX. Again we square both sides to get the following analog of equation (2.35): (a2 − c2 )X 2 + a2 Y 2 = a2 (a2 − c2 ). Replacing X by (x − h) and Y by (y − k), we get (2.38)

(a2 − c2 )(x − h)2 + a2 (y − k)2 = a2 (a2 − c2 ).

When h = k = 0, we get the previous equation (2.35); namely, (a2 − c2 )x2 + a2 y 2 = a2 (a2 − c2 ). At this point, two comments should be made. The ﬁrst is that, as in the case where the foci lie on the x-axis symmetrically around O, the equation (2.38) of the ellipse is also an equation of degree 2 in x and y. This can be seen by expanding equation (2.38); namely, (a2 − c2 )x2 + a2 y 2 + 2h(c2 − a2 )x + (−2a2 k)y + {(h2 − a2 )(a2 − c2 ) + a2 k2 } = 0. This is a special case of (2.15) on page 96 if we let A = (a2 − c2 ), B = 0, C = a2 , D = 2h(c2 − a2 ), E = −2a2 k, and F = (h2 − a2 )(a2 − c2 ) + a2 k2 . We also observe that, in this case, we have (compare with (2.31) on page 102) (2.39)

C>A>0

and 2

E2 D2 + − F > 0. 4A 4C

2

E 2 2 2 The last inequality is because D 4A + 4C − F = a (a − c ), by a simple computation, 15 and we know a > c > 0. A second comment is that we can multiply both sides of equation (2.38) by 1/(a2 (a2 − c2 )) to get the following analog of equation (2.36):

(2.40)

(x − h)2 (y − k)2 + = 1. a2 a2 − c2

Thus we see that the graph of (2.40) is exactly the translated image of the graph of (2.36) under the translation T (x, y) = (x + h, y + k) (see page 9). Acivity. Check the preceding assertion that the graph of (2.40) is the translated image of the graph of (2.36) under T . √ As in equation (2.37), we will rewrite (2.40) by letting b = a2 − c2 . Then we have (y − k)2 (x − h)2 + = 1, and 0 < b < a. (2.41) 2 a b2 This is the normal √ form of the equation of an ellipse whose foci lie on a horizontal line. Since c = a2 − b2 , we can directly read oﬀ from equation (2.41) that its graph is an ellipse with center (h, k), foci (h − c, k) and (h + c, k), and semimajor axis a. 15 The reason for the strange-looking expression of −F +D 2 /(4A)+E 2 /(4C) will be apparent when we get to equation (2.43) below, which is part of the proof of Theorem 2.19 on page 107. What we want to make clear is that this expression is not some clever concoction we make up but, rather, something handed to us naturally when we try to understand ellipses more deeply.

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

107

We are in a position to single out which equation of the form of (2.15) on page 96 has as its graph an ellipse whose foci lie on a horizontal line. In view of (2.39), it suﬃces to consider the following special case of equation (2.15): Ax2 + Cy 2 + Dx + Ey + F = 0 (2.42) where A, C, . . . , F satisfy (2.39). We will prove that, indeed, the graph of (2.42) is an ellipse with its foci lying on a horizontal line. To this end, rewrite the equation as D E A x2 + x + C y 2 + y = −F. A C Completing the squares gives 2 2 E2 D E D2 + − F. (2.43) A x+ +C y+ = 2A 2C 4A 4C We can bring this equation to normal form in the following way. Let K be the 2 E2 number on the right; i.e., K = D 4A + 4C − F . By (2.39), K > 0. Dividing through the preceding equation by K gives 2 2 A C D E + = 1. x+ y+ K 2A K 2C Now let (2.44)

a=

K , A

b=

K , C

h=

−D , 2A

and

k=

−E . 2C

Since C > A > 0 by (2.39), we have a > b > 0. Thus (2.42) becomes (x − h)2 (y − k)2 + = 1. a2 b2 This is an equation in the form of equation (2.41), and therefore the graph of (2.42) is an ellipse with center (h, k), semimajor axis√a, and foci (h − c, k) and (h + c, k), where a, b, h, and k are as in (2.44) and c = a2 − b2 . We have therefore proved:. Theorem 2.19. Let F1 and F2 be, respectively, two points (h−c, k) and (h+c, k) lying on the horizontal line y = k, where c > 0. Then the ellipse with foci F1 and F2 and semimajor axis a for an a > c is the set of points (x, y) satisfying the equation (x − h)2 (y − k)2 + = 1. a2 a2 − c2 Conversely, the graph of a second degree equation in two variables, Ax2 + Cy 2 + Dx + Ey + F = 0

where A, C, . . . , F satisfy (2.39)

is an ellipse with center (h, k), semimajor axis √ a, and foci (h − c, k) and (h + c, k), where a, b, h, k are given by (2.44) and c = a2 − b2 .

108

2. QUADRATIC FUNCTIONS AND EQUATIONS

It remains to make a few comments about ellipses whose foci lie on a vertical line. Let the two foci F1 and F2 be (0, c) and (0, −c) on the y-axis (instead of the x-axis). The ellipse consisting of all the points P = (x, y) satisfying |P F1 |+|P F2 | = 2a (0 < c < a) can be shown, as in equation (2.36) on page 104, to be the graph of the following y2 x2 equation: 2 + 2 = 1, where 0 < b < a, and b a √ b = a2 − c2 . For example, the graph of 43 x2 + 15 y 2 = 2 is an ellipse with foci on the y-axis because the equation is equivalent to x2 3 2

(0, a )

F1

P

( b, 0)

O

F2

y2 = 1, 10

+

and, of course, the denominators of the two terms on the left satisfy case, a = 10 and c = 10 − and (0, − 17 2 ). 2

2

3 2

=

17 2 .

3 2

< 10. In this Therefore the foci of this ellipse are (0, 17 2 )

When the foci of an ellipse lie on an arbitrary vertical line, x = h, so that its center is (h, k), for some constants h and k, let the foci be (h, k + c) and (h, k − c), where c > 0. Then the equation of the ellipse consisting of all the points P = (x, y) satisfying |P F1 | + |P F2 | = 2a (0 < c < a) is the following: (x − h)2 (y − k)2 + =1 2 b a2 √ where, again, 0 < b < a and b = a2 − c2 . This is the normal form of the equation of an ellipse whose foci lie on a vertical line. This ellipse is the translated image of 2 2 the ellipse with equation xb2 + ya2 = 1 under the translation T (x, y) = (x + h, y + k). (See page 9.) Expanding equation (2.45) yields

(2.45)

a2 x2 + (a2 − c2 )y 2 + (−2a2 h)x + (−2(a2 − c2 )k)y + {(a2 − c2 )(k2 − a2 ) + a2 h2 } = 0. This is a special case of equation (2.15) on page 96 where A = a2 , B = 0, C = a2 − c2 , D = −2a2 h, E = −2(a2 − c2 )k, and F = (a2 − c2 )(k2 − a2 ) + a2 h2 . In this case, observe that (2.46)

D2 E2 + − F > 0. 4A 4C

A > C > 0 and 2

2

E 2 2 2 The latter inequality is because D 4A + 4C − F = a (a − c ) > 0. This leads us to consider the graphs of equations of degree 2 in two variables of the form

Ax2 + Cy 2 + Dx + Ey + F = 0

where A, C, . . . , F satisfy (2.46).

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

109

We can prove as in Theorem 2.19 that these graphs are ellipses with foci lying on a vertical line. The following theorem summarizes the ﬁndings: Theorem 2.20. Let F1 and F2 be, respectively, two points (h, k+c) and (h, k−c) lying on the vertical line x = h, where c, h, and k are constants and c > 0. Then the ellipse with foci F1 and F2 and semimajor axis a for some a > c is the set of points (x, y) satisfying the equation (y − k)2 (x − h)2 + = 1. a2 − c2 a2 Conversely, the graph of a second degree equation in two variables, Ax2 + Cy 2 + Dx + Ey + F = 0

where A, C, . . . , F satisfy (2.46), 2

2

E is an ellipse with foci lying on a vertical line. If we let K = D 4A + 4C − F and let K K −D −E a= , b= , h= , and k = , A C 2A 2C

then this is an ellipse√with center (h, k), semimajor axis a, and foci (h, k + c) and (h, k − c), where c = a2 − b2 . We leave the details of the proof of the theorem to an exercise (see Exercise 10 on page 115). Hyperbolas When two points F1 and F2 are given as in the case of the ellipse, i.e., |F1 F2 | = 2c, there is a parallel development that considers the collection of all points P = (x, y) that satisﬁes not the previous condition of |P F1 | + |P F2 | = 2a but the condition that either |P F1 | − |P F2 | = 2a or −2a. More precisely, let F1 and F2 be two points in the coordinate plane so that |F1 F2 | = 2c > 0. Let a be a positive number so that 0 < a < c. Then the collection of all the points P that satisﬁes either |P F1 | − |P F2 | = 2a or |P F1 | − |P F2 | = −2a is called a hyperbola with foci F1 and F2 .16 It is easily seen that such a hyperbola consists of two disjoint geometric ﬁgures corresponding to |P F1 | − |P F2 | = 2a and |P F1 | − |P F2 | = −2a, respectively; these are called the two branches of the hyperbola. We usually abbreviate the deﬁnition by saying that a hyperbola is the collection of all the points P satisfying |P F1 | − |P F2 | = ±2a. The requirement that 0 < a < c shows up in the following way. Let foci F1 and F2 lie on a horizontal line as shown below, and let |F1 F2 | = 2c. The condition that 0 < a < c implies that the two branches of the hyperbola intersect the segment F1 F2 at Q and Q , respectively, so that |QQ | = 2a and |F1 Q| = |F2 Q |. Then the left branch consists of all the points satisfying |P F1 | − |P F2 | = −2a while the right branch consists of all the points satisfying |P F1 | − |P F2 | = 2a.

16 It will be seen to be a consequence of the triangle inequality (Theorem G36 on page 247) that if this collection of points P is to consist of more than two points, then necessarily a < c. See Exercise 10 on page 253.

110

2. QUADRATIC FUNCTIONS AND EQUATIONS

2a

F1

ʹ

F2

There are critical diﬀerences between ellipses and hyperbolas. Relegating the discussion of these diﬀerences to the next subsection, we will concentrate on their similarities here. As in the case of ellipses, if the line joining the foci of a hyperbola is neither horizontal nor vertical, then the mixed term in the equation deﬁning the hyperbola will be nonzero. Therefore, we will henceforth restrict ourselves to only those hyperbolas whose foci lie on a horizontal or a vertical line. With this understood, the preceding discussion for ellipses carries over almost verbatim to hyperbolas. In particular, if the foci lie on the x-axis, so that F1 = (−c, 0) and F2 = (c, 0), then the hyperbola will be the graph of the following equation of degree 2 in two variables: (2.47)

(c2 − a2 )x2 − a2 y 2 − a2 (c2 − a2 ) = 0.

Despite the formal similarity of equation (2.47) with the corresponding equation (2.35) for ellipses on page 104, we will give the derivation of (2.47) because this derivation has some unsuspected subtleties. Thus we are given |P F1 |−|P F2 | = ±2a. We ﬁrst take up |P F1 | − |P F2 | = 2a. With P = (x, y), we have

Therefore,

(x + c)2 + y 2 = (x − c)2 + y 2 + 2a.

Since both side are positive, this equation is equivalent to the equality of the squares of both sides; i.e., (x + c)2 + y 2 = (x − c)2 + y 2 + 4a (x − c)2 + y 2 + 4a2 . After simplifying, this is equivalent to (2.48)

cx − a2 = a

(x − c)2 + y 2 .

Since the right side of (2.48) is positive, so is the left side. Thus, once again, the equation (2.48) is equivalent to the equality of the squares of both sides. After simplifying, we immediately obtain (2.47). Next, suppose |P F1 | − |P F2 | = −2a. We rewrite this equation as 2a + |P F1 | = |P F2 |, and therefore

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

111

Both sides being positive, this equation is equivalent to the equality of the squares of both sides: 4a2 + 4a (x + c)2 + y 2 + (x + c)2 + y 2 = (x − c)2 + y 2 . After simplifying, we get (2.49)

a

(x + c)2 + y 2 = −cx − a2 .

Since the left side is positive, so is the right side. Therefore, equation (2.49) is equivalent to the equality of the squares of both sides. After simplifying, we again obtain (2.47). The reason we went through the proof of (2.47) in such detail is that, unlike the proof of equation (2.35) for ellipses on page 104, the proof of (2.47) has to deal with two separate cases. But by a stroke of good fortune, however, both cases lead to the same equation, namely, (2.47). Let it be understood that several of the proofs about hyperbolas below will each have to deal with two separate cases, and this fact will not be mentioned again. In addition, you may also be puzzled by the fact that the right side of (2.49) is positive until you realize that equation (2.49) holds only for those P = (x, y) that are constrained to satisfy |P F1 | − |P F2 | = −2a. A similar remark also applies to the fact that the left side of (2.48) is positive. Intuitively, equation (2.48) only applies to points (x, y) on the right branch of the hyperbola, and equation (2.49) only applies to points (x, y) on the left branch of the hyperbola. It remains to point out that equation (2.47) is a special case of equation (2.15) on page 96, where A = (c2 −a2 ), B = 0, C = −a2 , D = E = 0, and F = a2 (c2 −a2 ). Note that A and C have opposite signs; i.e., one of A and C is positive and the other negative. Dividing both sides of this equation (2.47) by a2 (c2 − a2 ), we obtain

Letting b =

√ c2 − a2 , we have

x2 y2 − = 1. a2 c2 − a2

x2 y2 − = 1. a2 b2 This is the normal form of the equation of a hyperbola whose foci are symmetrically placed on the x-axis with respect to the origin. The key feature of (2.50) is that, because the right side of (2.50) is positive, no point of the form (0, y) can satisfy (2.50) and therefore the hyperbola is disjoint from the y-axis. Also observe that (±a, 0) lie on the hyperbola. (2.50)

Example. The graph of x2 − 4y 2 = 1 is seen to be a hyperbola with foci on the xaxis when the equation is written in the form of (2.50): y2 x2 − 2 = 1. 1 1 2

The points (±1, 0) are on the hyperbola. Consider now the hyperbola whose points satisfy |P F1 | − |P F2 | = ±2a, where F1 = (h − c, k) and F2 = (h + c, k) are points on the horizontal line y = k, where h, k are constants and we have 0 < a < c as usual.

112

2. QUADRATIC FUNCTIONS AND EQUATIONS

P = r(x, y) @ @ @ @r r q (h, k) F1 = (h − c, k) F2 = (h + c, k) Then the hyperbola is the graph of the following equation: (2.51)

(c2 − a2 )(x − h)2 − a2 (y − k)2 = a2 (c2 − a2 ),

where 0 < a < c.

The proof of equation (2.51) is entirely analogous to the proof of equation (2.38) on page 106. By expanding (2.51), we see that the hyperbola is the graph of the following equation of degree 2 in two variables: (c2 − a2 )x2 − a2 y 2 − 2(c2 − a2 )hx + 2a2 ky (2.52)

+{(c2 − a2 )h2 − a2 k2 − a2 (c2 − a2 )} = 0.

In the notation of equation (2.15) on page 96, equation (2.52) is the special case of (2.15) where A = c2 − a2 , B = 0, C = −a2 , D = −2(c2 − a2 )h, E = 2a2 k, and F = (c2 − a2 )h2 − a2 k2 − a2 (c2 − a2 ). In terms of these values of A, C, . . . , F , we have the following inequalities about the coeﬃcients of the equation of a hyperbola whose foci lie on a horizontal line: E2 D2 + − F > 0. (2.53) A > 0, C < 0, and 4A 4C 2

2

E 2 2 2 The last inequality is because D 4A + 4C − F = a (c − a ) > 0. To see how such an inequality arises naturally, compare equation (2.43) and the comments surrounding it. Incidentally, by dividing both sides of equation (2.51) by a2 (c2 − a2 ), we obtain

(x − h)2 (y − k)2 − =1 2 a b2 √ where b = c2 − a2 . This is the normal form of the equation of a hyperbola whose foci are placed symmetrically on the horizontal line y = k with respect to the point (h, k). This hyperbola is the translated image of the hyperbola of equation (2.50) under the translation T (x, y) = (x + h, y + k). (See page 9.) It is now time to revisit the old question: what kind of equation of degree 2 (see equation (2.15) on page 96) will have a hyperbola as its graph? On account of (2.53), we may restrict ourselves to the following special case: Ax2 + Cy 2 + Dx + Ey + F = 0

where A, C, . . . , F satisfy (2.53).

Observe that, as far as A and C are concerned, the important thing is that they have opposite signs (see page 111 for the deﬁnition). However, there is no harm in requiring A to be positive, because we can simply multiply both sides of the equation by (−1) otherwise. That said, the reasoning used in the proof of Theorem 2.19 on page 107 easily leads us to the proof of the second half of the following theorem. Theorem 2.21. Let F1 and F2 be, respectively, two points (h−c, k) and (h+c, k) lying on the horizontal line y = k, where c > 0. Then a hyperbola, with foci F1 and F2 , consisting of all the points P = (x, y) satisfying |P F1 | − |P F2 | = ±2a (0 < a < c), is the graph of the equation (x − h)2 (y − k)2 − 2 = 1. 2 a c − a2

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

113

Conversely, the graph of a second degree equation in two variables, Ax2 + Cy 2 + Dx + Ey + F = 0

where A, C, . . . , F satisfy (2.53),

is a hyperbola with foci (h − c, k) and (h + c, k), where h = −D 2A and k = c is deﬁned as follows: let K K E2 D2 + − F, a= , b= . K= 4A 4C A −C √ Then c = a2 + b2 .

−E 2C

, and

In an exercise (Exercise 13 on page 115), you will be asked to supply the details of the proof of the second half of the theorem. Finally, we will wrap up the discussion of the similarities between ellipses and hyperbolas by brieﬂy addressing the case of a hyperbola with both foci lying on a vertical line. Thus let the foci be F1 and F2 , where |F1 F2 | = F1 2c (c > 0). Let 0 < a < c, and we consider the hyperbola of all the points P = (x, y) so that Q |P F1 | − |P F2 | = ±2a. Since a < c, the two branches of the hyperbola intersect the vertical 2a segment F1 F2 at two points Q and Q so that |QQ | = 2a and |F1 Q| = |F2 Q |. The point Q beQʹ longs to the upper branch satisfying |P F1 | − |P F2 | = −2a, and the point Q belongs to the lower F2 branch satisfying |P F1 | − |P F2 | = 2a. When the foci lie symmetrically on the y-axis with respect to the origin as F1 = (0, c) and F2 = (0, −c) (c > 0), the hyperbola of points satisfying |P F1 | − |P F2 | = ±2a (0 < a < c) is the graph of the equation, y2 x2 + =1 b2 a2 where b2 = c2 − a2 and a, b > 0. This is the normal form of the equation of a hyperbola whose foci are symmetrically placed on the y-axis with respect to the origin. From (2.54), it is obvious that the x-axis is disjoint from the hyperbola because no point of the form (x, 0) can be a solution of (2.54). We also have the analog of Theorem 2.20 on page 109 for hyperbolas. −

(2.54)

Theorem 2.22. Let F1 and F2 be, respectively, two points (h, k+c) and (h, k−c) lying on the vertical line x = h, where c, h, and k are constants and c > 0. Then the hyperbola with foci F1 and F2 consisting of all the points (x, y) satisfying |P F1 | − |P F2 | = ±2a (0 < a < c) is the graph of the equation (y − k)2 (x − h)2 + = 1. c2 − a2 a2 Conversely, the graph of a second degree equation in two variables, −

Ax2 + Cy 2 + Dx + Ey + F = 0 where A, C, . . . , F satisfy A > 0,

C < 0,

and

E2 D2 + − F < 0, 4A 4C

114

2. QUADRATIC FUNCTIONS AND EQUATIONS

−E is an ellipse with foci (h, k + c) and (h, k − c), where h = −D 2A and k = 2C , and c is deﬁned as follows: let E2 −K K D2 + − F, a= , b= . K= 4A 4C A C (Keep √ in mind that K is now a negative number and that C is also negative.) Then c = a2 + b2 .

Because the computations and the reasoning are quite similar, we will leave the proofs of these assertions to an exercise (see Exercise 14 on page 115.) Let us conclude with an observation about ellipses and hyperbolas. We know that all circles are similar to each other and that all parabolas are similar to each other (Exercise 4(c) on page 95). However, not all ellipses are similar to each other and not all hyperbolas are similar to each other. It can be shown, for example, that two ellipses ax2 + by 2 = 1 and cx2 + dy 2 = 1 are similar if and only if ab = dc or a d b = c. The history of conic sections in the development of mathematics provides an object lesson on the importance of fundamental scientiﬁc research—the kind of research that seems to be detached from reality. We are not certain what led to the discovery of these curves by the Greeks around the fourth century BC, but very likely it was inspired by what we call nowadays “recreational mathematics”. The Greeks were fascinated by the game of doubling the volume of a given cube using only straightedge and compass (see Section 7.3 on pp. 325ﬀ. for a fuller discussion). Now the only curves that can be drawn with a straightedge and compass are lines and circles, and since all attempts at a solution had failed up to that point, they started to look for curves that they could use, other than lines and circles, to help them solve the puzzle. That was when they started to experiment with the curves that could be obtained by intersecting a right circular cone with a plane, i.e., the conic sections. The credit of discovering the conic sections is generally given to Menaechmus (circa 380–320 BC), who was the ﬁrst to show that if parabolas instead of circles are used, then the problem of doubling the volume of a given cube becomes solvable. See page 95 of [Boyer-Merzbach]. That seems to have been the beginning of the geometric study of conic sections. But in spite of the great work of Apollonius, the conic sections found no substantial applications within mathematics and languished for about eighteen centuries until 1609, when Johannes Kepler (1571–1630) announced his monumental discovery that the orbits of the planets around the sun are ellipses. See [Katz, Chapter 10]. From the perspective of Newtonian mechanics, any motion around the sun traces out conic sections, and the elliptic orbits of the planets tell only part of the story. Planets are celestial objects held captive by the sun’s gravitational ﬁeld, but in theory, those celestial objects that can escape the gravitational pull of the sun (and thereby get out of the solar system never to return) will trace out hyperbolic or parabolic orbits. In October of 2017, an interstellar visitor, ‘Oumuamua—together with its hyperbolic orbit around the sun—was oﬃcially identiﬁed as one such (see [Wiki-‘Oumuamua]), and it is expected to stay in the solar system for roughly another 20,000 years before exiting.17 Nowadays, the fundamental role of conic sections in mathematics 17 On November 16, 2019, a second interstellar visitor was spotted by the Hubble Space Telescope, and it is a comet! See [Wiki-2I/Borisov].

2.3. GRAPHS OF EQUATIONS OF DEGREE 2

115

and in the sciences is clearly understood, and their history furnishes a resounding argument that, even in our quest to understand the world around us, there will always be a place for the pursuit of knowledge for its own sake.

Exercises 2.3. (1) (i) What is the graph of (x + y − 2)(3x − y + 1) = 0? Explain. (ii) What is the graph of (x + 1)2 − 3y 2 = 0? (iii) What is the graph of (x2 − y 2 ) + y − x = 0? (2) (a) Find the center and the radius of the circle which is the graph of 3x2 +3y 2 +6x−9y−37 = 0. (b) Do the same for 5x2 +5y 2 −8x+4y−61 = 0. (3) Let P be the parabola that is the graph of equation (2.16) on page 97. Let T be the translation so that T (x, y) = (x + h, y + k), and let P = T (P). Prove directly that P is the graph of equation (2.17) on page 98. (4) (i) Show that the graph of y = −2(x−1)2 +4(x−1)−3 is a parabola. What are its focus and directrix? (ii) Show that the graph of 3y 2 − x − 1 = 0 is a parabola. What are its focus and directrix? (5) (a) Is the graph of the equation 3x2 + 4y 2 + x − y − 80 = 0 a circle or an ellipse? If it is a circle, ﬁnd its center and radius. If it is an ellipse, ﬁnd its foci and semimajor axis. (b) Do the same for 4x2 + 3y 2 + x − y − 80 = 0. (6) (i) Find the foci and the semimajor axis of the ellipse deﬁned by −2x2 − 3y 2 + x − y + 10 = 0. (ii) Do the same for 4x2 + y 2 − 8x + y + 1 = 0. (7) Given an ellipse with foci at (−c, 0) and (c, 0) and passing through (45, 0) and (0, 23). What is c? (8) Show that the graph of x = a(y − k)2 + b(y − k) + c, where a, b, c, and k are constants and a = 0, is a parabola. What are its focus and directrix? (9) Show that the graph of x = ay 2 + by + c, where a, b, c are constants (a = 0), is a parabola so that 2 1 b its focus is ( 4ac−b + 4a , − 2a ), 4a 2 1 − 4a , and its directrix is the vertical line deﬁned by x = 4ac−b 4a 4ac−b2 b its vertex is ( 4a , − 2a ). (10) Write out the details of the proof of Theorem 2.20 on page 109. (11) Consider the set of all triangles ABC so that their perimeter is a ﬁxed constant K and so that the length of one side, |BC|, is also a ﬁxed constant a. (i) Describe all the possible positions of the third vertex A. (ii) What is the largest possible distance of A from the line LBC ? (12) (a) Determine the foci of the hyperbola deﬁned by 2x2 − 5y 2 + 4x − 1 = 0 and sketch the graph. What are the points of intersection of the hyperbola with the line joining the foci? (b) Do the same for 2x2 −5y 2 +4x+10y+2 = 0. (13) Prove the second half of Theorem 2.21 on page 112. (14) (i) Give the details of the proof of the ﬁrst part of Theorem 2.22 on page 113. (ii) Do the same for the second part of Theorem 2.22. (15) (i) Prove that if the graph of Ax2 + Cy 2 + Dx + Ey + F = 0 is a parabola, then exactly one of A and C is 0. (ii) Prove that if the graph of Ax2 + Cy 2 + Dx + Ey + F = 0 is a hyperbola, then AC < 0.

116

2. QUADRATIC FUNCTIONS AND EQUATIONS

2.4. The concept of an asymptote The main objective of this section is to point out a critical diﬀerence between hyperbolas and ellipses: the former have asymptotes. We will give a deﬁnition of an asymptote as best we can without the use of limits. If we look at the graph of a hyperbola such as x2 − 4y 2 = 1, we will notice that it gets “very close” to the lines y = 12 x and y = − 12 x when it is far from the origin. We will now explain the meaning of two graphs of equations being “very close” by the use of functions. The explanation will be somewhat informal because the full explanation requires the concept of limits of functions.18 To begin with, observe that the hyperbola is not the graph of a function because it does not satisfy the vertical line rule. However, suppose we concentrate on the part of the hyperbola in the upper half-plane {y ≥ 0}; then we are looking at x2 − 4y 2 = 1,

with y ≥ 0.

We get y 2 = 14 (x2 − 1). If x2 − 1 ≥ 0, or equivalently, if |x| ≥ 1, we can solve for y √ to get y = + 12 x2 − 1 (using the fact that y ≥ 0; the “+” sign is for emphasis). In order to express the graph of this equation as the graph of a function, we introduce a standard notation: we will denote the ray (see page 354 for the deﬁnition) consisting of all the numbers ≤ a (where a is some number) by (−∞, a], and the ray of all the numbers ≥ b (where b is some number) by [b, ∞). In this notation, the collection of all the numbers satisfying |x| ≥ 1 is therefore the union of two semi-inﬁnite intervals, (−∞, −1] ∪ [1, ∞). Then the part of the hyperbola in the upper half-plane is the √ graph of the function h : (−∞, −1] ∪ [1, ∞) → R so that h(x) = + 12 x2 − 1. There are now two cases to consider: x ≥ 1 and x ≤ −1. Case I. Suppose x ≥ 1. Then what we want to show is that the part of the hyperbola in the ﬁrst quadrant is “very close” to the line y = 12 x. Now the hyperbola √ above [1, ∞) is the graph of the function h : [1, ∞) → R so that h(x) = 12 x2 − 1, and the graph of y = 12 x above [1, ∞) is the graph of the linear function g(x) = 12 x. By deﬁnition, the hyperbola x2 − 4y 2 = 1 is “very close” to the line y = 12 x if (2.55)

|h(x) − g(x)| is arbitrarily small for all suﬃciently large x.

The meaning of the last phrase will be further elucidated below. We ﬁrst simplify |h(x) − g(x)|: 1 1 1 x2 − 1 − x = x2 − 1 − x . |h(x) − g(x)| = 2 2 2 √ √ Of course, when x is very large, there x2 − 1 and x2 . √ √ is little diﬀerence between Therefore when x is very large, x2 − 1 − x is roughly x2 − x = x − x = 0. So it stands to reason that when x is very large, |h(x) − g(x)| is roughly equal to 0, which is essentially the claim in (2.55). However, we can make the argument more 18 See

the appendix of Section 6.1 in [Wu2020c].

2.4. THE CONCEPT OF AN ASYMPTOTE

117

√ convincing by using a standard trick to convert the diﬀerence x2 − 1 − x into a form that makes its behavior more transparent when x is large: 1 2 ( x − 1 − x) = 2

=

√ 1 2 x2 − 1 + x ( x − 1 − x) √ 2 2 x −1+x √ √ 1 ( x2 − 1 − x)( x2 − 1 + x) √ . 2 ( x2 − 1 + x)

The purpose of the trick is to get the numerator inside the absolute value to be √ free of square roots. We can see this if we let β = x2 − 1. Then the numerator becomes (β − x)(β + x) = β 2 − x2 = (x2 − 1) − x2 = −1. Therefore, we have

−1 1 . |h(x) − g(x)| = √ 2 2 x − 1 + x

√ Since x ≥ 1 > 0, x2 − 1 + x ≥ 0 + x ≥ 1, so the denominator inside the absolute value is already positive. Thus, we may remove the absolute value symbols: 1 1 1 1 √ |h(x) − g(x)| = ≤ · . 2 2 2 x x −1+x As x gets suﬃciently large, x1 gets arbitrarily small; a fortiori, so does |h(x) − g(x)|. This proves (2.55). Therefore when x ≥ 1, the hyperbola x2 − 4y 2 = 1 in the upper half-plane is, in this sense, very close to the line y = 12 x. Case II. Suppose x < −1. In this case, we are looking at the part of the hyperbola in the second quadrant and we want to show that the hyperbola x2 − 4y 2 = 1 there is “very close” to the graph of the line y = − 12 x. Now the hyperbola in the second √ quadrant is the graph of the function h : (∞, −1] → R deﬁned 1 by h(x) = 2 x2 − 1, and the line lying above (−∞, 1] is the graph of the linear function f (x) = − 12 x. Then, the meaning of the hyperbola x2 − 2y 2 = 1 being “very close” to the line y = − 12 x is that (2.56)

|h(x) − f (x)| is arbitrarily small for all suﬃciently large |x|.

The reasoning that follows is similar to the preceding case, so we can be more brief. We have 1 1 |h(x) − f (x)| = x2 − 1 + x = (−x) − (−x)2 − 1 . 2 2 Because x is negative, |x| = −x. Therefore, |h(x) − f (x)| =

1 |x| − (|x|2 − 1) . 2

118

2. QUADRATIC FUNCTIONS AND EQUATIONS

Therefore, |h(x) − f (x)| =

=

=

2 − 1) (|x| |x| + 1 (|x| − (|x|2 − 1) 2 2 |x| + (|x| − 1) 1 1 2 (|x| + (|x|2 − 1) 1 2

1 (|x| + (|x|2 − 1)

≤

1 1 · . 2 |x|

It follows that when |x| is suﬃciently large, |h(x) − f (x)| gets arbitrarily small, which is exactly (2.56). In any case, the conclusion is that when |x| get suﬃciently large, the hyperbola x2 − 2y 2 = 1 gets very close to the two lines 1 y=− x 2

and

y=

1 x 2

on (−∞, −1] and [1, ∞), respectively. If a function f is deﬁned on [x0 , ∞) (for some constant x0 ) and for some constants a and b, |f (x) − (ax + b)| is arbitrarily small as |x| gets suﬃciently large,19 then we say the line (deﬁned by) y = ax + b is an asymptote of the graph of f on [x0 , ∞).20 In this language, the line y = 12 x is an asymptote of the hyperbola in the upper half-plane of f on [1, ∞). One deﬁnes an asymptote of the graph of a function on (−∞, x0 ] in a similar manner and concludes likewise that the line y = − 12 x is an asymptote of the hyperbola in the upper half-plane on (−∞, −1]. If one looks at the lower half-plane, one would come to exactly the same conclusion, namely that the two lines y = 12 x and y = − 12 x are asymptotes of the hyperbola on (−∞, −1] and [1, ∞), respectively. Taking both half-planes into account, we conclude that the hyperbola x2 − 4y 2 = 1 has two asymptotes; namely y = ± 12 x. The reasoning in this example is perfectly general. For a given hyperbola 2 2 deﬁned by (x−k) − (y−) = 1 (where a, b > 0) it is not diﬃcult to prove that the a2 b2 lines y − = ± ab (x−k) are asymptotes of the hyperbola (see Exercise 2 immediately following). Thus each hyperbola always has two asymptotes. Exercises 2.4. (1) What are the asymptotes of the following hyperbolas: (i) −2x2 +3y 2 = 12. (ii) 4x2 − 2y 2 − 4x − 1 = 0. (iii) x2 − 3y 2 + 2x + 12y − 11 = 0. 2 2 (2) Prove that a hyperbola (x−k) − (y−) = 1 (where a, b > 0) always has a2 b2 two asymptotes, namely the lines y − = ± ab (x − k). 19 While not strictly necessary, we use |x| rather than x because this deﬁnition will remain valid as is in the following discussion of the behavior of f on (−∞, x0 ]. 20 The precise deﬁnition of an asymptote is given in the appendix of Section 6.1 in [Wu2020c].

2.4. THE CONCEPT OF AN ASYMPTOTE

119

√ √ (3) (i) Prove √ directly √ that if the foci of a hyperbola are F1 = ( 2, 2) and F2 = (− 2, − 2), √then the hyperbola consisting of points P so that √ |P F1 | − |P F2 | = ±2 2 is the graph of xy = 1. (Thus c = 2 and a = 2 in the notation of page 109. This is a departure from our usual insistence on having the foci lying on either a horizontal line or a vertical line.) (ii) Explain why the x-axis is an asymptote of this hyperbola.

CHAPTER 3

Polynomial and Rational Functions After such a detailed study of linear and quadratic functions, one would expect another long chapter on cubic functions (polynomial functions of degree 3) and yet another on quartic functions (polynomial functions of degree 4), etc. The fact that this does not happen is a consequence of the fact that the theory loses its simplicity when the degree of the polynomial exceeds 2. Recall that the study of quadratic functions was greatly facilitated by the method of completing the square. There is no tool of comparable power and simplicity for polynomial functions of degree exceeding 2. Analogs of the quadratic formula for quadratic equations continue to hold for cubic and quartic equations,1 but they are unwieldy and therefore not particularly useful. It is a famous theorem of Abel and Galois that for equations of degrees exceeding 4, no analog of the quadratic formula exists.2 Consequently, we know far less about arbitrary polynomial functions. All we can do is to give a general discussion of the most basic properties of polynomial functions and rational functions. This chapter deviates from earlier chapters in one respect: although the statements of its theorems are entirely elementary, their proofs require a slight knowledge of basic facts from calculus about limits and derivatives so that the exposition here can no longer claim to be a logical consequence of what is in the preceding chapters. This deviation is justiﬁed by the fact that the theorems of this chapter expand students’ understanding of polynomials and deserve a place in the school classroom. As often happens in the upper levels of school mathematics, it is sometimes necessary to make use of some advanced theorems without proof. For example, in Section 5.3 on page 196, we will make use of the fact that every polynomial of degree n has exactly n (complex) roots (the fundamental theorem of algebra). Fortunately, the proofs for the advanced theorems that are needed in this chapter will become accessible as soon as we get to Chapters 2 and 6 of [Wu2020c]. 3.1. Some basic facts about polynomials In this section, we prove some basic theorems about polynomials with the help of some theorems from calculus, such as the intermediate value theorem and Rolle’s theorem. We will prove that every odd-degree polynomial function must have a zero, in contrast with the fact that, for example, the polynomial function f (x) = x4 + 1 clearly has no zeros. We will also show that a polynomial function of degree n cannot have more than n zeros. 1 For

cubic equations, see Exercise 6 on page 203. any book on Galois theory for a proof, e.g., Chapter 3 of [Hadlock]. N. H. Abel (1802– ´ Galois (1811–1832) was French. Theirs are among the most revered 1829) was Norwegian and E. names in mathematics, despite the fact that they both died in their twenties, due to poverty and a senseless duel, respectively. 2 See

121

122

3. POLYNOMIAL AND RATIONAL FUNCTIONS

Intermediate value theorem and odd-degree polynomial functions (p. 122) The number of zeros of a polynomial function (p. 126) Intermediate value theorem and odd-degree polynomial functions Our ﬁrst concern is with the phenomenon that some quadratic functions have no zeros. For example, x2 + 5, or more subtly 5x2 − 32x + 71. For that matter, the polynomial x26 + 5 also has no zero because x26 = (x13 )2 ≥ 0 (the square of a number being always ≥ 0), so that x26 + 5 ≥ 0 + 5 = 5 > 0. Against this background, the following theorem acquires signiﬁcance. Recall that a function of the form f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 for ﬁxed numbers an , . . . , a0 (an = 0) is called a polynomial function (see page 4) and n is called its degree. Of course we will soon get sloppy (like everybody else, because we can) and will sometimes call a polynomial function simply a polynomial. We also call an the leading coeﬃcient of f . For a later need, we also introduce the terminology of a monic polynomial (function) to be one whose leading coeﬃcient is 1.

Theorem 3.1. An odd-degree polynomial function must have a zero.

This theorem will also be seen to be a consequence of Theorem 5.8 on page 200. Theorem 3.1 should be (like all theorems in mathematics) taken literally: it guarantees that there is one zero for an odd-degree polynomial, but it does not preclude the possibility that an odd-degree polynomial may have many zeros. For example, the 7th-degree polynomial (x − 3)(x6 + 1) has exactly one zero at x = 3, but the 5th-degree polynomial h(x) = (x − 1)(x − 2)(x − 3)(x − 4)(x − 5) has ﬁve zeros. The proof of Theorem 3.1 requires the following special case of the intermediate value theorem (IVT).3 For its statement, we ﬁrst recall some standard notation: we denote all the numbers x satisfying a ≤ x ≤ b by [a, b], and we call this a closed interval (see page 10). By an open interval (a, b), we mean all the numbers x between a and b; i.e., a < x < b.4 Then the IVT states that if a polynomial function g deﬁned on the closed interval [a, b] satisﬁes g(a) = g(b), then for any number M between g(a) and g(b), there is an x0 ∈ (a, b) so that g(x0 ) = M . This theorem is entirely plausible because, graphically, it says the graph of g must “cross” the horizontal line y = M if g(a) < M < g(b) or g(a) > M > g(b).

3 To

be proved in full generality in Section 6.2 of [Wu2020c]. the closed interval [a, b] diﬀers from the open interval (a, b) in having two endpoints. It will be seen in Chapters 2 and 6 of [Wu2020c] that those two endpoints make all the diﬀerence in the world when it comes to the convergence of sequences. If you are puzzled by the fact that two points could make such a diﬀerence, you will have to ﬁnd the reason in the more general context of the topology of Euclidean spaces; [Rosenlicht] would be a good starting point for this purpose. 4 Thus

3.1. SOME BASIC FACTS ABOUT POLYNOMIALS

123

Y g(b) M graph of g g(a) X a x0 b The IVT is an example of some theorems in K–12 that students should know about because they are a basic part of the school curriculum, but which—because of their advanced nature—are usually not proved. Compare the discussion of FASM (Fundamental Assumption of School Mathematics) on page 4 and in Section 2.7 of [Wu2020a], which give a general guiding principle on how to cope with the arithmetic of irrational numbers within K–12; namely, simply treat them formally as if they were rational numbers.5 The reason IVT is relevant to Theorem 3.1 is clear: to prove that a polynomial function f has a zero, one way is to ﬁnd a large positive number c so that on [−c, c], f (c) > 0 and f (−c) < 0. Then because f (−c) < 0 < f (c), IVT implies that for some x0 ∈ (−c, c), f (x0 ) = 0. We will also need a simple inequality. Lemma 3.2. For any numbers a and b1 , . . . , bk , a + b1 + · · · + bk ≥ a − (|b1 | + |b2 | + · · · + |bk |). Proof. We know that for each bi , −|bi | ≤ bi ≤ |bi |. In particular, bi ≥ −|bi |, for each i = 1, 2, . . . , k. Thus by (B) of Section 2.6 in [Wu2020a] (recalled on p. 356 of this volume), we have a + b1 + · · · + bk ≥ a − |b1 | − |b2 | − · · · − |bk | = a − (|b1 | + |b2 | + · · · + |bk |). The lemma follows. Before giving the general proof of Theorem 3.1, let us prove a special case to get a feel for the general argument to come. Thus consider f (x) = x3 + 123x2 − 456x − 789. The idea is that for a very large positive c, f (c) > 0 because c being large, c3 completely dominates the other terms insofar as they involve at most c2 . For the same reason, the number f (−c) = −c3 +123c2 +456c−789 must be negative because −c3 is so “large in the negative direction” that the other terms involving at most c2 will hardly matter. IVT then yields the desired zero for this f , as explained above. Such an intuitive argument is good because it serves to convince us that Theorem 3.1 must be true, but the intuitive argument has yet to be converted into real mathematics because we have to use reasoning to push the argument to be beyond the shadow of a doubt. One reason the proof of Theorem 3.1 is worth learning is that, in the process, we get to see—at least in one case—how to convert intuition to a proof. The key step is to rewrite f as 123 −456 −789 + + . f (x) = x3 1 + x x2 x3 5 The

precise statement of FASM is given on page 351 of this volume.

124

3. POLYNOMIAL AND RATIONAL FUNCTIONS

Then let

123 −456 −789 + + x x2 x3 3 so that f (x) = x g(x). The reason for isolating g is that Lemma 3.2 will lead to a proof that g(x) > 0 when the absolute value |x| is suﬃciently large. Once this is done, then for a suﬃciently large positive c, we would have g(c) > 0 and g(−c) > 0. Consequently, g(x) = 1 +

f (c) = c3 g(c) > 0

and

f (−c) = −c3 g(−c) < 0.

By an earlier remark, IVT implies that f has a zero. Let us therefore prove that g(x) > 0 when the absolute value |x| is suﬃciently large. By Lemma 3.2, we have that for all x, 123 −456 −789 + + g(x) ≥ 1 − x x2 x3 (3.1)

= 1−

123 456 789 + 2+ 3 |x| |x| |x|

.

We may assume that |x| > 1, so that |x|3 > |x|2 > |x|. By the cross-multiplication algorithm, 1 1 1 . < < 3 2 |x| |x| |x| This implies 456 789 456 789 and , < < |x|2 |x| |x|3 |x| so that 123 456 789 123 456 789 + 2+ 3 < + + . |x| |x| |x| |x| |x| |x| By (A) on page 356, we have 123 456 123 456 789 789 123 + 456 + 789 + 2+ 3 > − + + . − = − |x| |x| |x| |x| |x| |x| |x| It then follows from (3.1) that for all x so that |x| > 1, (3.2)

g(x) > 1 −

123 + 456 + 789 . |x|

1 Let a number c (positive or negative) satisfy |c| (123 + 456 + 789) < 12 . By (A) on page 356, we have 1 1 − (123 + 456 + 789) > − . |c| 2 By (3.2), the following holds for such a c:

g(c) > 1 −

1 1 = . 2 2

Now let such a large c be positive; then f (c) = c3 g(c) > c3 · 12 > 0 and f (−c) = −c3 g(−c) < −c3 · 12 < 0 (we have used (E) on page 356). By IVT, this implies that for some x0 ∈ (−c, c), f (x0 ) = 0 and the proof of this special case of Theorem 3.1 is complete. We are now ready to tackle the general case.

3.1. SOME BASIC FACTS ABOUT POLYNOMIALS

125

Proof of Theorem 3.1. Let the polynomial function be f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 for some ﬁxed constants an , . . . , a0 (an = 0), where n is odd. We have f (x) = an f0 (x), where an−1 n−1 an−2 n−2 a0 f0 (x) = xn + x + x + ···+ . an an an Because an = 0, we see that f (k) = 0 for some number k if and only if f0 (k) = 0. The diﬀerence between f and f0 is that f0 is a monic polynomial. Therefore it suﬃces to prove that a monic odd-degree polynomial such as f0 (x) always has a zero. Thus we may let an = 1; i.e., f is now f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 . We will prove that any suﬃciently large c will satisfy f (c) > 0 and f (−c) < 0. Once this is done, by applying IVT to f on the closed interval [−c, c] with f (−c) < 0 < f (c), we see that for some x0 ∈ (−c, c), f (x0 ) = 0, as desired. To ﬁnd such a c, we rewrite f as an−1 an−2 a0 + 2 + ··· + n . f (x) = xn 1 + x x x For ease of discussion, we let an−2 an−1 a0 def g(x) = 1 + + 2 + ··· + n. x x x Then we have f (x) = xn g(x). We will concentrate on getting a handle on g(x). For any x so that |x| > 1, we have, by Lemma 3.2, a a n−1 an−2 0 g(x) ≥ 1 − + 2 + ···+ n x x x (3.3)

= 1−

|an−1 | |an−2 | |a0 | + + ···+ n |x| |x|2 |x|

.

Since |x| > 1, we have |x|n > |x|n−1 > · · · > |x|2 > |x| (see (D) on page 356). By the cross-multiplication algorithm, 1 1 1 1 1 1 , , ··· , . < < < |x|n |x| |x|n−1 |x| |x|2 |x| Since |an−2 |, . . . , |a0 | are all ≥ 0, we also have |a0 | |a0 | , ≤ n |x| |x|

|a1 | |a1 | , ≤ n−1 |x| |x|

··· ,

|an−2 | |an−2 | . ≤ 2 |x| |x|

Therefore, |an−1 | |an−2 | |a0 | |a0 | |an−1 | |an−2 | + + + ··· + . + ··· + n ≤ 2 |x| |x| |x| |x| |x| |x| This implies |an−1 | |an−2 | |an−1 | |an−2 | |a0 | |a0 | − + + + ···+ + ··· + n ≥ − . |x| |x|2 |x| |x| |x| |x|

126

3. POLYNOMIAL AND RATIONAL FUNCTIONS

Now (3.3) becomes, for all x so that |x| > 1, g(x) ≥ 1 − = 1−

|an−1 | |an−2 | |a0 | + + ···+ |x| |x| |x|

1

· |an−1 | + |an−2 | + · · · + |a0 | . |x|

Let K = |an−1 | + |an−2 | + · · · + |a0 |. We choose a positive number c so large that c > 1 and Kc < 12 . Then, K 1 1

· |an−1 | + · · · + |a0 | = < c c 2 so that

1 1

1 · |an−1 | + · · · + |a0 | > 1 − = . c 2 2 In exactly the same manner, we get g(c) ≥ 1 −

1 1 1 · |an−1 | + · · · + |a0 | > 1 − = . | − c| 2 2

g(−c) ≥ 1 − Altogether, we get g(c) > f (c) = cn g(c) > cn ·

1 2

and g(−c) > 12 . Consequently,

1 1 > 0 and f (−c) = −cn g(−c) < −cn · < 0. 2 2

Now we know that 0 is a number between f (−c) and f (c). By IVT, there is some number x0 ∈ (−c, c) so that f (x0 ) = 0. Theorem 3.1 is proved. The number of zeros of a polynomial function Our next goal is to generalize the fact that a quadratic function has at most two zeros. The precise statement is: Theorem 3.3. A polynomial function of degree n has at most n (distinct) zeros. We will use calculus to prove this theorem here, but a purely algebraic proof will be given in Section 5.1 on page 184. Both proofs make use of mathematical induction (see Section 1.7 on page 57). Proof of Theorem 3.3. First recall a basic tool from calculus, Rolle’s theorem: Let f be a continuous function deﬁned on a closed interval [a, b], so that it is diﬀerentiable in the open interval (a, b) and so that f (a) = f (b). Then there is a number c in c ∈ (a, b), i.e., a < c < b, so that the derivative of f at c satisﬁes f (c) = 0.6 Of course, polynomial functions are diﬀerentiable everywhere. We are trying to prove that the following statement Sn is true for each positive integer n: Sn : A polynomial function of degree n has at most n zeros. 6 We

will give a proof of Rolle’s theorem in Section 6.4 of [Wu2020c].

3.1. SOME BASIC FACTS ABOUT POLYNOMIALS

127

We use mathematical induction. First we check S1 : does a polynomial function of degree 1 have at most 1 zero? Yes, this is easy to see; e.g., it is proved in Section 6.2 of [Wu2020a] that such a function has exactly 1 zero. Next, suppose Sk is true; then we must prove that Sk+1 is also true. So let f be a polynomial function of degree k + 1, and let x1 < x2 < x3 < · · · < xm be any m of its zeros arranged in increasing order. By Rolle’s theorem, in each of the (m − 1) intervals (x1 , x2 ), (x2 , x3 ), . . . , (xm−1 , xm ), there is at least one zero of the derivative f of f . Therefore, the number of zeros of f ≥ (m − 1). Now f is a polynomial function of degree k (since the degree of f is k + 1). Because we are assuming that Sk is true, k ≥ the number of zeros of f . Hence k ≥ (m − 1), which is the same as m ≤ (k + 1). This says that if we take any m zeros of f , then this m is at most k + 1. In particular, the total number of zeros of the polynomial function f of degree k + 1 is at most k + 1. This proves Sk+1 , and thus Theorem 3.3 as well. The line of reasoning in the preceding proof allows us to draw some qualitative conclusions about the graphs of polynomial functions. Suppose we are told that the following graph of a polynomial function g crosses the x-axis only at three points as shown and that g has degree 5. Should we believe it?

Er r F

O r C Ar

Br

r D

Observe that Theorem 3.3 does not yield much information about this question, because the graph of g crosses the x-axis three times and therefore g appears to have 3 zeros. This would be consistent with g being of degree 5. Furthermore, every horizontal line appears to intersect the graph at no more than 3 points, and this means the function g(x) + c has at most 3 zeros; this would also be consistent with the polynomial g(x) + c being of degree 5, and therefore also consistent with g itself being of degree 5. But let us look at the derivative g . The tangents to the graph of g are horizontal at the six points A, B, . . . , F , and so g has at least 6 zeros. Since g is a polynomial function, the degree of g must be at least 6, according to Theorem 3.3. The degree of g then must be at least 7. So it cannot be 5. We can informally say that the graph of g has “peaks” at A, C, E, and “valleys” at B, D, F . It is intuitively clear that the number of zeros of g is at least the total number of “peaks” and “valleys” in the graph of g, so the preceding reasoning leads to the informal conclusion that the degree of a polynomial function is at least equal to 1 plus the total number of “peaks” and “valleys” in its graph. This can be rigorously proved by the use of calculus.

128

3. POLYNOMIAL AND RATIONAL FUNCTIONS

Exercises 3.1. (1) Approximately where does the graph of the function x3 + 2x2 − 1 meet the x-axis? (2) Prove that if two polynomial functions are equal, then they have the same degree and their coeﬃcients are pairwise equal. (3) Show that if f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 , where n is a positive integer and each ai ∈ R, then given any M > 0, there is a c > 1 so that f (c) > M . (4) Let f (x) = x3 − 1321x2 − 2890x − 694. Find a number c so that for all x > c, f (x) > 106 . (5) Assume a polynomial function f (x) = x55 −98765x53 −789x5 −1. Prove: If M is a positive number (no matter how large), there is a positive number c so that f (c) > M and so that f (−c) < −M . (6) Assume a polynomial function f (x) = an xn +an−1 xn−1 +· · ·+a1 x+a0 for some ﬁxed constants an , . . . , a0 (an > 0), and assume n is odd. Prove: If M is a positive number (no matter how large), there is a positive number c so that f (c) > M and so that f (−c) < −M . (7) A function f : R → R is said to be even if f (−x) = f (x) for all numbers x, odd if f (−x) = −f (x). (a) Give an example of a polynomial function that is even. (b) Do the same for an odd function. (c) Prove that every function from R to R is equal to the sum of an even function and an odd function. (You will need a knowledge of calculus for the next two parts.) (d) Give an example of a function that is even (respectively, odd) that is not a polynomial. (e) Apply part (c) to the function ex : what is the even function in this case, and what is the odd function? 2 +bx+c , where a, b, c, A, and B are (8) Let f be the function f (x) = axAx+B constants and a > 0, A > 0. Prove that for any given M > 0, there is a t > 0 so that f (t) > M . 3.2. Descartes’ rule of signs There are very few general theorems about polynomial functions, but Descartes’ rule of signs is one of them. Because it yields some information—even if not much—about the presence or absence of positive or negative zeros of a polynomial function, it is worth knowing. This section describes this rule and brieﬂy discusses some of its applications. The main theorem of this section is a classical result due to Ren´e Descartes (1596–1650), a codiscoverer of analytic geometry (the other codiscoverer is Pierre Fermat (1601–1665)).7 For its statement, we will have to deﬁne the multiplicity of a zero of a function. The motivation is the consideration of a quadratic function such as f (x) = (x − 1)2 . Yes, 1 is a zero of f , but clearly, we sense intuitively 7 Descartes’ mathematical output is contained in the third appendix of his magnum opus on philosophy, Discours de la m´ ethode . . . , published in 1637. His rule of signs is given in the third part of the appendix. It is here that Descartes ﬁnally codiﬁed our present system of symbolic notation, but—at least in France—his fame as a philosopher probably exceeds his fame as a mathematician. Incidentally, it is not likely that Descartes had a proof of his rule of signs, though he probably had a heuristic argument. The reason is that its proof does require some knowledge of the derivative of a function, and Descartes died before Newton and Leibniz brought calculus to the world.

3.2. DESCARTES’ RULE OF SIGNS

129

that although 1 is also a zero of g(x) = (x − 1)(x − 2), the nature of these two zeros is diﬀerent. In some sense, 1 should count as a double zero of f because f (x) = (x − 1)(x − 1), and the “roots” of f are therefore 1 and 1. There are various ways to make sense of this intuitive feeling; one way is to resort to the use of the derivative. Notice that f (x) = 2(x − 1) and therefore f (1) = 0, whereas since g (x) = (x − 1) + (x − 2), g (1) = 0. The precise deﬁnition is that a number r is said to be a zero of multiplicity k of a function f if r is a zero of the ﬁrst k − 1 derivatives of f , but not a zero of the k-th derivative (here the function f is regarded as the 0th derivative of f ). A zero of multiplicity 2 is called a double zero. A zero of multiplicity 1 is called a simple zero. In the preceding examples, 1 is a simple zero of g and a double zero of f . Recall that a polynomial function is said to be monic if its leading coeﬃcient is 1. For the rest of this section, let a monic polynomial function f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be given. The importance of having a monic polynomial in the present discussion is that its leading coeﬃcient is positive. This will streamline the ensuing discussion, as we shall see. In the following, we will be counting the number of times the coeﬃcients of f change signs (from positive to negative or from negative to positive), and we need to set up the convention for this purpose that the zero coeﬃcients are ignored. Thus the coeﬃcients of both h(x) = x6 − x2 + 3x + 6 and g(x) = x3 − x2 + 3x + 6 change signs 2 times, although the coeﬃcients of h are, strictly speaking, 1, 0, 0, 0, −1, 3, 6. In this context, the only thing that matters is that the coeﬃcients of h go from a positive number 1 (and bypassing the string of three 0’s) to a negative number −1 and then back to a positive number 3. We can now state Descartes’ theorem. Theorem 3.4 (Descartes’ Rule of Signs). Let C+ be the number of times the coeﬃcients of a monic polynomial function f change signs, and let Z+ be the number of positive zeros of f , counting multiplicity. Then Z+ ≤ C+ , and the two integers Z+ and C+ are both even or both odd. Moreover, if we set g(x) = f (−x), let C− be the number of times the coeﬃcients of g change signs, and let Z− be the number of negative zeros of f , counting multiplicity. Then Z− ≤ C− , and the two integers Z− and C− are both even or both odd. Before we worry about proving the theorem, let us see if it is good for anything. The goal in this kind of theorem is to give information about the presence or absence of the positive or negative roots of a polynomial function. From this point of view, this theorem is not ideal because it oﬀers an upper bound of Z+ in terms of C+ without also oﬀering a lower bound. Nevertheless, some information is better than no information at all. To reinforce this idea, we give two examples. Consider f (x) = x6 − 2x − 1. We have C+ = 1. Since Z+ ≤ C+ , we get Z+ ≤ 1. But since C+ is odd, so is Z+ . Thus Z+ is an odd whole number ≤ 1, which means Z+ = 1. Next, with g(x) = f (−x), we have g(x) = x6 + 2x − 1. Hence Z− ≤ C− = 1 and, as before, Z− = 1. We conclude that f (x) = x6 − 2x − 1 has one positive zero and one negative zero. As a second example, suppose we are trying to graph g(x) = x3 − x2 + 7x − 11. We notice that g(−x) = −x3 − x2 − 7x − 11, so that

130

3. POLYNOMIAL AND RATIONAL FUNCTIONS

C− = 0. By the theorem, Z− = 0. This tells us that we can aﬀord to pay more attention to the behavior of the graph of g on the right half-plane of the y-axis. The proof of this theorem is quite intricate and may not be suﬃciently instructive to merit spending class time on it. A complete proof will be posted on the author’s homepage, https://math.berkeley.edu/~wu/. However, it is possible to explain by an example how Descartes might have conjectured such a theorem, and we will proceed to do that. Let f be a cubic polynomial function which is the product of its linear factors; i.e., f (x) = (x − a)(x − b)(x − c). We have f (x) = x3 − (a + b + c)x2 + (ab + ac + bc)x − abc. We now verify Descartes’ rule of signs for such a cubic function. First suppose that Z+ = 3; i.e., all the zeros a, b, c are positive and the four coeﬃcients 1,

−(a + b + c),

(ab + ac + bc),

−abc

obviously alternate in signs. Thus also C+ = 3. In this case, Z+ = C+ . Next, suppose Z+ = 2. Since there is no diﬀerence between a, b, or c for this purpose, we may assume that a ≤ 0 and b, c > 0. If a = 0, then the coeﬃcients of f are 1, −(b + c), bc, 0. Thus C+ = 2. If however a < 0, then abc < 0 so that −abc > 0. Since 1 > 0, the coeﬃcients of f are positive,

−(a + b + c),

(ab + ac + bc),

positive.

There are two cases to consider: a + b + c > 0 and ≤ 0. If a + b + c > 0, then C+ = 2 regardless of whether ab + ac + bc is positive or negative or zero. If a + b + c ≤ 0, then the coeﬃcients of f become positive,

positive or 0,

(ab + ac + bc),

positive.

We now show that ab + ac + bc < 0, which implies that C+ = 2. If not, then ab + ac + bc ≥ 0, so that bc ≥ (−a)(b + c). But since a + b + c ≤ 0, b + c ≤ (−a) and therefore bc ≥ (−a)(b + c) ≥ (b + c)(b + c) = (b + c)2 which is impossible because bc ≥ (b + c)2 implies bc ≥ b2 + 2bc + c2 (recall that b and c are positive). Next, suppose Z+ = 1. We may let c > 0 and a, b ≤ 0 without any loss of generality. We have to show C+ is 1 or 3. First, if a = b = 0, then the coeﬃcients of f become 1, −c, 0, 0 and clearly C+ = 1. If both a and b are negative, then −abc < 0 and the coeﬃcients of f are 1, −(a + b + c), (ab + ac + bc), negative. It is easy to verify that no matter what the signs of −(a + b + c) and (ab + ac + bc) may be, C+ is either 1 or 3, and we are done again. Therefore we may assume one of a, b is negative and the other 0. Let us say a = 0 and b < 0. The coeﬃcients of f now become 1, −(b + c), bc, 0. Since bc < 0, C+ = 1 no matter what −(b + c) may be. Finally, suppose Z+ = 0 and we must prove that C+ is 0 or 2. In fact C+ = 0, and this will be left as an exercise.

3.3. RATIONAL FUNCTIONS

131

We have shown at least in the case of a cubic polynomial function that Z+ ≤ C+ always holds and the two integers are both even or odd. Similar experimentations then lend credence to the fact that Descartes’ rule of signs is true in general. Exercises 3.2. (1) Let f be the cubic polynomial function f (x) = (x − a)(x − b)(x − c), where a, b, c are numbers. Prove directly the following special case of Descartes’ rule of signs: if Z+ = 0, then also C+ = 0. (2) Discuss the nature of the zeros of the following polynomial functions: (a) f (x) = x7 + 2x6 − 8x3 + 3x + 1. (b) g(x) = x4 + x2 + 17x − 2. (c) h(x) = x3 + 12x + 5. (3) (a) Use Descartes’ rule of signs to determine the number of positive and negative zeros of x3 + x2 − x − 1. (b) Can you directly locate the zeros of this polynomial?

3.3. Rational functions The interest in rational functions, as far as school mathematics is concerned, is twofold: they are only one step away from polynomials and are therefore worth knowing, and surprisingly, their graphs display features that are genuinely diﬀerent from those of polynomials—they have asymptotes—and are therefore instructive for that reason. The goal of this section is to explain the latter phenomenon by making informal use of the concept of a limit. The functions that come closest to polynomial functions are the rational functions, which are the quotients of polynomial functions; i.e., f is a rational function if there are polynomial functions p(x) and q(x), so that for each number x, f (x) is the division p(x)/q(x). Because we cannot divide by 0, it is understood that the domain of f is outside the (at most) n zeros of q(x), where n is the degree of q (Theorem 3.3 on page 126). Both in form and in substance, rational functions are to polynomial functions as rational numbers are to whole numbers. Mathematical Aside: Recall that Q is the quotient ﬁeld of the integral domain Z. By the same token, the ﬁeld of rational functions is the quotient ﬁeld of the integral domain of polynomial functions. The simplest rational function is probably the “reciprocal function” r(x) = x1 . The domain of r(x) is all the nonzero numbers. This function already exhibits the new features of the graphs of rational functions that we wish to emphasize; namely, these graphs generally have a horizontal asymptote as well as a “vertical asymptote”. We have already discussed nonvertical asymptotes in Section 2.4 on p. 116. We now explain intuitively what it means for a vertical line to be an “asymptote” of the graph of a function.8 We say the line x = c is a vertical asymptote of the graph of a function f if |f (s)| gets arbitrarily large when either s decreases to c (s > c) or when s increases to c (s < c) or both. The y-axis is an 8 The

precise deﬁnition of an asymptote is given in the appendix of Section 6.1 in [Wu2020c].

132

3. POLYNOMIAL AND RATIONAL FUNCTIONS

1 asymptote of r(x) because |r(x)| = |x| so that |r(x)| gets arbitrarily large if |x| is arbitrarily small. This can be seen from the following graph of r(x):

Recall from page 118 that a nonvertical line represented by y = mx + k is an asymptote of the graph of a function f if the diﬀerence |f (x) − (mx + k)| goes to 0 as x gets arbitrarily large in either the positive direction or the negative direction. In the case of r(x), the x-axis (which is the graph of y = 0) is an asymptote 1 1 and certainly |x| goes to 0 when x gets arbitrarily large in because |r(x) − 0| = |x| the positive or negative direction. x . After Let us look at a slightly diﬀerent rational function g(x) = x2 +x−2 factoring the denominator, we rewrite g as x . g(x) = (x − 1)(x + 2) This expression of g lets us know that the domain of g is the set of all the numbers not equal to −2 or 1. In this case, we are going to show that the vertical lines x = −2 and x = 1 are asymptotes, as is the x-axis. The fact that the x-axis is an asymptote follows from a computation that makes use of the cancellation law (see page 356 or Section 2.5 of [Wu2020a]): g(x) =

x · x12 x = = x2 + x − 2 (x2 + x − 2) x12 1+

1 x 1 2 x − x2

.

We see that as x gets bigger and bigger in either the positive or negative direction, the denominator gets close to 1 while the numerator gets close to 0. As the x-axis is the graph of the equation y = 0, 1 x −→ 0 = 0 |g(x) − 0| = 1 1+ 1 − 2 x

x2

where the right arrow “−→” indicates that the fraction gets close to 01 . So according to the deﬁnition, the x-axis is an asymptote. The preceding reasoning actually proves that if a rational function g(x) is a quotient q(x)/p(x), where p(x) and q(x) are polynomials so that deg p(x) > deg q(x), then the x-axis is an asymptote of the graph of g(x). See Exercise 10 in Exercises 2.3 of [Wu2020c]. So far our discussion has been mainly about nonvertical asymptotes of the graph of a function. It is time to take a serious look at some vertical asymptotes. Consider the vertical lines x = 1 and x = −2 and the function x . g(x) = (x − 1)(x + 2)

3.3. RATIONAL FUNCTIONS

133

Observe that as x gets closer and closer to either 1 or −2 (but not equal to either), the absolute value |g| of g gets bigger and bigger, as shown:

Let us present the detailed argument for the line x = −2, for instance. First let x get close to −2 from the right; i.e., x > −2. Write x = (−2) + s, where s is small but positive. In particular, 0 < s < 1, so that 1 < 2 − s and 0 < 3 − s < 3. Then 1 1 · (3 − s) < (2 − s) · 3, which implies 2−s 3−s > 3 , by the cross-multiplication algorithm, so that 1 2−s −2 + s 1 1 = · . g(x) = > (−3 + s)s s 3−s s 3 This clearly shows that g(x) gets arbitrarily large when s decreases to 0. If on the other hand x < −2, write x = −2 − s, where 0 < s < 1. Then 1 2+s −2 − s =− g(x) = . (−3 − s)(−s) s 3+s Now 2 < 2 + s while 3 + s < 4 so that,again by the cross-multiplication algorithm, 2+s 2 1 1 1 2+s 1 1 1 1 3+s > 4 = 2 . Since s > 0, we have s 3+s > s · 2 . Therefore, g(x) < − s · 2 and therefore, 1 1 |g(x)| = −g(x) > · . 2 s As x gets closer and closer to −2, s gets increasingly small (because x = −2 − s) and therefore 1s gets large without bound. Thus |g(x)| gets arbitrarily large as x approaches −2 regardless of whether from the right or the left of −2. Hence the vertical line x = −2 is an asymptote of g. Finally, we give an example of a rational function with an asymptote that is neither horizontal nor vertical. Consider the rational function 2x2 . h(x) = x+4 The fact that the vertical line x = −4 is an asymptote will be ignored in this discussion because we have already seen how to verify this. We will concentrate instead on the fact that h has a nonvertical asymptote. The key to the existence of a nonvertical asymptote is the fact that the degree of the numerator of h exceeds the degree of the denominator by 1. The intuitive explanation of why this is so is quite clear: for the graph of h to be “almost straight” far away from the origin, h should be almost a nonconstant linear function far away from the origin. If the degree of the numerator of h is m and the degree of the denominator is n, then intuitively h behaves like a polynomial of degree m − n far away from the origin. Therefore

134

3. POLYNOMIAL AND RATIONAL FUNCTIONS

unless m − n = 1, h will not be anything like a nonconstant linear polynomial far away from the origin and its graph will have no asymptote. For the case at hand, the numerator of h is quadratic and the denominator is linear, so the graph of h will likely have an asymptote. It remains to show how to get the asymptote. The idea is to make use of division-with-remainder for quadratic polynomials (Theorem 2.10 on page 84). We divide the numerator 2x2 by the denominator x + 4 to get quotient 2x − 8 and remainder 32: 2x2 = (2x − 8)(x + 4) + 32. Notice that the quotient (2x − 8) is necessarily a linear polynomial because the diﬀerence in degree between numerator and denominator is 1. The presence of x + 4 in the numerator of h now leads to a cancellation, as in the following: 32 (2x − 8)(x + 4) + 32 = (2x − 8) + . h(x) = x+4 x+4 This therefore implies that for x > 0, 32 → 0 as x gets arbitrarily large. |h(x) − (2x − 8)| = x + 4 We see that the desired asymptote is the graph of the quotient 2x−8 of the division of 2x2 by x + 4, as shown:

Exercises 3.3. (1) (i) Find the asymptotes of

2x2 +5 x+2

and sketch its graph.

2

+5 . (ii) Do the same for 2x2−x −x and sketch its graph. (2) (i) Find the asymptotes of 4x−3

(ii) Do the same for

x2 −1 x2 −4 .

+2 (3) Find the asymptotes of f (x) = −3x x+2 , and explain why they are asymptotes. Discuss the behavior of the graph of f around the asymptotes and sketch the graph of f . 2 −2x+5 , and explain why they are (4) Find the asymptotes of f (x) = −x2x−1 asymptotes. Discuss the behavior of the graph of f around the asymptotes and sketch the graph of f . 4 (5) Can the rational function x2x+1 have a nonvertical asymptote? Explain. 2

CHAPTER 4

Exponential and Logarithmic Functions Besides polynomials and rational functions, the other standard so-called elementary functions are the exponential and logarithmic functions and the trigonometric functions. We will deal with trigonometric functions in Chapter 1 of [Wu2020c]. The objects of immediate interest are the exponential and logarithmic functions. There are two ways to approach these two classes of functions. The more mathematically satisfying way starts with the logarithmic function and then deﬁnes the exponential function as the inverse function of the logarithm; this is the approach that will be given in Chapters 6 and 7 of [Wu2020c]. In this approach, every concept is unambiguously deﬁned and the validity of the laws of exponents becomes completely transparent. Unfortunately, this approach requires the use of calculus and therefore ignores the needs of the general public which has to make practical use of the exponential and logarithmic functions (e.g., the Richter scale for the magnitude of an earthquake) regardless of whether they ever take a course in calculus or not. For this reason, a second approach to these functions is to deﬁne the exponential functions on rational numbers and pretend that they are already deﬁned on all real numbers (this should be reminiscent of FASM; see page 351). Then the logarithm is deﬁned as the inverse function of the exponential function. This approach—especially the way it is presented in TSM1 —has glaring pitfalls but is also blessed with the unrivaled advantage of being elementary. Let us discuss these issues with some care. The overriding diﬃculties with the second approach are that (1) in deﬁning the rational exponents of a positive number α, one must be very precise in diﬀerentiating between a heuristic argument that justiﬁes a deﬁnition and a bona ﬁde proof and (2) one must convey the idea that, while the laws of exponents appear to be mere number facts, they are actually properties of the exponential function. It is the function that provides the proper context for an understanding of these laws. The general lack of precision in TSM almost guarantees that point (1) will lead to disastrous consequences. Indeed, the heuristic arguments for deﬁning 30 = 1 and √ 5 5/2 deﬁning 3 as ( 3) are routinely mistaken by students and teachers for proofs √ so that the equalities α0 = 1 and α5/2 = ( α)5 (for any positive α) have generally come to be regarded as theorems. As for (2), the lack of eﬀort in TSM to place the laws of exponents for rational exponents in the √context of exponential functions leaves students with the impression that writing ( 3)5 as 35/2 is just another parlor game about changing the notation for numbers. This situation is not improved by the fact that the laws of exponents for rational exponents are stated in TSM without any reasoning, not even for the ubiquitous 1 See

page xi for the deﬁnition of TSM. 135

136

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

identity2 that (4.1)

√ √ √ n n n a b = ab

for all positive numbers a and b and for all positive integers n. It is a fact that the proofs of these “laws of exponents” for rational exponents—with a few minor exceptions such as (4.1)—are simply too intricate for school mathematics and, therefore, should be assumed on faith.3 But in the context of the general absence of reasoning in TSM, these laws become just another conﬁrmation of students’ suspicion that the only way mathematics can be learned is by rote memorization. The deﬁnitions of rational exponents in the second approach to exp x and log x mentioned above are meant to ease the introduction of exponential functions, e.g., 3x for all real numbers x. The idea is that if students can make sense of the function 3x for all rational x, then they can at least formally extrapolate 3x to all irrational x such as 3π . Unfortunately, because TSM cannot make sense of 3x even for rational x and cannot handle irrational numbers in general (witness the absence of any mention of FASM), the exponential functions end up being thoroughly mysterious objects in TSM. We want to salvage this approach because its elementary character allows for the introduction of the logarithmic function to school mathematics. It is clear, however, we need to minimize some of the liabilities that we have just pointed out for the purpose of a good school mathematics education. We propose a compromise solution. We will introduce the exponential functions ﬁrst, but we will do so by stating Theorem 4.1 on page 141 from the outset. We are following the tradition in school mathematics of making judicious use of a few advanced theorems (such as the fundamental theorem of algebra on page 197) to illuminate school mathematics when absolutely necessary. From the vantage point of Theorem 4.1, it is now possible to survey the landscape of rational exponents with coherence and simplicity. On the basis of this theorem, we can prove all we need to know about rational exponents, usually with relative ease. To compensate for the lack of a proof for Theorem 4.1, we will provide ample motivation for it and also show how to apply it eﬀectively to solve problems and to deﬁne the logarithm. These will be the main objectives of this chapter.

4.1. An interpolation problem The goal of this section is to introduce a new class of functions, the exponential functions, by way of a new phenomenon: interpolation. To understand what interpolation is all about, consider the following question: if the values of a function are known on a small collection of numbers such as the whole numbers, is there a “reasonable” way to make sense of the values of this function on all of R? The common mathematical formulation of this question is whether we can interpolate this function from the whole numbers to the real numbers. We will work this out for the special case of the exponential function. The phenomenon of interpolation (p. 137) The main theorem on exponential functions (p. 139) 2A

self-contained proof of (4.1) can be found on page 143. for example, pp. 184–191 of [Wu2010c]. The proofs of the laws of exponents—in one fell swoop—will be the high point of Chapter 7 in [Wu2020c]. 3 See,

4.1. AN INTERPOLATION PROBLEM

137

The phenomenon of interpolation To understand the need for interpolation, let us consider a standard problem. Often we get a function C(n) that is deﬁned only when n is a positive integer or whole numbers and, for various reasons, we would like C(n) to make sense even when n is an arbitrary real number. In fact, we came across exactly one on page 6. In that case we had a function, C(n) = 65n (dollars), that represents the cost of buying n books. Although the function C(n) is a function deﬁned on the positive integers, it can be easily “extended” to be a function F (x) deﬁned on the real line, F : R → R, by letting F (x) = 65x for all real numbers x. In that case, we got oﬀ easy because all we had to do was to replace the letter “n” by the letter “x”. We say F “extends” C because, while F (x) makes sense for every real number x, it “stays close to” C in the precise sense that, for any positive integer n, F (n) = C(n) (both are equal to 65n). Therefore F is not diﬀerent from C other than the fact that it can assign a number F (x) to any number x regardless of whether x is a positive integer or not. In so doing, of course the function F loses contact with √ the reality of √ book-buying because there is no such thing as the cost of buying 5 4 books (= F ( 5)). But as functions √ go, F is far easier to handle than C. For example, one can multiply F by 5 and not worry√about books, but one would really not want to do that to C because how would 5 · C(3) come up at the cash register? A practical reason for why F is preferable is that the graph of C(n) is a collection of dots, and dots are tiresome to draw: 585 455 325 195 65

r

r

r

r

r

r

r

r

1 3 5 7 9 On the other hand, the graph of F (x) is a line and, by any measure, a line is more drawable and more recognizable than a sequence of dots: r r 455 r r 325 r r 195 r r 65 1 3 5 7 9 585

We call the function F an interpolation of the function C, in the sense that F is not only deﬁned on R but also coincides with C wherever C is deﬁned; i.e., for all 4 Mathematical Aside: From an advanced standpoint, C(n) is hard to handle because we cannot apply the common tools of calculus to C(n).

138

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

positive integers n, F (n) = C(n). The graph of F is now seen to “connect the dots” in the graph of C. This is a useful and intuitive interpretation of interpolation: the process of connecting the dots of a given graph. More precisely, the graph of F is obtained from the graph of C by connecting the dots in the graph of the latter by linear segments; the fact that these separate segments together form a line is just an accident. If the interpolated function F of a given function C has the property that the graph of F is obtained from the graph of C by joining the consecutive dots in the latter by line segments, we call F the linear interpolation of C. Obviously there are many other ways to interpolate C, and the following graph shows one among an inﬁnite number of possibilities: 585 455

r r 325 r r 195 r r 65 1 3 5

r

7

r

9

The preceding example of interpolating the function C(n) = 65n by the function F (x) = 65x is trivial as far as interpolation goes; it was obvious how to get it done. The reason we bring up the concept of interpolation is that we have a much more interesting collection of functions deﬁned on the positive integers awaiting “suitable” interpolations. Let us start at the beginning. Unless stated to the contrary, α, β, γ will stand for positive numbers, and m, n, k, for positive integers. Recall that, by deﬁnition, α2 = αα, α3 = ααα, α4 = αααα, . . . , and in general, αn = αα · · · α

(n times).

The positive integer n is called the exponent or power of αn , and α is called the base of αn . One also speaks of αn as raising α to the n-th power. Here are the most basic facts concerning positive integer exponents: (E1)

αm αn = αm+n ,

(E2)

(αm )n = αmn ,

(E3)

(αβ)m = αm β m ,

where m, n are any positive integers. The proofs of (E1)–(E3) are obvious. For example, if m = 5 and n = 3, then (E1) holds because 8 5+3 α5 α3 = ααααα

=α = α . ααα = αααααααα 5

3

8

4.1. AN INTERPOLATION PROBLEM

139

Likewise, (E2) holds because (α3 )5

= (α3 )(α3 )(α3 )(α3 )(α3 ) = (ααα)(ααα)(ααα)(ααα)(ααα) = α5×3 = α3×5 .

The general proof for arbitrary m and n is no diﬀerent for both (E1) and (E2). The same is true of (E3). The formal simplicity and elegance, not to mention usefulness, of (E1)–(E3) must have inspired in people long ago the wish that these identities would be true not only for the positive integers but for all numbers. So our goal here is to make sense of αx for an arbitrary real number x, so that the following analogs of (E1)– (E3), called laws of exponents, continue to hold for all real numbers s and t: (E4)

αs αt = αs+t ,

(E5)

(αs )t = αst ,

(E6)

(αβ)s = αs β s .

Coincidentally, we should note that nature also has plenty of say in this matter because it seems to be just as eager to make sense of αx for any number x. For example, starting with 95 grams of the radioactive element radium, if we try to predict its mass m(t) after t years—where t is a fraction—it is a well-established fact that the following formula for m(t) is an excellent approximation to the observed data: m(t) = 95 · 2−t/1590 where 1590 is what is called the half-life of radium (the time it takes the initial mass to decay down to half the amount). Obviously, the experimental veriﬁcation of this formula is only for a relatively small number of fractional values of t, but it is natural to expect that the formula holds also for all values of t. It means that t of the number 2 for any t. Let we have to ﬁrst make sense of the exponent − 1590 us do that. The main theorem on exponential functions We begin by taking stock of where we stand. Let α be a positive number; then we have seen that α determines a function deﬁned on all positive integers so that if n is a positive integer, the function assigns αn to n. Thus the domain of deﬁnition of the function n → αn is the positive integers. What we now look for is an interpolation of this function αn to a function αx deﬁned on R, so that it assigns to any number x a positive number αx and so that (E4)–(E6) are valid.5 In this light, we see that (E4)–(E6) are statements about the function αx rather than an accidental collection of statements about numbers in a strange notation.6 Observe also that if n is a positive integer, the number αn retains the unambiguous meaning of α · α · · · α (n times). 5 Out of respect for tradition, we continue to denote the interpolated function by the same notation, αx , and rely on the symbol x (instead of n) to indicate that it is the interpolation of the original function αn deﬁned on the positive integers. 6 According to TSM, (E4)–(E6) are merely number facts. Wrong.

140

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

In order to get an idea of the diﬃculty of getting such an interpolation, let us look at an example. Let α = 2 and we use the linear interpolation (see page 138 for the deﬁnition) of the function n → 2n ,

for all positive integers n,

to see if it satisﬁes (E4). For the sake of clarity, let us denote the function that is the linear interpolation of 2n by F (x) (instead of 2x ); thus for each positive integer n, we have F (n) = 2n . In this notation, (E4) becomes the statement that F (s)F (t) = F (s + t); our pressing concern at this point is whether or not the equality F (s)F (t) = F (s + t) holds for all real numbers s and t. For instance, could F (s)F (t) = F (s + t) be true for s = t = 1.5? That is, is it true that F (1.5)F (1.5) = F (1.5 + 1.5)?

Because we are using the linear interpolation, the graph of the function F (x) (for all x in R) to the right of (1, 0) is given by the union of the slanted line segments above. Therefore the point (1.5, F (1.5)) is, by deﬁnition, the midpoint of the segment joining (1, 2) to (2, 4), as shown. The value of F (1.5) is thus the length of the vertical segment above 1.5, which is the average of the lengths of the vertical segments above 1 and 2 (see Exercise 1 on page 142), and the latter are, respectively, 2 and 4 because 21 = 2 and 22 = 4. Since 12 (2 + 4) = 3, we see that if we use the linear interpolation, we have F (1.5) = 3. It follows that F (1.5)F (1.5) = F (1.5 + 1.5), because the left side is 3 · 3 = 9 while the right side is F (3) = 23 = 8. This is the ﬁrst hint that the desired interpolation of the function 2n cannot be something simplistic like the obvious linear interpolation. There is in fact an additional diﬃculty: we cannot ﬁx an α and interpolate the function n → αn in isolation, one function at a time, but must interpolate all of them, for all α, “simultaneously”. To explain what this means, recall that the interpolated function, x → αx , must also satisfy (E5). The statement of (E5), as is, can be misleading, and we want to rephrase it to bring out the subtlety. Because we only need to verify (E5) one s at a time, let us ﬁx such an s and verify (E5) for every t. To this end, let β = αs . Now consider the interpolation of the function n → β n ; then (E5) is really a statement about the relationship between this particular interpolated function, x → β x , and the interpolation of the function n → αn , namely, the equality β t = αst

for every real number t.

4.1. AN INTERPOLATION PROBLEM

141

When stated like this, (E5) is seen to be about an inﬁnite number of equalities between pairs of interpolated functions α(sx) and β x (because as s changes, so does β = αs ). This is what we meant by “simultaneous interpolations” at the beginning of this paragraph. Having pointed out the nontrivial nature of these interpolations, we will bring closure to this whole discussion by quoting a theorem that gives us everything we want from such an interpolation. At ﬁrst glance, the theorem seems to address only (E4) but not (E5) or (E6), but implicit in this theorem is the fact that, among the three identities (E4)–(E6), only (E4) is truly basic while (E5) and (E6) are consequences of (E4). All this will be explained in Section 7.3 of [Wu2020c], but for now, we will assume this theorem (a proof will be given in Section 7.3 in [Wu2020c]). Theorem 4.1. Let a positive constant α be given. Then there exists a unique function αx : R → R so that: (A) αx is continuous, and for all positive integers n, αn has the usual meaning; i.e., αn = α · α · · · α (n times). (B) For all (real) numbers s and t, αs αt = αs+t . We remark that in (A), the term “continuous” is used because, without it, the theorem is false. Since continuity will be used only in a peripheral way in this volume (see pp. 157ﬀ. and 159), and since continuous functions will be the subject of Sections 6.1–6.2 in [Wu2020c], we will not spend any time on deﬁning continuity here. Condition (B) is just (E4). The fact that (E5) and (E6) follow (B) of Theorem 4.1 (i.e., (E4)), at least for rational values of s and t, will be proved in the next section on pp. 147ﬀ.; the validity for all s and t will follow from the density of Q in R, proved in Section 2.4 of [Wu2020c]). The function αx in the theorem is known as an exponential function with base α. Note that we are allowing α to be an irrational number from the beginning; we really have no choice because, when α = 3 and s = 1/2 in (E5), for instance, we √ will be called upon to make sense of (31/2 )t , and 31/2 = s is irrational because of Theorem 3.9 of [Wu2020a] (quoted on page 358 of this volume). We give a graphic illustration of the interpolation of the function 3n (n being a positive integer) by 3x . Here is the graph of 3n ; it is a sequence of dots:

142

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

The graph of 3x is seen to connect the dots of the graph of 3n :

In the next two sections, we will explore the consequences of Theorem 4.1. In summary, the message from this section that you cannot aﬀord to forget is that the introduction of real-number exponents of a positive number is a consequence of our desire to interpolate the function n → αn deﬁned on the positive integers n. Exercises 4.1. (1) Let ABCD be a trapezoid so that AD is parallel to BC (i.e., the line LAD is parallel to the line LBC ). Let E and F be the midpoints of AB and CD, respectively. Prove that EF is parallel to BC and |EF | = 12 (|AD|+|BC|). (2) Assume that the function n → 3n is interpolated by 3x as in Theorem 4.1. (a) If you are told that 30.25 = 1.3 approximately, what is an approximate value of 30.5 ? (No calculator.) (b) What is an approximate value of 90.5 ? (No calculator.) (3) For a given positive constant α, suppose there is a continuous function F : R → R so that (A) F (1) = α and (B) for all s, t ∈ R, F (s)F (t) = F (s + t). Assume Theorem 4.1 on page 141. Prove that F (x) = αx for all x.

4.2. Rational exponents Granting the existence of a positive n-th root of a positive number, we√will√prove √ its uniqueness and use it to prove the ubiquitous identity that n αβ = n α n β for all positive numbers α and β. Then we will use Theorem 4.1 of the last section to determine the values of the rational exponents of a positive number and prove (E5) and (E6), at least for rational exponents. The section concludes with the comprehensive deﬁnition of a number expression and a heuristic explanation for the diﬃculty of deﬁning 00 . Existence and uniqueness of the n-th root (p. 143) The rational exponents of a number (p. 144) Proofs of (E5) and (E6) (p. 147) Miscellaneous comments (p. 151)

4.2. RATIONAL EXPONENTS

143

Existence and uniqueness of the n-th root We will begin with some general facts about the n-th root of a positive number. We will assume the following fundamental theorem, which will be proved in Section 2.5 of [Wu2020c]. Theorem 4.2. Given a positive number α and a positive integer n, there exists one and only one positive number γ so that γ n = α. Given a positive number α and a positive integer n, a positive number γ is said to be a positive n-th root of α if γ n = α. In this terminology, Theorem 4.2 asserts that the positive n-th root of a positive number √ exists and is unique. The standard notation for the √ positive n-th root of α is n α. We can slightly extend the deﬁnition by deﬁning n 0 = 0 for a positive integer n because the only number β that satisﬁes β n = 0 is 0 (see Exercise 9 on page 152). Note that the case n = 2 is distinguished and the positive second root √ √ is called the positive square root; 2 α. the notation for the positive square root is α rather than the more elaborate √ √ n Also note that, as a matter of deﬁnition, α is always positive. Therefore 4 = 2, and never −2. The positive third root of α is traditionally called its positive cube root. Remarks. A few comments will help to clarify this theorem. (1) First of all, the positivity of α and γ in Theorem 4.2 is essential: (i) if α = −4 and n = 2, for example, the theorem is false as there can be no such γ (see (2.1) on page 64), and (ii) if α = 4 and n = 2 but γ is not required to be positive, then the theorem is again false because γ 2 = 4 for both γ = 2 and γ = −2. (2) The uniqueness part of the theorem, which says that there is at most one such γ, is not diﬃcult to prove; an exercise with hints (Exercise 1 on page 152) yields one proof, but the uniqueness will also be a simple consequence of Lemma 4.5 on page 148. However, the existence part (that for any positive α and for any positive integer n, there exists a γ so that γ n = α) is the source of the diﬃculty; as mentioned previously; this existence statement will be proved in Section 2.5 of [Wu2020c]. (3) But are we really saying that it is√diﬃcult to ﬁnd the square root of some√ thing as simple as 2? Indeed, what would 2 be? On the computer, one can get 2 as a ﬁnite decimal with any number of decimal digits as one wishes, yet the square of any of these ﬁnite decimals will almost equal 2 but not exactly. For example, the decimal 1.41421356237 has a square equal to 1.9999999999912458800169.

√ It is close to 2 but not 2. You can get as many decimal digits for 2 from the computer as you want, and the result will not change: it will always be almost but not quite. It is therefore reassuring to have Theorem 4.2. As illustration of the power of the uniqueness assertion in Theorem 4.2, we will give a proof of identity (4.1) on p. 136; namely, for any two positive numbers α and β and for any positive integer n, √ n αβ = n α n β. (4.2)

144

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

This identity is particularly useful when n = 2, which takes the form of √ αβ = α β. In fact, this special case was already used in the study of quadratic functions (equation (2.2) on page 65). √ √ The proof of (4.2) goes as follows: we are trying to show that n α n β is equal to the √ positive n-th root of αβ. According to Theorem 4.2, it suﬃces to show that √ n α n β raised to the n-th power is equal to αβ. To this end, we simply compute n

√ √ √ √ n α n β = ( n α n β)( n α n β) · · · · ( n α n β) .

n

According to Theorem 2 in the appendix of Chapter 1 of [Wu2020a] (quoted on page 358), the right side is equal to

√ √ √ n n α n α··· n α β n β··· n β .

By deﬁnition of the symbol

√ n

n

n

, this is equal to αβ. Altogether, we have n

√ n α n β = αβ.

By the preceding italicized remark, this proves (4.2). We should point out the following consequence of (4.2) that is an equally useful fact. Let α, β > 0 and let n be a positive integer. Then √ n α α n = √ (4.3) . n β β √ The proof of (4.3) can be obtained by simply replacing n β in the preceding argu1 ment with √ . We can leave the details as an exercise (see Exercise 5 on p. 152). n β There is another proof, however. By equation (4.2), α α n = n β· . β· n β β √ The right side is evidently equal to n α. Thus, √ α n β· n = n α. β But this is equivalent to (4.3). The rational exponents of a number At this point, we can re-engage with Theorem 4.1 on p. 141. We will show how this theorem determines the values of β x when x is rational. Let α be a ﬁxed positive number in the ensuing discussion.7 First, we recall part (B) of the theorem: αs αt = αs+t

(4.4)

for all s and t.

Our ﬁrst claim is that (4.5) 7 See

α0 = 1. page 146 for an explanation of why we restrict the base α to be a positive number.

4.2. RATIONAL EXPONENTS

145

Indeed, letting s = 0 and t = 1 in (4.4), we obtain α0 α1 = α1 . By part (A) of Theorem 4.1, this is equivalent to α0 α = α. Multiplying both sides by α1 , we get (4.5). Equation (4.5) has the important consequence that αx > 0 for every x.

(4.6)

To prove this, we ﬁrst show that αx/2 = 0 for every x. This is because, by (4.4), we have αx/2 · α−x/2 = α0 = 1 x/2 nor α−x/2 can be 0. Such being the case, and therefore neither α αx = αx/2 · αx/2 = (αx/2 )2 > 0 by (2.1) on page 64. Thus (4.6) is proved. Equation (4.5) allows us to draw another conclusion about the relationship between αx and α−x . Let s = x and let t = −x in (4.4); then we get αx · α−x = α0 , so that αx · α−x = 1. Therefore, 1 (4.7) α−x = x for every x. α One consequence of (4.7) is that if we know the value of αx when x is a fraction, then we know the value of αx for all rational numbers x. It may be worthwhile to consolidate our gains at this point. Let R+ denote the positive real numbers. For a positive number a, Theorem 4.1 says that ax is a function from R to R. But (4.6) says we can be more precise: we actually have ax : R → R+ ; i.e., the function ax is always positive. Moreover, the graph of ax has a kind of “inverted symmetry” with respect to the y-axis in the sense that if at a positive number t, at is equal to B, then at the mirror reﬂection −t of t on the x-axis with respect to the origin, the value of a−t is equal to 1/B. The natural next step in our quest to understand Theorem 4.1 is to investigate the value of αm/n , where m and n are positive integers. We claim that for any positive integer n, √ 1 (4.8) α n = n α. To prove (4.8), we observe that equation (4.4) has a simple generalization: for any n numbers s1 , s2 , . . . , sn , we have (4.9)

αs1 · αs2 · · · αsn = αs1 +s2 +···+sn .

The proof makes use of Theorem 2 in the appendix of Chapter 1 in [Wu2020a] (recalled on page 358 of this volume) and mathematical induction; it is simple enough to be left as an exercise (Exercise 6 on page 152). On the basis of (4.9), we have 1 1 1/n 1/n n +···+ n = α1 = α. α · α1/n

· · · · α = α n

This means

n α1/n = α.

Thus α1/n is a number that, when raised√to the n-th power, is equal to α. By the uniqueness part of Theorem 4.1, α1/n is n α. This proves equation (4.8). We pause to make two observations about equation (4.8). The ﬁrst is that, from our standpoint, this equation is the reason we use the exponential notation α1/n to

146

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

√ denote the n-th root of α: the value of the function αx happens to equal n α when 1 x = n . In TSM, the laws of exponents are presented as “number facts”, thereby creating the impression√that the exponential notation is just a game people play when they are tired of n α and want to write it diﬀerently. A second observation is that, as a result of equation (4.8), we can restate the identity (4.2) as follows: for any positive α and β, (αβ)1/n = α1/n β 1/n .

(4.10)

Observe that (4.10) is a special case of (E6) on page 139 when s = 1/n. We are ﬁnally in a position to determine the value of αx when x is a fraction m n:

(4.11)

αm/n =

√ m n α

for all fractions

m . n

First consider the special case of 45/3 ; we will show that it is equal to the cube root of 4 raised to the ﬁfth power, because 45/3

1

1

1

1

1

= 43+3+3+3+3 1 1 1 1 1 = 43 · 43 · 43 · 43 · 43 √ 3 = ( 4)5 (by (4.8)).

(by (4.9))

The proof in general of (4.11) is basically the same as in the special case: αm/n

1

1

1

= α n + n +···+ n 1 1 1 n n n = α (by (4.9)) · α · · · α m √ (by (4.8)). = ( n α)m

Thus equation (4.11) is proved. In summary, we have from (4.5), (4.7), and (4.11) that for any positive number α and any positive integers m and n,

√ m ⎧ m/n n α = α , ⎪ ⎪ ⎪ ⎨ α0 = 1, (4.12) ⎪ ⎪ ⎪ ⎩ −m/n 1 α = (√ n α)m . The list in (4.12) gives the values of αx whenever x is a rational number. This may be the appropriate moment to reﬂect on why we want the base α of αx to be positive. Suppose α were negative; what could α1/2 be? According to (4.8), it should be a number whose square is the negative number α. But by (2.1) on page 64, the square of every nonzero (real) number is positive, so α1/2 would not even make sense in this case. The same diﬃculty will persist for α1/n when n is even and α is negative. Therefore for αx to make sense for every rational number x, we have to require α to be positive. Of course, when n is an odd integer, then α1/n —being the number whose n-th power is equal to α—makes sense regardless of whether α is positive or negative. For example, (−27)1/3 = −3. See Exercise 8 on page 152.

4.2. RATIONAL EXPONENTS

147

Proofs of (E5) and (E6) Our next goal, the main goal of this section, is to prove the following basic properties of the exponential function: Theorem 4.3. Let α > 0. Then, (4.13)

(αs )t = αst

for all rational numbers s and t.

Theorem 4.4. Let α, β > 0 and let s be a rational number. Then (4.14)

(α · β)s = αs · β s

for all rational numbers s.

Notice that Theorems 4.3 and 4.4 are the special cases of (E5) and (E6), respectively, where s and t are rational numbers (rather than real numbers in general). However, if we make use of a standard theorem about continuous functions (Theorem 6.6 in Section 6.1 of [Wu2020c]), then Theorems 4.3 and 4.4 will in fact be suﬃcient to prove (E5) and (E6) in general. Notice also that equation (4.10) is the special case of Theorem 4.4 when s = 1/n. We ﬁrst prove Theorem 4.3. It is not likely that—due to its length and intricacy—you will ever present the complete proof of this theorem in a high school classroom. However, the proof is interesting and instructive because it will prove, along the way, some very useful facts. We begin by pointing out two special cases of Theorem 4.3 that are already known to us. By equations (4.11) and (4.8), we have m (4.15) (α1/n )m = αm/n for all fractions . n This is the special case of Theorem 4.3 when s = n1 and t = m for positive integers m and n. Next, we show that

−1/n m = α−m/n . (4.16) α This is then a special case of Theorem 4.3 when s = − n1 and t = m for positive integers m and n. The reason for (4.16) is that, by (4.7), α−m/n =

1 . αm/n

By (4.15), we have α−m/n =

1 (α1/n )m

.

Therefore, α−m/n =

1 (α

1/n

) · (α

) · · · · (α1/n )

1/n

.

m

By the product formula for rational quotients (see p. 356),8 , we have α−m/n =

1 1 1 × 1/n × · · · × 1/n . 1/n α α α

m

8 Together

with FASM.

148

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

By (4.7) again, this implies −1/n −1/n −1/n m α−m/n = α · α−1/n ) .

· · · · α = (α m

This is exactly (4.16), as desired. We will also need the following basic fact about n-th powers: Lemma 4.5. Let α and β be two positive numbers. If for a positive integer n, αn = β n , then α = β. It is clear that the uniqueness part of Theorem 4.2 follows immediately from this lemma. We will deduce Lemma 4.5 from the following useful lemma about the inequality of n-th powers. Lemma 4.6. Let α and β be two positive numbers. If α < β, then αn < β n for any positive integer n. Conversely, if for some positive integer n, αn < β n , then α < β. Note that the hypothesis of the positivity of α and β in Lemma 4.6 is crucial. For instance, −2 < 1 but (−2)2 > 12 , or, −3 < −2 while (−3)4 > (−2)4 . Postponing the proof of Lemma 4.6 for the moment, let us ﬁrst give the proof of Lemma 4.5. Proof of Lemma 4.5. Suppose αn = β n for a positive integer n. To show α = β, it suﬃces to show—by virtue of the trichotomy law (see page 359)—that it is impossible to have either α < β or α > β. So suppose α < β. Lemma 4.6 implies that αn < β n , and this contradicts the hypothesis that αn = β n . Similarly, the assumption that α > β leads to αn > β n , which also contradicts the hypothesis that αn = β n . The proof of Lemma 4.5 is complete. The preceding proof of Lemma 4.5 is valuable because its argument using Lemma 4.6 has many applications. However, one should be aware of a second proof of the lemma that is even shorter. See Exercise 1 on page 152. We will now retroactively prove Lemma 4.6. Proof of Lemma 4.6. Suppose α < β. We will prove αn < β n for all positive integers n by induction on n. This assertion being trivially true for n = 1, suppose it has been shown to be true for n − 1. We will prove that it is true for n. Since the assertion is true for n − 1, we have αn−1 < β n−1 . Since α > 0, we multiply both sides of this inequality by α to get αn < αβ n−1 . But β n−1 > 0, so multiplying both sides of the inequality α < β by β n−1 gives αβ n−1 < β n . Combining this inequality with the previous one then gives αn < αβ n−1 < β n . The induction is complete and we see that αn < β n for all n. We now prove the converse; i.e., if αn < β n for some positive integer n, then α < β. As in the proof of Lemma 4.5, we will appeal to the trichotomy law (see p. 359) among numbers (see page 359) and prove instead that it is impossible to have

4.2. RATIONAL EXPONENTS

149

either α = β or α > β, so that α < β is the only possibility. Let us ﬁrst rule out α = β: in this case, clearly αn = β n for any positive integer n, which contradicts the hypothesis that αn < β n . Now suppose α > β, then the ﬁrst half of this proof shows that αn > β n , again contradicting the hypothesis. Therefore, only α < β is possible. This proves Lemma 4.6. Corollary of Lemma 4.6. Let α and β be two positive numbers. If α < β, then α1/n < β 1/n for all positive integers n. Conversely, if for some positive integer n, α1/n < β 1/n , then α < β. The proof of the corollary will be left as an exercise (Exercise 7 on page 152). We now head towards the proof of Theorem 4.3. We ﬁrst prove a special case of the theorem. Lemma 4.7. For any positive number α and for any rational number s and any positive integer k,

s k = αsk . α Observe that this lemma diﬀers from Theorem 4.3 only in the fact that the k in the lemma is a positive integer rather than an arbitrary rational number. Proof of Lemma 4.7. If s = 0, both sides of the above equation are equal to 1 (see (4.12)). We may therefore assume that s = 0. Suppose s > 0. Then s = m/n for some positive integers m and n. Then we have to prove (4.17)

(αm/n )k = αkm/n

for all positive integers k.

The left side is equal to ((a1/n )m )k by (4.11) on page 146, which is equal to (a1/n )km by (E2) on page 138. The last is equal to akm/n by (4.11) again, and this proves (4.17). Now consider the case of s < 0. Let s = −m/n for some positive integers m and n. Then we have to prove (α−m/n )k = α−km/n Now let γ = α

−1/n

for all positive integers k.

. Then by (4.16), we have (α−m/n ) = (α−1/n )m = γ m

and

α−mk/n = (α−1/n )mk = γ mk . Therefore, what we have to prove can be rephrased in terms of γ as (γ m )k = γ km .

Since this follows from (E2) on page 138, the lemma is completely proved. We are now in a position to give the Proof of Theorem 4.3. The plan of the proof is to simply apply Lemmas 4.5 and 4.7 over and over again. We have to prove that, for α > 0, (αs )t = αst

for all rational numbers sand t.

If t = 0, both sides are equal to 1 (see (4.5) on p. 144). Henceforth, we may assume t = 0. First consider the case of t > 0. Then t = m/n for some positive integers m and n. What we must prove is then (4.18)

(αs )m/n = αsm/n

for any rational number s.

150

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

In order to prove (4.18), Lemma 4.5 implies that it suﬃces to prove the equality of the n-th power of both sides of (4.18). But by Lemma 4.7, the n-th power of the left side is ((αs )m/n )n = (αs )(m/n)n = (αs )m while the n-th power of the right side is (αsm/n )n = α(sm/n)n = αsm . Therefore we have to prove (αs )m = αsm

for any rational number s.

But this is exactly the statement of Lemma 4.7. Thus the theorem is proved in case t > 0. Consider now the case of a negative t. In this case, t = −m/n for some positive integers m and n. Thus we have to prove

s −m/n α = αs·(−m/n) for all rational numbers s. Because s(−m/n) = −(sm/n), equation (4.7) (on page 145) implies that this is equivalent to proving 1 1

m/n = α(sm/n) . s α However, this follows immediately from equation (4.18). The proof of Theorem 4.3 is complete. We can now give the proof of Theorem 4.4. We must prove equation (4.14); i.e., (α · β)s = αs · β s

for all rational numbers s.

If s = 0, there is nothing to prove as both sides will be equal to 1 (see (4.5) on p. 144). Therefore we may assume s = 0. If s > 0, then s = m/n for some positive integers m and n. In this case, we have to prove (4.19)

(α · β)m/n = αm/n · β m/n .

By Lemma 4.5, it suﬃces to prove that the n-th power of both sides of equation (4.19) are equal. Now the n-th power of the left side of (4.19) is equal to (α · β)m , according to Lemma 4.7. On the other hand, the right side of (4.19) when raised to the n-th power is equal to

m/n m/n n α ·β = (αm/n )n (β m/n )n according to (E3) on page 138. Using Lemma 4.7 once more, we see that the right side of the preceding equation is equal to αm · β m , which is equal to (α · β)m , according to (E3) again. Thus both sides of equation (4.19) are indeed equal when raised to the n-th power. Equation (4.19) is proved. It remains to prove (4.14) when s < 0. In this case, s = −(m/n) for some positive integers m and n. Therefore we must prove (4.20)

(α · β)−(m/n) = α−(m/n) · β −(m/n) .

By (4.7) on p. 145, this is equivalent to proving 1 1 1 = m/n · m/n . m/n (α · β) α β

4.2. RATIONAL EXPONENTS

151

By applying the product formula for rational quotients (see p. 356) to the right side,9 this in turn is equivalent to proving 1 1 = m/n m/n . (α · β)m/n α ·β According to (4.19), this equality is correct. The proof of Theorem 4.4 is complete.

Miscellaneous comments We conclude this section with two comments. First, we bring closure to the deﬁnition of a number expression, or expression for short, in a given collection of numbers x, y, . . . , w. This concept was ﬁrst deﬁned in Section 6.1 of [Wu2020a] and was later expanded on page 14 of this volume. Now we know that on the semiinﬁnite interval [0, ∞) consisting of all numbers ≥ 0, there is deﬁned a function √ which associates with each t the number n t (n is a positive integer). Therefore, we arrive at the ﬁnal form of the deﬁnition: A number expression in a given collection of numbers x, y, . . . , w is a number obtained from these x, y, . . . , w, from a collection of speciﬁc real numbers, and from the values of a given collection of functions f , . . . , h at these x, y, . . . , w using a speciﬁc combination of arithmetic operations (i.e., +, −, ×, ÷) and the operation of taking the positive n-th root of a positive number. √ For example, 4 5x2 + x + 8 − 2x3 is an expression in the number x. Finally, we end this section by informally addressing the issue of why we do not consider 0r for all numbers r (this is the case of α = 0 in αr ); in particular, we do not deﬁne 00 . The reason is not quite cut-and-dry, as we shall see. Suppose αr is deﬁned for all α ≥ 0, and suppose it satisﬁes (E4)–(E6). First of all, if n is a positive integer, then of course 0n = 0 · 0 · · · 0, which is 0. So√0n = 0 for a positive 1 integer n. Still with n as a positive integer, 0 n would be n 0 as before, so that 1 m 1 0 n = 0 (see Exercise 9). If m is also a positive integer, 0 n = (0 n )m = 0, by (E5). Thus 0t = 0 for all positive rational numbers t. We now consider 00 . Assuming that this is a well-deﬁned number, we will show that it would lead us into some uncomfortable situations. By (E4), we have 0r 0s = 0r+s for all r ≥ 0 and s ≥ 0. Letting r = s = 0 yields 0 0 0 0 = 00+0 . This means 00 00 = 00 . 0 If we let k = 0 , this says k must satisfy k2 = k, so that k2 − k = 0, and therefore k(k − 1) = 0. Either k = 0 or k − 1 = 0 (see Corollary 1 of Theorem 2.9 in [Wu2020a], quoted on page 356—and FASM). In other words, 00 = 0

or

00 = 1.

Suppose 00 = 1. Because of (E4), we have 01 · 0−1 = 01−1 , i.e., 0 · 0−1 = 1, so that no matter what 0−1 may be, 0 = 1, a contradiction. Therefore 00 cannot be 9 Together

with FASM.

152

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

equal to 1 and we conclude 00 = 0. At this point, we have to invoke the concept of limit.10 If the general deﬁnition of αr for all nonnegative α (i.e., allowing α to also equal 0) is at all “reasonable”, we would “expect” that lim α0 = 00 ,

α→0+

where α → 0+ has the usual meaning of letting α converge to 0 from the right (i.e., α > 0). So on the one hand, we have 00 = 0, and on the other, we know that for α > 0, α0 = 1. Therefore lim 1 = 0,

α→0+

which means 1 = 0. Again impossible. Our conclusion is therefore that, in the consideration of αr , if we also allow α to be 0, then the behavior of the symbol αr will not live up to our normal expectations of such a symbol (being a continuous function of α and r). This is why we restrict the symbol αr to positive values of α. Note that the preceding reasoning for not considering 0r depends on some vague philosophical considerations such as how we “expect” αr to behave as a function of α for a ﬁxed r. This kind of vague “guiding principle” is also an integral part of doing mathematics. Exercises 4.2. In the exercises below, α, β are positive numbers. (1) Let n be a positive integer. Show that there is at most one positive number β, so that β n = α by making use of the identity in Section 6.1 of [Wu2020a]: xn+1 − y n+1 = (x − y)(xn + xn−1 y + xn−2 y 2 + · · · + xy n−1 + y n ) for all numbers x and y and for any positive integer n. (2) (a) Prove (α2 − β 2 )6 (α + β)−5 = (α + β)(α − β)6 . (b) Simplify (α3 − β 3 )7.3 (α2 + αβ + β 2 )−7.3 . √ x2 ? (3) What is the function f : R → R so that f (x) = √ (4) Solve: (a) x − 6 x + 8 = 0. √ (b) x5 + 715 = 0. (c) x2 − 363 = 0. (d) x − x − 1 = 0. (5) Prove equation (4.3) on page 144 by imitating the argument that proves equation (4.2) on page 143. (6) Prove equation (4.9) on page 145. (7) Prove the corollary to Lemma 4.6 on page 148. (8) Let n be an odd positive integer. Use the Theorem 4.2 on page 143 to prove that given any x ∈ R (x not being necessarily positive), there is a unique t ∈ R so that tn = x. (9) Prove that if β is a nonzero rational number, then β n = 0 for any positive integer n. (Hint: Use mathematical induction.)

10 See

Chapter 2 of [Wu2020c].

4.3. EXPONENTIAL FUNCTIONS

153

(10) Without using (E4)–(E6), but using equation (4.8) on page 145, explain 1 1 why (αm ) n = (α n )m for all positive integers m, n. (11) Without using (E4)–(E6), but using equation (4.11) on page 146, prove m 1 1 that (α n ) m = α n . (12) Write a careful proof of (αr β s )t = αrt β st for all rational numbers r, s, and t. (13) Let n be a positive integer. Using the deﬁnition of the n-th root of a 1 positive number, give a direct proof that n α1 = √ n α. (14) Let α be a positive number. (a) Making Theorem 4.3, prove that √ use of √ m n mn for all positive integers m and n, α = α. (b) Without using Theorem 4.3, give a direct proof of the same.

4.3. Exponential functions The purpose of this section is to get acquainted with exponential functions αx . First, we point out the obvious fact that these functions are the multiplicative analogs of linear functions without constant term. Then we prove a sequence of basic properties of exponential functions and collect them together into a comprehensive theorem, Theorem 4.9. Generalities (p. 153) Basic properties of exponential functions (p. 156) The theorem (p. 160) Generalities Fix an α > 0. Furthermore, if α = 1, αx is the constant function 1 and is hardly interesting. For this reason, we assume for the rest of this section that α > 0 and α = 1. By Theorem 4.1 on p. 141, we have a function αx deﬁned on R so that αx > 0 for all x (see equation (4.6) on page 145). We may therefore write αx : R → (0, ∞), where, as before, R+ denotes all the numbers > 0. Recall that αx is called an exponential function with base α (page 141). First, we point out the obvious fact that this function αx is a very natural function, being the multiplicative analog of the linear function without constant term, g(x) = αx, in the following sense. Let x be a positive integer n. Then g(n) = αn = nα = α + α + · · · + α + α .

n

If we replace addition with multiplication, then the right-hand side becomes n α · α · α · · · α = α . n

Therefore at least on the positive integers n, αx is the direct multiplicative analog of the linear function g(n). These functions αx and g(x) share another property: a

154

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

linear function is determined by its values at two distinct points (because its graph is a line and a line is determined by two of its points), and the same is true of exponential functions (see Exercise 8 on page 162). We will continue to assume that (E4)–(E6) on page 139 are valid for all numbers r, s and for all positive numbers α and β; i.e., (E4)

αs αt = αs+t ,

(E5)

(αs )t = αst ,

(E6)

(αβ)s = αs β s .

Of course (E4) comes with Theorem 4.1. Although Theorems 4.3 and 4.4 on page 147 only prove (E5) and (E6) for rational numbers s and t, we will nevertheless assume (E5) and (E6) for all numbers s and t (see the comment that immediately follows Theorem 4.4 on page 147). Having said that, we take note of the fact that each time we consider a function αx , we have to be careful whether α > 1 or α < 1; these two kinds of functions exhibit distinct behaviors. See Theorem 4.9 on page 160 and Exercise 4 on page 161. Before proceeding further, let us at least graph the function 2x (compare page 140). This means we will plot many points of the graph of the exponential function 2x where x is a rational number (see especially the meaning of rational exponents on pp. 144ﬀ). First of all, we should certainly get the points (n, 2n ) when n = −1, 0, 1, 2, 3, 4. These are (−1, 0.5), (0, 1), (1, 2), (2, 4), (3, 8), (4, 16).

It would be reasonable to add at least the following points. (The calculator comes in handy at this point.) We round oﬀ the y-coordinates to the ﬁrst decimal place: √ √ (−0.5, √12 ) = (−0.5, 0.7), (0.5, 2) = (0.5, 1.4), (1.5, ( 2)3 ) = (1.5, 2.8). √ √ √ (2.5, ( 2)5 ) = (2.5, 5.7), (3.5, ( 2)7 ) = (3.5, 11.3), (4.5, ( 2)9 ) = (4.5, 22.6).

4.3. EXPONENTIAL FUNCTIONS

155

Because the right end looks a little sparse, let us throw in a few more points to (literally) ﬁll in the picture. Again, we round oﬀ the y-coordinates to the ﬁrst decimal place: √ 4 (2.75, ( √ 2)11 ) = (2.75, 6.7), 4 (3.75, ( 2)15 ) = (3.75, 13.5),

√ 4 (3.25, ( √ 2)13 ) = (3.25, 9.5), 4 (4.25, ( 2)17 ) = (4.25, 19).

It should be quite clear at this point what the graph of 2x looks like. In the course of plotting these points, we also get to see, in a concrete way, the advantage of having the precise determination of 2x when x is a rational number (see equations (4.7) and (4.11) on pp. 145 and 146, respectively). We will return to the shape of the graph of an exponential function in general on page 160.

156

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Basic properties of exponential functions We will make a sequence of small observations—labeled (A) to (G) ((G) is on page 159)—about the function αx and then summarize them in a comprehensive theorem, Theorem 4.9 on page 160. We begin with the deﬁning properties of αx : (A) α0 = 1. (B) αs αt = αs+t for all numbers s and t. For (A), see equation (4.5) on page 144. (B) is the second part of Theorem 4.1 on page 141. Next, we formally deﬁne a function F to be increasing if, for any two numbers x1 and x2 in its domain, x1 < x2 implies F (x1 ) < F (x2 ); we also say F is nondecreasing if, for any two numbers x1 and x2 in its domain, x1 < x2 implies F (x1 ) ≤ F (x2 ). (Caution: There is no uniformity in the literature about this particular piece of terminology. What we call an “increasing function” is sometimes called a “strictly increasing function”, and what we call a “nondecreasing function” would then be called an “increasing function” in that context.) Note that an increasing function is automatically injective in the sense of the deﬁnition in Section 1.1 (page 5); i.e., F (x1 ) = F (x2 ) implies x1 = x2 . We will prove this by contradiction. If x1 = x2 , then either x1 < x2 or x1 > x2 (see the trichotomy law on page 359). If the former, then F (x1 ) < F (x2 ), and if the latter, then F (x1 ) > F (x2 ). In any case, F (x1 ) = F (x2 ), a contradiction. Thus x1 = x2 and F is injective. Unless stated otherwise, we will concentrate on those exponential functions αx where α > 1 in the remainder of this section. (C) If α > 1, then αx is an increasing function. Recalling that R+ stands for the positive real numbers, we therefore have an injective function αx : R → R+ . We cannot give a complete proof of (C) without going more deeply into the real numbers and without making use of limits, but we will present an argument that is “almost good enough” and then complete it—as a side remark—using advanced ideas. We will try to separate what is proved and what is plausible but not quite proved. We ﬁrst prove a lemma that is of independent interest. Lemma 4.8. Let 0 < γ < 1 < α. Then for any fraction

m n

> 0,

γ m/n < 1 < αm/n . Proof. First assume 1 < α, and we have to show that 1 < αm/n for any nonzero fraction m n . Since 1 < α, the corollary of Lemma 4.6 on page 148 implies that 11/n < α1/n . But 11/n = 1 because both 1 and 11/n have the property that their n-th power is 1, i.e., (11/n )n = 1 and (1)n = 1, so the uniqueness of the n-th root of 1 implies that 11/n = 1. Hence from 11/n < α1/n , we conclude 1 < α1/n . By Lemma 4.6, we see that 1m < (α1/n )m . By Theorem 4.3, (α1/n )m = αm/n . Therefore 1 < αm/n .

4.3. EXPONENTIAL FUNCTIONS

157

The proof of γ m/n < 1 if 0 < γ < 1 can be obtained by repeating the same argument. An alternate argument is the following: let α = 1/γ; then 1 < α. Therefore the preceding paragraph implies that 1 < αm/n . By (E5) on page 154, αm/n = (γ −1 )m/n = γ −(m/n) =

1 . γ m/n

Therefore 0 < γ < 1 implies 1

1, αx is almost increasing in the precise sense that if s and t are rational numbers and s < t, then αs < αt . To this end, let t − s = r; then r > 0. Since r is a rational number, r is equal to a fraction m n for some positive integers m and n and therefore αr > 1, by Lemma 4.8. We have t = s + r, so that αt

= αs+r = αs αr > αs · 1 = αs .

(by (E4)

Hence αt > αs , as desired. The complete proof of (C) requires that we show something more; namely, if α > 1 and s and t are real numbers so that s < t, then αs < αt . In the following indented passages—which can be omitted if one so wishes—we show how to complete the argument by the use of the concepts of limit and continuity. With α > 1, our goal is to prove that for any real numbers s and t, not necessarily rational, s < t implies αs < αt . For this purpose, we will appeal to Section 2.4 in [Wu2020c]. Step 1. For any number s > 0, αs ≥ 1. We use the density of the rational numbers in R (Section 2.4 in [Wu2020c]) to produce a sequence of rational numbers {sn } so that s < sn and sn −→ s. Since the sn ’s are positive (because sn > s > 0), Lemma 4.8 implies that αsn > 1. Therefore, by the continuity of αx , αs = lim αsn ≥ 1. n→∞

Step 2. For any number s > 0, αs > 1. By the density of the rational numbers in R, we can choose a rational number r so that 0 < r < s. 0

r

s

We have s = (s − r) + r. Because s − r > 0, Step 1 implies that αs−r ≥ 1. Because r is a positive rational number, Lemma 4.8 implies that αr > 1. Therefore, using (E4), we have αs = α(s−r)+r = αs−r αr ≥ 1 · αr > 1. Thus αs > 1, as desired.

158

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Step 3. αx is an increasing function. Let s and t be any two numbers so that s < t, and we must prove αs < αt . We have t = (t − s) + s, so that by (E4), αt = αt−s αs . Because t − s > 0, Step 2 implies αt−s > 1. Thus, αt = αt−s αs > 1 · αs = αs . So αs < αt , as desired. The proof of (C) is complete. Before we proceed to the next two assertions ((D) and (E) below), we pause to take note of the fact that knowing αx is increasing does not mean αx must “increase to inﬁnity” in the sense of getting bigger and bigger without bound as x gets arbitrarily large. For example, here is the graph of an increasing function which does not get arbitrarily large as x gets larger and larger. (For those who are curious, this is the graph of the arctangent function, which will be deﬁned in Section 1.8 of [Wu2020c].)

What we want to show is that it is nevertheless the case that if α > 1, then αx does get arbitrarily large as x gets arbitrarily large. We formulate this assertion carefully in two steps, as follows. (D) If α > 1, then given any M > 0, there is a positive integer n so that αn > M . We claim that if b > 0, then (1 + b)n > 1 + nb for all integers n ≥ 2. This can be proved by using the binomial theorem (see page 203), but it is simpler to prove it by mathematical induction on n. The claim is clearly true if n = 2. Suppose the claim is true for n. Then we must prove that it is also true for n + 1; i.e., we must prove (1 + b)n+1 > 1 + (n + 1)b. This is so because, using the induction assumption, (1 + b)n+1 = (1 + b)n (1 + b) > (1 + nb)(1 + b) = 1 + (1 + n)b + nb2 > 1 + (n + 1)b. Thus the claim is true for all integers n ≥ 2. Now since α > 1, we have α = 1 + β for some positive β. Therefore, αn = (1 + β)n > 1 + nβ.

4.3. EXPONENTIAL FUNCTIONS

159

Thus if we choose n to be large, for example n > M β , then we can guarantee that n α > M . (The existence of such a large integer n will be proved in Section 2.4 of [Wu2020c].) (E) If α > 1, then “αx goes to inﬁnity as x goes to inﬁnity.” More precisely, given any M > 0, there is an n > 0 so that for all x > n, αx > M . This is an immediate consequence of (C) and (D). (F) Let α > 1. Given any y0 > 1, there is an x0 > 0 so that αx0 = y0 . The proof of this fact depends on two assumptions: one is that the function αx is continuous (see part (A) of Theorem 4.1 on p. 141) and the other is that the intermediate value theorem is valid (see page 122). It should be pointed out that in Sections 7.4 and 6.2 of [Wu2020c], respectively, these assumptions will be proved. For now, this way of ﬁnding an x0 so that αx0 is equal to a given number is an idea worth learning, so we will just concentrate on explaining this idea. By (E), there is an n so that αn > y0 . Thus on the interval [0, n], we have α0 = 1 < y0 whereas αn > y0 , so that α0 < y0 < αn . graph of αx q

αn M q

y0 1

x0

O

n

The intermediate value theorem (see page 122) now guarantees that there is some x0 , so that 0 < x0 < n and αx0 = y0 . Still with α > 1, we want to know whether a given number y0 > 0 can be written as αx0 for some x0 . If y0 > 1, then (F) shows that this is indeed the case. If y0 = 1, of course we already know that 1 = α0 . What about positive numbers y1 which are less than 1? The answer is also aﬃrmative, as the following shows. (G) Let α > 1. Given any positive y1 so that y1 < 1, there is an x1 < 0 so that αx1 = y1 . To see this, consider the number y0 = 1/y1 , which is > 1 because y1 < 1. By (F), there is an x0 > 0 so that αx0 = y0 . From equation (4.7), we have α−x0 =

1 1 = . x 0 α y0

But 1/y0 = y1 . Therefore, α−x0 = y1 , and (G) is proved by letting x1 = −x0 . Together, (F) and (G) say that the function αx : R → R+ is surjective (see page 5 for the deﬁnition). This is the right place to mention the diﬀerence in the use of the two words “injective” and “surjective”. In general, given any function

160

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

g : D → T , the question “Is it injective?” is independent of what T is. This is because it makes sense to inquire, given any two points x1 and x2 in D, whether g(x1 ) = g(x2 ) regardless of what T may be. However, the question of whether g is surjective or not depends critically on T . To make this point explicit, consider the function at hand, αx : R → R+ . We have just seen that this function is surjective. Now since R+ is a subset of the larger interval (−1, ∞), we can also write αx as αx : R → (−1, ∞). Now the function αx : R → (−1, ∞) is deﬁnitely not surjective because − 12 lies in (−1, ∞) but there is no x ∈ R so that g(x) = − 12 because, by (4.6) on page 145, g(x) > 0 for every real number x. Thus αx : R → R+ is surjective but αx : R → (−1, ∞) is not.

The theorem We can now summarize (A)–(G) in one comprehensive theorem.

Theorem 4.9. Let α > 0 and α = 1; then the function αx has the following properties: (i) α0 = 1. (ii) αs+t = αs · αt for all real numbers s and t. (iii) If α > 1, then αx is an increasing function which is bijective from R to R+ .

If α > 0 but α < 1, then the analog of Theorem 4.9 continues to hold for such an α; see Exercise 4 on page 161. We note that when α < 1, the function αx becomes decreasing rather than increasing (see page 161 for the deﬁnition of decreasing). We can easily sketch the graph of αx for α > 1: it is increasing and its y-values include every positive number as x ranges through R. Moreover, we also know the value of αx at two points: α0 = 1 and α1 = α. Thus, for example, the graph of f (x) = 2x looks like this:

The straight line in the picture is the graph of y = x; it is usually called the diagonal. We note that the graph of 2x is entirely above the diagonal, but we should not assume that this is true for the graph of αx for every α > 1. For example, the graph of 1.3x , while looking in general terms similar to that of 2x ,

4.3. EXPONENTIAL FUNCTIONS

161

actually crosses the graph of y = x at two points, as shown:

One can see this directly: the question is whether the function h(x) = 1.3x − x is ever negative on [0, ∞). A simple computation shows that while h(0) = 1 and h(8) > 0.15, we also have h(2) = −0.31 and h(7) < −0.73. So it is not surprising that h is negative on an interval containing [2, 7]. There is a counterpart of Theorem 4.9(iii) for the case of α < 1, but we leave it to an exercise (Exercise 4 on page 161). The graph of αx when α < 1 would look completely diﬀerent because αx is now decreasing. The graph of ( 12 )x , for example, is the reﬂection of the graph of 2x across the y-axis (see Exercise 6 below).

graph of ( 21 )x

graph of 2x

We conclude this section with the reminder that the complete proofs of Theorem 4.9 and its counterpart for α < 1 depend on the following theorems that will be proved in Sections 7.3 and 6.2 of [Wu2020c], respectively: Theorem 4.1 (page 141) and the intermediate value theorem (page 122). Exercises 4.3. (1) Solve for x: (a) 42x − 3 · 4x − 4 = 0. (b) 22x − 5 · 2x − 24 = 0. (2) If an algebra student comes to you and asks why αs = αt implies s = t, where α > 0 but α = 1, and s, t are numbers, how would you explain this to her? (3) Let s be any positive rational number and let α, β > 0. Prove that (a) α < β if and only if αs < β s and (b) α = β if and only if αs = β s . (4) (a) A function F : I → R (where I is a segment in R) is said to be decreasing, if x1 < x2 for any x1 , x2 ∈ I, then F (x1 ) > F (x2 ). Show that a decreasing function is injective. (b) Show that if 0 < α < 1, then the function f : R → R+ deﬁned by f (x) = αx is decreasing. (c) Show that the function f of part (b) is bijective.

162

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(5) For any positive number β and for any rational number r, prove that r 1 1 = r. β β (6) Show that the graphs of the two functions αx and (1/α)x are symmetric with respect to the y-axis. (7) Prove that if 0 < β < 1, then for all rational numbers r and s so that r < s, β r > β s . (8) (i) Show that if f (x) = ax and g(x) = bx are two exponential functions (a and b are positive numbers not equal to 1) and, at two distinct points p and q on R, f (p) = g(p) and f (q) = g(q), then a = b. (ii) Express part (i) in terms of the graphs of exponential functions. What geometric fact does this remind you of? (9) Let αx be a given exponential function (α is positive and not equal to 1). Fix a constant t. For each x, the slope of the line joining the two points (x, αx ) and (x + t, αx+t ) then deﬁnes a new function F (x), so that F (x) =

αx+t − αx . (x + t) − x

Prove that for some constant K, F (x) = Kαx . (Thus the slope grows or decays the same way αx itself does.) 4.4. Logarithms Theorem 4.9 on page 160 and Exercise 4 on page 161 together show that the function f : R → R+ so that f (x) = αx is bijective for any positive α = 1. Therefore f has an inverse function (see page 6). This inverse function is the classical logarithm with base α (or more correctly, the logarithmic function with base α). This section explores the basic properties of the logarithm and its graph, and gives a brief historical account of its origin. Inverse functions (p. 162) The deﬁnition and basic properties of logα x (p. 164) Historical origin of the logarithm (p. 166) Graph of logα x (p. 169) Change of base formula (p. 172) Inverse functions One cannot discuss the exponential function and its inverse function—the logarithm (to be introduced presently)—in any meaningful way without a careful discussion of the general concept of inverse function itself. There is all the more reason to do so because TSM perennially glosses over the meaning of inverse function when the logarithm is introduced. Let us begin with some generalities about inverse functions (see page 6). Recall that if I and J are two (ﬁnite or inﬁnite) intervals on the number line and a function F : I → J is given, we say a function G : J → I is an inverse function of F (and F is an inverse function of G) if G(F (x)) = x for every x ∈ I and F (G(t)) = t for every t ∈ J. Symbolically, G is an inverse function of F if G ◦ F = idI and F ◦ G = idJ , where idI and idJ are the identity functions of I and J, respectively.

4.4. LOGARITHMS

163

We remark that an inverse function G of F : I → J, if it exists, must be unique. Indeed, if G0 is another inverse function of F : I → J, then we will prove G = G0 . By deﬁnition of equal functions (see page 3), we have to show that for any t ∈ J, G(t) = G0 (t). To this end, since G is an inverse function of F , we have F (G(t)) = t. Since by assumption G0 is also an inverse function of F , we have F (G0 (t)) = t. Thus F (G(t)) = F (G0 (t)). Since F is injective, we conclude that G(t) = G0 (t), as desired. It follows that, from now on, if F : I → J has an inverse function, then we may speak of the inverse function of F : I → J. We also recall that a function F : I → J has an inverse function G : J → I if and only if F is bijective (see Exercise 6 on page 15). If we unravel the deﬁnition of bijectivity, then assuming F : I → J is bijective, we can describe explicitly how the inverse function G : J → I of F is deﬁned. Thus given t ∈ J, we have to deﬁne G(t) ∈ I. By the bijectivity of F : I → J, there is a unique number x ∈ I so that F (x) = t for the given t ∈ J. Then we deﬁne G(t) to be the number x ∈ I so that F (x) = t. To show that the G so deﬁned is the inverse function of F : I → J, we ﬁrst observe that, from the deﬁnition of G, we have immediately F (G(t)) = F (x) = t, for any t ∈ J. Schematically, we have

Fs(x) = t

J

F

I

G

s

x = G(t) To also show that G(F (x)) = x for every x ∈ I, let F (x) = t, with t ∈ J. Note that this x ∈ I is the only number in I that is mapped to t because F is injective. The above deﬁnition of G now says that G(t) is this number x ∈ I. Consequently, G(F (x)) = G(t) = x, as desired. Intuitively, the inverse function G : J → I of a given function F : I → J is easy to describe: if F moves a point x of I to a point t of J, then what G does is to move t right back to x. This is because if F (x) = t as given, then of course G(F (x)) = G(t), but since G(F (x)) = x for every x ∈ I, the left side of the equality G(F (x)) = G(t) is just x so that x = G(t). This is exactly the claim that G moves t back to x. Similarly, for the same F and G, if we look at F as the inverse function of G, then if G moves a point t of J to x of I, F will move x back to t, for the reason that G(t) = x implies F (G(t)) = F (x), which implies t = F (x) (since F (G(t)) = t for all t). However, none of this would make any sense unless we know F : I → J is bijective to begin with, and this is where TSM lets students down: TSM does not emphasize that the bijectivity of F is a pre-condition for the discussion of the inverse function of F . Whether we like it or not, inverse functions are an integral part of mathematics and, without knowing it, we have been using inverse functions all along. Take, for example, the two functions g√: R → R√and h : R → R deﬁned by g(x) = x3 for every x ∈ R and h(x) = 3 x, where 3 x denotes the unique number t so that t3 = x (the existence of such a t is essentially guaranteed by Theorem 4.2 on page 143, but see Exercise 8 on p. 152). It is now trivial to verify that h(g(x)) = x and g(h(x)) = x for every x ∈ R. Another example is the function : [0, ∞) → [0, ∞)

164

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

√ and the function s : [0, ∞) → [0, ∞) deﬁned by (t) = t and s(t) = t2 for all t ≥ 0. Note that in this case the domain of deﬁnition of is [0, ∞) rather than all of R as the square root of a negative number is not a real number (see Section 5.2 on page 189), and the domain of deﬁnition of s has to be restricted from R to [0, ∞) to ensure that s is injective (as otherwise s(t) = s(−t) = t2 ). Other common examples of inverse functions can be easily cited. For example, ﬁx a nonzero number c and deﬁne functions F : R → R and G : R → R by F (x) = cx and G(x) = x/c for any x ∈ R. Then again, F (G(x)) = x and G(F (x)) = x for all x ∈ R. Incidentally, the fact that F and G are each other’s inverse function is the substance of the common but erroneous statement that “multiplication and division are inverse operations”. The correct statement is: If c is a ﬁxed nonzero number, then dividing by c and multiplying by c are inverse operations. Another example of inverse functions is furnished by the pair h : R → R and k : R → R by h(x) = x + c and k(x) = x − c for any x ∈ R, where c is again a ﬁxed constant. The simple checking that these are indeed inverse functions will be left to the reader. The deﬁnition and basic properties of logα x Now we go back to the exponential function αx : R → R+ , where α is a ﬁxed positive constant = 1. For the sake of notational clarity, we will denote the function αx by f , so that f (x) = αx for any number x. Because f is bijective, it has an inverse function g : R+ → R. Since f and g are each other’s inverse function, we have g(f (x)) = x for every x ∈ R and f (g(t)) = t for every t ∈ R+ (see page 163). Since f (x) = αx , we have (4.21)

g(αx ) = x

and

αg(t) = t

for every x ∈ R and for every t ∈ R+ . As is well known, this g(t) is denoted by logα t in the literature,11 and is called the logarithm of t with base α (or sometimes, to base α). The number α of logα is the base of the logarithm, and the base is always understood to be = 1. In Chapter 7 of [Wu2020c], we will formally introduce the number e, e = 2.71828 . . ., called the base of the natural logarithm. The corresponding exponential function ex is distinguished by its properties in terms of diﬀerentiation and integration (see Chapter 7 of [Wu2020c] again). Its inverse function, loge t, is simpliﬁed to log x without a subscript by universal consent in mathematics; in the sciences and engineering, however, this log x is denoted by ln x. Now we go back to logα t in general for a positive α = 1. We have (4.22)

logα : R+ → R is a bijection

and (4.21) becomes, for every x ∈ R and for every t ∈ R+ , (4.23)

logα (αx ) = x

and

αlogα t = t.

It follows immediately from the ﬁrst equality in (4.23), by setting x = 1, that (4.24)

logα α = 1.

11 You may have already noticed the notational anomaly: by tradition, we write log t rather α than logα (t).

4.4. LOGARITHMS

165

Moreover, we also have 1 = − loga t for all t > 0. t The proof of the ﬁrst part of (4.25) is simple enough: logα 1 = logα α0 = 0. For the proof of the second part, we ﬁrst observe two basic properties of logα t:

(4.25)

(4.26) (4.27)

logα 1 = 0

and

logα st logα tx

logα

= logα s + logα t, = x logα t.

We will approach these counterparts of (E4) and (E5) (see page 139) about αx , by transferring all arguments to αx itself, which we know a little better. We will exploit the fact that the function αx is injective, so that, to show u = v for real numbers u and v, it suﬃces to show—by virtue of the deﬁnition of injectivity—that αu = αv . Thus, instead of directly proving (4.26), we will prove αlogα st = αlogα s+logα t . To this end, observe that, by the second identity in (4.23), the left side is αlogα st = st. But by (E4) (page 139), the right side is also equal to st because αlogα s+logα t = αlogα s αlogα t = st. Therefore the two sides are equal and (4.26) is proved. We apply the same method to prove (4.27). Thus we prove instead x

αlogα t

= αx logα t .

The left side is tx (by (4.23)) while the right side, by (E5), is x

αx logα t = αlogα t = tx . Therefore (4.27) is valid. Now we can give the simple proof of the second equality of (4.25). By (4.26), we have 1 1 = logα 1 = 0. logα t + logα = logα t · t t (4.25) is now completely proved. Finally, we note that if α > 1, the fact that αx is increasing (Theorem 4.9 on page 160) is reﬂected also in the behavior of logα ; namely, logα t is also increasing. We prove this fact by using similar ideas. Let 0 < s < t, and we must show logα s < logα t. Let u = logα s and v = logα t. Then what we want is u < v. We begin by proving something that seems to be a bit less; namely, (4.28)

αu < αv .

This is true because, by (4.23), the left side is s and the right side is t, and the fact that s < t is guaranteed by assumption. That said, we claim that (4.28) implies u < v. Indeed, suppose not, then by the trichotomy law (see p. 359), either u = v or v < u. The former implies αu = αv , which contradicts (4.28), and the latter implies αv < αu (because αx is an increasing function), which also contradicts (4.28). So u < v after all and we have proved that logα t is increasing. We collect together the main points of the preceding discussion about logα into a comprehensive theorem, which is the counterpart of Theorem 4.9 on page 160.

166

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Theorem 4.10. Let α > 0 and α = 1; then the inverse function logα : R+ → R of the exponential function αx : R → R+ has the following properties: (i) logα 1 = 0, and logα α = 1. (ii) logα st = logα s+logα t for all s, t > 0, and logα tx = x logα t for all t > 0 and for all real numbers x. (iii) If α > 1, then logα x is an increasing function of x which is bijective from R+ to R. The corresponding fact for (iii) that logα is decreasing when 0 < α < 1 is left as an exercise (Exercise 4 on page 173). Historical origin of the logarithm Part (ii) of Theorem 4.10, to the eﬀect that logα converts a multiplication st to an addition logα s + logα t, is historically of great importance. For the purpose of illustration, let us take α to be 3 and, at the risk of oversimpliﬁcation, explain the original impetus for the introduction of the logarithm by minimizing all the technicalities about inverse functions and laws of exponents. Conﬁning ourselves for the moment only to numbers which can be expressed as a positive integer power of 3, i.e., 3(= 31 ), 9(= 32 ), 27(= 33 ), 81(= 34 ), 243(= 35 ), 729(= 36 ), . . . , we note that each such number is identiﬁed uniquely with its exponent as a power of 3; i.e., once 3 is ﬁxed, then 9 is uniquely identiﬁed with 2 (because 9 = 32 ), 27 is uniquely identiﬁed with 3 (because 27 = 33 ), 81 is uniquely identiﬁed with 4 (because 81 = 34 ), 177147 is uniquely identiﬁed with 11 (because 177147 = 311 ), etc. Let us call 2 the logarithm of 9, 4 the logarithm of 81, 11 the logarithm of 177147, etc. More brieﬂy, we write log 9 = 2,

log 81 = 4,

log 729 = 6,

log 177147 = 11,

...,

log 3n = n.

The obvious fact that 3m 3n = 3m+n for integers m and n then leads to log(3m 3n ) = log 3m+n = m + n = log 3m + log 3n . Let us highlight this conclusion: (4.29)

log(3m 3n ) = log 3m + log 3n

for all positive integers m and n.

(This is of course a special case of Theorem 4.10(ii).) For example, we have (4.30)

log(243 × 177147) = log 243 + log 177147

because 243 = 35 and 177147 = 311 . Let us use equation (4.30) to illustrate how the logarithm simpliﬁes computations. To ﬁnd the product of 243 × 177147, we can directly multiply, of course. But before the age of calculators and computers, computations of this sort were excruciatingly tedious. Four centuries ago, the Scottish scholar John Napier12 (1550–1617) made the observation that, via the exponents of a ﬁxed number such as 3, log converts multiplication to addition (in the sense of equation (4.29)) and can be used systematically to simplify computations, provided a good table of logarithms in base 3 has been compiled. To get such a table, one computes—once and for all—as many 12 Napier was a colorful personality. He seemed to be more interested in theology than mathematics; he was virulently anti-Catholic. His work in mathematics was to him a hobby.

4.4. LOGARITHMS

167

of the powers of 3 as one can manage, such as the following: n 1 2 3 4 5 6 7 8

3n 3 9 27 81 243 729 2187 6561

n 3n 9 19 683 10 59 049 11 177 147 12 531 441 13 1 594 323 14 4 782 969 15 14 348 907 16 43 046 721

n 17 18 19 20 21 22 23 24

3n 129 140 163 387 420 489 1 162 261 467 3 486 784 401 10 460 353 203 31 381 059 609 94 143 178 827 282 429 536 481

To make such a table useful, one interchanges the columns of the above table so that the powers of 3 come ﬁrst, with the corresponding logarithms coming second. So a table of logarithms would look something like this:13 k log k 3 1 9 2 27 3 81 4 243 5 729 6 2187 7 6561 8

k log k 19 683 9 59 049 10 177 147 11 531 441 12 1 594 323 13 4 782 969 14 14 348 907 15 43 046 721 16

k log k 129 140 163 17 387 420 489 18 1 162 261 467 19 3 486 784 401 20 10 460 353 203 21 31 381 059 609 22 94 143 178 827 23 282 429 536 481 24

Now, back to the multiplication of 243×177147. Instead of doing the multiplication directly, Napier would look up the logarithms of 243 and 177147, and they are 5 and 11. Then equation (4.30) implies that log(243 × 177147) is 5 + 11 = 16. Therefore equation (4.30) told him that log(243 × 177147) = 16. Now, according to the above table, log 43046721 = 16. He knew that distinct numbers would have distinct logarithms, so he concluded that 243 × 177147 = 43046721. Thus by compiling a comprehensive table of logarithms (for this ﬁxed number 3) once and for all, he could replace a multiplication such as 243 × 177147 by the addition log 342 + log 177147. (There is a brief hint of how Napier might have been led to this discovery in Exercise 6 of Exercises 1.4 of [Wu2020c].) This discussion may seem to be too restrictive to be of interest, because it appears to be applicable only to numbers that are exactly powers of 3. However, consider the product 239 × 355628. Now, 239 is not among the entries in the above table and neither is 355628. But 239 is close enough to 243 (which is in the table), and 355628 is in between the two entries 177147 and 531441 in the table. Since 13 Napier

had already compiled a quite extensive table of the logarithms of numbers.

168

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

the midpoint between 177147 and 531441 is 354294, which is suﬃciently close to 355628, we may approximate log 354294 by the midpoint of 11 (= log 177147) and 12 (= log 531441). Therefore we approximate log 354294 by 11.5. Consequently (we use “≈” to denote “approximately equal to”), (4.31)

log(239 × 355628) = log 239 + log 355628 ≈ 5 + 11.5 = 16.5.

Since 16.5 is not in the table, we again approximate. From the table, log 43046721 = 16, and log 129140163 = 17, so an intuitive guess is that the number midway between 43046721 and 129140163 would have a logarithm of approximately 16.5. Since the midpoint between 43046721 and 129140163 is 86093442, a reasonable guess is that log 86093442 ≈ 16.5. Comparing with (4.31), we see that 239 × 355628 ≈ 86093442. The exact value of 239 × 355628 being 84995092, the percentage error14 is 1098350 86093442 − 84995092 = ≈ 1.3%. 84995092 84995092 Not bad at all. We have just indicated how a table of logarithms may be used to obtained approximate values of the multiplication of arbitrary positive numbers. Now if 100% accuracy is not the goal of such computations (and remember that in a science such as astronomy, especially in the 17th century, its measurements were very crude to begin with and therefore approximate answers were all that astronomers were looking for), then a table of logarithms would be an invaluable aid in scientiﬁc computations. By reﬁning the table of logarithms (in our present context, this could mean, for example, getting the values of log3 t when t is a positive integer rather than just an integer power of 3) and by tweaking the methods of approximations in the use of such a table, the logarithm was made into a precise computational tool (and basically the only one) in the days of hand calculations. Napier’s “ﬁxed number” (the base of the logarithm) was not 3 to be sure. Contrary to common misconception, it was not e either (where e is the base of the natural logarithm), but it was more or less 1/e. However, the principle is the same. Historically, 10 was the base that was adopted after Napier’s initial discovery, and the log in that case is called the common logarithm,15 denoted by log10 or, in school mathematics, just log. Nowadays, the logarithm in base e, i.e., loge , is almost the only logarithm used in advanced mathematics and science; it is called the natural logarithm and, as noted in the last subsection, it is simply written

14 The

ratio of the absolute error to the correct value, in percents. was the modiﬁcation of Napier’s logarithm by the English mathematician John Briggs (1561–1630). Briggs’s four day trip from London to Edinburgh in 1615 gave birth to the account of the legendary meeting that “almost one quarter of an hour was spent, each beholding other with admiration, before one word was spoken.” 15 This

4.4. LOGARITHMS

169

as log x in mathematics.16 But no matter what base is used for the logarithm, the reduction of multiplication to addition by the use of logarithm was a tremendous saving of labor and, for over three centuries, tables of logarithms saved scientists countless hours in computations with data. A statement attributed to PierreSimon Laplace (1749–1827)17 says that the discovery of logarithms “by shortening the labors doubles the life of the astronomer.” Many of our students no longer know about the historical background of the logarithm and have come to regard it merely as another function they need to learn in order to pass exams. It would be far better, however, if this part of mathematical history would be kept alive in the school classroom because it is not only about the evolution of an important mathematical idea but, ultimately, also about a key development in human culture. With the advent of computer software and calculators, one may legitimately ask why one should bother to learn about the logarithm now. There are many reasons, among them the following two. One is that since multiplication is still more complicated than addition, if there is a concrete way to convert multiplication into addition, in the sense of part (ii) of Theorem 4.10 on page 166, i.e., log st = log s + log t

for all positive s, t,

then it is automatically worth knowing. The other reason is that both exponential functions and logarithms ﬁgure prominently not just in mathematics but in science as a whole. You may recall from calculus, for example, that trying to solve the simplest of diﬀerential equations arising from growth or decay problems, y = y, leads to the exponential function ex . Observe also that the Richter scale for measuring the energy of an earthquake is formulated in terms of the common logarithm. There is no way one can avoid either. Graph of logα x Our next objective is to get to know the graph of logα a bit. Let α > 1. Theorem 4.9(iii) on page 160 says that logα is an increasing function which is a bijection from R+ to R. The “increasing” property implies that the graph rises as we move to the right. The “bijective” property implies that logα t goes to inﬁnity as t goes to inﬁnity (see (E) on page 159 for this terminology). This is because logα is in particular surjective, so that given any number M no matter how large, there is a t0 > 0 so that logα t0 = M , and since logα is increasing, we see that logα t > M for all t > t0 . Consequently, the graph of logα on (t0 , ∞) is above the horizontal line y = M . Still with α > 1, the fact that for a positive t, t being very small implies x1 is very large, and the fact that logα t = − logα 1t (see (4.25)) implies that logα t goes to minus inﬁnity as t goes to 0, in the sense that if K < 0 is given, then there is a suﬃciently small positive number t so that logα t < K. 16 Engineers

and most scientists denote the natural logarithm by ln. mathematician and friend of Napoleon. However, his position in mathematics is independent of that friendship, to be sure. He was regarded as the natural successor to Newton in celestial mechanics, and he also made landmark contributions to the theory of probability. The Laplace equation and the Laplace transform are still with us everywhere. 17 French

170

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

If 0 < α < 1, then logα t is decreasing and goes to minus inﬁnity as t goes to inﬁnity. We leave this as an exercise (Exercise 6 on page 173). We summarize all the information on the graph of logα for α > 1 as follows: logα 1 = 0, logα α = 1, the graph of logα x climbs arbitrarily high as x increases, and it has the negative y-axis as an asymptote (see Section 2.4 on pp. 116ﬀ.) as x gets close to 0. So the following general picture should not come as any surprise:

Analogously, a typical graph of logα x when 0 < α < 1 is the following:

We now come to a critical piece of information about the graph of logα . Recall that we call the line deﬁned by the equation y = x the diagonal (see page 160) Our main claim is the following. Theorem 4.11. The reﬂection with respect to the diagonal of the graph of αx is the graph of logα x. You may wish to review the deﬁnition of a reﬂection on page 354 at this point. It also follows from this theorem that the reﬂection of the graph of logα x with respect to the diagonal line y = x is the graph of αx . In the graph below, the upper left curve is the graph of αx , and the lower right one is logα x. The diagonal y = x is also shown. (It was pointed out in the preceding section that for some numbers α close to 1, e.g., α = 1.3, the graph of αx would intersect the diagonal at two points so that the disjointness of the three graphs in the following picture should not be considered to be typical.)

4.4. LOGARITHMS

171

The proof of the theorem requires a preliminary observation. Let E be the graph of αx and let L be the graph of logα x. We claim that (4.32)

a point (p, q) is in E ⇐⇒ the point (q, p) is in L.

Take a point (p, q) ∈ E; we will show that (q, p) ∈ L. By deﬁnition of the graph E, (p, q) ∈ E implies q = αp . Taking logα of both sides, we get p = logα q, so that (q, p) = (q, logα q) which is obviously in L. The proof of the converse is identical: if (q, p) ∈ L, then p = logα q so that αp = q, and (p, q) = (p, αp ) ∈ E. The claim is proved. Proof of Theorem 4.11. We want to prove that the reﬂection of E, which is the graph of ax , with respect to the diagonal line y = x is equal to L, the graph of loga x. Recall that if Λ is the reﬂection with respect to a line L, then for any point (a, b) not lying on L, the reﬂected point Λ(a, b) is characterized by the fact that L is the perpendicular bisector of the segment joining (a, b) to Λ(a, b). In view of (4.32), if we can prove that the diagonal is the perpendicular bisector of the segment joining an arbitrary point (a, b) (a = b) to the point (b, a), then the reﬂection across the diagonal maps E to L. (b, a) r @ @ @ r @(t, t) @ @

@r (a, b)

O

We claim that the diagonal intersects the segment joining (a, b) to (b, a) at the midpoint of the segment. The fact that the diagonal intersects the line joining (a, b) to (b, a) is easy to see because these lines have slopes 1 and −1, respectively, and are therefore not parallel (see Theorem 6.17 of [Wu2020a], recalled on page 359). To prove that the segment joining (a, b) to (b, a) intersects the diagonal requires

172

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

some ﬁnesse, however. For simplicity, we do so indirectly. Let t = 12 (a + b), then the point (t, t) belongs to the diagonal. It is easy to see that the line joining (t, t) to (a, b) coincides with the line joining (t, t) to (b, a) as they both have slope −1 (see Theorem 6.9 of [Wu2020a], recalled on page 359). Thus the points (a, b), (t, t), and (b, a) are collinear. Moreover, by the distance formula (see page 356), the distance from either (a, b) or (b, a) to (t, t) is (a − t)2 + (b − t)2 . Thus (t, t) is the midpoint of the segment joining (a, b) to (b, a). This shows that the diagonal intersects the segment joining (a, b) to (b, a) precisely at the midpoint of the segment, as claimed. But, as noted, the slope of the line joining (a, b) to (b, a) is −1 while the slope of the diagonal is 1. By Theorem 6.18 of [Wu2020a] (recalled on page 359), these lines are perpendicular. Hence the diagonal is the perpendicular bisector of the segment joining (a, b) to (b, a) after all. The proof of the theorem is complete. One can draw some obvious conclusions from the theorem. One is the fact, already observed above, that the negative y-axis is an asymptote of the graph of logα if α > 1 whereas the positive y-axis is an asymptote of the graph of logα if 0 < α < 1. This is because the negative x-axis is an asymptote of the graph of αx if α > 1 and the positive x-axis is an asymptote of the graph of αx if 0 < α < 1, so that upon taking the reﬂection across the diagonal, we have the corresponding statements about the graph of logα . Consequently, logα x goes to minus inﬁnity as x goes to 0 if α > 1, and it goes to inﬁnity as x goes to 0 if 0 < α < 1. Theorem 4.11 also gives us a diﬀerent understanding of the fact that if α > 1, the increasing function logα x goes to inﬁnity as x goes to inﬁnity. So suppose this is not true. Then there is a positive number M so that logα x < M for all x ∈ R. This would mean that the graph of logα lies below the horizontal line y = M . After reﬂection across the diagonal, this becomes the statement that the graph of αx lies to the left of the vertical line x = M . Since αx is deﬁned on all of R, this is a contradiction. The reasoning used in the proof of Theorem 4.11 is actually valid in a broader context and serves to prove the following general theorem. We will leave the details of the proof to Exercise 9 on page 174. Theorem 4.12. Let f : I → J and g : J → I be each other’s inverse function, where I and J are (ﬁnite or inﬁnite) intervals on the number line. Then their graphs are each other’s reﬂection with respect to the diagonal.

Change of base formula We conclude with a standard question in school algebra: given a, b > 0 and a, b = 1, how are loga x and logb x related for a positive number x? Thus let loga x kx = logb x where the subscript x in kx indicates that this number is a priori dependent on x. We want an explicit value of kx in terms of a, b, and perhaps x. This can be done in more than one way. Still with the given x, we let x = bu for some u (we can do that because the function f (x) = bx is surjective!). Then using logb b = 1, we get loga x = loga bu = u loga b

and

logb x = logb bu = u logb b = u

4.4. LOGARITHMS

173

so that

u loga b = loga b. u The surprise is that kx turns out to be a constant and, in particular, independent of x. Therefore loga x = (loga b) logb x for all x > 0. kx =

This is known as the change of base formula for logarithms. (Do not try to commit this formula to memory! Rederive it each time you need it.) If we let x = a in the change of base formula, we obtain immediately loga b logb a = 1. This fact can also be proved directly. See Exercise 11 immediately following. In any case, we can also write the change of base formula as loga x =

logb x . logb a

Pedagogical Comments. The change of base formula is something worth knowing, of course, but it is also a technical piece of information that is not often needed. TSM and, unfortunately, standardized tests tend to exaggerate its importance, the same way they exaggerate the importance of the factoring of quadratic polynomials with integer coeﬃcients. Such pedagogical errors should be avoided. End of Pedagogical Comments. Exercises 4.4. (1) Very roughly, the Richter (magnitude) scale measures the energy released by an earthquake and is deﬁned to be log10 (A/A0 ), where A is the energy of the quake and A0 is a ﬁxed constant. (i) Suppose two earthquakes have energies A1 and A2 and they register as 7 and 8 on the Richter scale, respectively. How much bigger is A2 compared with A1 ? (ii) If two other quakes have energies A3 and A4 and they register as 6 and 9 on the Richter scale, respectively, how much bigger is A4 compared with A3 ? (2) Solve for √ x in each of the following: (a) x + 2 x − 15 = 0. (b) 5−2x − 3 · 5−x + 2 = 0. (c) loga x + loga (x − 1) = 2, where a > 0 but a = 1. (3) (a) Sketch the graph of f (x) = log2 (x+7) by specifying the domain, where it crosses the x-axis, and as many points as you can. (b) Sketch the graph of g(x) = log3 |x|. Specify its domain, where it crosses the x-axis, and as many points as you can. (4) If a function F : I → R (where I is a segment in R) is decreasing and surjective, then according to Exercise 4 on page 161, F is injective and therefore bijective. Let G : R → I be the inverse function of F . Show that G is also decreasing. log8 x for all positive x. (5) Discuss the function h(x) = log2 x (6) Prove directly that if 0 < α < 1, then logα t goes to minus inﬁnity as t goes to inﬁnity.

174

4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(7) (a) Let α > 0 but α = 1, and let x > 0. How does logα x compare with log α1 x? (b) For a, b > 0 but a, b = 1, solve loga (logb x) = 0 and loga (logb x) = 1. (8) Let a, b be positive numbers so that a < b but a, b = 1. Is logb logb a always deﬁned? If not, exactly when is it deﬁned? Is it positive or negative? (9) Give a proof of Theorem 4.12 on page 172. (10) Give a diﬀerent derivation of the change of base formula on page 172 for logarithms by letting x = av for some number v. (11) Give a direct proof that, for a, b > 0 and a, b = 1, loga b logb a = 1. (Hint: Express a in terms of b, for example.) (12) Recall that an annual interest rate of x percent compounded n times annually means that if the calendar year is divided into n equal periods (so that each period has 365 n days), anaccount of q dollars at the

beginning of a period will earn an amount of x/n 100 q dollars at the end of the period. (a) Derive a formula which gives the amount of money in a savings account at the end of k years, where k is a positive integer, if the initial deposit in the account is P dollars, the annual interest rate is x percent compounded n times annually, and no money is ever withdrawn from the account. (b) Let D be a number greater than the P of part (a). How many years (the number of years understood to be an integer) will it be before the amount in the account of part (a) is at least D dollars? Express your answer in terms of P , D, n, x, and log. (Note: You can use a logarithm with any base.) (13) With an annual interest rate of 4.5% compounded (1 time) annually, how long will it take an initial deposit of $7,123,567,021 to double its value?

CHAPTER 5

Polynomial Forms and Complex Numbers Up to this point, the algebra we have been doing may be called generalized arithmetic, because the symbols employed in all the expressions we have come across all stand for numbers (review Section 6.1 of [Wu2020a] if necessary). It follows that all our computations have been with numbers. This kind of algebra is more or less the algebra of the period from al-Khwarizmi (780–850) to 1800 (see, e.g., [Bashmakova-Smirnova], Chapters 4–6). After 1800, the work of Gauss, Abel, Galois, and others slowly transformed algebra into an abstract study of structure. School algebra, at some point, must also give an introduction to such abstract considerations. The study of polynomials provides the most natural venue to showcase the transition from algebra as generalized arithmetic to formal algebra. The main topics to be taken up in this chapter, complex numbers, the fundamental theorem of algebra, and the binomial theorem, will serve to illustrate this transition. It will be seen that this more abstract perspective even sheds new light on the algebra of polynomials that we have come to know up to this point. 5.1. Polynomial forms This section introduces the formal concept of a polynomial form—an abstraction version of a polynomial—and proves its most basic properties. This is our ﬁrst excursion into formal algebra, and the extra step of formalization will be justiﬁed by the gain in conceptual clarity in subsequent sections. Among the theorems to be proved in this section, we single out the rational roots theorem, a theorem usually badly taught and therefore often misunderstood in TSM. The deﬁnition and arithmetic of polynomial forms (p. 175) Why polynomial forms (p. 177) The division algorithm (p. 180) Roots of polynomial forms (p. 184) The rational roots theorem (p. 186) The deﬁnition and arithmetic of polynomial forms So far, a polynomial is just a sum of multiples of powers of a number x. Now we change our point of view by considering sums of multiples of powers of a symbol X, f (X) = an X n + an−1 X n−1 + · · · + a1 X + a0 where ai ∈ R for i = 0, 1, . . . , n and an = 0. (We hasten to add that this condition an = 0 does not preclude the possibility that for some i < n, ai = 0.) To avoid confusion with the polynomials we have been working with thus far, we will call 175

176

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

such an f (X) a polynomial form but will continue to refer to n as its degree, the an , . . . , a0 as its coeﬃcients, and an as its leading coeﬃcient. The term a0 will also be referred to as its constant term. As a matter of convention, we will also omit the writing of a term if its coeﬃcient is 0. Thus 2X 2 + 0X + 3 will be abbreviated to 2X 2 + 3. By the above deﬁnition of degree for a nonzero polynomial form, the degree of a constant polynomial form such as f (X) = 5 is 0. The zero polynomial g(X) = 0 is troublesome (we will explain later why it is troublesome). One can simply ignore it by not assigning it a degree (nobody says we must), or assign it a degree of −∞ as a matter of expediency. Because we have never encountered this kind of purely formal (symbolic) mathematics, a few comments are necessary to reassure you as to what this is all about. We will address the need for this kind of formal mathematics in the subsection Why polynomial forms on page 177 and, in fact, implicitly in the rest of this chapter. For now, let us address the simpler issue of how to do it correctly. What we have deﬁned — a polynomial form — is just a formal expression, in other words, a piece of notation, no more and no less. We are not saying at all that this is something you should have known1 or that once you see this, you would know how to add or multiply two polynomial forms. We may have some guesses as to how we should proceed, but in mathematics, if something is not known, i.e., not yet deﬁned, we do not pretend that we already know bits and pieces of it, at least not in a formal sense. In particular, the operations of adding and multiplying polynomial forms will have to be deﬁned presently; i.e., we have to specify how to compute with them even if the temptation is great to assume that they can be computed like polynomial functions. So for now, all that matters is that you recognize a polynomial form when you see one. Our next goal is to learn how to perform arithmetic operations on them. We ﬁrst take up the simpler problem of clarifying what it means to say that two polynomial forms are equal. By deﬁnition, two polynomial forms are equal if the corresponding coeﬃcients of the same power of X in the two forms are pairwise equal. In particular, two equal polynomial forms must have the same degree, because they must have the same leading coeﬃcient. Note that the concept of the equality of polynomial forms is a matter of deﬁnition and is not subject to psychological interpretations. This is another piece of information that should help clarify the current concern about the meaning of the equal sign in the mathematics education research literature. As a general guiding principle, the arithmetic operations we are going to deﬁne on polynomial forms should be entirely consistent with the same operations if in each f (X) we replace the symbol X by a number x. In other words, the deﬁnitions of the arithmetic operations on polynomial forms are guided, closely, by those on polynomials.2 We will indicate the replacement of X by x in any polynomial form g(X) simply by the notation g(x). Of course, g(x) is nothing but an ordinary polynomial in the number x.

1 But of course if you have taken a course in abstract algebra, you would recognize that a polynomial form is nothing but an element of the polynomial ring R[x]. 2 Mathematical Aside: We are implicitly exploiting the fact that there is an isomorphism from the polynomial ring R[x] to the ring of polynomial functions over R.

5.1. POLYNOMIAL FORMS

177

We specify how to do these arithmetic operations among polynomial forms in the most obvious way possible, which is to deﬁne addition and multiplication among polynomial forms by treating X as if it were a number, so that, at least formally, dealing with polynomial forms does not introduce any surprises. The idea is so simple that, in place of the most general deﬁnition possible (which would require the use of symbolic notation not appropriate to school mathematics), it suﬃces to indicate what is intended by a simple example in each case. Thus let g(X) = a2 X 2 + a1 X + a0

and

h(X) = b3 X 3 + b2 X 2 + b1 X + b0 ,

where the ai ’s and bj ’s are numbers and a2 = 0, b3 = 0. Then by deﬁnition, their sum is g(X) + h(X) = b3 X 3 + (a2 + b2 )X 2 + (a1 + b1 )X + (a0 + b0 ) and their product is g(X)h(X) = (a2 b3 )X 5 + (a1 b3 + a2 b2 )X 4 + (a0 b3 + a1 b2 + a2 b1 )X 3 +(a0 b2 + a1 b1 + a2 b0 )X 2 + (a0 b1 + a1 b0 )X + a0 b0 . In other words, the product of two polynomial forms is obtained by multiplying out all possible terms and then collecting like terms by their powers. It is immediately seen that, because the additions and multiplications among the coeﬃcients are associative, commutative, and distributive, so are the addition and multiplication of polynomial forms. The multiplication of polynomial forms has the desirable property that if we denote the degree of a polynomial form f (X) by deg f (X), then for all nonzero f (X), g(X), (5.1)

deg(f (X)g(X)) = deg f (X) + deg g(X).

The simple proof will be left as an exercise (see Exercise 1 on page 188). Now the problem of assigning a degree to 0 arises in this situation: do we want to say, without restriction, that the equality in (5.1) is correct for all f (X) and g(X) regardless of whether they are zero or not? If so, let f (X) = 0 and let g(X) = X. Then by (5.1), deg 0 = deg 0 + 1, and we see that deg 0 cannot be an integer. For this reason, we assign to 0 the degree of −∞ if we insist that this equality be valid without restriction. An equally prudent way to deal with the situation is simply to not assign a degree to 0 and restrict the equality (5.1) on degrees to be one among nonzero polynomials. Why polynomial forms Why polynomial forms instead of just polynomials? This question cannot be answered satisfactorily within the conﬁne of school mathematics. The idea roughly is that in a polynomial f (x), x always plays the primary role and the coeﬃcients always play a subordinate role. However, we will soon come upon situations where it is necessary, conceptually, to disengage x from the coeﬃcients altogether. Thus, the set of all polynomial forms in X with real numbers as coeﬃcients is denoted by R[X], to be called the polynomial ring over R. By singling out the symbol X and the coeﬃcients in this manner, we open the way to change R to some other number systems (such as the complex numbers C which will be discussed in Section

178

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

5.2 or the integers Z) or to allow X to assume other values distinct from those of the coeﬃcients.3 This is an example of abstraction and generality at work in advanced algebra. We mention in passing that if we ﬁx a number b, then we can deﬁne a function from the set of polynomial forms R[X] to R by assigning to each f (X) the number f (b). This function has nice properties, which are easily veriﬁed, such as the fact that it assigns to each f (X) + g(X) the number f (b) + g(b) and assigns to each f (X)g(X) the number f (b)g(b). This is an example of what is called a ring homomorphism from R[X] to R, and homomorphisms are basic to any algebraic discussions in advanced mathematics. In a limited way, we can illustrate the advantage of the abstraction by considering the problem of division among polynomials. Imitating the fact that the division of polynomial functions leads to rational functions, we introduce a rational form as a formal expression of the type f (X) , g(X)

where f (X) and g(X) are polynomial forms, g(X) = 0.

By “formal” expression, we mean that f (X)/g(X) as yet has no meaning other than that we have created this symbol. As in the case of polynomial forms, we will have to say what can be done with rational forms, as we do now. First, we agree to identify every polynomial form f (X) with the rational form f (X) 1 . For example, 0 0 is identiﬁed with 1 . The set of all rational forms in X is denoted by R(X), to be called the ﬁeld of ration forms over R.4 We are going to show that, in terms of formal structure, R(X) is to R[X] as the rational numbers Q is to the integers Z. We begin by deﬁning what it means for two rational forms to be equal. f2 (X) Assume given fg11 (X) (X) and g2 (X) ; then by deﬁnition, (5.2)

f1 (X) f2 (X) = g1 (X) g2 (X)

means

f1 (X)g2 (X) = f2 (X)g1 (X).

In other words, we rely on the cross-multiplication algorithm in fractions as a guide to deﬁne the equality of rational forms. Again, note that there is no ambiguity at all about the meaning of equality between rational forms. As a consequence, 0 every rational form g(X) is equal to 0, and the analog of “equivalent fractions” is automatically valid among rational forms, in the sense that for all polynomial forms f (X), g(X), h(X), we have (5.3)

f (X) h(X)f (X) = . g(X) h(X)g(X)

Why? Because clearly f (X)(h(X)g(X)) = g(X)(h(X)f (X)), and so by (5.2), the (X) = 0 if and only if f (X) = 0. two rational forms are equal. Also observe that fg(X) 3 Mathematical Aside: As an example of the latter, one can allow X to be a square matrix of a ﬁxed dimension, and such a polynomial occurs naturally, for example, in the Cayley-Hamilton theorem in linear algebra. 4 Thus R[X] is the collection of all polynomial forms and R(X) is the collection of all rational forms. The notations are completely unsatisfactory, in the same way that using [a, b] and (a, b) to denote the closed and open intervals with endpoints a and b is completely unsatisfactory.

5.1. POLYNOMIAL FORMS

179

We deﬁne addition and multiplication in R(X) by imitating what we know about rational quotients (see Section 2.5 in [Wu2020a]): f1 (X) f2 (X) + g1 (X) g2 (X)

f1 (X)g2 (X) + f2 (X)g1 (X) , g1 (X)g2 (X)

=

f1 (X) f2 (X) f1 (X)f2 (X) · = . g1 (X) g2 (X) g1 (X)g2 (X) But now we have to show that these operations are well-deﬁned, in the following sense. Take the case of the multiplication of rational forms. Suppose f1 (X) F1 (X) f2 (X) F2 (X) = and = ; g1 (X) G1 (X) g2 (X) G2 (X) then for the above deﬁnition to make sense, we must be able to show that f1 (X)f2 (X) F1 (X)F2 (X) = . g1 (X)g2 (X) G1 (X)G2 (X) The same goes for the deﬁnition of addition. The routine proofs will, however, be left as an exercise (Exercise 1 on page 188). Observe that, once addition and (X) (X) (X) (X) = fg(X) and 1· fg(X) = fg(X) multiplication have been deﬁned, we again have 0+ fg(X) for any f (X) and g(X). (X) assumes the We now observe that in the ﬁeld of rational forms R(X), fg(X) property of the division of f (X) by g(X), in the sense that the following holds in R(X): (5.4)

g(X) ·

f (X) = f (X). g(X)

This simple statement calls for two explanations: (1) why is equation (5.4) correct and (2) in what sense is equation (5.4) a “division” in R(X)? We do (1) ﬁrst: g(X) ·

f (X) g(X)

g(X) f (X) · 1 g(X) g(X) · f (X) = 1 · g(X) = f (X), =

where the last step makes use of the “equivalent fractions” property of rational forms. As to (2), what equation (5.4) says is that if we denote the element f (X) g(X) of R(X) by K(X), then we would have the following equality in R(X): f (X) = K(X) · g(X). Comparing this with the interpretation of rational numbers as divisions of integers (e.g., Section 2.5 of [Wu2020a]), we see why K(X) would be called “the division of f (X) by g(X)”. (X) has a multiplicative inverse Note also that every nonzero rational form fg(X) g(X) f (X) .

This then completes the analogy that the set of rational forms R(X) is to the set of polynomial forms R[X] as the rational numbers Q are to the integers Z.

180

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

If we were constrained to discuss rational functions rather than rational forms, (x) each time then we would have to deal with the domain of a rational function fg(x) we face it: this domain is not the number line but only the part of the number line outside the zeros of g(X). When we have to add or multiply a few rational functions in one sitting, the situation can get awkward fast. But for rational forms, (X) is not a function and there is no such awkwardness because a rational form fg(X) g(X) has no zeros. The division algorithm The analogy of polynomial forms with whole numbers leads us to consider the analog of division-with-remainder among whole numbers in the context of polynomial forms. This is the following important division algorithm. Theorem 5.1 (Division algorithm for R[X]). Let g(X) be a nonzero polynomial form in R[X]; then for any polynomial form f (X) in R[X], there are unique polynomial forms Q(X) and r(X), also in R[X], so that (5.5)

f (X) = Q(X)g(X) + r(X)

and r(X) is either 0 or deg r(X) < deg g(X). This Q(X) in (5.5) is called the quotient and r(X) the remainder of the division algorithm of f (X) by g(X) As usual, g(X) is called the divisor. This algorithm is a direct generalization of Theorem 2.10 on page 84. Compared with the ordinary long division of f (X) by g(X) taught in high school, this algorithm is more precise in terms of the concepts involved. For example, we bring out the precise deﬁnition of the remainder (its degree is smaller than the degree of the divisor) and the uniqueness of the quotient and the remainder. To motivate the proof, let us do a concrete example and then analyze the procedure to see what is really going on. For deﬁniteness, take g(X) = 2X − 5, and we approach the division algorithm as an attempt to ﬁnd ways to divide all polynomial forms by g(X). We can do it systematically. First consider all 0-degree polynomial forms f (X); i.e., f (X) = c where c ∈ R and c = 0. This case is trivial because c = 0 · g(x) + c gives the division algorithm with quotient 0 and remainder c. Then we go on to forms of degree 1; let us say f (X) = aX + b, where a = 0. The long division is then 2X − 5

1 2a

aX + b aX − 52 a b + 52 a

Note that what we did was to ﬁnd the number a2 so that, when it multiplies the divisor 2X − 5, this produces the polynomial form aX − 52 a whose leading term (i.e., the term of the highest degree in the polynomial form) aX matches the leading term of the dividend aX + b. Hence, in the subsequent subtraction we only get a degree-0 polynomial b + 52 a. This process is expressed by the equality 5 a (aX + b) − (2X − 5) = b + a . 2 2

5.1. POLYNOMIAL FORMS

181

Rewriting it slightly then yields the division algorithm for this case: 5 a aX + b = (2X − 5) + b + a 2 2 where the quotient is a2 and the remainder is b + 52 a. Next, we do the division of all quadratic polynomial forms f (X) by g(X), then all cubic ones. Suppose that has been done. The next challenge is to obtain the division algorithm of the division of a degree-4 polynomial form by g(X). Consider f (X) = 3X 4 − X 2 + 12 X + 8. Again we do the ﬁrst step of the long division: 2X − 5

3 3 2X

3X 4 3X 4 −

− X2 +

1 2X

+ 8

− X2 +

1 2X

+ 8

15 3 2 X 15 3 2 X

Again, the strategy is to multiply the divisor 2X − 5 by 32 X 3 so as to produce 3 4 a polynomial form 3X 4 − 15 2 X whose leading term 3X matches the leading term 1 4 2 of the dividend 3X − X + 2 X + 8. Therefore in the subsequent subtraction, we 1 3 2 get a polynomial form 15 2 X − X + 2 X + 8 whose degree is smaller than that of f (X). This process is described by 3 3 1 1 15 3 X (2X − 5) = X − X 2 + X + 8. 3X 4 − X 2 + X + 8 − 2 2 2 2 Now the fact that the right-hand side is a cubic becomes signiﬁcant: we already know how to divide all cubic polynomial forms by 2X − 5. Therefore we get 15 2 71 15 3 359 1 1923 2 X −X + X +8= X + X+ . (2X − 5) + 2 2 4 8 16 16 Combining the two equations, we get 3 3 1 4 2 X (2X − 5) 3X − X + X + 8 − 2 2 15 2 71 359 1923 X + X+ . = (2X − 5) + 4 8 16 16 Transposing the second term on the left and applying the distributive law, we obtain the division algorithm for the division-with-remainder of the 4th-degree polynomial form 3X 4 − X 2 + 12 X + 8 by g(X): 3 3 15 2 71 359 1 1923 4 2 X + X + X+ . 3X − X + X + 8 = (2X − 5) + 2 2 4 8 16 16 71 359 1923 2 Here the quotient is 32 X 3 + 15 4 X + 8 X + 16 and the remainder is 16 . We can see from the preceding example that, for the division algorithm to be valid, it is essential that beyond the associative, commutative, and distributive laws, the coeﬃcients have the property that every nonzero number has a multiplicative inverse.5 For example, we had to multiply 2X − 5 by 32 X 3 , which can be made 5 Mathematical Aside: In the language of abstract algebra, the critical requirement is that the coeﬃcients are elements of a ﬁeld and both R and C (see Section 5.2 on p. 189) are ﬁelds.

182

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

explicit as (2−1 × 3)X 3 . Observe also that, in essence, we have made use of mathematical induction in passing from the division of a 4th-degree polynomial form to that of a cubic polynomial form. The general proof follows this reasoning exactly. For purely technical reasons, we have to use a slightly diﬀerent version of mathematical induction (compare the principle of mathematical induction on page 58). Let Pn be a statement about the positive integer n for each n. The principle of complete induction states: suppose both of the following hold for the collection {Pn }: (CI1) P1 is true. (CI2) Pk+1 is true if P1 , P2 , . . . , Pk are true. Then Pn is true for all n. We leave the reason for the validity of the principle of complete induction to an exercise (Exercise 6 on page 188). The proof of the division algorithm can now be given. Proof of Theorem 5.1. We begin by concentrating on the existence of Q(X) and r(X) as asserted and leave the question of their uniqueness to the end. We ﬁrst dispose of the trivial cases in Theorem 5.1. First of all, we may assume that f (X) is nonzero, because if f (X) = 0, then we simply let Q(X) = r(X) = 0. Next, we may assume that both g(X) and f (X) have degree ≥ 1. Indeed, if the degree of g(X) is 0, then g(X) = c ∈ R and c = 0 so that 1 f (X) c + 0. f (X) = c We may let Q(X) = ( 1c f (X)) and r(X) = 0, and we are done. Therefore we will assume that deg g(X) ≥ 1. Then if the degree of f (X) is 0, there would be nothing to prove because we may then let Q(X) be 0, the remainder r(X) be f (X) itself, and we have deg r(X) = deg f (X) = 0 < 1 ≤ deg g(X). Henceforth we may therefore assume that the degrees of f (X) and g(X) are both positive. Thus we ﬁx a g(X) with degree ≥ 1, and for this g(X), we shall prove Theorem 5.1. This means we have to prove that for any polynomial form f (X) of degree equal to 1, 2, 3, . . . , we can ﬁnd polynomial forms Q(X) and r(X) satisfying equation (5.5) and so that either r(X) = 0 or deg r(X) < deg g(X). We will employ complete induction on the degree of f (X) for the proof. We begin with the base case of the degree of f (X) being equal to 1. If degree g(X) is 1, let g(X) = aX + b and also let f (X) = cX + d (a, c = 0). Since bc c , cX + d = (aX + b) + d − a a we may let Q(X) = ac and r(X) = d − bc a and we are done. If, however, the degree of g(X) is at least 2, then f (X) = 0 · g(X) + f (X), and we may let Q(X) = 0 and r(X) = f (X) because this r(X) has degree less than the degree of g(X) and we are done again. Thus if degree f (X) is 1, the theorem is proved.

5.1. POLYNOMIAL FORMS

183

Still with the ﬁxed g(X) of degree ≥ 1, we will prove the theorem for the case of an f (X) of degree > 1. We employ complete induction on the degree of f (X). Suppose the theorem has been proved for any polynomial form f (X) of degree 1, 2, . . . , k. We will prove the theorem when deg f (X) = k + 1. Let f (X) = cX k+1 + f1 (X), where c = 0 and f1 (X) is a polynomial form of degree ≤ k. In case deg g(X) > k + 1, then f (X) = 0 · g(X) + f (X) shows that letting Q(X) = 0 and r(X) = f (X) would do. We may therefore assume from now on that deg g(X) ≤ k + 1. Let g(X) = aX m + g1 (X), where a = 0, m ≤ k + 1, and g1 (X) is a polynomial form of degree ≤ m − 1. Then c c (5.6) f (X) − X (k+1)−m g(X) = f1 (X) − X (k+1)−m g1 (X). a a (Notice that this step requires the existence of the multiplicative inverse of a.) Let h(X) be the right-hand side; i.e., c h(X) = f1 (X) − X (k+1)−m g1 (X). a Note that the degree of h(X) is at most k because the degree of the second term on the right is ≤ ((k + 1) − m) + (m − 1) = k. We do not know that the degree of h(X) is exactly k; it could be 1, 2, . . . , or k. Nevertheless, since we are using complete induction, we know that there are polynomial forms Q (X) and r(X) so that h(X) = Q (X)g(X) + r(X) where deg r(X) < deg g(X). Hence, from (5.6), we have c f (X) − X (k+1)−m g(X) = h(X) = Q (X)g(X) + r(X). a Transposing the second term of the left-hand side and applying the distributive law, we have c X (k+1)−m + Q (X) g(X) + r(X). f (X) = a Letting Q(X) be ac X (k+1)−m + Q (X) and observing that deg r(X) < deg g(X), we are done as far as existence is concerned. To prove the uniqueness of Q(X) and r(X) in equation (5.5) with the requirement that either r(X) = 0 or deg r(X) < deg g(X), let f (X) = Q(X)g(X) + r(X) = Q1 (X)g(X) + r1 (X) where Q(X), Q1 (X), r(X), and r1 (X) are polynomial forms and deg r(X), deg r1 (X) < deg g(X). This implies that {Q(X) − Q1 (X)}g(X) = {r1 (X) − r(X)}. Suppose Q(X) = Q1 (X); then Q(X) − Q1 (X) is a polynomial of degree ≥ 0 and therefore {Q(X)−Q1 (X)}g(X) is a polynomial form of degree ≥ deg g(X). But the degree of the polynomial form on the right-hand side, {r1 (X)−r(X)}, is < deg g(X). This is a contradiction. Therefore Q(X) = Q1 (X), so that 0 = {r1 (X) − r(X)}, and hence r(X) = r1 (X) as well. The proof of Theorem 5.1 is complete. Acivity. Why is mathematical induction as deﬁned in Section 1.7 (pp. 57ﬀ.) not good enough for proving Theorem 5.1?

184

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

The basic importance of the division algorithm for R[X] can be appreciated if we recall what one can do with the division-with-remainder for integers. Review Chapter 3 of [Wu2020a] at this point if necessary. We will leave a few such applications of the division algorithm for R[X] to the exercises. It may be mentioned that, in advanced mathematics, both the integers Z and R[X] are examples of Euclidian domains essentially because they both possess a division algorithm. Roots of polynomial forms What we will discuss next are the implications of the division algorithm on the nature of the roots of a polynomial equation f (X) = 0, which are, by deﬁnition, the real numbers c so that f (c) = 0 (remember that here we are discussing a polynomial form f (X) in R[X]). For convenience, we also refer to c as a root of f (X) in R[X]. We say a polynomial form h(X) is a factor of another polynomial form f (X) if f (X) = h(X)g(X) for some g(X) ∈ R[X]. In this case, we also say h(X) divides f (X). In symbols, h(X)|f (X). Theorem 5.2 (Factor theorem). A number c is a root of f (X) ∈ R[X] ⇐⇒ (X − c) is a factor of f (X). Proof. By the division algorithm, f (X) = Q(X)(X − c) + k for some Q(X) ∈ R[X] and for some k ∈ R. In particular, f (c) = Q(X) · 0 + k. Therefore c is a root of f (X) ⇐⇒ k = 0 in the division algorithm f (X) = Q(X)(X − c) + k. The theorem follows. We can now give a purely algebraic proof of Theorem 3.3 on page 126, to the eﬀect that a polynomial f (x) of degree n has at most n distinct zeros.6 Indeed, given f (x), consider the corresponding polynomial form f (X) obtained by replacing each number x by the symbol X. We will prove that any polynomial form of degree n has at most n distinct roots. Let r1 , r2 , . . . , rk be any k distinct roots of f (X). At this point, we cannot assert that these are all the distinct roots of f (X) because we do not as yet know that the total number of distinct roots of f (X) is ﬁnite. However, a knowledge of this ﬁniteness is not necessary, as the following argument shows. By the factor theorem, we have f (X) = (X − r1 )f1 (X) for a polynomial form f1 (X) of degree n − 1. Letting X = r2 , we get 0 = f (r2 ) = (r2 − r1 )f1 (r2 ). Since the ri ’s are distinct, r2 − r1 = 0. Thus f1 (r2 ) = 0 and r2 is a zero of the polynomial form f1 (X). Applying the factor theorem to f1 (X), we get f1 (X) = (X − r2 )f2 (X) for a polynomial form f2 (X) of degree n − 2. Therefore, f (X) = (X − r1 )(X − r2 )f2 (X). Continuing this way, we get f (X) = (X − r1 )(X − r2 ) · · · (X − rk )fk (X) for a polynomial form fk (X) of degree n − k. This says that no matter what k is, k ≤ n, and Theorem 3.3 is proved. Recall that in Chapter 3, Theorem 3.3 was proved with the help of calculus. 6 To the extent that we have been dealing with the real numbers R exclusively up to this point, all these zeros are understood to be real numbers.

5.1. POLYNOMIAL FORMS

185

Next, we give a preliminary reﬁnement of Theorem 3.3, which is Theorem 5.3 immediately following. (The ultimate reﬁnement will be given in Theorem 5.7 on page 200.) To this end, we have to introduce a new concept. By the factor theorem, if r is a root of a polynomial form f (X), then (X − r)|f (X); therefore there is a largest positive integer m(r) so that (X − r)m(r) |f (X). This integer m(r) is called the multiplicity of the root r. In view of the factor theorem, the multiplicity m(r) of r may also be characterized as follows: m(r) is the positive integer with the property that if we write f (X) = (X − r)m(r) g(X) for a polynomial form g(X), then g(r) = 0. If m(r) = 1, then we say r is a simple root (or simple zero), and if m(r) = 2, then we say r is a double root (or double zero). With the use of calculus, one can show that this deﬁnition of multiplicity for the root (zero) of a polynomial is the same as the one oﬀered in Section 3.2; this will be Exercise 8 in Exercises 6.3 of [Wu2020c]. A root r of multiplicity 3, for instance, means f (X) = (X − r)3 f1 (X) = (X − r)(X − r)(X − r)f1 (X) for some polynomial form f1 (X), so that f1 (r) = 0. Therefore, this r is a root of f (X) three times rather than just once. This is the intuitive meaning of multiplicity. Now the theorem we are after is the following. Theorem 5.3. A polynomial form in R[X] of degree n has at most n (real) roots, counting multiplicity. Proof. It is essentially the same as the preceding argument. Since we now know that the total number of distinct roots of f (X) is at most n, we may let r1 , . . . , rk be all the distinct roots of f (X). Then as before, writing m(1) for the multiplicity m(r1 ) of r1 , we have f (X) = (X − r1 )m(1) f1 (X), where f1 (X) is a polynomial form of degree n − m(1) so that f1 (r1 ) = 0. Letting X = r2 in the preceding equation, we get 0 = f (r2 ) = (r2 − r1 )m(1) f1 (r2 ). Since the ri ’s are distinct, r2 − r1 = 0. Thus f1 (r2 ) = 0 and r2 is a root of f1 (X). Let us say the multiplicity of r2 as a root of f1 (X) is m(2). This means f1 (X) = (X − r2 )m(2) f2 (X) for a polynomial form f2 (X) of degree n − m(1) − m(2) so that f2 (r2 ) = 0. Therefore, (5.7)

f (X) = (X − r1 )m(1) (X − r2 )m(2) f2 (X).

We pause to observe that (♠)

m(2) is in fact the multiplicity of r2 as a root of f (X).

To prove this, suppose the multiplicity of r2 as a root of f (X) is so that f (X) = (X − r2 ) g(X) for a polynomial form g(X). Thus g(r2 ) = 0. Now in equation (5.7), let us write h(X) for the polynomial form (X − r1 )m(1) f2 (X); then we also have f (X) = (X − r2 )m(2) h(X), where h(r2 ) = 0.

186

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

Therefore (X − r2 ) g(X) = (X − r2 )m(2) h(X). If = m(2), let us say < m(2). Then 1 1 (X − r2 ) g(X) = (X − r2 )m(2) h(X). (X − r2 ) (X − r2 ) By equation (5.4) on page 179, we get g(X) = (X − r2 )(m(2)−) h(X) with m(2) − > 0. Letting X = r2 , the right side is 0, and therefore g(r2 ) = 0, contradicting g(r2 ) = 0. Thus the assumption < m(2) cannot hold and we must have > m(2). But then a similar argument, using h(r2 ) = 0, shows that the assumption > m(2) cannot hold either. Therefore = m(2) after all, and m(2) is indeed also the multiplicity of the root r2 in f (X). This proves (♠). Now we go back to equation (5.7) and write f2 (X) = (X − r3 )m(3) f3 (X), where f3 (X) is a polynomial form of degree n − m(1) − m(2) − m(3) and f3 (r3 ) = 0. Moreover, we also know by the same reasoning as in the preceding paragraph that m(3) is the multiplicity of r3 as a root of f (X). Continuing this way, we get f (X) = (X − r1 )m(1) (X − r2 )m(2) · · · (X − rk )m(k) fk (X) where fk (X) is a polynomial form of degree n−m(1)−m(2)−· · ·−m(k), the m(i)’s are positive integers, and each m(i) is the multiplicity of ri as a root of f (X). In particular, n − m(1) − m(2) − · · · − m(k) ≥ 0, so that n ≥ m(1) + m(2) + · · · + m(k). The proof of Theorem 5.3 is complete. The rational roots theorem It is not easy to ﬁnd the roots of a polynomial form. However, if the coeﬃcients of a polynomial form are all integers, then it is easy to detect whether it has any rational root, i.e., any root that is a rational number. This is most easily seen from a simple example. Consider the quadratic polynomial form g(X) = 35X 2 −11X −6. Its roots are 35 and − 27 , because it is easy to see that g( 53 ) = g(− 27 ) = 0. How could we have guessed these roots? We could, if we had noticed that the numerators 3 and 2 of the roots are related to the constant term 6 of g, and the denominators 5 and 7 are related to the leading coeﬃcient 35 of g(X). More precisely, by the factor theorem above (see also Vi`ete’s theorem on page 83), we know 3 2 g(X) = X − X+ k 5 7 for a degree 0 polynomial form k (i.e., k is a constant). Comparing the coeﬃcients of the X 2 term on both sides, we get k = 35 and therefore g(X) = (5X − 3)(7X + 2) = 5 × 7 X 2 + (10 − 21)X − 3 × 2 . This expression of g(X) explicitly displays the fact that the numerators 3 and 2 of the roots are divisors of the constant term 6 of g(X), and the denominators 5 and 7 of the roots are divisors of the leading coeﬃcient 35 of g(X). This then suggests, if we are looking for roots of g(X) that are rational numbers, that we should look

5.1. POLYNOMIAL FORMS

187

among those k so that k is a divisor of the constant term 6 and is a divisor of the leading coeﬃcient 35. Thus the only possibilities for the rational roots of g(X) are 1 2 3 6 ±1, ±2, ±3, ±6, ± , ± , ± , ± , 35 35 35 35 2 3 6 1 2 3 6 1 ± , ± , ± , ± , ± , ± , ± , ± . 5 5 5 5 7 7 7 7 A simple trial and error then yields the two roots 35 and − 27 above. The following theorem says that this is no accident but is a general phenomenon. Before stating the theorem, however, we observe that some caution is called for. 9 9 is a root of g(X) = 35X 2 − 11X − 6, because 15 = 35 , but clearly the Indeed, 15 9 numerator 9 of 15 is no divisor of the constant term 6, and likewise the denominator 9 15 of 15 does not divide the leading coeﬃcient 35. This suggests that we should use only rational numbers k (where k and are integers) in lowest terms, in the sense that the whole numbers |k| and || are relatively prime. (This deﬁnition is therefore consistent with the deﬁnition of a fraction in lowest terms (recalled on page 353).) Theorem 5.4 (Rational roots theorem). Let f (X) = an X n + an−1 X n−1 + · · · + a1 X + a0 be a polynomial form so that all its coeﬃcients ai (i = 0, 1, . . . , n) are integers. Then any rational root k of f (X) (where k and are integers), if it is in lowest terms, must satisfy k|a0 and |an . We give a word of caution: this theorem says nothing about roots which are not rational. For example, the rational roots of g(X) = X 2 + X − 3 must be among ±1 and ±3, but none of the four is a root of g(X) because g(±1) = 0 and g(±3) = 0 (see the factor theorem). The roots are, instead, the pair of irrational numbers: √ 1 (−1 ± 13). 2 Because the divisibility properties among whole numbers, especially the Key Lemma (Lemma 3.4) of [Wu2020a] (see page 357), are usually not taught in middle school, or at least usually not taught well, the rational roots theorem in algebra courses in high school often ends up being taught by rote without any reasoning. As a result, students get the misconception that this theorem is a test for whether a polynomial has any real roots. No. We repeat: this is only a test for the existence of rational roots of a polynomial with integer coeﬃcients. Please make this point absolutely clear to your students when you teach. Proof. We are given that f ( k ) = 0, so we have, by deﬁnition, n n−1 k k k an + an−1 + · · · + a1 + a0 = 0. Multiplying both sides by n , we get an kn + an−1 kn−1 + · · · + a1 kn−1 + a0 n = 0. We can rewrite the preceding equation in two diﬀerent ways: (5.8)

k(an kn−1 + an−1 kn−2 + · · · + a1 n−1 ) = (−a0 )n

and (5.9)

(an )kn = (−)(an−1 kn−1 + · · · + a1 kn−2 + a0 n−1 ).

188

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

In (5.8), k divides the left side and therefore must also divide the right side. Thus |k| divides |a0 |||n . But by hypothesis, |k| and || are relatively prime, and therefore |k| and ||n are also relatively prime. By the Key Lemma (Lemma 3.4) in [Wu2020a] (recalled on page 357), |k| must divide |a0 |; i.e., k divides a0 . By arguing in exactly the same way with (5.9), we conclude that must divide an . The proof is complete. Exercises 5.1. (1) (a) Show that for all nonzero polynomial forms f (X), g(X), deg(f (X)g(X)) (X) is equal to = deg f (X) + deg g(X). (b) Show that a rational form fg(X) 0 if and only if f (X) = 0. (c) Show that the deﬁnitions of adding and multiplying rational forms in R(X) are well-deﬁned. (2) (a) Find all the roots of 2x3 − x2 − 3x − 1. (b) Find all the roots of 2x4 − 7x3 − 29x2 + 86x + 56. A B (3) (i) Show that the rational form X3X−3 2 −X−2 can be written as X+1 + X−2 for some A, B ∈ R. Explain your steps carefully. (ii) Show that the rational 1 , for a, b ∈ R, where a and b are distinct, can always be form (X−a)(X−b) A B for some A, B ∈ R. Explain your steps carefully. written as X−a + X−b (4) Let the notation be as in Theorem 5.1 on page 180. (a) Apply the division algorithm to the polynomial forms f (X) = 3X − 1 and g(X) = 7X 5 − 2X 3 +X 2 − 12 X −1. (b) Do the same for f (X) = 7X 5 −2X 3 +X 2 − 12 X −1 and g(X) = 3X − 1. (c) Do the same for f (X) = 53 X 6 − 52 X 4 + 76X and g(X) = 2X 2 − 3. (5) A polynomial form p(X) is said to be the GCD of the polynomial forms f (X) and g(X) if it is the highest-degree polynomial form that divides both f (X) and g(X). (Note that the GCD so deﬁned is necessarily indeterminate up to a nonconstant multiple.) Use ideas from Chapter 3 of [Wu2020a] to ﬁnd the GCD of f (X) = X 3 − X 2 + 3X − 3 and g(X) = X 6 + 3X 4 + X 2 + 3. (Don’t forget that you have to prove your answer is correct.) (6) Explain why the principle of complete induction announced on page 182 is correct. (7) Consider the set of all polynomial forms whose coeﬃcients are restricted to be integers. This set is usually denoted by Z[X]. Would the division algorithm be valid for Z[X]? In other words, is the following correct? Given any polynomial forms f (X) and g(X) in Z[X], there are unique polynomial forms Q(X) and r(X) in Z[X] so that f (X) = Q(X)g(X) + r(X) where r(X) is either 0 or has a degree < the degree of g(X). If so, prove it. If not, describe the ways in which it might fail. (8) Let the Fibonacci sequence (Fn ) be deﬁned for all whole numbers n by F0 = 0, F1 = 1, and for all n ≥ 2 by Fn = Fn−1 + Fn−2 . Use the principle of induction to prove Cassini’s identity: Fn2 − Fn+1 Fn−1 = (−1)n+1 for all n ≥ 1.

5.2. COMPLEX NUMBERS

189

5.2. Complex numbers We introduce complex numbers in this section, prove their basic properties, and prove the complex quadratic formula for quadratic polynomials with complex coeﬃcients. Basic properties of complex numbers (p. 189) The square roots of a complex number (p. 191) The quadratic formula in C[X] (p. 193) Basic properties of complex numbers Every real number is, by deﬁnition, nothing but a point on the x-axis (which is identiﬁed with the number line in this discussion) of the coordinate plane. The fact that we can add and multiply real numbers may be interpreted as the possibility of conferring an algebraic structure on the points on the x-axis so that they can be added and multiplied. We now go a step further by showing how to add and multiply any two points in the coordinate plane. The resulting collection of points is called the complex numbers, to be denoted by C. Here are the details. Given two points (a, b) and (c, d) in the (coordinate) plane, where a, b, c, d are understood to be real numbers in this discussion until stated otherwise, we have to say what it means to add (respectively, multiply) them to get a third point (x, y) so that the associative, commutative, and distributive laws hold. We ﬁrst do it algebraically. By deﬁnition, (5.10) (5.11)

(a, b) + (c, d) (a, b) · (c, d)

= (a + c, b + d), = (ac − bd, ad + bc).

The fact that these two operations now satisfy the associative, commutative, and distributive laws will be left as an exercise (see Exercise 1 on page 196). We usually omit the dot “·” in the writing of multiplication in (5.11). The most basic concern about these deﬁnitions has to do with the fact that the number line is embedded in C as its x-axis, and therefore these deﬁnitions of addition and multiplication on C, when applied to points in Q—the rational numbers on the x-axis—will then specify a new way to add and multiply rational numbers. Since the arithmetic operations on Q are already well established (see Chapter 2 of [Wu2020a]), the concern is whether the new deﬁnitions agree with the original ones on Q. If not, there would be chaos. Fortunately they do. (In the language of extension as introduced in Section 1.2 of [Wu2020a], the deﬁnitions of the arithmetic operations on C will now be shown to be extensions of those on R.) Thus let a and c be two rational numbers; then as rational numbers we have the usual sum and product, a + c and ac. Now, when regarded as complex numbers, a and c become (a, 0) and (c, 0), and therefore a + c and ac become (a + c, 0) and (ac, 0). The question is, do the preceding deﬁnitions yield (a, 0) + (c, 0) = (a + c, 0) and

(a, 0)(c, 0) = (ac, 0),

respectively? An inspection immediately reveals that this is indeed what we get if b = d = 0 in (5.10) and (5.11). We therefore conclude that the deﬁnitions of addition and multiplication in C are extensions of the original deﬁnitions of the same in Q.

190

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

This deﬁnition of addition has a geometric interpretation. For a parallelogram with vertices O, (a, b), and (c, d) as shown, the point (a, b) + (c, d) is the fourth vertex of the parallelogram (Exercise 2 on page 196, where some hint is provided).

O

(c, d) r

s(a, b) + (c, d)

r (a, b)

The geometric interpretation of multiplication is even more important, but it must wait for the discussion of trigonometry in Section 1.6 of [Wu2020c]. The next task is to show how to relate this deﬁnition of complex numbers to their usual representation as “x + iy” and the standard computations with this representation. We do so via a sequence of observations (i)–(ix), and a main conclusion is that every nonzero complex number has a unique multiplicative inverse. See (viii). (i) We have just observed that a real number t (on the x-axis) is identiﬁed with the point (t, 0) in the plane and that the deﬁnitions of addition and multiplication in (5.10) and (5.11) yield the equalities (a, 0) + (c, 0) = (a + c, 0) and (a, 0)(c, 0) = (ac, 0), respectively. Consequently, the addition and multiplication of two real numbers a and c then have the usual meaning whether regarded as two real numbers or as two complex numbers (a, 0) and (c, 0). From now on, for two real numbers a and c, we identify them with (a, 0) and (c, 0), respectively, so that a + c and ac become identiﬁed with (a, 0) + (c, 0) and (a, 0)(c, 0), respectively. (ii) We will agree to denote the point (0, 1) by the symbol i. Then, (a, b) = a + ib for all complex numbers (a, b), because a + ib

= (a, 0) + (b, 0)(0, 1) = (a, 0) + (0, b) (by (5.11)) = (a, b) (by (5.10)).

The real numbers a and b are called the real and imaginary parts of a + ib. If the real part of a complex number is 0, the number ib is called a pure imaginary number. (iii) (a + ib)(c + id) = (ac − bd) + i(ad + bc). This is a direct translation of (5.11) in the notation of “i” using (ii). (iv) i2 = −1. This follows from (iii) by letting a = c = 0 and b = d = 1. (v) A complex number a + ib is equal to zero, by deﬁnition, if a = b = 0. This deﬁnition is consistent with the deﬁnition that a point (a, b) in the plane is equal to the origin (i.e., O) if and only if a = b = 0. (vi) Introduce the conjugate of a complex number z as follows: if z = a+ib, then the conjugate z of z is, by deﬁnition, z = a − ib. Also introduce the modulus or absolute value |z| of z by √ |z| = zz.

5.2. COMPLEX NUMBERS

191

Then a trivial computation using (iii) gives |a + ib| = a2 + b2 . Thus for any z ∈ C, the modulus |z| is a nonnegative real number and |z| > 0

if and only if z = 0.

If a is a nonzero real number and z is a complex number, then the product a1 z makes sense and it is usually simply written as az . With this understood, we have z (5.12) z = 1 if z = 0. |z|2 Indeed, this is nothing more than a rewriting of zz = |z|2 . (vii) For complex numbers z and w, (5.13)

z + w = z + w,

zw = z w,

|z| = |z|,

and

|zw| = |z| · |w|.

These are simple computations (Exercise 3 on page 196). Note also that the real numbers are exactly those complex numbers w which satisfy w = w. (viii) Every nonzero complex number z has a unique multiplicative inverse z −1 , in the sense that zz −1 = 1. Indeed, (5.12) implies that it suﬃces to let z −1 =

z . |z|2

The fact that if z is another complex number and zz = 1, then z = z −1 , will be left as an exercise (see Exercise 4 on 196). (ix) Assume complex numbers w, z, with z = 0. Then we deﬁne w divided by ) to be the complex number c so that w = cz. This deﬁnition is z (in symbols, w z the same as the deﬁnitions for the division of fractions and the division of rational numbers given in Section 1.5 and Section 2.5 of [Wu2020a], respectively. Since zz −1 = 1 by (viii), we see that c = c · 1 = c · zz −1 = (cz)z −1 = wz −1 . It follows that

w = wz −1 . z

The square roots of a complex number The goal of the rest of this section is to prove the quadratic formula for quadratic equations with complex coeﬃcients. Anticipating the presence of the square root in the quadratic formula, we ﬁrst make sense of square roots in the context of complex numbers. We say a complex number (x + iy) is a square root of the complex number (a + ib) if (x + iy)2 = a + ib. We will need the following simple but basic fact. Lemma 5.5. Every nonzero complex number has exactly two distinct complex square roots. Remark. Once we have the polar form of a complex number, we will generalize this lemma by exhibiting, for every nonzero complex number, its n distinct complex n-th roots (see Section 1.6 of [Wu2020c]).

192

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

Proof of Lemma 5.5. Let the given nonzero complex number be (a + ib). Then (x + iy) is a square root of (a + ib) if and only if (5.14) (5.15)

x2 − y 2 2xy

= a, = b.

These are simple consequences of the fact that (x + iy)2 = (x2 − y 2 ) + i(2xy). Now the proof of the lemma breaks up into two cases. Case 1: b = 0. Thus we are given a nonzero real number a and must show that a has exactly two distinct square roots. Since a = 0, either a > 0 or a < 0. First suppose a > 0. If (x + iy) is a square root of a, then (5.15) implies that 2xy = 0 so that at least one of x and y is 0. Now x = 0 because x = 0 would imply (by (5.14) a = −y 2 ≤ 0, a contradiction. Therefore y = 0; this implies that any square root √ of a has to be a real number x. Then Lemma 2.1 on page 64 implies that x = ± a. We pause to take stock of where we are: we have just shown that if a real number a has a complex√root, then the complex root is actually a real √ number that either of ± and it must be one of ± a. A priori, this does not say √a is a √ square root of a. But of course this is easy to check: (± a)2 = a so that ± a are indeed square roots √ of a. Altogether, we have shown that the only square roots of a positive a are ± a. If a < 0, then x + iy being a square root of a implies that 2xy = 0, by (5.15). This time, y = 0 because y = 0 would imply by (5.14) that a = x2 ≥ 0, contradicting a < 0. Thus x = 0. Consequently, any square root of a is a pure imaginary number iy. But (iy)2 = a implies −y 2 = a, which implies √ in turn that y 2 = −a; since −a > 0, the preceding paragraph implies that y = ± −a. It follows that any square root of a negative a is √ √ iy = i(± −a) = ±i −a. √ 2 It remains to check the simple fact that, √ indeed, (±i −a) = a. It follows that all the square roots of a negative a are ±i −a. This proves the lemma in case b = 0. Case 2: b = 0. In this case, if x + iy is a square root of a + ib, then 2xy = b = 0 by (5.15). Therefore x = 0 and y = 0, and we get (5.16)

y=

b . 2x

Now according to (5.14), we have x2 − y 2 = a and therefore, by (5.16), we have x2 −

b2 = a. 4x2

Multiplying both sides by 4x2 and transposing, we get 4(x2 )2 − 4a(x2 ) − b2 = 0. This is a quadratic equation in x2 , and the quadratic formula yields √ 1 4a ± 16a2 + 16b2 2 = (a ± a2 + b2 ). x = 8 2 But note that for any a ∈ R and any b = 0, a − a2 + b2 < 0 and a + a2 + b2 > 0.

5.2. COMPLEX NUMBERS

Since x = 0, x2 > 0. Hence we get x2 = and consequently,

193

1

a + a2 + b2 2

1

a + a2 + b2 . 2 Recalling (5.16), we see that if x + iy is a square root of a + ib, then b 1

and y = a + a2 + b2 , x=± 2 2x or, in more detail, 1

b x=± and y = ±

a + a2 + b2 √ . 2 1 a + a2 + b2 2 x=±

2

It remains to show that if b = 0 and x + iy denotes either of the two complex numbers ⎫ ⎧ ⎬ ⎨ 1 b (5.17) ± (a + a2 + b2 ) + i , √ ⎭ ⎩ 2 2 1 (a + a2 + b2 ) 2

2

then (x + iy) = a + ib. As we have seen, such is the case if and only if we can verify both (5.14) and (5.15). Since the computation that is required for such a veriﬁcation is entirely straightforward, we will leave it as an exercise (Exercise 7 on page 196). Thus we have shown that the two numbers in (5.17) are all the possible square roots of a + ib in case b = 0. The proof of Lemma 5.5 is complete. We will follow the common practice of denoting either of the two square roots of a + ib in (5.17) by √ a + ib. We emphasize, however, that this is a defective √ notation. Unlike the case of the square root of positive numbers, the notation a + ib here is ambiguous because it does not specify which of the two square roots of a + ib is meant. Once we have done enough trigonometry in [Wu2020c], we will revisit the issue of the square roots of a complex number and prove more generally that, for any positive integer n, any complex number z has n distinct n-th roots; i.e., there are distinct complex numbers w1 , w2 , . . . , wn so that win = z for i = 1, 2, . . . (see Section 1.6 of [Wu2020c]). The quadratic formula in C[X] We can now tackle quadratic equations in general. Recall that we did not completely solve all real quadratic equations ax2 + bx + c = 0, where a, b, c are real numbers. We got solutions only when the discriminant b2 − 4ac is greater than or equal to 0. With complex numbers at our disposal, we can now solve all such quadratic equations even when a, b, c are complex numbers regardless of what the discriminant may be. Consider quadratic polynomial forms aX 2 + bX + c, where a, b, c are complex numbers. For convenience, let the set of all polynomial forms with

194

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

coeﬃcients in C be denoted by C[X]; this is the polynomial ring over C. As usual, a complex number r is a root (or zero) of the polynomial form F (X) ∈ C[X] if F (r) = 0. We claim that the same quadratic formula now provides the roots of all complex quadratic polynomial forms in C[X]; i.e., given any (aX 2 + bX + c) ∈ C[X] with a = 0, its roots are −b ±

(5.18)

√ b2 − 4ac . 2a

Notice that the quadratic formula, insofar as it gives both roots √ without favoring one over the other, is immune to the ambiguity of the meaning of b2 − 4ac (recall from the preceding subsection: we do not know from this notation which of the two square roots of b2 − 4ac is meant). The proof of this claim is identical to the case of R[X] and may be succinctly given as follows. (Caution: The succinctness of the exposition is predicated on your willingness to painstakingly verify that each of the logical equivalences below, i.e., “⇐⇒”, is indeed an equivalence; i.e., “A ⇐⇒ B” is a shorthand for two assertions, namely, A =⇒ B and B =⇒ A.) Let x be a complex number. Then because ax2 + bx + c = 0 ⇐⇒ ax2 + bx = −c, we have ax + bx + c = 0 ⇐⇒ 2

b 2 a x + x = −c a

⇐⇒

c b 2 x + x =− a a

⇐⇒

b2 b c b2 2 − . x + x+ 2 = 2 a 4a 4a a

The left side of the last equality is equal to (x + 2 −4ac to b 4a . Therefore, 2 ax + bx + c = 0 ⇐⇒ 2

But b2 − 4ac = 4a2

b x+ 2a

b 2 2a ) ,

2 =

while the right side is equal

b2 − 4ac . 4a2

2 √ ± b2 − 4ac . 2a

Therefore, ax2 + bx + c = 0 if and only if (5.19)

√ b ± b2 − 4ac . x+ = 2a 2a

b After transposing 2a to the right in (5.19), we see that (5.18) does provide all the 2 roots of (aX + bX + c), as claimed.

5.2. COMPLEX NUMBERS

195

For cubic polynomial forms, there is an analog of the quadratic formula, called the Cardano formulas,7 which exhibit the roots of a cubic in terms of its coeﬃcients. See [Osler] and also Exercise 6 on page 203. However, these are more unwieldy than the quadratic formula and are therefore less useful. There are even more complicated formulas for the roots of quartic (i.e., 4th-degree) polynomial forms. For polynomial forms of degree ≥ 5, the famous theorem of Abel and Galois says that there are no such general formulas for their roots. We now give a more leisurely discussion of the notational ambiguity of the square root of a complex number. Recall that for a positive number a, the symbol √ a denotes an unambiguous number, namely, the positive number so that its square is a. As pointed out on page 65 (and also page 143), this uniqueness leads to the useful identity (2.2) stated on page 65, to the eﬀect that √ √ √ a b = ab, valid for all positive a and b. If now a and b are complex numbers, so that we can no longer specify the choice of their “positive” square roots, this identity √ √becomes −1 −1. false if taken literally. For example, if a = b = −1, the left side would be √ 2 We naturally take −1 to be i, and therefore √ the left side becomes ii = i = −1. But the right side, which is (−1)(−1) = 1, is just 1 if taken to have its usual √ √ meaning. Then the equality −1 −1 = (−1)(−1) would assert −1 = 1, which is absurd. Naturally, this equality is √ not wrong if we make judicious choices of the square roots roots. One way is to look at 1 as indicating one of the two of 1. Then √ √ square we may take it to be −1, in which case the validity√of −1 −1 = (−1)(−1) is restored. Another√way√is to look at one of the two −1, let ﬁrst, as −i √ us√say the but not i. Then −1 −1 = (−i)i = 1, and the equality −1 −1 = (−1)(−1) would again be valid. But the need to choose the correct square root on a caseby-case basis is contrary to the spirit √ of a general formula, so there is no glory in √ √ asserting that the identity a b = ab is also valid for complex numbers a and b, “provided judicious choices √are made about square roots in each instance”. √ √ The failure of a b = ab, as is, underscores the importance of the uniqueness of the square root of positive numbers. In a broader context, it also underscores the importance of having proofs of theorems. In this case, we have the satisfaction of √ √ √ knowing exactly why the identity a b = ab is true for positive a and b and why it breaks down in general for complex numbers. Mathematical Aside: Having conferred algebraic structures on a line and a plane to make them into the real numbers R and the complex numbers C, respectively, it is natural to inquire if we can go on to do the same for all Euclidean spaces of higher dimensions, Rn for n ≥ 3. The answer is no, at least not if we insist on conferring the structure of a ﬁeld on Rn . We now know that if we are willing to give up the commutativity of multiplication, then we can do it for R4 (the quarternions), 7 Gerolamo Cardano (1501–1576) of Italy published these formulas in his Ars Magna in 1545. He was a famous physician of his day and was a colorful personality. The so-called Cardano formulas have a complicated history. They were ﬁrst discovered by Scipione del Ferro (1465–1526) but Cardano publicized them together with the quartic formulas of Lodovico Ferrari (1522–1565), thereby making them known to the whole world. Ars Magna became very inﬂuential, not the least because it made systematic use of negative numbers and also made a ﬁrst attempt at using complex numbers.

196

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

and if we are also willing to give up the associativity of multiplication, then we can do it for R8 (the Cayley numbers or the octonions). The proof that there are no other possibilities beyond these four cases requires the modern arsenal of algebraic topology. See the entry [Division Algebra] from Wolfram MathWorld for a summary. Exercises 5.2. (1) Show that the addition and multiplication of complex numbers satisfy the associative, commutative, and distributive laws. (2) Verify the geometric interpretation of the addition of complex numbers on page 190. (Caution: This looks easy when you think only in terms of two points in the ﬁrst quadrant, but the parallelogram interpretation is supposed to be valid for any two points. Try to use Lemma 6.17 of [Wu2020a]—quoted on page 359—for example.) (3) For complex numbers z and w, show that z + w = z + w,

zw = zw,

and

|zw| = |z| · |w|.

(4) Show that the multiplicative inverse of a nonzero complex number is unique. (5) Show that the distance between two complex numbers w and z, considered as two points in the plane, is given by |w − z|. (6) Write each of the following in the form a + ib for some real numbers a and 2 − i2 4 + i 15 1+i . (b) 3 . (c) 1 b: (a) 1. 1−i 2+i 3 − i4 (7) Prove that the number in (5.17) on page 193 is a square root of a + ib. (8) (a) Write the square roots of i in the form a + ib for some real numbers a and b. (b) Do the same for 2 − i3. √ √ (9) Find all the real numbers x so that x + 5 − 2 x − 1 = 0. (10) Solve: (a) 9X 2 + i = 0. (b) 3X 2 − 3X + i = 0. (11) If a and b are real numbers and a + ib is a root of iX 2 − 2X + (2 + i), what could a and b be? (12) Prove the following generalization of the theorem of Vi`ete (Lemma 2.9 on page 83): let f (x) = ax2 + bx + c be a complex quadratic polynomial form, and let r1 and r2 be the zeros of f (see (5.18) on p. 194). Then r1 + r2 = − ab and r1 r2 = ac . 5.3. Fundamental theorem of algebra This section gives a precise statement of the fundamental theorem of algebra and deduces a few immediate consequences. It will be seen that the division algorithm for complex polynomial forms plays a crucial role. We also specialize the fundamental theorem of algebra to real polynomial forms to show that the latter is always a product of linear and quadratic real polynomial forms. We saw in the last section that, as a result of the quadratic formula, every quadratic polynomial form with complex coeﬃcients always has two roots (counting multiplicity), provided we allow the roots to be in C. One of the more remarkable theorems in mathematics is that a corresponding fact holds for polynomial forms in C[X] of all degrees. The basic reason is the following.

5.3. FUNDAMENTAL THEOREM OF ALGEBRA

197

Fundamental Theorem of Algebra. Every F (X) ∈ C[X] has a complex root. Please note that this theorem does not merely say that every polynomial with real coeﬃcients has a complex root. Rather, it says that every polynomial with complex coeﬃcients also has a complex root. Intuitively, this says that although we get complex numbers from real numbers by looking for roots of polynomials with real coeﬃcients, we will not get a number system larger than C by looking for roots of polynomials with complex coeﬃcients because C already contains all such roots. This is one reason why complex numbers are distinguished. The proof of this theorem involves some advanced concepts no matter how well one disguises them; the most natural proof uses the theory of holomorphic functions of one variable, and the exposition in Chapter 9 of [Knopp] is as good as any. (There is a very terse proof phrased entirely in terms of real variables in Section 3.8 of [Stillwell], and there is also one using algebraic topology in Section 5 of Chapter XI in the classic treatise of [Eilenberg-Steenrod].) It will not be a proﬁtable use of our time to attempt a proof here. We will assume the truth of this theorem and concentrate instead on its applications. The following consequences of the fundamental theorem of algebra belong to the most basic part of school mathematics. You would do well to not only know how to use them to solve problems about polynomials, but also to learn why they are true, assuming of course the fundamental theorem of algebra. We observe above all else that the fundamental theorem of algebra says every real polynomial form f (X) ∈ R[X] has a complex root. Contrast this with the polynomial form X 8 + 2, which has no real roots at all. Now we push this theorem to its logical conclusion. Theorem 5.6. Every F (X) in C[X] of positive degree is equal to a product of complex linear polynomial forms, and these linear forms are unique in the sense that if F (X) = c(X − r1 )(X − r2 ) · · · (X − rn ) and F (X) = c (X − t1 )(X − t2 ) · · · (X − tm ) for complex numbers c, c , r1 , . . . , rn , t1 , . . . , tm , then c = c , m = n, and after re-indexing the t’s if necessary, we also have ri = ti for i = 1, . . . , n. It follows that each F (X) ∈ C[X] of degree n (n > 0) has exactly n roots in C and these roots are unique up to rearrangement. (This should remind you of the uniqueness part of the fundamental theorem of arithmetic in Section 3.2 of [Wu2020a].) To prepare for the proof, we ﬁrst observe that the division algorithm (page 180) is also valid for C[X]:8 Given any polynomial forms f (X) and g(X) in C[X], with g(X) = 0, there are unique polynomial forms Q(X) and r(X) in C[X] so that f (X) = Q(X)g(X) + r(X) where r(X) is either 0 or has a degree < the degree of g(X). 8 As we pointed out earlier, the need to consider polynomials with complex (rather than real) coeﬃcients is one reason for introducing the concept of polynomial forms.

198

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

The proof is identical to the case of R[X] on page 182, because, as already remarked in that proof, it depends only on the fact that (in addition to the associativity, commutativity, and distributivity of the addition and multiplication in C) every nonzero element of C has a multiplicative inverse. Once we have the division algorithm for C[X], the proof of the factor theorem (see page 184)—originally stated for R[X]—also extends to the case of C[X]. Thus we have: A complex number r is a root of F (X) ∈ C[X] ⇐⇒ (X − r) is a factor of F (X); i.e., F (X) = (X − r)G(X) for some G(X) ∈ C[X]. Proof of Theorem 5.6. Given F (X) ∈ C[X], we ﬁrst prove the existence of the requisite complex linear forms. The fundamental theorem of algebra guarantees the F (X) has a root r1 ∈ C. By the factor theorem, F (X) = (X − r1 )F1 (X) for some F1 (X) ∈ C[X]. In turn, F1 (X) has a root r2 in C, so that F1 (X) = (X − r2 )F2 (X) for some F2 (X) ∈ C[X]. Therefore, F (X) = (X − r1 )(X − r2 )F2 (X). We repeat with F2 (X), so that after n − 2 more steps, where n = deg F (X), we arrive at F (X) = (X − r1 )(X − r2 ) · · · (X − rn )Fn (X) for some Fn (X) ∈ C[X]. Now the degree of the right side is n + deg Fn (X), while the degree of the left is of course n. Thus deg Fn (X) = 0 and Fn (X) is some (nonzero) complex number c. Thus (5.20)

F (X) = c(X − r1 )(X − r2 ) · · · (X − rn )

for some c ∈ C. (By equating the coeﬃcients of both sides, we see that c is in fact the leading coeﬃcient of F (X).) This proves the existence part of Theorem 5.6. Next, we prove the uniqueness of the linear forms (X − r1 ), . . . , (X − rn ) in (5.20). Thus suppose we also have (5.21)

F (X) = c (X − t1 )(X − t2 ) · · · (X − tm )

for some complex numbers c , t1 , . . . , tm and some positive integer m. We will prove that c = c , m = n, and after re-indexing the t’s if necessary, we also have ri = ti for i = 1, . . . , n. First, it follows immediately from (5.20) and (5.21) that c(X − r1 )(X − r2 ) · · · (X − rn ) = c (X − t1 )(X − t2 ) · · · (X − tm ). Because the equality of two polynomial forms implies they have the same degree, we have m = n (see page 176). But the equality of two polynomial forms also implies that the coeﬃcients of X n in the two forms are equal (see page 176 again); thus also c = c . Therefore we have (5.22)

(X − r1 )(X − r2 ) · · · (X − rn ) = (X − t1 )(X − t2 ) · · · (X − tn ).

Now if there is an rk (for an integer k so that 1 ≤ k ≤ n) on the left side of (5.22) that is not equal to any of the ti ’s on the right, then by setting X = rk in (5.22), we get 0 on the left but a nonzero number on the right, a contradiction. Hence, every rk on the left side of (5.22) is equal to one of the ti ’s on the right. For the same reason, every tk on the right side of (5.22) is equal to one of the ri ’s on the

5.3. FUNDAMENTAL THEOREM OF ALGEBRA

199

left. It follows that there is a collection of distinct r1 , r2 , . . . , rp for some integer p ≤ n, so that every one of the n complex numbers {r1 , r2 , . . . , rn } is equal to one of the r1 , r2 , . . . , rp , and every one of the n complex numbers {t1 , t2 , . . . , tn } is also equal to one of the same collection r1 , r2 , . . . , rp . Now consider r1 . Of course, it may happen that the total numbers of rj ’s on the left side of (5.22) equal to r1 is a(1) (a(1) being a positive integer) while the total numbers of rj ’s on the right side of (5.22) equal to r1 is b(1) (b(1) being a positive integer), and a(1) = b(1). We now prove that this does not happen. To this end, we rewrite the left side of (5.22) as (X − r1 )a(1) G(X),

where G(X) is a polynomial form and G(r1 ) = 0.

In other words, G(X) is a product of (X − r2 )’s, (X − r3 )’s, . . . , and (X − rp )’s. Similarly we may rewrite the right side of (5.22) as (X − r1 )b(1) H(X),

where H(X) is a polynomial form and H(r1 ) = 0;

i.e., H(X) is a product of (X − r2 )’s, (X − r3 )’s, . . . , and (X − rp )’s. Therefore (5.22) now becomes (5.23)

(X − r1 )a(1) G(X) = (X − r1 )b(1) H(X),

where G(r1 ) = 0 and H(r1 ) = 0 and a(1) and b(1) are positive integers. We claim that a(1) = b(1). Suppose not, and we will deduce a contradiction. Without loss of generality, we may assume a(1) < b(1); let us denote b(1) − a(1) by c(1). Then multiplying both sides of (5.23) by 1/(X −r1 )a(1) yields (by virtue of equation (5.3) on page 178): G(X) = (x − r1 )c(1) H(X). But if X = r1 , then the left side is nonzero whereas the right side is 0 (because c(1) > 0). The contradiction proves that a(1) = b(1). What is true for r1 is true for r2 , r3 , . . . , rp ; i.e., each of r2 , r3 , . . . , rp appears the same number of times in both sides of (5.22). It follows that, after a rearrangement of the ti ’s if necessary, we have r1 = t1 , . . . , rn = tn . This proves the uniqueness of the linear forms (X − r1 ), . . . , (X − rn ) in (5.20) and hence also Theorem 5.6. We can deﬁne the multiplicity of a root r of a complex polynomial form f (X) ∈ C[X] as in the case of R[x]; i.e., it is the largest positive integer m(r) so that (X − r)m(r) |f (X). Then as in Theorem 5.3 on page 185, we have the following immediate consequence of Theorem 5.6. Corollary of Theorem 5.6. If F (X) ∈ C[X] has degree n, then it has exactly n complex roots r1 , r2 , . . . , rn , counting multiplicity, and these roots are unique. If c is the leading coeﬃcient of F (X), then F (X) = c(X − r1 )(X − r2 ) · · · (X − rn ). Of course, if n = 2, then the quadratic formula for complex coeﬃcients given in the last section says more in this case: if f (X) = aX 2 + bX + c, then the two roots are, explicitly in terms of the coeﬃcients, √ −b ± b2 − 4ac . 2a If n = 3 or 4, we also get explicit formulas in terms of the coeﬃcients, cumbersome as these may be. (In Exercises 4 to 7, we give a guided tour of the derivation of the

200

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

formulas for n = 3, the Cardano formulas. The ideas for deriving the formulas for n = 4 are not essentially diﬀerent from the case of n = 3; see pp. 191–194 of [Rotman].) For a general polynomial of degree greater than 4, we now know that no such general formula in terms of the coeﬃcients can exist; for a detailed account of the history of this development that culminated in the work of Abel and Galois (see footnote on page 121) in the early nineteenth century, see [Tignol]. The corollary furnishes an example of what is known in mathematics as an existence theorem: it guarantees that there must be n complex roots without exhibiting explicitly what they are. It is still a problem of great interest in current research in numerical analysis how to devise eﬃcient (inﬁnite) procedures to get at these roots in the general case. As we observed in the proof of Theorem 5.6, the n roots r1 , . . . , rn need not be distinct. After grouping together the same roots, we clearly have the following theorem. Theorem 5.7 (Theorem on roots with multiplicity). A polynomial form F (X) ∈ C[X] of degree n has k distinct roots r1 , r2 , . . . , rp , p ≤ n, so that F (X) = c(X − r1 )m(1) (X − r2 )m(2) · · · (X − rp )m(p) where c is the leading coeﬃcient of F (X), the m(i)’s are positive integers, and m(1) + · · · + m(p) = n. The numbers r1 , . . . , rp and the positive integers m(1), . . . , m(p) are unique. Theorem 5.7 is clearly a sharpened version of Theorem 5.3 on page 185. The reasoning that proves (♠) on page 185 also shows that each m(i) is in fact the multiplicity of the root ri , for i = 1, . . . p. Finally, we consider an f (X) ∈ R[X] of degree n. Since R[X] ⊂ C[X], f (X) is also in C[X] and, as such, it is equal to a product of n linear polynomial forms in C[X]. Since f (X) is a real polynomial form, it would be desirable to have an expression of f (X) as a product of “simple” real polynomial forms. The example of x2 + 1 ∈ R[X] shows that we cannot expect such an f (X) to be a product of real linear polynomial forms. The next theorem shows what is the best possible under the circumstances. Theorem 5.8. Every f (X) ∈ R[X] is equal to a product of real linear polynomial forms and real quadratic polynomial forms that have no real roots. Note that if f (X) is a quadratic polynomial form aX 2 + bX + c with real coeﬃcients a, b, c, then Theorem 5.8 follows from the quadratic formula: if the discriminant b2 − 4ac is ≥ 0, then f (X) is the product of the following real linear forms: √ √ −b − b2 − 4ac −b + b2 − 4ac X− . f (X) = a X − 2a 2a If the discriminant is negative, then f (X) is itself a real quadratic polynomial form which has no real roots. A particularly simple consequence of Theorem 5.8 is that an odd-degree polynomial form with real coeﬃcients must have at least one real root, because one of the factors guaranteed by Theorem 5.8 must be a real linear form. This fact was already proved in Theorem 3.1 on page 122. Consequently, a cubic polynomial form g can be factored completely into a product of linear (possibly complex)

5.3. FUNDAMENTAL THEOREM OF ALGEBRA

201

polynomial forms if one can guess one real root r, for then the quadratic factor in the factorization of g can be further factored into a product of (possibly complex) linear forms by appealing to (5.18) on page 194 and Theorem 5.6. The following example gives more details of this line of reasoning. Example. Assume a cubic polynomial form f (X) = X 3 + X 2 − 4X + 6 in R[X]. By the rational roots theorem on page 187, we ﬁnd by trial and error that −3 is a root. Therefore by the factor theorem on page 184, f (X) = (X + 3)g(X) for some quadratic polynomial form g(X). Using the procedure of the division algorithm (page 180 in Section 5.1), we ﬁnd that g(X) = X 2 − 2X + 2. (Of course the uniqueness part of Theorem 5.8 gives an a priori guarantee that g(X) is a real polynomial form.) By the quadratic formula, the roots of g(X) are 1 ± i. Thus by Theorem 5.6, g(X) = (X − (1 − i))(X − (1 + i)), and therefore, f (X) = (X + 3)(X − (1 − i))(X − (1 + i)). As mentioned in connection with the corollary to Theorem 5.6, how to ﬁnd, explicitly, the linear and quadratic polynomial forms guaranteed by Theorem 5.8 for a general real polynomial form of arbitrary degree is still a challenge. Proof of Theorem 5.8. We divide the proof into three steps. Step 1. If β is a complex root of f (X), so is β. Indeed, let f (β) = 0. If f (X) = an X n + · · · + a1 X + a0 , then each ai is a real number by hypothesis. Therefore 0 = f (β) = an β n + · · · + a1 β + a0 = an (β)n + · · · + a1 β + a0

(by (5.13) on p. 191)

= an (β) + · · · + a1 β + a0

(each ai is real).

n

One recognizes that the last sum is f (β). Thus f (β) = 0. Step 2. (X − β)(X − β) is a real quadratic polynomial form for every β ∈ C. This is because (X − β)(X − β) = X 2 − (β + β)X + ββ and β + β and ββ = |β|2 are both real. Step 3. We prove Theorem 5.8 for an f (X) ∈ R[X] by the principle of complete induction (see page 182) on the degree of f (x). If the degree of f (X) is 1, then there is nothing to prove. So suppose we know that all real polynomial forms of degree 1, 2, . . . , n have the property asserted in Theorem 5.8. Let f (X) be a real polynomial form of degree n + 1, and we have to prove that f (X) has the same property. If f (X) has a real root r, then by the factor theorem, there is a real polynomial form g(X) so that f (X) = (X −r)g(X). Since the degree of g(X) is n, the induction hypothesis implies that g(X) is a product of real linear polynomial forms and real quadratic polynomial forms that have no real roots. Thus the same is true of f (X) and we are done. Therefore, we may assume that f (X) has no real roots. Let β be a complex root of f (X); so β is not real. By Step 1, β is also a root of f (X). By Step 2, the quadratic polynomial form (X − β)(X − β), to be denoted by g(X), has real coeﬃcients; i.e., g(X) ∈ R[X]. By the division algorithm of f (X) by g(X) (see page 180), f (X) = Q(X)g(X) + (aX + b)

202

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

for some real polynomial forms Q(X) and the remainder aX + b. If X = β or β, both f (X) and Q(X) are equal to 0. Thus the linear form ax+b, as a form in C[X], has two distinct roots, β and β. This contradicts the corollary of Theorem 5.6 on page 199 unless a = b = 0. Hence f (X) = Q(X)g(X). But the degree of Q(X) is n − 1, so again the induction hypothesis implies that Q(X) is a product of real linear polynomial forms and real quadratic polynomial forms that have no real roots. Since a real root of Q(x) would be a real root of f (X) (because f (X) = Q(X)g(X)) and since we are assuming that f (X) has no real roots, it follows that Q(X) has no real roots. Therefore Q(X) is a product of real quadratic polynomial forms that have no real roots. The same is therefore true of f (X). The proof of Theorem 5.8 is complete. Exercises 5.3. (1) (a) Write the quadratic polynomial form 5X 2 − X + 2 as a product of two linear polynomial forms in C[X]. (b) Write X 4 − 4X 3 + 5X 2 − 4X + 4 as a product of linear and quadratic polynomial forms in R[X]. (c) Write 3X 3 − 8X 2 − 2X + 4 as a product of real linear polynomial forms. (2) Given that −1 + i is a root of 25X 3 + 44X 2 + 38X − 12, factor this cubic polynomial form into a product of a real linear polynomial form and a real quadratic polynomial form. (3) What is the minimum degree √ √with real coeﬃcients having √ of a polynomial the following roots: 5, i5, 3 − i, − 3 + i, 3 + i, 1 + i √13 ? (4) (i) Given a cubic equation x3 +ax2 +bx+c = 0, where a, b, c are constants, let y be another symbol so that x = y + k, where k is a constant to be chosen. Show that with the choice of k = − a3 , the equation in x becomes an equation in y without a second degree term; i.e., y 3 + Ay + B = 0, for some constants A and B. (ii) Generalize part (i) to the following: a polynomial equation of any degree n ≥ 2 can be rewritten as a polynomial equation without a term of degree n − 1. (You may use the binomial theorem of the next section if you wish.) (5) (i) Prove that for any numbers a and b, the following is an identity: (a − b)3 + 3ab(a − b) = a3 − b3 .

(5.24)

(ii) Assume a cubic equation x3 + mx = n, where m and n are constants. Suppose we can choose numbers a and b so that 3ab = m, a3 − b3 = n. Then prove that (a − b) is a solution of x3 + mx = n. (iii) To solve the m . With this choice of b understood, show equations in (5.24), let b = 3a that (a, b) solves (5.24) if and only if a is a solution of (a3 )2 − n(a3 ) −

m 3 3

= 0.

(iv) Finally, prove that if a is a cube root of a solution of the quadratic m 3 equation y 2 − ny − ( m 3 ) = 0 and if b is so chosen that b = 3a , then (a − b) 3 is a solution of x + mx = n.

5.4. BINOMIAL THEOREM

203

(6) (i) Let x3 + mx = n be given. Make use of Exercise 5(iv) to show that if a+ and a− are the cube roots n 2 m 3 n 2 m 3 3 n 3 n + − + or a− = + , a+ = 2 2 3 2 2 3 then each of

a+ −

m 3a+

and

a− −

m 3a−

is a solution of the original cubic x3 + mx = n. (ii) Make use of Exercise 4(i) to write down the Cardano formulas for the roots of a general cubic equation x3 + ax2 + bx + c = 0. (7) In Exercise 6(i), each of the cube roots a+ and a− has three choices, corresponding to the fact that a nonzero number has three distinct cube roots. Therefore the totality of these cube roots in m m and a− − a+ − 3a+ 3a− appear to generate six solutions of the cubic equation x3 + mx = n. Does this contradict the corollary to Theorem 5.3 on page 199? (Be careful!) (8) Show that the two deﬁnitions of the multiplicity of the zero of a polynomial on page 129 and page 199 are equivalent.

5.4. Binomial theorem The binomial theorem gives an explicit expansion of (X + Y )n for two symbols X and Y and for any positive integer n; it generalizes the well-known identity (X + Y )2 = X 2 + 2XY + Y 2 . This theorem is more than a theorem in algebra because it also ﬁgures prominently in probability and statistics, among other ﬁelds. In the second subsection of this section, we give a brief indication why this is so. The binomial theorem and Pascal’s triangle (p. 203) The binomial coeﬃcients (p. 207) The binomial theorem and Pascal’s triangle The binomial theorem is about the expansion of (X + Y )n , where X, Y are both symbols and n is a positive integer. In this case, we assume that the addition and multiplication of the symbols are associative, commutative, and distributive; e.g., X + Y = Y + X, XY = Y X, X(X + Y ) = X 2 + XY , etc. The main theorem we are after is the following. Theorem 5.9 (Binomial theorem). For each positive integer n, n n n X n−1 Y + X n−2 Y 2 + · · · + XY n−1 + Y n , (X + Y )n = X n + 1 2 n−1 where the binomial coeﬃcients for whole numbers 0 ≤ k ≤ n are deﬁned by n n! = k k!(n − k)!

204

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

and the n factorial n! of a whole number n is deﬁned by 0! = 1, and if n > 0, then n! = 1 · 2 · 3 · · · (n − 1) · n. Note that the binomial coeﬃcients are always positive, and it is simple to check that they satisfy for all whole numbers 0 ≤ k ≤ n the identity n n (5.25) = . k n−k It will be seen that in fact they are all integers (see for example the identity (5.26) immediately following or the interpretation of these coeﬃcients in terms of counting on page 207). Mathematical Aside: The added generality of allowing X, Y in the binomial theorem to be any two symbols which add and multiply in accordance with the expected associative, commutative, and distributive laws is important. This theorem is applicable in a wide range of situations, e.g., when X, Y are commuting matrices or elements of a commutative ring. There are many proofs of this theorem. The usual proof is by induction on n. In this case, the key is the following identity: n n n+1 (5.26) + = . k−1 k k The proof of this identity is Exercise 11 in Exercises 1.3 of [Wu2020a]. We leave the induction proof of the binomial theorem as an exercise but will give an alternate proof based on counting. First, an observation about the binomial coeﬃcients. Note that the identity (5.26) on binomial coeﬃcients is the basis of the construction of Pascal’s triangle. As is well known, the latter is the triangle in which the n-th row gives the coeﬃcients of the binomial expansion (X + Y )n−1 and which is constructed, row by row, so that the two numbers on the extreme left and extreme right are simply 1 and 1 and so that each of the remaining numbers in each row is the sum of the two numbers in the preceding row immediately above it. More precisely, consider the ﬁrst 6 rows: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Thus the 6-th row, 1 5 10 10 5 1 is obtained from the preceding row 1 4 6 4 1 by ﬁrst writing down the two numbers at both ends as 1 and 1, and then each of 5, 10, 10, 5 is the sum of the two numbers immediately above it: 5 = 1 + 4,

10 = 4 + 6,

10 = 6 + 4,

5 = 4 + 1.

The reason that this is the correct way to write down all the coeﬃcients of the binomial expansions (X + Y )n as n runs through 1, 2, . . . is the following. Let us

5.4. BINOMIAL THEOREM

205

scrutinize how the coeﬃcients of (X + Y )5 are obtained from those of (X + Y )4 . By the binomial theorem, the coeﬃcients of (X + Y )4 are 4 4 4 4 4 , , , , 0 1 2 3 4 and the coeﬃcients of (X + Y )5 are 5 5 5 , , , 0 1 2

5 , 3

5 , 4

5 . 5

Of course the numbers at both ends are just 1 and 1: 5 5 = = 1. 0 5 According to identity (5.26), the remaining four numbers in the latter row are obtained from the former by 5 4 4 5 4 4 = + , = + , 1 0 1 2 1 2 4 5 4 4 5 4 . = + , = + 4 3 2 3 4 3 Evaluating the binomial coeﬃcients, we obtain an explanation of the construction of the 6-th row of the Pascal triangle 5 = 1 + 4, 10 = 6 + 4,

10 = 4 + 6, 5 = 4 + 1.

Schematically, that is

1

1 4 6 4 1 @ @ @ @ R @ @ R @ R @ R 5 10 10 5

1

Of course the reasoning is perfectly general and works for any row of the Pascal triangle. Finally, we are ready to prove the binomial theorem. Proof of the binomial theorem. We will analyze in detail the proof of 5 5 5 5 5 5 4 3 2 2 3 (X + Y ) = X + X Y + X Y + X Y + XY 4 + X 5 . 1 2 3 4 Once this is understood, the proof of the general case will become an afterthought. To better explain the reasoning of this special case, we consider the seemingly more complicated expansion (X1 + Y1 )(X2 + Y2 )(X3 + Y3 )(X4 + Y4 )(X5 + Y5 ) where the Xi and Yj are symbols for i, j = 1, . . . , 5, which we assume to satisfy the associative, commutative, and distributive laws when added or multiplied. After we have obtained the expansion of (X1 + Y1 ) · · · (X5 + Y5 ), then by letting Xi = X and Yj = Y for all i, j, we get back the original expansion (X + Y )5 . The reason for looking at the more complicated situation will be apparent presently.

206

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

It is straightforward to see that (X1 + Y1 )(X2 + Y2 )(X3 + Y3 )(X4 + Y4 )(X5 + Y5 ) is equal to the sum of all all all all all all

the the the the the the

terms terms terms terms terms terms

containing containing containing containing containing containing

only the X’s and exactly 1 of the Y ’s exactly 2 of the Y ’s exactly 3 of the Y ’s exactly 4 of the Y ’s only the Y ’s.

and and and and

For deﬁniteness, let us ﬁnd out the number of terms which contain exactly three Y ’s. The terms which contain, for example, Y1 , Y3 , and Y5 would have to take X2 from the factor (X2 + Y2 ) and X4 from the factor (X4 + Y4 ). Since we assume the multiplication of all the X’s and Y ’x is commutative, there is thus exactly one such term, namely, X2 X4 Y1 Y3 Y5 . Similarly, there is only one term which contains Yi , Yj , Yk for any given triple of indices i, j, k. Therefore, to determine the number of terms which contain exactly three of the Y ’s, it suﬃces to determine how many distinct ways there are of writing down Yi Yj Yk where i, j, k are distinct indices running through 1, 2, . . . , 5. For ease of keeping count for the moment, let us ﬁrst assume that the order of Yi , Yj , Yk in each Yi Yj Yk matters, so that Yi Yj Yk is considered to be diﬀerent from Yk Yi Yj or Yj Yk Yi . Let E be the set of all the Yi Yj Yk , where i, j, k are distinct and the order Yi , Yj , Yk matters. We want the number of elements in E. The reasoning is well known. Once we write down a Yi (and there are 5 possibilities for the value of i), the number of possibilities for the value of j is limited to 4 because we cannot repeat the already chosen value of i. Furthermore, once i, j have been written down, the number of possibilities for the value of k is down to 3 because we cannot repeat either of the chosen values of i and j. Thus there are exactly 5 × 4 × 3 elements in E. But now, we remember that multiplication among symbols is commutative so that among the elements in E, many are the same. For example, Y1 Y2 Y3 = Y2 Y3 Y1 = Y3 Y1 Y2 = Y1 Y3 Y2 = · · · . Thus we should identify all the Yi Yj Yk ’s in E which have the same indices i, j, k, regardless of order. There are 3! of these because, again, if we enumerate them one by one, the ﬁrst index has 3 possibilities, the second has only 2 (since it cannot repeat the ﬁrst), and there is only 1 possibility for the third index (because it cannot repeat the already chosen ﬁrst two). This means that the set E of 5 × 4 × 3 elements is partitioned into equal groups of 3! elements, where all the elements in each group are equal to each other because multiplication is commutative. The total number of such groups is therefore 5 5 × 4 × 3 × 2! 5! 5×4×3 = = = . 3 3! 3!2! 3!2! Therefore the total number of distinct ways of writing down YiYj Yk , where the indices i, j, k are distinct and their order does not matter, is 53 . Consequently, the total number of terms containing exactly 3 of the Y ’s in the expansion of (X1 + Y1 )(X2 + Y2 )(X3 + Y3 )(X4 + Y4 )(X5 + Y5 )

is also 53 .

5.4. BINOMIAL THEOREM

207

A little reﬂection reveals that this reasoning is perfectly general. The total number of terms containing exactly k (0 ≤ k ≤ 5) of the Y ’s in this expansion is

5 . k Finally, we let X1 = X2 = X3 = X4 = X5 = X and Y1 = Y2 = Y3 = Y4 = Y5 = Y . Then we get the binomial expansion of (X + Y )5 as 5 5 5 5 X 4Y + X 3Y 2 + X 2Y 3 + XY 4 + X 5 . X5 + 1 2 3 4 The reasoning when (X + Y )5 is replaced by (X + Y )n for any positive integer n is no diﬀerent. The proof of the binomial theorem is complete. The binomial theorem is one of the foundational results in mathematics. It has numerous applications in other scientiﬁc disciplines, e.g., in ﬁnite probability. The most substantial application we have made of this theorem in these volumes is the fact that if a > 1, then an becomes arbitrarily large as the positive integer n becomes arbitrarily large (see (d) on page 158). We recall this argument: let a = 1 + b so that b > 0; then n 2 n 3 n n b + b + · · · + bn . a = (1 + b) = 1 + nb + 2 3 Since each term in the binomial expansion on the right is positive, we see that an > 1 + nb. If any positive number m is given, we now see why an > m for all suﬃciently large n. Indeed, by letting n be any positive integer exceeding m b , we would have an > 1 + m, as desired. Concerning the last assertion about letting “n be any positive integer exceeding m ”, we may take it on faith at this point even when b is irrational. But eventually b we will take a close look at this possibility; see the discussion in Section 2.4 of [Wu2020c], especially the remark after the proof of Theorem 2.13 (Archimedean Property) in [Wu2020c]. The binomial coeﬃcients We now give a brief illustration of how the binomial coeﬃcients arise naturally in a context other than the binomial theorem. Consider the following problem: In how many distinct ways can three balls of the same color be put into 5 numbered boxes if each box holds only one ball? We ﬁrst tackle the simpler problem of placing three balls of distinct colors — let us say red, white and blue — into 5 numbered boxes if each box holds only one ball. For this simpler problem, the usual explanation is this: the red ball has 5 choices (i.e., can be put into any of the 5 boxes), the white ball then has only 4 choices (i.e., can be put into any of the remaining 4 boxes) and then the blue ball has only 3 choices. So the answer is that there are 5 × 4 × 3 ways. This explanation is a bit too glib because it seems to be biased against both the blue and white balls in that these balls are both given fewer choices than the red ball; if so, is this an accurate way of keeping track of the counting? Let us do it diﬀerently. Call each placement of the r (red), w (white), and b (blue) balls into the 5 boxes an event. Our problem is to count the total number

208

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

of events. We do so by assigning each event a code consisting of three distinct integers each of which being one of {1, 2, 3, 4, 5}: the ﬁrst integer (from the left) is the box number of the r ball, the second integer is the box number of the w ball, and ﬁnally the last integer is the box number of the b ball. For example, suppose the event consists of a w ball in box 1, r ball in box 3, and b ball in box 4, as shown:

W 1

2

R

B

3

4

5

Then the code of this event is 314. It is straightforward to verify that there is a bijection between the total number of events and the total number of codes. Hence, instead of counting the total number of events, it suﬃces to count only the total number of codes by listing them systematically, without repetition. Suppose the ﬁrst integer is 3, as above. Then the second integer cannot be 3 because each box holds only one ball. Thus the second integer can only be one of 1, 2, 4, and 5; i.e., there are only 4 possibilities for the second integer. The ﬁrst two integers of the code are therefore 31, 32, 34, or 35. In the case of 31 (as in 314), then the third integer can only be one of 2, 4, and 5; i.e., there are only 3 possibilities for the third integer. Similarly for 32, 34, and 35. Reasoning in this way, we arrive at the fact that the following are the total number of codes whose ﬁrst integer is 3:

3 XXX # c XXX # c XXX # c XXX # c XXX # c 31 32 34 35

312

J J

314

J

J J 315

321

J

J

J J J 324 325

J J

J

J

J 341 342 345

J

J

J

J

J 351 352 354

This listing explains why the total number of codes is 4 × 3 if the ﬁrst integer is 3. Now if the ﬁrst integer is not 3 but 1, 2, 4, or 5, then the same reasoning prevails. It follows that the total number of codes (where the ﬁrst integer can be 1, 2, 3, 4, or 5) is 5 × 4 × 3. This is the answer for the simpler problem where the balls are of distinct colors. Now we return to our original problem where the balls are all of the same color. We will use the case of distinct-colored balls as our starting point in order to make conclusions about the case of same-colored balls. In the above example of 314 for balls with colors R, W, and B, what happens now is that the following events will all be considered to be the same if the colors are removed from the balls:

5.4. BINOMIAL THEOREM

W

R

B

W

B

R

R

W

B

R

B

W

B

W

R

209

B R W 1 2 3 4 5 Of course we can see that there are 6 such events, but what is the reasoning? It is this: since we are staying with these colored balls in boxes 1, 3, and 4, we are asking: In how many distinct ways can three balls of distinct colors be put into 3 numbered boxes if each box holds only one ball? We have already done a similar problem above and know that the answer is 3 × 2 × 1 = 3!. This is then the number of events that will be indistinguishable from the event 314 if all the colored balls are replaced by balls of the same color. What is true for the event 314 is true for all other events. Thus if all the colored balls are now replaced by balls of the same color, then all the events of distinct-colored balls will be divided into equal groups of 3! events that will be indistinguishable from each other when the colors on the balls are removed. It follows that among the 5 × 4 × 3 events of distinct-colored balls, there are only 5×4×3 3! distinct events once the colors of the balls are removed. The answer to our original problem is therefore 5 5 × 4 × 3 × 2! 5! 5×4×3 = = = . 3 3! 3! × 2! 3! × 2!

We see how the binomial coeﬃcient 53 naturally emerges. The general case of placing k balls of the same color into n numbered boxes is similar and will be left as an exercise. Exercises 5.4. (1) Prove that there are 10! total numbers of seconds in 6 weeks. (2) This exercise basically asks for a proof of the binomial theorem by induction, but instead of the usual general argument, you are asked to write it up in the way speciﬁed below. Assume you know the following is true: 4 4 4 (X + Y )4 = X 4 + X 3Y + X 2Y 2 + XY 3 + Y 4 . 1 2 3 (You of course know that the coeﬃcients are actually 1, 4, 6, 4, 1, but you are not allowed to use this fact.) Assume also that you know that for all integers 0 ≤ k ≤ n, n n n n = 1, = n, and = 0 1 k n−k and

n n n+1 + = . k−1 k k

210

5. POLYNOMIAL FORMS AND COMPLEX NUMBERS

Then using these facts alone, prove 5 5 5 5 5 5 4 3 2 2 3 (X + Y ) = X + X Y + X Y + X Y + XY 4 + X 5 . 1 2 3 4 (3) Prove the following identity: for all whole numbers p, k and for all positive integers n, n−1 p + k n + p = . p n−1 k=0

(This is one of the identities discovered by the Chinese mathematician Chu Shih-Chieh (= Zhu Shijie) around 1300; it was also used by Fermat 1 to evaluate 0 xp dx around 1630. The identity can also be generalized to all real values of p and is then known as one of the Chu-Vandermonde identities. See pp. 59–60 of [Askey].) (4) Prove that n n n n n + + + ···+ + = 2n . 0 1 2 n−1 n (5) (a) Suppose you have k balls of distinct colors and n numbered boxes 1, 2, . . . , n, where k ≤ n. In how many distinct ways can the k balls be put in these n boxes if each box can hold only one ball? (b) Suppose you have k balls of the same color and n numbered boxes 1, 2, . . . , n, where k ≤ n. In how many distinct ways can the k balls be put in these n boxes if each box can hold only one ball? (6) (a) Which of the following is bigger? (1.01)48 or 1.48 ? (NO calculators.) (b) Which of the following is bigger? (1.01)25 or 1.28 ? (NO calculators.) (7) (a) Suppose you have three balls of distinct colors and ten boxes numbered 1, 2, . . . , 10 and each box holds only one ball. In how many distinct ways can the balls be put in these boxes so that the three balls are in adjacent boxes? (b) With the assumption as in (a), the probability of placing A , where A is the balls in adjacent boxes is by deﬁnition the quotient B the number of distinct ways the three balls can be put in adjacent boxes and B is the number of distinct ways the balls can be put in the ten boxes. Compute this probability. (8) (a) A whole number is said to be a ﬁve-digit number if it has ﬁve digits and if the leading digit on the left is nonzero. How many ﬁve-digit numbers with no repeated digits are there? (Be careful.) (b) The probability of a ﬁve-digit number having no repeated digits is by deﬁnition the A , where A is the total number of ﬁve-digit numbers with no quotient B repeated digits and B is the total number of all ﬁve-digit numbers. If you write down a ﬁve-digit number at random, what is the probability that it has no repeated digits? (c) What is the probability that a seven-digit number has no repeated digits? (The meaning of this “probability” can be extrapolated from part (b).) (9) How many distinct ways are there of painting 30 numbered boxes so that 12 of them will be red, 7 of them white, 3 of them blue, 2 of them pink, and 6 of them purple?

5.4. BINOMIAL THEOREM

211

(10) (a) Explain why, with an initial deposit of P dollars and an interest rate of r (think of r as 5%, for instance) compounded n times annually, the amount of money in the account after k years is P (1+ nr )nk (see Exercise 12 on page 174) for the relevant deﬁnitions). (b) Explain why this amount is always greater than the amount in the account after k years if the interest rate is the simple interest rate r (i.e., compounded only once a year). (c) Explain why, if m and n are positive integers and m < n, r mk ) < P (1 + nr )nk , where r is assumed to be positive. In other P (1 + m words, the more often the interest is compounded during the year, the more interest there will be at the end of the year. (For part (c), you will have to use calculus or something similar.) (11) Let X be a symbol. Prove that (X − 1)2 is a factor of the polynomial form X n − n(X − 1) − 1 for

any integernn > 1 by applying the binomial theorem to expand X n as (X − 1) + 1 . (This is essentially Exercise 8 on page 61.)

CHAPTER 6

Basic Theorems of Plane Geometry This chapter is devoted to a systematic exposition of plane Euclidean geometry that was started in Chapters 4 and 5 of [Wu2020a]. It gives a fairly comprehensive introduction to the geometry of the triangle and the circle. The ﬁrst task is to prove the SSS criterion of congruence for triangles; together with the SAS and ASA criteria (see page 218), SSS completes the collection of basic facts that standard high school geometry textbooks use to prove most of the geometric theorems. The reader may wish to at least skim through Chapter 8 at this point for a general orientation on some of the key issues surrounding the teaching of high school geometry. In Section 6.3 on page 229, we will further elaborate on the pedagogical ramiﬁcations of having all three criteria—SAS, ASA, and SSS—available. The next task is to ﬁll in the two gaps that were purposely left open in [Wu2020a]; we are referring to the missing proofs of the following theorems. The ﬁrst, the fundamental theorem of similarity (FTS), is a basic fact already used, implicitly and explicitly, all through Chapters 5–6 of [Wu2020a] and Chapter 1 of this volume. For the statement of FTS, we recall some standard notation: if A and B are two points in the plane, AB (respectively, LAB ) will denote the segment (respectively, the line) joining them, and RAB will denote the ray on the line LAB issuing from A that contains B. By abuse of language, we also say two segments DE and BC are parallel, DE BC, if their corresponding lines are parallel; i.e., LDE LBC . Theorem G10 (FTS). Let ABC be given, and let D, E be points on the rays RAB and RAC , respectively, with neither being equal to A. If |AE| |AD| = = r, |AB| |AC| then DE BC and

|DE| |BC|

= r.

A

A

@ @ @ @ @E D @ @C B

@ @ @ @ @ @ @C B @ @E D @ r>1

r C > B. In symbols, A ∗ C ∗ B. This deﬁnition is independent of the particular number line structure on (see Section 4.1 of [Wu2020a]). Observe that A ∗ B ∗ C if and only if C ∗ B ∗ A. The segment AB is the collection of all the points C between A and B together with A and B themselves.

6.1. REVIEW

215

For the next assumption, we recall that the plane is said to be the disjoint union of three sets U , V , and W if they are disjoint and if their union is the whole plane. (L4) (Plane separation) A line L separates the plane into two nonempty subsets, H+ and H− , called the half-planes of L. The half-planes H+ and H− satisfy the following two properties: (i) The plane is the disjoint union of H+ , H− , and L, and the half-planes H+ and H− are convex. H− H+ L (ii) If two points A and B in the plane belong to diﬀerent half-planes, then the line segment AB must intersect the line L. s q B H+ L (L5) To each pair of points A and B of the plane, we can assign a number dist(A, B), called the distance between A and B so that (i) dist(A, B) = dist(B, A) and dist(A, B) ≥ 0. Furthermore, dist(A, B) > 0 ⇐⇒ A = B. (ii) Given a ray with vertex O and a positive number r, there is a unique point B on the ray so that dist(O, B) = r. (iii) Let O and A be two points on a line L so that dist(O, A) = 1, and let O and A be the 0 and 1 of a number line on L (as in (L3)). Then for any two points P and Q on L, dist(P, Q) coincides with the length of the segment P Q on this number line. (iv) If A, B, C are collinear points and C is between A and B, then dist(A, B) = dist(A, C) + dist(C, B). The next assumption introduces the concept of the degree of an angle. Recall that an angle ∠AOB is a region in the plane rather than two rays with a common vertex (see page 351). More precisely, we are given that LOA and LOB intersect at O. Then ∠AOB is either the intersection of the closed half-plane1 of LOA containing B and the closed half-plane of LOB containing A (this angle is usually referred to as the convex angle) or the complement of this intersection together with the two rays ROA and ROB (this is usually referred to as the nonconvex angle). Next, we need the concept of adjacent angles. We say two angles ∠AOC and ∠COB, with a common side ROC , are adjacent angles with respect to ∠AOB if C belongs H−

1 See

q A

page 352 for a deﬁnition.

216

6. BASIC THEOREMS OF PLANE GEOMETRY

to ∠AOB (as a region in the plane; let it be stated explicitly that in this case, ∠AOB can denote either the convex angle or the nonconvex angle), and ∠AOC and ∠COB are subsets of ∠AOB. For example, if ∠AOB is understood to denote the convex angle, then in context, ∠AOC has to be the convex (shaded) subset on the left rather than the nonconvex subset indicated by the arc on the right. A

A

O

C

O

C B

B

On the other hand, if ∠AOB is understood to denote the nonconvex angle, then in context (looking at the drawings below), ∠AOC has to be the shaded subset on the left rather than the nonconvex subset indicated by the arc on the right.

A

C

A O

C

O

B

B

Adjacent angles ∠AOC and ∠COB (with respect to ∠AOB) are the analogs— among angles—of segments AC, CB so that A, B, C are collinear and C is between A and B. (L6) To each angle ∠AOB, we can assign a number |∠AOB|, called its degree, so that (i) 0 ≤ |∠AOB| ≤ 360◦ , where the small circle ◦ is the abbreviation of “degree”. (ii) Given a ray ROB and a number x so that 0 < x < 360 and x = 180, let one of the two closed half-planes of the line LOB be speciﬁed. Then there is a unique ray ROA lying in the speciﬁed closed half-plane of LOB so that |∠AOB| = x◦ , where ∠AOB denotes the convex angle if x < 180 and the nonconvex angle if x > 180. (In the following picture, the speciﬁed closed half-plane is the closed upper half-plane of LOB .)

A

O

x°

A

x° B

O B

6.1. REVIEW

217

(iii) |∠AOB| = 0◦ ⇐⇒ ∠AOB is the zero angle; |∠AOB| = 180 ⇐⇒ ∠AOB is a straight angle; |∠AOB| = 360◦ ⇐⇒ ∠AOB is the full angle at O. (iv) If ∠AOC and ∠COB are adjacent angles with respect to ∠AOB, then ◦

|∠AOC| + |∠COB| = |∠AOB|. Assumption (L6) deserves a special comment: part (i) asserts the existence of angles of any degree from 0 to 360, and this existence assumption is ultimately what guarantees the existence of a rotation of any degree around a given point. (L7) The basic isometries (rotations, reﬂections, and translations) have the following properties: (i) A basic isometry maps a line to a line, a ray to a ray, and a segment to a segment. (ii) A basic isometry preserves lengths of segments and degrees of angles. (L8) (Crossbar axiom) Given a convex angle AOB, for any point C in ∠AOB not equal to O, the ray ROC intersects the segment AB (indicated as point D in the following ﬁgure). A @ @D O ``` C ``` ``` @ ``@ ` @ ``` B The geometric theorems We now list the theorems whose numbers are prefaced by the letter “G” (G = Geometry). They come in two groups. There are ﬁfteen in the ﬁrst group: G1–G9, G12–G15, and G18–G19. What these ﬁfteen theorems have in common is the fact that their proofs depend only on the concept of congruence (= ﬁnite composition of basic isometries). The concept of similarity (= ﬁnite composition of congruences and dilations) is never used in their proofs and therefore their validity has been ﬁrmly established. We also recall that Theorems G1–G3 are needed to show that the concept of reﬂection is well-deﬁned. (The fact that rotation is well-deﬁned is not in question, and the fact that translation is well-deﬁned follows from the parallel postulate and assumptions (L3) and (L4).) On the other hand, the validity of the remaining nine theorems, G10–G11, G16–G17, and G20–G24, does depend on the validity of Theorem G10, which is as yet unproven.2 Theorem G1. Let O be a point not contained on a line L, and let be the rotation of 180◦ around O. Then maps L into a line parallel to itself; i.e., (L) L. As remarked earlier (page 214), this theorem guarantees that given a point P not lying on a given line L, there exists a line passing through P and parallel to L. 2 The proof of Theorem G25 given in Section 5.3 of [Wu2020a] depends on Theorem G10. However, in Exercise 1 on page 227, you will be asked to give a proof of Theorem G25 that makes use of the concept of congruence only. So Theorem G25 does not depend on Theorem G10.

218

6. BASIC THEOREMS OF PLANE GEOMETRY

Theorem G2. Two lines perpendicular to the same line are either identical or parallel to each other. Theorem G3. A transversal of two parallel lines that is perpendicular to one of them is also perpendicular to the other. Theorem G4. Opposite sides of a parallelogram are equal. Theorem G5. Let be a line which is neither parallel to LAB nor equal to LAB . Then the translation TAB maps to a line parallel to itself. Theorem G6. (a) Every congruence is an isometry; it preserves lines and the degrees of angles, and it is also a bijection. (b)The inverse of a congruence is a congruence. (c) Congruences areclosed under composition in the following sense: if F and G are congruences, so is F ◦ G. Theorem G7. If for two triangles ABC and A B C , |∠A| = |∠A |,

|∠B| = |∠B |,

|∠C| = |∠C |

|AB| = |A B |,

|AC| = |A C |,

|BC| = |B C |,

and

then ABC ∼ =” stands for “is congruent to”). = A B C (“∼ Theorem G8 (SAS). Assume two triangles ABC and A B C so that |∠A| = |∠A |, |AB| = |A B |, and |AC| = |A C |. Then the triangles are congruent. Theorem G9 (ASA). Assume two triangles ABC and A B C so that |AB| = |A B |, |∠A| = |∠A |, and |∠B| = |∠B |. Then the triangles are congruent. Theorem G12. Let O be a point on a line L, and let be the rotation of 180◦ around O. Then interchanges the two half-planes of L. Theorem G13. Let L and L be two lines meeting at a point O, and P , Q (respectively, P , Q ) are points lying on opposite half-lines of L (respectively, L ) determined by O. Then |P O| = |OQ| and |P O| = |OQ | ⇐⇒ P P QQ is a parallelogram. L @ " " " @ " P @ " Q " @ " " @O " " @ "" @ @Q P " " @ " @

L

Theorem G14. A quadrilateral is a parallelogram ⇐⇒ it has one pair of sides which are equal and parallel.

6.1. REVIEW

219

Theorem G15. Let ABC be given, and let D and E be midpoints of AB and AC, respectively. Then DE BC and |BC| = 2|DE|. A @ @ @ @E D @ @ @ @C B Theorem G18. Alternate interior angles of a transversal with respect to a pair of parallel lines are equal. The same is true of corresponding angles. Theorem G19. If the alternate interior angles of a transversal with respect to a pair of distinct lines are equal, then the lines are parallel. The same is true of corresponding angles. As mentioned above, the validity of the next ten theorems hinges on the validity of the ﬁrst (FTS). Theorem G10 (FTS). Let ABC be given, and let D, E be points on the rays RAB |AE| and RAC , respectively, neither equal to A or B. If |AD| |AB| = |AC| and their common value is denoted by r, then DE BC

and

|DE| = r. |BC|

A

A

@ @ @ @ @E D @ @C B r1

Theorem G11 (FTS*). Let ABC be given, and let D be a point on the ray RAB not equal to A or B. Let the line parallel to BC and passing through D intersect the line LAC at E. Then E lies in the ray RAC and |AD| |AE| |DE| = = . |AB| |AC| |BC| Theorem G16. Dilations map segments to segments. More precisely, a dilation D maps a segment P Q to the segment joining D(P ) to D(Q). Moreover, if the line LP Q does not pass through the center of the dilation D, then the line LP Q is parallel to the line containing D(P ) and D(Q).

220

6. BASIC THEOREMS OF PLANE GEOMETRY

Theorem G17. Let D be a dilation with center O and scale factor r. Then: (a) D is a bijection. In fact its inverse is the dilation with the same center O but with a scale factor 1/r. (b) For any segment AB, |D(AB)| = r|AB|. (c) D maps rays to rays, angles to angles, and preserves degrees of angles. Theorem G20. Given two triangles ABC and A B C , their similarity, i.e., ABC ∼ A B C , is equivalent to the following equalities: |∠A| = |∠A |,

|∠B| = |∠B |,

|∠C| = |∠C |

and |AB| |AC| |BC| = = . |A B | |A C | |B C | Theorem G21 (SAS for similarity). Assume two triangles ABC and A B C . If |∠A| = |∠A | and |AB| |AC| = , |A B | |A C | then ABC ∼ A B C . Theorem G22 (AA for similarity). Two triangles with two pairs of equal angles are similar. Theorem G23 (Pythagorean theorem). If the lengths of the legs of a right triangle are a and b and the length of the hypotenuse is c, then a2 + b2 = c2 . Theorem G24 (Converse of Pythagorean theorem). If triangle ABC satisﬁes |CB|2 + |CB|2 = |AB|2 , then |∠C| = 90◦ . Theorem G25 (HL). Two right triangles with equal hypotenuses and a pair of equal legs are congruent. 6.2. SSS and ﬁrst consequences The goal of this section is to prove the standard theorem that two triangles whose three pairs of corresponding sides are equal must be congruent (SSS). The three theorems, SAS (Theorem G8), ASA (Theorem G9), and now SSS, form the cornerstone of the discussion of triangles in the high school geometry curriculum. Because the proof of SSS requires only Theorems G26 and G27 and because the proofs of these two theorems in turn require only assumptions (L1)–(L8), the proofs of these three basic theorems (SAS, ASA, and SSS) can be given right after the proof of Theorem G7 if we so desire. This has a bearing on the teaching of geometry, as we will explain in the next subsection, Section 6.3.3 Characterization of the perpendicular bisector of a segment (p. 221) Applications of the characterization (p. 223)

3 We should add that Theorems G26 and G27 are of independent interest and should not be regarded as things to be discarded as soon as SSS is proved. On the contrary, this material has to be done regardless of whether we want to prove the SSS theorem or not.

6.2. SSS AND FIRST CONSEQUENCES

221

Characterization of the perpendicular bisector of a segment First, we give some deﬁnitions. Recall that a triangle is isosceles if it has (at least) two equal sides. If in ABC, |AB| = |AC|, we will, for convenience, refer to A as the top vertex and ∠A as the top angle. The angles ∠B and ∠C are the base angles, and side BC is the base. For a general triangle, the line passing through A and perpendicular to BC is called the altitude on BC or sometimes the altitude from A. Sometimes, the segment AD, where D is the point of intersection of this perpendicular line with BC, is also called the altitude from A. This is a typical example of the same term being used in school mathematics for two diﬀerent purposes. A QQ Q Q Q Q Q Q Q B D C

A a Qa Qaa Q aa Q aa Q aa Q aa Q aa Q Q a D B C

There is no such ambiguity about the next deﬁnition, however. The segment joining a vertex to the midpoint of the opposite side is by deﬁnition a median of the triangle. Theorem G26. (a) Isosceles triangles have equal base angles. (b) In an isosceles triangle, the perpendicular bisector of the base, the angle bisector of the top angle, the median from the top vertex, and the altitude on the base all coincide. In the statement of part (b) of the theorem, “coincide” is, strictly speaking, the wrong word to use. The correct statement should be that the perpendicular bisector of the base, the angle bisector of the top angle, the median from the top vertex, and the altitude on the base all lie on the same line. (Recall that the angle bisector is a ray and the median is a segment, while the other two are lines.) This kind of abuse of language is, however, common in the subject and in general does no harm. To put assertion (b) in context, observe that in a general triangle, the lines containing the following objects are distinct: the angle bisector AK of angle ∠A, the median AA and the altitude AD on BC, and the perpendicular bisector P A of the side BC opposite A. (See the picture below.) It is therefore remarkable that in the case of an isosceles triangle, all four lines issuing from the top vertex would collapse into one line.

A P

B

D K Aʹ

C

222

6. BASIC THEOREMS OF PLANE GEOMETRY

Proof. Let |AB| = |AC| in ABC, and let the angle bisector of the top angle ∠A intersect the base BC at K.4 It is immediately seen that ABK ∼ = ACK on account of SAS (page 218). Therefore, |∠B| = |∠C| because corresponding angles of congruent triangles are equal. This proves part (a), and part (b) also follows easily from this congruence. However, it is instructive to give a second proof without using SAS but using a reﬂection directly. Such a proof is more transparent because it shows a little more of why the theorem is true. Therefore let Λ be the reﬂection across the angle bisector LAK . Since |∠BAK| = |∠KAC| and since reﬂections preserve the degrees of angles, Λ maps the ray RAB to the ray RAC by virtue of part (ii) of assumption (L6) (page 216). Let Λ(B) = B . Since A is on the line of reﬂection, Λ(A) = A so that |AB| = |Λ(AB)| = |AB |. By hypothesis, |AB| = |AC|, and therefore |AC| = |AB |. Since C and B are both on the same ray RAC , we have B = C (see the uniqueness statement in (L5)(ii) on page 215); i.e., Λ(B) = C. Now it is also true that Λ(K) = K and Λ(A) = A because K and A lie on the line of reﬂection of Λ. Since Λ maps segments to segments, Λ(BK) = CK and Λ(BA) = CA, and since Λ also maps a ray to a ray, we have Λ(∠B) = ∠C. Thus |∠B| = |∠C| because Λ preserves degrees of angles. This proves part (a).5 For part (b), observe that since LAK is the line of reﬂection and Λ(B) = C, Λ(∠AKB) = ∠AKC

and Λ(BK) = CK.

Therefore |∠AKB| = |∠AKC|, and since ∠BKC is a straight angle, we see that |∠AKB| = |∠AKC| = 90◦ . Since also |BK| = |CK|, LAK is the perpendicular bisector of BC. Since LAK is, by construction, also the angle bisector of ∠A, every statement in part (b) follows. The proof is complete. The following characterization of the perpendicular bisector of a segment is a direct consequence of Theorem G26. It will be found to be extremely useful. Theorem G27. The perpendicular bisector of a segment is the set of all points equidistant from the two endpoints of the segment. Proof. Given a segment BC, let its midpoint be M and its perpendicular bisector be . Furthermore, let E be the set of all the points K in the plane equidistant from the two endpoints B and C; i.e., |KB| = |KC|. We have to prove that = E.

K H HH HH HH r HC B M r D 4 The

217).

fact that the angle bisector of ∠A intersects BC is implied by the crossbar axiom (page

5 Note that one can use SAS to prove ABK ∼ ACK to arrive at the same conclusion = that |∠B| = |∠C|. A similar remark applies to the proof of Theorem G27.

6.2. SSS AND FIRST CONSEQUENCES

223

We ﬁrst show ⊂ E. Let D ∈ , i.e., let D be a point on the perpendicular bisector of BC, and we have to show that D ∈ E, i.e., D is equidistant from B and C. By the deﬁnition of reﬂection, the reﬂection Λ across maps B to C and D to D, so that Λ(BD) = CD and, therefore, |BD| = |Λ(BD)| = |CD|. In other words, D is equidistant from B and C. Consequently, D ∈ E and ⊂ E. Conversely, we show E ⊂ . Let K ∈ E, i.e., let K be equidistant from B and C, and we must show K lies on the perpendicular bisector of BC. Now M ∈ , and if K = M , there is nothing to prove. So let K = M . Then KBC is isosceles with |KB| = |KC|, and therefore by Theorem G26(b), the median KM of KBC is also the perpendicular bisector of BC. Thus K ∈ , and E ⊂ , as desired. Applications of the characterization As a ﬁrst application of Theorem G27, we prove the SSS criterion of triangle congruence, which is the main goal of this section. Theorem G28 (SSS). Assume triangles ABC and A B C . If the three sides are pairwise equal, i.e., |AB| = |A B |, |AC| = |A C |, and |BC| = |B C |, then the triangles are congruent. Proof. As in the case of the SAS and ASA theorems, we break up the proof into three steps, going from a special case to the most general. Case I. The two triangles satisfy in addition, A = A and B = B. Case II. The triangles satisfy in addition, A = A . Case III. The general case. Case I. In this case, there are two possibilities. (1) C and C are in opposite half-planes of LAB . (2) C and C are in the same half-plane of LAB . If (1) holds, then the fact that |AC| = |A C | and |BC| = |B C | implies that A and B are each equidistant from C and C . By Theorem G27 (and, of course, also (L1) on page 214), LAB is the perpendicular bisector of CC . Generically, we have the following pictures:

A = A

C HH HH HH HH B =B @ @ @ @ C

C H @HH @ H @ HH HH A = [email protected] B =B C

Let Λ be the reﬂection across LAB ; then by the deﬁnition of a reﬂection (see page 354), Λ(C ) = C while Λ(A ) = A = A and Λ(B ) = B = B. Therefore Λ(A B C ) = ABC, and the triangles are congruent, as desired. If (2) holds instead, then we claim C = C , so that ABC = A B C . To prove the claim, again let Λ be the reﬂection across LAB . Let Λ(A ) = A0 ,

224

6. BASIC THEOREMS OF PLANE GEOMETRY

Λ(B ) = B0 , and Λ(C ) = C0 . Thus, Λ(A B C ) = A0 B0 C0 . Of course, A0 = A and B0 = B, but now C0 and C are on opposite half-planes of LAB . The preceding discussion of possibility (1) implies that Λ(A0 B0 C0 ) = ABC. But Λ ◦ Λ = identity, so ABC = Λ(A0 B0 C0 ) = Λ(Λ(A B C )) = A B C , exactly as claimed, and Case I is completely proved. Case II. The two triangles have one pair of corresponding vertices in common, say, A = A . A = A qHH HH HH H

B HH

B

Let be the rotation around A that brings the ray RAB to the ray RAB . Because |AB| = |A B | by hypothesis, (B ) = B by the uniqueness statement of (L5)(ii) (see p. 215). Therefore if we denote (A ), (B ), and (C ) by A0 , B0 , and C0 , respectively, then (A B C ) = A0 B0 C0 and the triangles ABC and A0 B0 C0 have two pairs of corresponding vertices in common; namely, A = A0 and B = B0 . By Case I, there is a congruence ϕ so that ϕ(A0 B0 C0 ) = ABC. Hence ABC = ϕ(A0 B0 C0 ) = ϕ( (A B C )) = (ϕ ◦ )(A B C ). Since ϕ ◦ is a congruence, the triangles ABC and A B C are congruent. Case III. Finally, we deal with the general case. Let T be the translation that maps A to A, and denote T (A ), T (B ), and T (C ) by A0 , B0 , and C0 , respectively. Then T (A ) = A and A0 B0 C0 = T (A B C ). But A0 B0 C0 and ABC now have a pair of corresponding vertices in common because A0 = A. By Case II, there is a congruence ϕ so that ϕ(A0 B0 C0 ) = ABC. Therefore ABC = ϕ(A0 B0 C0 ) = ϕ(T (A B C )) = (ϕ ◦ T )(A B C ). Since ϕ ◦ T is a congruence (Theorem G6(c) on page 218), we have proved that two triangles ABC and A B C are congruent when they have three pairs of equal sides. The proof of Theorem G28 is complete. As an immediate application of SSS, we give another characterization of a parallelogram among quadrilaterals. Corollary 1. A quadrilateral is a parallelogram ⇐⇒ its opposite sides are equal. Acivity. Prove Corollary 1. Corollary 2. A rhombus is a parallelogram. We close this section with two other applications of the characterization of the perpendicular bisector of a segment. The ﬁrst is the converse of Theorem G26(a). Theorem G29. If two angles of a triangle are equal, then the triangle is isosceles. More precisely, if in ABC, |∠B| = |∠C|, then |AB| = |AC|.

6.2. SSS AND FIRST CONSEQUENCES

225

Proof. Let M be the midpoint of BC and let be the perpendicular bisector of BC passing through M . If A lies on , then Theorem G27 on page 222 implies that |AB| = |AC| and we are done. So suppose A does not lie on and we will deduce a contradiction. Since B and C lie on opposite half-planes of , A lies on either the half-plane containing B or the half-plane containing C. Without loss of generality, we may assume that it is the latter. Then A and B lie on opposite half-planes of . By assumption (L4)(ii) on page 215, the segment AB intersects at some point A0 . Observe that A0 = A and A0 = B.

A A0 l HH ll HH Hl Hl Hl r HC B M

By Theorem G27 again, we have |A0 B| = |A0 C|, and by Theorem G26(a) on page 221, we also have |∠ABC| = |∠A0 CB|. By the hypothesis that |∠ABC| = |∠ACB|, we obtain |∠A0 CB| = |∠ACB|. Observe that the segment AA0 lies in the same half-plane of LBC . We now face the familiar situation that we have two equal angles ∠A0 CB and ∠ACB with a common side RCB and A, A0 lie in the same half-plane of their common side LCB . Part (ii) of assumption (L6) on page 216 shows RCA0 = RCA . Since A and A0 are the intersections RCA ∩ LAB

and

RCA0 ∩ LAB ,

respectively, we have A = A0 and hence A lies on . This contradiction completes the proof of Theorem G29. Pedagogical Comments. In a school classroom, if a proof is demanded of the geometrically obvious fact that “the segment AA0 lies in the same half-plane of LBC ”, one can explain this as follows. Suppose A and A0 lie in opposite half-planes of LBC ; then assumption (L4)(ii) would again imply that AA0 contains a point D of LBC . Therefore the line LAA0 contains a point D of LBC , and D is between A and A0 . However, LAA0 already contains a point of the line LBC , namely, the point B. Since A0 is between A and B, B is not between A and A0 (see Lemma 4.4 of [Wu2020a] stated on page 357 of this volume). It follows that the line LAA0 contains two distinct points of LBC , namely, B and D. Thus LAA0 and LBC coincide (by assumption (L1)) and A, B, and C are collinear. Hence ABC is not a triangle. Contradiction. However, we suggest that proofs of this nature are better discussed privately than in a classroom setting. End of Pedagogical Comments. We now oﬀer an alternate proof of Theorem G29 that sheds light on the concept of congruence, but one that may be too subtle for beginners. Surprisingly, it was

226

6. BASIC THEOREMS OF PLANE GEOMETRY

ﬁrst discovered some seventeen centuries ago. Note that there are three other proofs of Theorem G29; see Exercise 4 on page 237, Exercise 5 on page 244, and Exercise 4 on page 253. Second proof. Recall the ASA theorem (Theorem G8): Assume two triangles ABC and A B C . If |BC| = |B C |,

|∠B| = |∠B |,

|∠C| = |∠C |,

then there is a congruence ϕ so that ϕ(ABC) = A B C . Now with ABC given so that |∠B| = |∠C|, we deﬁne a new triangle A B C so that A = A,

B = C,

and

C = B.

Then with this deﬁnition, these two triangles ABC and A B C are easily seen to fulﬁll the conditions of ASA; namely, |BC| = |B C |, |∠B| = |∠B |, and |∠C| = |∠C |.

B

A \

\

\ \

\ C

B = C

A = A \ \ \

\

\ C = B

Hence there is a congruence ϕ so that ϕ(ABC) = A B C , i.e., so that ϕ(A) = A, ϕ(B) = B , and ϕ(C) = C . Therefore, ϕ(AB) = A B , and we get |AB| = |A B |. But by deﬁnition, A B = AC, so we obtain |AB| = |AC|. The proof is complete. This remarkable proof is due to Pappus, who lived roughly six centuries after Euclid (see p. 254 of [Euclid-II]). Pappus did not have the precise concept of a congruence as a transformation of the plane that we now do; in those days, people talked in terms of “applying a triangle on another” or “placing a point on another”. In such an environment, Pappus’s proof must have raised some eyebrows. Acivity. Deﬁne a triangle to be equiangular if all three angles are equal. Prove: (a) An equilateral triangle is equiangular. (b) An equiangular triangle is equilateral. The next application of Theorem G27 is the remarkable fact that there is a circle that passes through the vertices of any given triangle. Theorem G30. (i) The three perpendicular bisectors of the sides of a triangle meet at a point which is equidistant from all three vertices, called the circumcenter of the triangle. (ii) There is one and only one circle that passes through the vertices of a given triangle, called the circumcircle of the triangle. Proof. Let the triangle be ABC, as shown, and let M , N be the midpoints of BC and AC, respectively. Also let the perpendicular bisectors of BC, AC be 1 and 2 , respectively.

6.2. SSS AND FIRST CONSEQUENCES

227

1 A 2 l % % l l% rN l % l % r lC B % M % % r% %O % First observe that 1 and 2 are distinct. Indeed, if they were the same line, then LAC and LBC —being two lines perpendicular to the same line—must be parallel (by Theorem G2 on page 218); this contradicts the fact that LAC and LBC intersect at C. So 1 and 2 are distinct after all. Next, we claim that 1 and 2 must intersect. If not, 1 2 , and we shall deduce a contradiction. Observe that LBC ⊥ 1 (i.e., LBC is perpendicular to 1 ), so Theorem G3 on page 218 implies that LBC ⊥ 2 . But LAC ⊥ 2 , so Theorem G2 on page 218 implies that LBC LAC , contradicting the fact that LBC and LAC intersect at C. Hence 1 and 2 must intersect, let us say, at a point O. Since O lies on the perpendicular bisector of BC, |OB| = |OC| (Theorem G27). Similarly, O lies on the perpendicular bisector of AC and we have |OA| = |OC|. The two equalities imply |OA| = |OB| = |OC|; i.e., O is equidistant from A, B, and C. The fact that O is equidistant from A and B implies, by Theorem G27 again, that O also lies on the perpendicular bisector of AB. As O already lies on the perpendicular bisectors of BC and AC, this proves part (i). Next, we prove part (ii). We ﬁrst prove that there is a circle passing through A, B, and C. Indeed, with the point O as above, let the circle with center O and radius |OA| be denoted by K. Then K is the sought-after circle because it must pass through B and C, on account of |OA| = |OB| = |OC|. We proceed to show that K is unique. Suppose there is another circle K passing through A, B, and C, and we will show that K coincides with K. Let the center of K be O . Since O is by deﬁnition equidistant from B and C, O lies on the perpendicular bisector 1 of BC (by Theorem G27). For exactly the same reason, O also lies on the perpendicular bisector 2 of AC. Therefore {O } = 1 ∩ 2 = {O}, and the center of K is just O. Since K passes through A, the radius of K is also |OA|. Thus K = K and the proof of the theorem is complete. Remark. Part (ii) of the preceding theorem is expressed in colloquial terms as “three points determine a circle”.What is meant is that given three noncollinear points, there is one and only one circle passing through these points, namely, the circumcircle of the triangle with these points as vertices. Exercises 6.2. (1) (a) Give a proof of Theorem G25 (page 220) that uses Theorem G26 (page 221) but does not use the Pythagorean theorem. In other words, prove the following without invoking the Pythagorean theorem: Two right triangles with equal hypotenuses and one pair of equal legs must be congruent. (b) If a ray issuing from the vertex of an angle has a point which is equidistant from both sides of the angle (see page 353 for the concept of distance of a point from a line), prove that it is the angle bisector.

228

6. BASIC THEOREMS OF PLANE GEOMETRY

(2) Prove that if the circumcenter is equidistant from the three sides of a triangle, the triangle is equilateral. (3) (a) Show that the three altitudes of an equilateral triangle are equal (in length). (b) If one side of an equilateral triangle is a, what is the length of its altitude? (4) (a) School mathematics deﬁnes a kite to be a quadrilateral ABCD so that two pairs of adjacent sides are equal; i.e., |AB| = |BC| and |CD| = |DA|. Prove that the diagonals of a kite are perpendicular to each other. (b) Prove that each diagonal of a rhombus bisects its angles. (5) Prove: (a) A parallelogram is a rectangle ⇐⇒ its diagonals are equal. (b) If a diagonal of a rectangle bisects an angle, then the rectangle is a square. (6) Prove that the circumcenter of a right triangle is the midpoint of the hypotenuse. (7) Prove that if at one vertex of a triangle the median coincides with the angle bisector, then the triangle is isosceles. Precisely, let F be the midpoint of BC in ABC so that AF is the angle bisector of ∠A. Then |AB| = |AC|. (8) Let |AB| = |AC| in ABC. Now prove: (a) If D and E are the midpoints of AB and AC, respectively, then |BE| = |CD| and they meet at the altitude from A. (b) The altitudes from B and C to AC and AB, respectively, have the same length, and they meet at the altitude from A. (c) Let the angle bisector of ∠B (respectively, of ∠C) meet AC at E (respectively, AB at D), then |BE| = |CD| and they meet at the altitude from A. (9) Suppose in ABC the altitudes from B and C to AC and AB, respectively, have the same length. Prove that ABC is isosceles: |AB| = |AC|. (10) Let ABCD be a trapezoid with AD BC but AB not parallel to CD. (a) Prove that |AB| = |CD| if and only if |∠B| = |∠C|. (b) Prove that |AB| = |CD| if and only if |AC| = |BD|. (11) (a) Let ABCDEF be a convex hexagon so that AB and AF are equal, BC and F E are equal, CD and ED are equal, and AD bisects ∠F AB. Prove that the hexagon is symmetric with respect to LAD . (b) Is the assumption of convexity in (a) necessary? (12) Let ABCDE be a regular pentagon and let be the perpendicular line from A to CD. Prove that is the perpendicular bisector of CD and BE. (13) Let |∠B| = |∠C| in ABC, and let D ∈ AB and E ∈ AC so that |AD| = |AE|. Suppose BE and CD meet at a point F . Prove that the line LAF is the perpendicular bisector of BC. (14) Assume ABC so that |BC| = a, |AB| = c, and |AC| = b. Let the altitude from A to BC be AD so that B ∗ D ∗ C. Express |AD| in terms of a, b, and c. (Note two things: (1) This exercise makes sense since SSS implies that a, b, and c uniquely determine AD, so all that this exercise does is make the dependence of |AD| on a, b, and c explicit. The assumption B ∗ D ∗ C simpliﬁes the exercise a bit but is not absolutely necessary because the case that D lies outside the segment BC can be handled in a similar manner. (2) This exercise should have been given back in Section 5.3 of [Wu2020a], right after the Pythagorean theorem was proved. The reason we did not do so was due to the amount of algebraic computations that is needed.)

6.3. PEDAGOGICAL COMMENTS

229

6.3. Pedagogical comments We give a brief explanation of why the traditional axiomatic treatment of plane geometry should be avoided. A more sophisticated mathematical explanation will be given in Chapter 8. At this point, we have in our possession the three most basic theorems that are used almost everywhere at the beginning of high school geometry: SAS, ASA, and SSS. If you happen to be a geometry teacher in high school, you can now pick up any traditional geometry textbook and—by ignoring its initial development from axioms—use it to instruct your students for the remainder of the school year. Let us amplify on what this means. As will be explained in Chapter 8, there are very persuasive arguments why the use of axioms to begin a course in school geometry is pedagogically defective. We have so far avoided even the mention of axioms. Of course we listed the assumptions (L1)–(L8) that we would use, but we treated the matter rather informally. These are in fact our “axioms”, so to speak, but they are so reasonable that we simply take them for granted, which is as it should be. What makes the treatment of geometry in this book diﬀerent from other traditional treatments is that, from the beginning, we rely on the basic isometries— rotations, reﬂections, and translations—to carry the main weight of the argument at every step. It is essential to let students see that the basic isometries are the mainstay of Euclidean geometry. From an advanced standpoint, what distinguishes the Euclidean plane from other surfaces is exactly this abundance of basic isometries6 and the fact that it satisﬁes the parallel postulate. This information is unfortunately hidden from students in TSM.7 There is an additional reason for initiating the discussion of geometry by using basic isometries rather than the standard axioms of Euclidean geometry: one avoids the dreary proofs of theorems at the beginning of an axiomatic treatment that are both trite and—to students—utterly trivial. See, for example, the brief discussion of this issue on page 333 of Chapter 8. By contrast, note that our fourth theorem (Theorem G4) is about opposite sides of a parallelogram being equal, and one must admit that this fact is hardly trivial or boring. What makes such an approach even more attractive is the ability to “explain” why SAS, ASA, and SSS are true (Theorems G8, G9, and G28 on page 218 and page 223) right at the beginning. Because students have had hands-on experience with basic isometries (see the discussion in Section 4.5 of [Wu2020a]), such an explanation of SAS, ASA, and SSS tends to demystify what “congruence” means8 and therefore increase the learnability of school geometry. TSM does not regard the basic isometries as an essential component of either geometry or mathematics. Prospective teachers learning from the present volume therefore would have been handicapped by “going against the grain” but for the timely arrival of CCSSM. The teaching of geometry in grade 8 and beyond in the CCSSM is in fact aligned with the mathematical development described in Chapters 4 and 5 of [Wu2016a], Chapters 4, 5, and 6 of [Wu2020a], and Chapters 6 and 7 of this volume. Speciﬁcally, the CCSSM call for the use of rotation, translation, 6 Mathematical Aside: The Euclidean plane, the hyperbolic plane, and the sphere are the only simply connected homogeneous 2-dimensional Riemannian manifolds. 7 See page xi for a deﬁnition of TSM. 8 This statement is based on the author’s personal encounters with teachers.

230

6. BASIC THEOREMS OF PLANE GEOMETRY

and reﬂection to start the informal geometric discussion in middle school specify that the AA criterion for similarity (Theorem G22 on page 220) be used to support the deﬁnition of the slope of a line and promote the early proofs of SAS, ASA, and SSS in high school so that the more traditional approach to proofs of geometric theorems may be carried out thereafter. One should not infer from the opening paragraph of this section that the possibility of transitioning from the use of basic isometries at the beginning of the geometric discussion to the usual treatment in the standard geometry textbooks is an end devoutly to be wished. On the contrary, since these standard geometry textbooks are generically the embodiment of TSM, the quality of the mathematics in most of them leaves much to be desired. Nevertheless, we want to assure you that, at least from a practical point of view, this volume has the potential of giving you the freedom of choice. It is not a bad thing to have the assurance that learning a subject correctly can also reap practical beneﬁts. It is hoped that, in any case, the present chapter can be used as a reference for a mathematically correct treatment of plane geometry.

6.4. Proof of FTS This section proves the fundamental theorem of similarity, Theorem G10, for the case that the ratio r is a fraction. Observe that this proof could have been given right after Theorem G9. The extension of the proof to the case of an arbitrary positive number r will have to wait until Section 2.6 of [Wu2020c], where the new ingredient will be the consideration of limits which is independent of the geometric development in this chapter. For this reason, we will be using Theorem G10—and therefore also all its consequences—in full generality without fear of circular reasoning. One such consequence is the proof of the SSS criterion for triangle similarity, which is outlined at the end of the section. Theorem G10 (FTS). In ABC, let D, E be points on the rays RAB and RAC , respectively, so that D = B and E = C. If for some positive number r, |AE| |AD| = = r, |AB| |AC| then DE BC and

|DE| = r. |BC|

A

@ @ @ @ @E D @ @C B r1

6.4. PROOF OF FTS

231

Remarks. (1) Recall that the positive number r is called the scale factor. Since D = B and E = C, we have r = 1. This explains why, in the above pictures, there are only two cases: r < 1 or r > 1. (2) Once we have proved FTS when r is a fraction, a general reasoning involving limits given in Section 2.6 of [Wu2020c] will give a complete proof of FTS for the general case where r is any nonnegative real number. By the cross-multiplication algorithm, the equality to

|AD| |AB|

=

|AE| |AC|

is equivalent

|AD| |AE| |AD| |AE| = ; i.e., = . |AB| − |AD| |AC| − |AE| |DB| |EC| (Compare Exercise 1 in Exercises 5.1 of [Wu2020a].) Therefore we have the following corollary. Corollary. Let ABC be given, and let D, E be points on RAB and RAC , respec|AD| |AE| = |EC| , then DE BC. tively, so that D = A, B and E = A, C. If |DB| Pedagogical Comments. The proof of FTS is probably not something beginners would be able to construct all by themselves. In this streamlined form, it is the result of many trials and errors plus a lot of experience in proving theorems. If you can discover the proof by yourself, good, but if not, do not let anyone mislead you into thinking that because you did not construct the argument yourself, you do not really understand it. Those who know mathematics know only too well that all of us always learn from others about things we cannot do ourselves. A main purpose of education is to learn. For the case at hand, there is a lot one can learn from this proof. For example, it reinforces the message embedded in the proofs of Theorem 1.7 in Section 1.4 of [Wu2020a] (on the area of a rectangle), Theorem G9 (ASA) in Section 4.6 of [Wu2020a], and Theorem G28 (SSS) on page 223 (of this volume), namely, that one should always try to break up a complicated proof into simpler steps. Indeed, our ﬁrst step in the proof of FTS is to prove FTS for the special case that r is fraction of the form k1 for a positive integer k (i.e., |AB| = k|AD| and |AC| = k|AE|); see Lemma 6.1 immediately following. Here the surprise is that one can prove this geometric statement by mathematical induction on k. That done, the second step of proving FTS when r is an arbitrary fraction becomes straightforward. By Remark (2) above, this will bring to a close the proof of FTS. End of Pedagogical Comments. Proof of FTS (beginning). We ﬁrst prove the special case of FTS where the scale factor r is a unit fraction k1 for a positive integer k. Now k > 1 because, if k = 1, then from |AE| 1 |AD| = = =1 |AB| |AC| k we conclude that |AD| = |AB| and |AE| = |AC|, so that D = B and E = C. This contradicts the hypothesis that D = B and E = C. Thus k > 1. Let us now state clearly what we have to show: if k is a positive integer > 1 and if |AD| |AE| 1 = = , |AB| |AC| k 1 then DE BC and |DE| |BC| = k . Using the cross-multiplication algorithm, we may rephrase this special case of FTS in an equivalent form, as follows.

232

6. BASIC THEOREMS OF PLANE GEOMETRY

Lemma 6.1. Let k be any positive integer ≥ 2. Suppose in ABC we have points D and E in AB and AC, respectively, so that |AB| = k|AD| and |AC| = k|AE|. Then DE BC and |BC| = k|DE|. Before giving the proof of Lemma 6.1, we make some preliminary observations. If k = 2, Lemma 6.1 is just Theorem G15 (see page 219). To go beyond Theorem G15, we can try our hand at proving the special case of the lemma for k = 3 (i.e., r = 13 ). Because this special case already yields critical insight into the general case, we will prove it in detail. Thus we assume that in ABC we have |AB| = 3|AD|, |AC| = 3|AE|, and we shall prove DE BC and |BC| = 3|DE|. A

D

@ @ G F @E @ @ @ H @ @ @

B

C

The main idea is that since we already know the validity of the case k = 2 (which is Theorem G15 on page 219), we should exploit it. Thus, we will imitate the proof of the case k = 2 (i.e., Theorem G15) if at all possible.9 To this end, we will do as we did in the proof of Theorem G15, namely, extend DE along the ray RDE until |DF | = 3|DE|. Join CF . Now if we can prove DBCF is a parallelogram, then we certainly get DE BC. Moreover, we get |BC| = |DF | (see Theorem G4 on page 218) so that |BC| = 3|DE| in view of |DF | = 3|DE|. Thus the case of k = 3 would be proved if we can prove DBCF is a parallelogram. To prove DBCF is a parallelogram, we will appeal to Theorem G14 on page 218 by proving |BD| = |CF | and BD CF . To show |BD| = |CF |, let H and G be the midpoints of EC and EF , respectively. Since |AC| = 3|AE|, we have |EC| = 2|EA| so that |EH| = |EA|. Similarly, since |DF | = 3|DE|, we have |EF | = 2|ED| so that |EG| = |ED|. Thus the 180◦ rotation around E would map H to A and G to D and therefore would map HG to AD. A rotation being an isometry (see assumption (L7) on page 217), (6.1)

|HG| = |AD|.

Now in ECF , H and G being the midpoints of the sides EC and EF , Theorem G15 implies that |CF | = 2|HG|. By (6.1), we get (6.2)

|CF | = 2|AD|.

But |AB| = 3|AD| by hypothesis, so |BD| = |AB| − |AD| = 2|AD|. Combined with (6.2), we obtain the sought-after result: (6.3)

|BD| = |CF |.

Next, we will prove BD CF . This is because the 180◦ rotation around E maps HG to AD so that, by Theorem G1 on p. 217, HG AD. But we also have HG CF , for the reason that H and G are the midpoints of the sides EC and EF in ECF and therefore Theorem G15 implies that HG CF . Thus each of AD 9 For

the proof of Theorem G15 in question, see Section 5.1 in [Wu2020a].

6.4. PROOF OF FTS

233

and CF being parallel to HG, we have AD CF ,10 which is to say, BD CF . In view of (6.3), DBCF is a quadrilateral with a pair of sides (BD and CF ) which are parallel and equal. By Theorem G14 (page 218), DBCF is a parallelogram. By a previous remark, we are done with FTS in case n = 3, i.e., in case r = 13 . The preceding proof shows how a knowledge of the case k = 2 can be used to prove the case k = 3. This then suggests that we can use the knowledge of the case k = 3 to prove the case k = 4; i.e., knowing that the lemma is valid when k = 3, we can prove that if |AB| = 4|AD| and |AC| = 4|AE|, then DE BC and |BC| = 4|DE|. If we know anything about mathematical induction at all, we would get the idea at this point that doing mathematical induction on k would be one way to prove Lemma 6.1 in general. That said, we are ready for the formal proof. Proof of Lemma 6.1. We use induction on k. We can get the induction started with k = 2, which is just Theorem G15. Next, we assume the lemma is true for k (k ≥ 2) and proceed to prove that it is true for k + 1. Precisely, we assume that the lemma is valid for one value of k and will prove that, in triangle ABC, if D ∈ AB, E ∈ AC so that |AB| = (k + 1)|AD| and |AC| = (k + 1)|AE|, then (6.4)

DE BC

and

|BC| = (k + 1)|DE|.

Extend DE along the ray RDE to F so that |DF | = (k + 1)|DE|. Join CF . Here is a schematic representation of the situation. A @ G F D @E @ @ [email protected] @ @ @ @C B As in the case of k = 3, our goal is to prove that the quadrilateral DBCF is a parallelogram. Indeed, if this is true, then clearly DE BC (because DF BC). In addition, |BC| = |DF | by Theorem G4 on page 218. Since |DF | = (k + 1)|DE| by the construction of F , we get |BC| = (k + 1)|DE|. Thus, (6.4) is proved and so is the lemma. It remains to prove that DBCF is a parallelogram. To this end, we will appeal to Theorem G14 on page 218 by proving |BD| = |CF | and BD CF . Let G and H be points on EF and EC, respectively, so that |EG| = k1 |EF | and |EH| = k1 |EC|. Equivalently, we have |EF | = k|EG| and |EC| = k|EH|. By the induction hypothesis that the lemma is true for k, we get (6.5)

HG CF

and

|CF | = k|HG|.

10 By Lemma 4.3 in [Wu2020a] (stated on page 357 of the present volume), which says two distinct lines parallel to a third are parallel to each other.

234

6. BASIC THEOREMS OF PLANE GEOMETRY

Next, we claim that |HG| = |AD|. To see this, observe that the hypothesis that |AC| = (k + 1)|AE| implies that |EC| = |AC| − |AE| = k|AE|. Coupled with |EC| = k|EH|, we get |AE| = |HE|. Similarly, the fact that |DF | = (k + 1)|DE| and |EF | = k|EG| implies that |DE| = |GE|. Therefore the 180◦ rotation centered at E maps H to A and G to D, so that (HG) = AD. By Theorem G1 on page 217, we get AD HG

and

|AD| = |HG|

(the latter is because preserves lengths, by assumption (L7)). Now we can bring (6.5) into the discussion. From AD HG and CF HG, we conclude AD CF (two distinct lines parallel to a third are parallel to each other11 ), or equivalently, BD CF . From |AD| = |HG| and |CF | = k|HG|, we conclude |CF | = k|AD|. Since |BD| = |AB| − |AD| = (k + 1)|AD| − |AD| = k|AD|, we also get |CF | = |BD|. Thus the two sides BD and CF of DBCF are equal and parallel, and DBCF is a parallelogram as claimed. By a previous remark, this completes the proof of Lemma 6.1. Proof of FTS (conclusion). We can now prove FTS when the scale factor r is a fraction m n , where m and n are positive integers. Let D, E be points in the segments AB and AC, respectively, of triangle ABC so that D = B and E = C and |AE| m |AD| = = . |AB| |AC| n We must prove that DE BC and |DE| m = . |BC| n Take a point D on AB so that |AD | = |AE | = n1 |AC|, as shown: A

1 n |AB|

and a point E on AC so that A

@ @ @ D @E @E D @ @C B

@ @ @ D @E @ @ @C B @ @E D @ m>n

m |AB|

We also record for future reference the following theorem: Theorem G31 (SSS for similarity). Two triangles are similar if and only if their corresponding sides are proportional. More precisely, ABC and A B C are similar if and only if |AB| |AC| |BC| = = . |A B | |A C | |B C | The proof is very similar in spirit to the proof of Theorem G21 (SAS for similarity),13 as follows. We may assume that |A B | > |AB|, so that also |A C | > |AC|. On the segments AB and AC, let B0 and C0 be chosen so that |A B0 | = |AB| and |A C0 | = |AC|. We want to prove that A B0 C0 ∼ = ABC by using the SSS criterion, but we do not as yet know whether |B0 C0 | = |BC|. However, Theorem G10 (FTS) and the hypothesis of Theorem G31 show that such must be the case. The rest of the proof is then the same as the proof of Theorem G21. The details can be left as an exercise (Exercise 6 on page 237). Exercises 6.4. (1) Assume that Lemma 6.1 on page 232 has been proved for the case k = 3. Use this fact to write out a detailed proof of the lemma for the case k = 4. In Exercise 8 of Exercises 5.1 in [Wu2020a], we proved that FTS* implies FTS. We can therefore get a diﬀerent proof of FTS by ﬁrst proving FTS* (Theorem G11). The next two exercises outline such a proof. (2) Let k be any positive integer ≥ 2. Suppose in ABC we have a point D in AB so that |AB| = k|AD|, and suppose the line passing through D and parallel to BC intersects the side AC at a point E. Then prove that |AC| = k|AE| and |BC| = k|DE|. (This is the analog of Lemma 6.1. Prove it also by induction.) (3) Let ABC be given, and let D be a point on the ray RAB so that |AD|/|AB| = m n . Suppose a line parallel to BC and passing through D intersects RAC at E (see picture below). Prove that |AE| |DE| m = = . |AC| |BC| n 13 See

Section 5.3 of [Wu2020a].

6.4. PROOF OF FTS

237

(Hint: Use the preceding Exercise 2 and imitate the proof of FTS.) A A

@ @ @ @ @E D @ @C B

@ @ @ @ @ @ B @C @ D E @

(4) Let |∠B| = |∠C| in ABC. Let the angle bisector of ∠A meet BC at D. Then: (i) Prove that ABD ∼ = ACD by using the AA criterion for similarity. (ii) Use (i) to give a new proof of Theorem G29 on page 224. (5) Let R be a rectangle with sides of length a and b. Let the sides of R be divided into n equal segments, where n is a positive integer. Join the corresponding division points on the opposite sides of R; the case of n = 4 is shown: Q2 Or Q Q Q Q Q Qr P2 Q Q Q Q Q Q From the upper left corner O of R, let Pi be the i-th division point from O on the left vertical side and let Qi be the i-th division point on the upper horizontal side. The case of i = 2 is shown. Prove that, for every i, the vertical line from Qi and the horizontal line from Pi meet at the diagonal of R as shown. (6) Prove SSS for similarity (Theorem G31). (7) Let 1 , 2 , and 3 be three parallel lines. Suppose a transversal meets them at A, B, C, respectively, and suppose another transversal meets them at A , B , and C , respectively. Prove that |A B | |AB| = . |AC| |A C | (8) Let and be parallel lines, and let ABC and A B C be two triangles so that A, A both lie on and BC and B C are segments of equal length lying on . Let a line L parallel to pass through a point D on AB, so that L meets AC, A B , and A C at E, D , and E , respectively. Prove that |DE| = |D E |. (9) Let D and E be points on AB and AC of a triangle ABC so that |AD|/|AB| = |AE|/|AC|. Prove that BE and CD intersect on the median from vertex A.

238

6. BASIC THEOREMS OF PLANE GEOMETRY

(10) In the picture

A

N

B

K

M

L

C

ABC is an equilateral triangle and KLM N is a square with vertices on the sides of ABC. If AB has length s, what is the length of KL?

6.5. The angle sum of a triangle The goal of this section is to prove the foundational result that the sum of the degrees of the angles of a triangle is 180◦ . We will use the term the angle sum of a triangle (or any convex polygon) to mean the sum of the degrees of all the angles. This proof is more subtle than usually realized, and the plane separation assumption (L4) plays a critical role behind the scenes. The teaching of this “simple” fact in the average high school classroom therefore requires some pedagogical ﬁnesse. This angle sum theorem will be needed (in the form of the exterior angle theorem) for the proof in the next section that every isometry is a congruence. Theorem G32 (Angle sum theorem). The sum of the (degrees of the) angles of a triangle is 180◦ . We will give two proofs. The ﬁrst is classical, and the second, while also classical, is less well known. We will make some pedagogical comments after the second proof. The concept of adjacent angles (page 215) will play a crucial role in both proofs. If C is a point in ∠AOB, then ∠AOC and ∠COB are adjacent angles with respect to ∠AOB, as shown in the picture below, and we have (see (L6)(iv) on page 216): (6.8)

|∠AOC| + |∠COB| = |∠AOB|. O \ \ \ \ A \ C B

Since all the angles involved in the following proofs are convex, there will be no ambiguity in interpreting the terms in equation (6.8). We also recall that C is between A and B means that the point C lies in the segment AB but not equal to either endpoint (see page 214). In symbols, A ∗ C ∗ B. First proof. Assume ABC. Let M be the midpoint of the side AC and let be the rotation of 180◦ around M . Let D = (B).

6.5. THE ANGLE SUM OF A TRIANGLE

239

E A

B

s

s JJ JqM J J C

D

Since (LBC ) = LAD , Theorem G1 (page 217) implies that LBC LAD . By Theorem G12 (page 218), B and D are on diﬀerent sides of the line LAC . Therefore ∠DAC and ∠BCA are alternate interior angles with respect to the transversal LAC of the parallel lines LBC and LAD and, by Theorem G18 (page 219), they are equal. Next, on line LAB , let E be a point so chosen that B ∗ A ∗ E. For the transversal LAB of the parallel lines LBC and LAD , we argue in a similar way and conclude that ∠EAD and ∠ABC are corresponding angles and are therefore also equal. To summarize, we now have (6.9)

|∠DAC| = |∠BCA|,

|∠EAD| = |∠ABC|.

Next, we claim that D lies in ∠EAC. To this end, we need only prove that (i) D and E lie on the same side of LAC and (ii) D and C lie on the same side of LAE , which is the same as LAB . We ﬁrst prove (i). Since E ∗ A ∗ B by choice, E and B lie on opposite sides of LAC . We already took note of the fact that B and D are on diﬀerent sides of the line LAC . Therefore D and E lie on the same side of LAC , proving (i). To prove (ii), it suﬃces to observe that LCD = (LAB ) and therefore LAB LCD . It follows that the segment CD cannot contain any point of LAB (because the line LCD is disjoint from LAB ), which then implies—by virtue of (L4)(ii) on page 215—that (ii) is valid. Therefore D lies in ∠EAC, so that ∠DAC and ∠EAD are adjacent angles. By equation (6.8), we have (6.10)

|∠EAC| = |∠EAD| + |∠DAC|.

In addition, with respect to the straight angle ∠EAB that contains C, the angles ∠EAC and ∠CAB are adjacent angles and therefore by equation (6.8) again, (6.11)

|∠EAC| + |∠CAB| = |∠EAB| = 180◦ .

Putting equations (6.10) and (6.11) together, we get |∠EAD| + |∠DAC| + |∠CAB| = 180◦ . In view of equation (6.9), we obtain |∠ABC| + |∠BCA| + |∠CAB| = 180◦ . The proof of Theorem G32 is complete. Second proof. The key idea of this proof is the observation that the preceding proof simpliﬁes when ∠B is a right angle. We break up the proof into two steps. Step 1. The angle sum of a right triangle is 180◦ .

240

6. BASIC THEOREMS OF PLANE GEOMETRY

D Aq Q Q Q Q qM Q Q Q Q Q B C Let ABC be a right triangle with the right angle at B, and let M be the midpoint of the hypotenuse AC. If is the rotation of 180◦ around M , let D = (B). By Theorem G12 (page 218), B and D are in opposite half-planes of LAC . Thus AB, BC, CD, and DA cannot intersect each other except at the points A, B, C, and D. Consequently, ABCD is a quadrilateral (see the deﬁnition of a polygon on page 354). Next, since LAD = (LBC ) and LAB = (LCD ), Theorem G1 (page 217) implies that LAD LBC and LAB LCD . Since ∠B is a right angle, Theorems G2 and G3 (page 218) imply that ABCD is a rectangle and each of its angles is 90◦ . Therefore, (6.12)

|∠B| + |∠BCD| + |∠D| + |∠DAB| = 360◦ .

Moreover, being a congruence and (ABC) = CDA, we have ABC ∼ = CDA. The equality of corresponding angles yields (6.13)

|∠DAC| = |∠BCA|,

|∠BAC| = |∠DCA|.

Now we claim that C is in ∠DAB. To this end, we have to prove that (i) C and D lie in the same half-plane of LAB and (ii) B and C lie in the same half-plane of LAD . Since LCD LAB , the segment CD is disjoint from LAB and therefore, by (L4)(ii) on page 215, C and D do lie on the same side of LAB . This proves (i). The proof of (ii) is similar. Thus C is in ∠DAB and therefore ∠DAC and ∠CAB are adjacent angles with respect to ∠DAB. By equation (6.8), we have |∠DAB| = |∠DAC| + |∠CAB|. Similarly, we argue with respect to ∠BCD and get |∠BCD| = |∠BCA| + |∠ACD|. Substituting these values of |∠DAB| and |∠BCD| into equation (6.12) yields |∠B| + (|∠BCA| + |∠ACD|) + |∠D| + (|∠DAC| + |∠CAB|) = 360◦ . By interchanging the two underlined terms, we clearly have (angle sum of ABC) + (angle sum of CDA) = 360◦ .

6.5. THE ANGLE SUM OF A TRIANGLE

241

Since ABC ∼ = CDA, this implies that the two angle sums are equal. So (angle sum of ABC) = 180◦ . The proof of Step 1 is complete. Step 2. The angle sum of any triangle is 180◦ . Given triangle ABC, we may assume ∠ABC is not a right angle, as otherwise we would be back to Step 1. Let the altitude from vertex A to the opposite side BC meets the line LBC at a point D. By Lemma 4.4 of [Wu2020a] (see page 357 of this volume), there are three possibilities: B ∗ D ∗ C, B ∗ C ∗ D, and ﬁnally C ∗ B ∗ D, as shown: A \ n m\ \ \ \ \ B D C

B

s C

A f h D

A J @ [email protected] [email protected] [email protected] J @ J @ D B C

First, suppose B ∗ D ∗ C (see above picture on the left). We will show that the angle sum of ABC is 180◦ . To simplify the notation, we have given names to two of the angles having A as a vertex, as shown. Applying Step 1 to ABD and ACD in succession and noting that |∠ADB| = |∠ADC| = 90◦ , we have |∠B| + |∠n| = 90◦ , |∠C| + |∠m| = 90◦ . Adding the two equations, we get |∠B| + (|∠n| + |∠m|) + |∠C| = 180◦ . Since D is between B and C, clearly D is in ∠BAC and therefore ∠n and ∠m are adjacent angles with respect to ∠BAC. By equation (6.8), we see that |∠n|+|∠m| = |∠BAC|. The preceding equation then says that the angle sum of ABC is 180◦ . The other two possibilities are similar, so we only deal with the case of B ∗C ∗D (see the middle picture above). Again, for the sake of clarity, we have given names to two of the angles having A as a vertex and one of the angles having C as vertex, as shown. Applying Step 1 to right triangle ACD, we get (6.14)

|∠h| + |∠s| = 90◦ .

Next we apply Step 1 to the right triangle ABD, and we have (6.15)

|∠B| + |∠BAD| = 90◦ .

Subtracting equation (6.14) from equation (6.15), we get (6.16)

|∠B| + |∠BAD| − |∠h| − |∠s| = 0◦ .

Now since C is between B and D by assumption, C is in ∠BAD. Therefore, ∠f and ∠h are adjacent angles with respect to ∠BAD, so that by equation (6.8), we have |∠f | + |∠h| = |∠BAD|. Substituting this value of |∠BAD| into (6.16) gives (6.17)

|∠B| + |∠f | = |∠s|.

Finally, since B ∗ C ∗ D, we may take ∠BCD to be the straight angle that contains A and therefore ∠BAC and ∠CAD (= ∠s) are adjacent angles. By equation (6.8)

242

6. BASIC THEOREMS OF PLANE GEOMETRY

again, |∠s| = 180◦ − |∠BCA|. Substituting this expression of |∠s| into equation (6.17), we get |∠B| + |∠f | = 180◦ − |∠BCA|. But this is equivalent to the statement that the angle sum of ABC is 180◦ . The proof of Theorem G32 is complete. Pedagogical Comments. It must be admitted that neither proof is entirely straightforward if we insist on giving all the details. In high school, a judicious presentation of a few (and only a few) such precise proofs in a geometry course can serve the purpose of opening the eyes of beginning students to the meaning of rigor and precision. However, the practice of routinely giving complete proofs for every theorem—especially a theorem such as Theorem G32—would likely kill oﬀ most students’ curiosity or interest in geometry. We have to exercise some restraint. In terms of everyday teaching in the school classroom, it would be justiﬁable, in the ﬁrst proof, to trust the picture and omit the reasoning for equation (6.10) and simply say: Because E lies in ∠DAC, we have |∠DAC| = |∠DAE| + |∠EAC|. One can make a side remark afterward that it is possible to prove that E lies in ∠DAC but that the reasoning is tedious. The proof of the angle sum theorem in [Euclid-I] in fact relies on the picture in the same naive way without regard to precise deﬁnitions. One of the main incentives for mathematicians of the nineteenth century—including M. Pasch (Prussian, 1843–1930), G. Peano (Italian, 1858–1932), D. Hilbert (German, 1862–1943), etc.—to reset the foundations of geometry was in fact to eliminate such reliance on pictures (because, after all, pictures are not part of the explicit assumptions) and replace them by reasoning based on the explicit assumptions alone. The second proof looks long, but the argument is intuitive and is one that students can easily remember, especially if the detailed reasoning about which angles are adjacent angles is skipped in favor of an inspection of the pictures. There is no harm in explicitly pointing out to students which uninteresting details have been intentionally glossed over and why. Most students will welcome the opportunity to skip over boring details, and for the few who do not ﬁnd such details boring, you can explain to them separately all the attendant subtleties. In fact, the need to do the latter is one reason why this volume always tries to give complete proofs to the extent possible. It may be added that the idea in the second proof—that of reducing considerations of a general triangle to considerations of right triangles—is an attractive one.14 We will see more examples of this idea in action in the consideration of area in Section 4.5 of [Wu2020c]. Teaching school mathematics usually involves complicated trade-oﬀs between accessibility and mathematical accuracy (for an extended discussion, see [Wu2006]). The more mathematics you know, the better you will be able to decide which aspect of the mathematics under consideration can aﬀord to be suppressed, and the better you can approximate the right pedagogical decision. There is no better illustration of the need for such trade-oﬀs than the case of geometry instruction in high school (cf. [Wu2010a]). In this regard, see also the remarks preceding Theorem G15 in Section 5.1 of [Wu2020a] as well as the italicized remarks following the proof of Theorem G18 in Section 5.2 (also of [Wu2020a]). In Chapter 8 (page 331) below, 14 It

has been strongly advocated by R. A. Askey for school mathematics in recent years.

6.5. THE ANGLE SUM OF A TRIANGLE

243

we will go more into the details of why such pedagogical choices must be made in the teaching of geometry in high school. End of Pedagogical Comments. There is a variant of the angle sum theorem that is already implicit in the second proof (more precisely, equation (6.17)) and which is also useful for applications, so we will single it out. Given a triangle ABC, suppose D lies on the line LBC and B ∗ C ∗ D. Then ∠ACD is said to be an exterior angle of ABC at the vertex C. Similarly, if E lies on the line LAC so that A ∗ C ∗ E, then ∠BCE is, by deﬁnition, also an exterior angle of ABC, as shown: A L L L

L L

q LC D LLq E Since obviously these two exterior angles at the vertex C are equal (see Lemma 5.1 on page 357), sometimes we simply refer to either as the exterior angle of ABC at C without any fear of confusion. Either of the angles ∠A (i.e., ∠BAC) and ∠B (i.e., ∠ABC) is called an opposite interior angle of ∠ACD. Then the above-mentioned variant is the following. B

Corollary 1 (Exterior angle theorem). An exterior angle of a triangle is the sum of its opposite interior angles and is therefore bigger than either. Proof. We use the notation of the preceding ﬁgure. By the angle sum theorem, (|∠A| + |∠B|) + |∠ACB| = 180◦ . But by the deﬁnition of the exterior angle, ∠ACD and ∠ACB are adjacent angles with respect to the straight angle ∠BCD that contains A. Thus, |∠ACD| + |∠ACB| = 180◦ . Subtracting the second equation from the ﬁrst, we get |∠A| + |∠B| = |∠ACD|, which proves the corollary. There is another obvious corollary of the angle sum theorem that gives an alternate characterization of a parallelogram. We will leave the proof as an exercise (Exercise 1 on page 243). Corollary 2. A quadrilateral is a parallelogram ⇐⇒ its angles at opposite vertices are pairwise equal. Exercises 6.5. (1) Prove Corollary 2 above. (2) (a) Prove that in a triangle, there can be at most one obtuse angle. (b) Prove that all isosceles right triangles are similar to each other. (c) Let ABCD be a square and let E be a point in the half-plane of LAD not containing B so that ADE is an equilateral triangle. What is the degree of ∠ABE? (3) In the text, we deduced the exterior angle theorem from the angle sum theorem. Now prove that, conversely, the angle sum theorem is a consequence of the exterior angle theorem.

244

6. BASIC THEOREMS OF PLANE GEOMETRY

(4) Prove that in a 90-60-30 triangle (i.e., a right triangle whose acute angles have 30 and 60 degrees), the bisector, median, altitude, and perpendicular bisector from the vertex of the right angle are distinct lines. (5) (a) Prove that if both ∠B and ∠C in ABC are acute, then the altitude from A to LBC intersects the latter at a point D so that B ∗ D ∗ C. (b) Give a new proof of Theorem G29 on page 224 by—notation as in that theorem—dropping a perpendicular from the top vertex A and using part (a). (6) Suppose ∠A in ABC is obtuse. Prove that the altitude from B or C intersects the opposite side at a point not belonging to the side; more precisely, if the altitude from B to LAC intersects the latter at E, then C ∗ A ∗ E, and similarly for the altitude from C. (7) (a) Give a third proof of Theorem G26(a) on page 221 using the following idea: drop the perpendicular from A to LBC , meeting the latter at D. Show D ∈ BC and that triangles ABD and ACD are congruent. (b) Could we have given this proof back in Exercises 6.2? (8) Let the altitude from vertex A of ABC meet the segment BC at D and suppose |AD|2 = |BD| · |DC|. Prove that ∠A is a right angle. (9) (a) Find the angle sum of the (degrees of the) angles of a convex n-gon. (b) Deﬁning the exterior angles of a convex polygon as in the case of a triangle, ﬁnd the sum of the (degrees of the) exterior angles of a convex n-gon. (Caution: Your proofs should make clear why the assumption of convexity is necessary.) (10) (a) Let L be a line and let ρ be a rotation of θ degrees around a point O. Prove that if θ = 180 or −180, then ρ(L) and L intersect and the four angles formed by L and ρ(L) at the point of intersection have either θ degrees or 180 − θ degrees. (b) Let L1 and L2 be two lines intersecting at Q and let P1 and P2 be points on L1 and L2 , respectively, distinct from Q. Prove that there is a rotation ρ of θ degrees so that ρ(L1 ) = L2 and ρ(P1 ) = P2 and so that θ is the degree of one of the four angles at Q. 6.6. Characterization of isometries In this section, we ﬁll in a crucial gap, already mentioned several times in Chapters 4 and 5 of [Wu2020a], by proving that every isometry in the plane is a congruence (Theorem G39 on page 250).15 This proof is quite sophisticated as it brings in a new idea, which is to study the ﬁxed points of an isometry. The triangle inequality (p. 244) The concept of a ﬁxed point of a transformation (p. 247) Every isometry is a congruence (p. 250) The triangle inequality Up to this point, we still do not know if an isometry of the plane is a congruence, but this will change as soon as we can prove Theorem G39 on page 250. To this end, we ﬁrst bring closure to the discussion of distance which was started in Section 4.2 in [Wu2020a]. Theorem G36 below (page 247) complements assumption (L5)(iv) 15 The

proof of Theorem G39 as presented here is based on Chapter 6 of [Lang].

6.6. CHARACTERIZATION OF ISOMETRIES

245

(see page 215). Recall from the last section that the notation A ∗ C ∗ B means the point C lies in the segment AB but is equal to neither endpoint. A key ingredient in the proof that an isometry is a congruence is Theorem G36, which tells us that the sum of the lengths of two sides of a triangle is always bigger than the length of the third side. At ﬁrst sight, this fact seems too obvious to require a proof, but the proof is actually quite subtle. It depends on another seemingly obvious fact, for the statement of which we ﬁrst give a deﬁnition. In a given ABC, we say BC is the side facing ∠A, AB is the side facing ∠C, etc. Then we are going to prove that, of the two sides facing two given angles in a triangle, the side facing the larger angle is longer. One can prove this by a contradiction argument (see Exercise 3(b) on page 253), but it is far simpler to ﬁrst prove its converse, as follows. Theorem G33. In a triangle, the angle facing the longer side is larger. More precisely, if in triangle ABC, |AC| > |AB|, then |∠B| > |∠C|. A @ @ @ @ @ B hhhhhh [email protected] hh @C Proof. We are given that |AC| > |AB|. Thus there is a point D in segment AC so that |AB| = |AD|, as shown: A @ @ @ @ α β @D B hhhhhh [email protected] hh @

C

We claim that D is in ∠ABC (= ∠B; recall that an angle is a region in the plane), for the following reason. To show that A and D lie in the same half-plane of LBC , note that the line LAD intersects LBC at C. So if the segment AD intersects LBC at a point P , then P = C (because C ∈ AD) and therefore LAC = LBC since they have two distinct points in common (by (L1) on page 214). Contradiction. Thus AD is disjoint from LBC and, by (L4)(ii) on page 215, A and D lie in the same half-plane of LBC . Similarly, D and C lie in the same half-plane of LAD . Together, these two facts imply that D ∈ ∠ABC. Let ∠ABD be denoted by ∠α and ∠ADB be denoted by ∠β. Then ∠α and ∠DBC are adjacent angles with respect to ∠ABC and therefore, by equation (6.8) on page 238, |∠ABC| > |∠α|. Since ABD is isosceles, by Theorem G26(a), |∠α| = |∠β|. By the exterior angle theorem (page 243), |∠β| > |∠C|. Thus, |∠ABC| > |∠α| = |∠β| > |∠C|. Therefore |∠ABC| > |∠C|, and the proof of the theorem is complete. The theorem that we were after is the converse of Theorem G33, and this converse turns out to be a corollary of Theorem G33 itself. (Recall that the same

246

6. BASIC THEOREMS OF PLANE GEOMETRY

interesting phenomenon occurs in the case of the Pythagorean theorem and its converse. See Section 5.3 in [Wu2020a].) The simple proof (by contradiction) that Theorem G33 implies its own converse can be safely left to an exercise (Exercise 3 on page 253). Note that the same exercise outlines another proof of the converse. Corollary. In a triangle, the side facing the larger angle is longer. That is, if in ABC, |∠B| > |∠C|, then |AC| > |AB|. The following is traditionally referred to as “the shortest distance between two points is (the length of) a straight line (segment)”. We will have a comment after the proof. Theorem G34. The sum of the lengths of two sides of a triangle exceeds the length of the third. Proof. Let us prove that in ABC, |AB| + |BC| > |AC|, as shown: A @ @ @ @ @ α β@ B ``` γ@D ``δ` ``` @ ``` ```@ `` @C Now if |AB| ≥ |AC|, there would be nothing to prove. Therefore, we may assume |AB| < |AC|. Then there is a point D between A and C so that |AD| = |AB|. As in the proof of Theorem G33, D lies in ∠ABC. Let us denote ∠ABD by ∠α, ∠ADB by ∠β, ∠BDC by ∠γ, and ∠DBC by ∠δ. Since |AB| = |AD| and |AC| = |AD| + |DC|, we see that |AB| + |BC| > |AC| is equivalent to |AD| + |BC| > |AD| + |DC|. Therefore, it suﬃces to prove |BC| > |CD|. By the preceding corollary, it suﬃces in turn to prove |∠γ| > |∠δ|. Note that because |AB| = |AD|, Theorem G26 implies that |∠α| = |∠β|. Using this fact and by repeated use of the exterior angle theorem (page 243), we have |∠γ| > |∠α| = |∠β| > |∠δ|. That is, |∠γ| > |∠δ|. This completes the proof of |AB| + |BC| > |AC|. A slightly diﬀerent proof is the following. As before, we have |∠α| = |∠β|. Looking at the angle sum of ABD, we see that ∠β is an acute angle, because |∠α| + |∠β| + |∠A| = 180◦ implies 2|∠β| + |∠A| = 180◦ , which in turn implies 2|∠β| < 180◦ . Hence |∠β| < 90◦ . Therefore ∠γ is an obtuse angle. But in any triangle, there can only be one obtuse angle on account of the angle sum theorem. Therefore, looking at BCD, we see that ∠δ is acute, and |∠γ| > |∠δ|. Here is a third proof of |∠γ| > |∠δ|. Clearly |∠γ| + |∠β| = 180◦ . Since |∠β| = |∠α|, we get |∠γ| + |∠α| = 180◦ .

6.6. CHARACTERIZATION OF ISOMETRIES

247

By the angle sum theorem, the sum of the angles of ABC is 180◦ . Thus |∠ABC| < 180◦ , and therefore |∠δ| + |∠α| = |∠ABC| < 180◦ = |∠γ| + |∠α|. We conclude that |∠δ| < |∠γ| as before. Remark. We want to explain why Theorem G34 says something less than the common interpretation that “the shortest distance between two points is a straight line”. Recall from Section 4.2 of [Wu2020a] that a polygonal segmentis a ﬁnite collection of segments A1 A2 , A2 A3 , A3 A4 , . . . , An−2 An−1 , An−1 An , with the understanding that these segments need not be collinear and that there may be intersections among them. The length of a polygonal segment is by deﬁnition the sum of the lengths of the individual segments (see Sections 4.2 and 4.3 of [Wu2020c] for a fuller discussion of lengths of curves). What one can deduce from Theorem G34 is the less general statement that, among all polygonal segments, the straight line segment joining two given points has the shortest length (Exercise 5 on page 253). To show that, in fact, among all curves, this straight line segment still has the shortest length, calculus must be used (see, e.g., [O’Neill, exercise 11 on page 56], which contains a detailed guidance for its proof). Again, see Section 4.2 of [Wu2020c] for an indication of the subtleties involved. Theorem G35. For any three distinct points A, B, C in the plane, |AB| ≤ |AC| + |CB|. The equality |AB| = |AC| + |CB| holds ⇐⇒ A ∗ C ∗ B. Proof. If the three points are collinear, then Lemma 4.4 in [Wu2020a] implies that there are only three possibilities: A∗B∗C or B∗A∗C or A∗C∗B (see page 357 in this volume). One concludes from the following pictures that (i) |AB| < |AC|+|CB| if A ∗ B ∗ C or B ∗ A ∗ C and (ii) |AB| = |AC| + |CB| if A ∗ C ∗ B. A

q B

C

B

q A

C

A

q C B

If the three points are not collinear, then |AB| < |AC| + |CB| because of Theorem G34. The conclusions of the theorem now follow trivially. The following is just a reformulation of Theorem G35 in terms of distance. Theorem G36 (Triangle inequality). For any three distinct points A, B, C in the plane, it is always true that dist(A, B) ≤ dist(A, C) + dist(C, B). The case of equality in this inequality (i.e., dist(A, B) = dist(A, C) + dist(C, B)) holds if and only if A ∗ C ∗ B. Incidentally, while we will prove presently that an isometry is a congruence and therefore must map lines to lines and segments to segments (Theorem G39 on page 250), it is of some interest to note that the triangle inequality already suﬃces to give a direct proof of the fact that an isometry must map lines to lines and segments to segments (Exercise 1 on page 253). The concept of a ﬁxed point of a transformation The proof of the theorem that every isometry of the plane is a congruence (Theorem G39) requires one more new idea. We call a point P a ﬁxed point of a transformation F of the plane if F (P ) = P . Here are some obvious examples of

248

6. BASIC THEOREMS OF PLANE GEOMETRY

ﬁxed points. A transformation F is the identity transformation if and only if every point of the plane is a ﬁxed point of F . If Λ is a reﬂection across a line L, then the ﬁxed points of Λ are exactly all the points on L. If F is a rotation of nonzero degree centered at a point O, then O would be the only ﬁxed point. It turns out that the converse is “almost” true: if an isometry has exactly one ﬁxed point, it must be a rotation around the ﬁxed point, or such a rotation followed by a reﬂection across a line containing the ﬁxed point. (Because this fact is not essential to the proof of Theorem G39, we will leave its proof to Exercise 12 on page 254.) On the other hand, a translation along a nonzero vector has no ﬁxed points. How ﬁxed points arise in this connection requires a brief summary and motivation of the proof. Given an isometry F , we want to prove that it is a congruence. It may be possible to directly exhibit a ﬁnite number of basic isometries ϕ1 , . . . , ϕn so that F = ϕ1 ◦ · · · ◦ ϕn , but such a proof seems not to have been found yet. What we are going to do is to “guess” a congruence ψ and show that F = ψ. Accepting this idea, we may ask what this ψ could be. Recall that in our proofs of SAS, ASA, and SSS, a knowledge of how two triangles correspond to each other leads in each case to a congruence, from the whole plane to the whole plane, which realizes this correspondence between the triangles. Taking that as a hint, we proceed as follows. Take three noncollinear points A, B, C and let A = F (A), B = F (B), C = F (C). Because of the triangle inequality (Theorem G36), A , B , C are also noncollinear. Since F is an isometry, the triangles ABC and A B C are congruent because of SSS. Thus there is a congruence ψ so that ψ(ABC) = A B C ; i.e., ψ(A) = A , ψ(B) = B , ψ(C) = C . Our hope is to be able to show that F = ψ. Now, what we have is the fact that both F and ψ behave the same way at least on the vertices of ABC; namely, F (A) = A , ψ(A) = A ,

F (B) = B , F (C) = C , ψ(B) = B , ψ(C) = C .

We can look at this coincidence diﬀerently. Recall that ψ −1 is also a congruence (see Theorem G6 on page 218), and therefore ψ −1 ◦ F is an isometry and this isometry has three noncollinear ﬁxed points A, B, and C: (ψ −1 ◦ F )(A) = (A), (ψ −1 ◦ F )(B) = (B), and (ψ −1 ◦ F )(C) = (C). We are going to prove that an isometry of the plane with three noncollinear ﬁxed points is the identity transformation. So ψ −1 ◦ F = I, and this implies F = ψ because, by applying ψ to both sides of ψ −1 ◦ F = I, we get ψ ◦ ψ −1 ◦ F = ψ ◦ I, which implies F = ψ since ψ ◦ ψ −1 = I. So F is a congruence. It is certainly far from obvious that the concept of a ﬁxed point can be brought to bear on a proof that tries to show that an isometry is a congruence. Here then is a new idea that we can learn from this proof. This idea also motivates us to explore in the next two theorems the ramiﬁcations when an isometry has ﬁxed points (see also Exercises 11 and 12 on page 254). Theorem G37. If F is an isometry of the plane and if P , Q are distinct ﬁxed points of F , then every point on the line LP Q joining P to Q is a ﬁxed point of F . Proof. This proof actually tells you why the theorem is true. First, consider the case of a point A lying in the segment P Q, A not being equal to either of the ﬁxed points P and Q. Since P ∗ A ∗ Q, |P A| + |AQ| = |P Q| (Theorem G35). Let A = F (A). Thus F maps P to P , Q to Q, and A to A .

6.6. CHARACTERIZATION OF ISOMETRIES

249

We are going to prove that A = A. Since F is an isometry, it preserves distances. Therefore from |P A| + |AQ| = |P Q|, we get |F (P A)| + |F (AQ)| = |F (P Q)|, so that |P A | + |A Q| = |P Q|. According to Theorem G35 again, the latter implies P ∗ A ∗ Q. Thus both A and A are points in the segment P Q and, consequently, they both belong to the ray RP Q . Again, because F is an isometry, |P A| = |P A | because F (P A) = P A . By the uniqueness part of (L5)(ii) on page 215, A = A; i.e., F (A) = A. So A is a ﬁxed point of F for every point A in the segment P Q. It remains to examine what happens when A is on LP Q but A ∈ P Q. Without loss of generality, we may assume that A ∗ P ∗ Q. q A

q P

q Q

Then exactly the same argument shows that if A = F (A), A ∗P ∗Q and the ray RQP contains both A and A . We have |QA| = |QA |, because F (QA) = QA and F is an isometry. We conclude that A = A because of the uniqueness statement in (L5)(ii), and F (A) = A as before. The proof is complete. The following theorem is the backbone of the argument that every isometry is a congruence. Theorem G38. An isometry of the plane that has three noncollinear ﬁxed points is the identity transformation. Proof. Let the three ﬁxed points of the given isometry F be A, B, and C. We have to prove that F ﬁxes any point P not equal to any of these three points; i.e., F (P ) = P . Since Theorem G37 says that every point lying on the lines joining A, B, C is a ﬁxed point of F , we are done if P is on one of these three lines. So suppose P does not lie on any of these lines. By Theorem G37 again, it suﬃces to construct a line passing through P that contains two distinct points on the lines joining A, B, and C (which consist of ﬁxed points of F ). To this end, take a point N on line LAC distinct from A and C. If the line LP N is not parallel to line LBC , then they intersect at a point M ; M is on LBC and M is obviously distinct from N . Then line LP N is the desired line. B q @Aq @ @ @ M @ @ [email protected] @ P r @q C @ @ If, however, the lines LP N and LBC are parallel, then LP N and LAB cannot be parallel because, if they were, then LBC and LAB would be parallel to each other since they are both parallel to LP N (Lemma 4.3 of [Wu2020a]; see page 357 of this volume). This is a contradiction because LBC and LAB intersect at B. So LP N and LAB are not parallel after all, and we may assume that they intersect at

250

6. BASIC THEOREMS OF PLANE GEOMETRY

some Q. Now Q is on LAB and is obviously distinct from N . Thus, once again, LP N contains two distinct points on the lines joining A, B, and C. The proof of Theorem G38 is complete. Every isometry is a congruence Finally, we are in a position to prove the main theorem of this section. Theorem G39. Every isometry of the plane is a congruence. Proof. Let F be an isometry. Take three noncollinear points A, B, C, and let A = F (A), B = F (B), and C = F (C). Since F is an isometry, we have |AB| = |A B |, |AC| = |A C |, and |BC| = |B C |.

B

A A A A A A AC

C A @ @ @ @ @ @ B

Since A, B, C are not collinear, the triangle inequality implies that |AB| + |AC| > |BC|. Because F is an isometry, we have |A B | + |A C | > |B C |. Again, by the triangle inequality, we see that A , B , C are noncollinear and therefore A B C is a triangle. By SSS (page 223), this says that there is a congruence ϕ so that ϕ(A B C ) = ABC. Therefore ϕ maps A to A, B to B, and C to C. Now the composition ϕ ◦ F —which is of course an isometry—maps A to A, B to B, and C to C. By Theorem G38, ϕ ◦ F = I, where I is the identity transformation of the plane. Therefore ϕ−1 ◦ ϕ ◦ F = ϕ−1 ◦ I, which implies, in turn, that F = ϕ−1 because ϕ−1 ◦ ϕ = I. Since ϕ−1 is a congruence (by Theorem G6 on page 218), the proof is complete. Theorem G39 is not only conceptually important, but also technically important in geometrical applications because it has the following corollary. Corollary 1. A similarity with scale factor equal to 1 is a congruence. The reason is that if F is such a similarity, then for any two points A and B, with A = F (A) and B = F (B), we have |A B | = 1 · |AB|. Thus F is an isometry and therefore a congruence. With a little more eﬀort, we can squeeze a slightly more precise statement out of the preceding proof of Theorem G39. For this statement, we recall that the verbal description of a composition ◦ T is “the transformation T followed by the transformation ”, i.e., in the same order that the transformations are applied. Corollary 2. An isometry F is either a translation or a translation followed by a rotation or a translation followed by a rotation and then followed by a reﬂection.

6.6. CHARACTERIZATION OF ISOMETRIES

251

Proof of Corollary 2. Using the notation as in the preceding proof, we use −−→ a translation T along the vector AA to map A to A . Suppose T (B) = B and T (C) = C . Since we know from the theorem that F is a congruence, its inverse transformation F −1 exists and F −1 maps A to A, B to B, and C to C. Then clearly F −1 ◦ T has A, B, and C as ﬁxed points. By Theorem G38, F −1 ◦ T = I, which implies F ◦ F −1 ◦ T = F ◦ I so that T = F . We are done in this case. Suppose T (B) = B . Let B = T (B). Since T is an isometry, we have the equality of segments |A B| = |T (AB)| = |AB| = |A B |, so that |A B| = |A B |. This says B and B are equidistant from A and therefore lie on a circle centered at A . Therefore, there is a rotation around A so that (B) = B . Now we have (6.18)

( ◦ T )(A) = A

and ( ◦ T )(B) = B .

If already ( ◦ T )(C) = C , then ◦ T is a congruence that maps A to A , B to B , and C to C . It follows that F −1 ◦ ◦ T is an isometry that has three noncollinear ﬁxed points A, B, and C. By Theorem G38, F −1 ◦ ◦ T = I and therefore F ◦ F −1 ◦ ◦ T = F ◦ I. This implies ◦ T = F , and we are done again. Now suppose (6.18) holds but ( ◦ T )(C) = C . Write C for ( ◦ T )(C) in the remainder of this discussion. We claim that in this case C and C must be in opposite half-planes of LA B . What we will prove is that if C and C are in the same half-plane of LA B , then they would be the same point and that would be a contradiction. Thus let C and C be in the same half-plane. C C Z Z b b bZ bZZ b bZ bZ bZ bZ b b Z A B Because ABC ∼ = A B C (SSS), we have |∠A| = |∠B A C |, |∠B| = |∠A B C |, and |∠C| = |∠C |. Since T and preserve degrees of angles (see (L7) on page 217), we have |∠A| = |∠BAC| = |( ◦ T )(∠BAC)| = |∠B A C|. Thus |∠B A C | = |∠A | = |∠A| = |∠B A C|. As C and C are in the same half-plane, the rays RAC and RAC coincide by the uniqueness statement in assumption (L6)(ii) (page 216). Similarly, the rays RBC and RBC coincide. Therefore, C and C, being the intersections of RA C and RB C , RA C and RB C , respectively, also coincide, a contradiction. This proves our claim that if C = C , these two points have to be in opposite half-planes of LA B . Such being the case, now let Λ be the reﬂection across LA B . Then Λ(C)

252

6. BASIC THEOREMS OF PLANE GEOMETRY

and C are in the same half-plane, and the preceding argument will again show that Λ(C) and C coincide and therefore (Λ ◦ ◦ T )(C) = Λ(C) = C . But Λ(A ) = A and Λ(B ) = B , so we also have (Λ ◦ ◦ T )(A) = A

and (Λ ◦ ◦ T )(B) = B .

It follows that the composition (Λ ◦ ◦ T ) maps A to A , B to B , and C to C . In other words, F −1 ◦ Λ ◦ ◦ T has three noncollinear ﬁxed points A, B, and C and therefore must be the identity transformation I, by Theorem G38. Hence, we have F −1 ◦ Λ ◦ ◦ T = I, and the usual argument then leads to Λ ◦ ◦ T = F . Corollary 2 is completely proved. We emphasize once more the remarkable nature of Theorem G39: starting with the deﬁnition that an isometry is a transformation of the plane that preserves distance—and only distance—we have no reason to expect, a priori, that an isometry is surjective and maps a line to a line, much less the fact that it would preserve the degree of an angle. Theorem G39, however, implies that an isometry automatically has all these properties. Pedagogical Comments. The three theorems, Theorems G34–G36, proved earlier in this section, are absolutely basic; they should be part of any high school curriculum. However, it is more diﬃcult to make the same case for the main theorem of this section, Theorem G39 (every isometry of the plane is a congruence). About its central importance in school geometry, there is no doubt. Its conclusions are stunning; e.g., (1) if a transformation of the plane preserves distance, then it also preserves degrees of angles and (2) if a transformation of the plane preserves distance, then it is surjective. The ﬂy in the ointment is that the proof of Theorem G39 is not short, so each teacher will have to make a case-by-case decision about how to present this proof in class or, in fact, whether to prove Theorem G39 at all. End of Pedagogical Comments. Mathematical Aside: In Section 5.3 of [Wu2020a], we introduced a nested sequence of groups of transformations in the plane, each being a subgroup of the next: group of translations ⊂ group of congruences ⊂ group of similarities ⊂ group of bijections. Now Theorem G39 tells us that a transformation of the plane is a congruence if and only if it is an isometry. In particular, the group of congruences coincides with the isometry group of the plane. The preceding sequence can now be put in its ﬁnal form: group of translations ⊂ isometry group ⊂ group of similarities ⊂ group of bijections. We also recall that the group of translations is abelian (see the end of Section 6.6 in [Wu2020a]. Before leaving the topic of isometries, we should mention that Theorem G39 and its Corollary 2 can be sharpened to the following theorem. Theorem. Every isometry of the plane is equal to a composition of at most three reﬂections.

6.6. CHARACTERIZATION OF ISOMETRIES

253

However, since this is a theorem that is nice to know but has not been shown, thus far, to be indispensable to school mathematics, we will leave the proof to a ﬁle entitled, Isometries as compositions of reﬂections, to be posted on the author’s homepage https://math.berkeley.edu/~wu/. Exercises 6.6. (1) Use the triangle inequality (Theorem G36) to give a direct proof of the fact that an isometry maps lines to lines and segments to segments. (2) Prove that if P and Q are two points inside a circle C, then |P Q| ≤ 2r, where r is the radius of C. (3) (a) Prove the corollary of Theorem G33 on page 246 by using Theorem G33. (b) Give a direct proof of the same corollary without using Theorem G33, as follows. Using notation as in that corollary, let D be a point on the ray RAC so that |AB| = |AD|; show that D cannot lie outside the segment AC so that D has to lie in the segment AC. (4) Use the corollary of Theorem G33 on page 246 to give another proof of Theorem G29 on page 224. (5) Prove that among all polygonal segments joining two given points, the straight line segment joining them has the shortest length. (Don’t forget to take care of the possible self-intersections of the polygonal segment.) (6) In ABC let A be a point inside the triangle but that does not lie on a side of the triangle. Prove: (i) |∠BAC| < |∠BA C|. (ii) |BA | + |A C| < |BA| + |AC|. (7) Let L be a line in the plane and let P and Q be two points on the same side of L. Deﬁne a function F : L → R by F (A) = |P A| + |QA| for each point A ∈ L. Find the point A0 in L at which F achieves its minimum.16 (8) Recall that in a quadrilateral ABCD, the segments AC, BD are the diagonals of the quadrilateral. (a) Prove that the perimeter of a quadrilateral (i.e., the sum of the lengths of all its sides) is greater than or equal to the sum of the (lengths of the) two diagonals. (b) Prove that the sum of the diagonals of a convex quadrilateral is greater than half the perimeter. (9) (a) Prove that (the length of) a median in a triangle is ≤ half of the sum of the lengths of the two sides issuing from the same vertex as the median. (b) Prove that the sum of the medians of a triangle is less than the perimeter of the triangle (i.e., the sum of the lengths of the three sides). (c) Consider the following statement: “Let r be a fraction so that r < 12 ; then a median in a triangle is ≤ r times the sum of the lengths of the two sides issuing from the same vertex as the median.” Give an intuitive argument to show that this is false. (10) (i) Prove that if a, b, and c are the lengths of three sides of a triangle, then |a−b| < c. (ii) Let F1 and F2 be two points in the plane and let |F1 F2 | = c, where c > 0. Fix a positive number a and let H be the collection of all the points P satisfying either |P F1 | − |P F2 | = 2a or |P F1 | − |P F2 | = −2a. Prove that if a = c, H consists of two points collinear with F1 and F2 , but that if a > c, H is the empty set. 16 This implies that if L is a mirror and light is emitted from P and reaches Q and if light takes the shortest path, then light will go from P to A0 to Q. The ﬁrst person to announce this result was apparently Heron of Alexandria (circa 10 AD–circa 70 AD). We will meet Heron again in Section 4.5 of [Wu2020c].

254

6. BASIC THEOREMS OF PLANE GEOMETRY

(11) Prove that if an isometry of the plane has two distinct ﬁxed points P and Q, then it is either the identity transformation or a reﬂection across the line LP Q . (12) Prove that if an isometry of the plane has exactly one ﬁxed point O, then it is either a rotation around O, or the composition of a rotation around O and a reﬂection across a line containing O. (13) Let a1 , a2 , b1 , b2 be any four numbers. (a) Prove that (a1 − b1 )2 + (a2 − b2 )2 ≤ |a1 − b1 | + |a2 − b2 |. (b) Prove that equality holds in part (a) if and only if either a1 = b1 or a2 = b2 . (c) Find a second proof of the fact that the equality (a1 − b1 )2 + (a2 − b2 )2 = |a1 − b1 | + |a2 − b2 | implies either a1 = b1 or a2 = b2 .

6.7. Some basic properties of a triangle Three lines are said to be concurrent if they meet at a point. A main focus of the plane geometry of triangles is on the properties of the points of concurrency of special lines in a triangle and the interrelationships among these points. In this section, we introduce a few distinguished points of concurrency associated with a triangle. One of these points is already known to us: the circumcenter, which is the subject of Theorem G30 on page 226). In the ﬁrst subsection, we collect together the standard points of concurrency in a triangle. These points may seem to be unrelated to each other, but remarkably, three of them are actually collinear, and this is the Euler line, the subject of the second subsection. Finally, we prove a theorem—Ceva’s theorem—that brings coherence to most of these points, and we will explain this phenomenon in the last subsection. Orthocenter, centroid, and incenter (p. 254) The Euler line (p. 258) Ceva’s theorem (p. 261) Orthocenter, centroid, and incenter Given a triangle, the perpendicular bisectors of the three sides are concurrent (Theorem G30 on page 226); recall that this point is called the circumcenter of the triangle. It turns out that the three altitudes (see page 221), the three medians (see page 221), and the three angle bisectors are also concurrent. These are classical results that were known already to Euclid ([Euclid-I] and [Euclid-II]), but a ﬂurry of activities on this topic in the nineteenth century uncovered some new phenomena (the Lemoine point and the Brocard points) that still retain their charm after more than a hundred years.17 We only have space here to deal with the classical results. 17 A wonderful reference is the book [Altshiller-Court]. Buy it while you can, because there is no telling when it will go out-of-print again.

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

255

The concurrency of the three altitudes of a triangle is intimately related to the concurrency of the three perpendicular bisectors. This fact is forcibly brought out in the following proof of the concurrency of the altitudes.18 Theorem G40. The three altitudes of a triangle meet at a point, called the orthocenter of the triangle. Acivity. Where is the orthocenter of a right triangles? In view of Theorem G40 and Theorem G30 on page 226 on the concurrency of the three perpendicular bisectors of a triangle, one may ask how one goes about proving the concurrency of any three lines in general. The answer is that there is no one general method that will work all the time. For example, Theorem G40 can be proved in at least four diﬀerent ways: in addition to the proof below, the remark on page 260 after the proof of the Euler line gives a second proof, the discussion of concurrency on pp. 264ﬀ. following the proof of Ceva’s theorem furnishes a third proof, and Exercise 15 on page 295 outlines a fourth proof. They are all diﬀerent. One standard method for this purpose is to let two of the lines meet at a point and then prove that this point of intersection must lie on the third line; this was the method employed to prove Theorem G30, and this is also the idea behind the fourth proof of Theorem G40 outlined in Exercise 15 on page 295. This will also be, more or less, the method of proof of Theorem G41 (page 256) for the concurrency of the medians. A second standard method is to try to apply Ceva’s theorem, and this works quite often. Finally, one can always rely on ingenuity to save the day; see, e.g., the following proof of Theorem G40. Proof. Let the altitudes of ABC be AD, BE, and CF . Through each vertex, draw a line parallel to the opposite side, resulting in a triangle which we denote by A B C , as shown below. If the triangle is acute (see picture on the left), we have D ∈ BC, E ∈ AC, and F ∈ AB (see Exercise 6 on page 244). However if (let us say) ∠A is obtuse (see picture on the right), then we have E ∗ A ∗ C and F ∗ A ∗ B (again see Exercise 6 on page 244).

ʹ

ʹ

ʹ

ʹ

ʹ

ʹ

By construction, the quadrilaterals AC BC, ABCB are parallelograms. Therefore, by Theorem G4, |C A| = |BC| = |AB |. Thus A is the midpoint of C B . 18 Recall that an altitude of a triangle refers to the line passing through a vertex and perpendicular to its opposite side as well as the segment between a vertex and the point of intersection of the perpendicular with the (line containing the) opposite side.

256

6. BASIC THEOREMS OF PLANE GEOMETRY

Moreover, since AD ⊥ BC and BC C B , we also know that AD ⊥ C B (Theorem G3 on page 218)). It follows that LAD is the perpendicular bisector of C B . Similarly, LF C and LBE are perpendicular bisectors of A B and C A , respectively. By Theorem G30, LAD , LF C , and LBE meet at the circumcenter of A B C . The proof is complete. Remark. This proof—sometimes attributed to Gauss—of the concurrency of the three altitudes of a triangle is sophisticated and is not likely one that beginners would be able to think up by themselves. However, it is worth learning because it is conceptually the simplest. Two other proofs of this result on pp. 264 and 295 (mentioned above) are more routine. The next topic of discussion is the concurrency of the medians in a triangle; recall that a median is the segment joining a vertex to the midpoint of the opposite side (see page 221). The proof of concurrency is equally sophisticated, but we ask you to view it from the point of view of Theorem G15; such a vantage point should make the whole proof seem natural, at least with hindsight. Here is the main idea. We will ﬁx one median, say, the median mB from the vertex B, and then show that another median must intersect mB at the point G of mB whose distance from B is 2 3 the length of mB . This then implies that all three medians meet at the point G. This idea of proving concurrency by showing that all the lines must go through a speciﬁc point will be used again in the proof of the Euler line below (Theorem G43 on page 258). Theorem G41. The three medians of a triangle meet at a point G, the centroid of the triangle. On each median, the distance of G to the vertex is twice the distance of G to the midpoint of the opposite side.

A

Bʹ

Cʹ

G M

N

B

C

Proof. We ﬁx the median issuing from B, to be called BB , B being the midpoint of AC. Let G be the point on BB so that |BG| = 2|GB |. We claim that the two medians issuing from A and C will meet BB at this G. Once this is done, we will have proved that all three medians meet at G. We will only consider the case of the median CC issuing from C, as the case of the median from vertex A is entirely similar. We will prove something that is equivalent to the preceding assertion, namely, that the point of intersection G of the segments CC and BB satisﬁes |BG| = 2|GB |. To this end, consider ABC. By Theorem G15 (page 219), C B BC and |C B | = 12 |BC|. Next, consider GBC. If M and N are midpoints of the two sides GB and GC, respectively, then by Theorem G15 again, M N BC and |M N | = 12 |BC|. Therefore, from C B BC

and M N BC,

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

257

we conclude that C B M N because two distinct lines parallel to a third are parallel to each other (Lemma 4.3 of [Wu2020a]; see page 357 of this volume). Also |C B | = |M N | (= 12 |BC|). By Theorem G14 (page 218), M N B C is a parallelogram and therefore the diagonals M B and N C bisect each other (Theorem G13, page 218). Thus, |GB | = |M G|, but since M is the midpoint of BG, we have |BM | = |M G| = |GB |, which is equivalent to |BG| = 2|GB |. The proof of Theorem G41 is complete. Remarks. 1. The preceding proof makes use of the concept of congruence but not similarity. This fact is of some importance, pedagogically, because we would like to be able to prove the concurrence of the medians for school students before they learn about the more subtle concept of similarity. However, if we can indulge ourselves and are free to make use of similarity, then the proof can be slightly simpliﬁed, as follows. We let the medians BB and CC meet at G as before. Then (without bothering with M and N ) because the lines BC and C B are parallel, the triangles GB C and GBC have three pairs of equal angles (on account of Theorem G18 on page 219) and are therefore similar because of the AA criterion of similarity (page 220). Therefore |BG| |BC| = =2 |B G| |B C | and we are done. 2. One proves in physics that the centroid of a triangle is the center of gravity of the triangle, in the following sense. Let a triangle be made of material that is homogeneous (completely uniform) and of the same thickness; then one can balance this triangle by allowing it to rest on the centroid. Mathematical Aside: Note that BB and CC must meet, as a simple application of the crossbar axiom (page 217) would show. However, we will be somewhat sparing in mentioning this kind of intuitively obvious but technical detail at this stage, because the geometric content of the theorems becomes increasingly substantial and we have to concentrate on that instead. The next point of a triangle to be brought up is very similar to the circumcenter, except that instead of taking the perpendicular bisectors of the sides of the triangle, we take the bisectors of the angles. Theorem G42. The three angle bisectors of a triangle meet at a point, the incenter of the triangle. The incenter is equidistant from the three sides. The proof is so similar to that of Theorem G30 (page 226) that we will leave it as Exercise 1 on page 266 (compare Exercise 8(b) in Exercises 5.3 of [Wu2020a]). Recall that the angle bisector of an angle is a ray lying in the angle issued from the vertex of the angle. In outline, one ﬁrst proves the analog of Theorem G27, to the eﬀect that the bisector of an angle < 180◦ is the set of all points in the angle equidistant from both sides of the angle. The rest of the reasoning then follows that of Theorem G30. In case you have noticed the absence of any reference to the analog of part (ii) of Theorem G30, see Exercise 3 on page 294.

258

6. BASIC THEOREMS OF PLANE GEOMETRY

Remark. In an analogous way, one proves that the bisector of an angle of a triangle and the bisectors of the other two exterior angles meet at a point (Exercise 2 on page 266). This is called an excenter of the triangle. A triangle has three excenters. The Euler line Let us begin to explore the relationship between four of the points of concurrency deﬁned thus far: the circumcenter, the orthocenter, the centroid, and the incenter. (More points of interest of a triangle will be deﬁned later on, e.g., pp. 294, 300, and 302.) The ﬁrst is both surprising and not too diﬃcult to prove, and one wonders why it is not in all the school textbooks. It involves only three of the four points explicitly, but the fourth point (the incenter) will actually enter in an implicit but spectacular way. See the discussion at the end of Section 6.9 on page 304 below. Theorem G43 (Euler line). Let H be the orthocenter, G the centroid, and O the circumcenter of a triangle ABC. (i) If G = O, then H = G = O and ABC is equilateral. (ii) If G = O, then H lies on the ray ROG , H ∗G∗O, and |HG| = 2|GO|; i.e., the distance from the centroid to the orthocenter is twice the distance of the centroid to the circumcenter. Remark. The line LHO is called the Euler line19 of ABC. The classical literature appears to focus only on the line LHO , but there are times when we will be interested explicitly in the segment HO. For this reason, we will use Euler segment to refer to HO when the occasion calls for it. In this terminology, we may rephrase part (ii) of Theorem G43 as follows: If in ABC, G = O, then also H = O and G lies in the Euler segment so that |HG| = 2|GO|. It will be seen that another point of great interest, the center of the nine-point circle (see Theorem G62 on page 302) also lies on the Euler segment. Note that there is no “standard” way for proving that three given points are collinear. Sometimes one can show that the angle formed by the three points is a straight angle. Other times one can show that the line joining two of the three points is characterized by certain properties and then show that the third point possesses those properties. In special situations, one may even be lucky enough to invoke Menelaus’s theorem (see Exercise 14 on page 268) to show directly that the points are collinear. The following proof is a bit subtle and does not fall into any of these categories. The proof relies on a simple lemma, and we will begin with that. Lemma 6.2. Let P Q be a chord on a circle C. Then the perpendicular line from the center of C to P Q is the perpendicular bisector of P Q. Proof of lemma. Let the center of the circle C be O. The perpendicular line from O to P Q is then the altitude from the top vertex of the isosceles triangle P OQ to P Q and is therefore the perpendicular bisector of the base P Q, by Theorem G26(b) (page 221). The proof is complete.

O

P

Q

19 We have encountered Leonhard Euler (1707–1783) before in connection with Mersenne primes in Section 6.1 of [Wu2020a].

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

259

Proof of Theorem G43. We ﬁrst deal with the trivial case, case (i), where G = O. Because G cannot lie on AB, BC, or AC (see Theorem G41 on page 256), we may drop a perpendicular from G to a side, let us say, BC; let intersect BC at A .

A @ @ @ rG = [email protected] B

A

@ @

C

Since G = O, is also the perpendicular from the circumcenter O to BC; therefore is the perpendicular bisector of BC (by Lemma 6.2) and A is the midpoint of BC. Consequently, rewriting as LGA , we see that LGA is the perpendicular bisector of BC. Now AA is a median of ABC and G must lie on the AA . Thus the lines LAA and LGA have two points in common: G and A . By assumption (L1) (page 214), the lines coincide; i.e., LAA = LGA . It follows that LAA is the perpendicular bisector of BC. In particular, since A lies on this perpendicular bisector, |AB| = |AC| (by Theorem G27 on page 222). In like manner, we also get |BA| = |BC| and |CA| = |CB|. This proves that ABC is equilateral and part (i) is proved. Next, we prove part (ii). Let G = O. On the ray ROG , let H be the point so that H ∗ G ∗ O and |H O| = 3|GO|.

(6.19)

H q

Gq

Oq

We claim that H is the orthocenter of ABC. Assuming the claim and observing that the equality |H O| = 3|GO| is equivalent to the equality |H G| = 2|GO|, we see that we have proved Theorem G43. To prove the claim, it suﬃces to show that any altitude of the triangle contains the point H . Without loss of generality, we consider the altitude from the vertex A to BC and will prove that this altitude contains H . Now if H = A, then for sure the altitude from A contains H . Therefore we may henceforth assume that H = A and we have a triangle GAH . Let the perpendicular from O to BC intersect BC at A ; observe that A is the midpoint of BC (see Lemma 6.2). See the following pictures.

260

6. BASIC THEOREMS OF PLANE GEOMETRY

A Hʹ

G

O

A

Hʹ

G

B

Aʹ

C

B

Aʹ

C

O Thus G lies on the median AA and by Theorem G41 (page 256), |AG| = 2|A G|. From (6.19), we also have |H G| = 2|OG|. We conclude that |AG| |H G| = = 2. |A G| |OG| Since |∠AGH | = |∠A GO| because they are opposite angles at G (Lemma 5.1 in [Wu2020a]; stated on page 357 of the present volume), we get the triangle similarity AGH ∼ A GO on account of the SAS criterion for similarity (Theorem G21 on page 220). Therefore |∠H AG| = |∠OA G|. Now the fact that H ∗ G ∗ O guarantees that O and H lie in opposite half-planes of LAA . Thus ∠H AG and ∠OA G are alternate interior angles20 of the transversal LAA of the two lines LOA and LAH . Hence LOA LAH because of the equality of alternate interior angles (Theorem G19 on page 219). Since LOA ⊥ LBC , we conclude that LAH ⊥ LBC (Theorem G2 on page 218). So LAH is the altitude from A after all and H lies in the altitude from A. This proves the claim and hence, by a previous remark, also Theorem G43. Remark. Implicit in this proof is a second proof of the concurrency of the altitudes of a triangle(Theorem G40). Indeed, all three altitudes pass through the point H deﬁned above (6.19) on page 259. Pedagogical Comments. This is such an important theorem—because it shows the internal coherence of the geometry of the triangle—that we should ﬁnd ways to let students see at least the main ideas of the proof. One virtue of this proof is that it allows students to appreciate the detailed knowledge they have gathered in Theorem G41 (page 256) about the centroid. A drastic suggestion (in case the lack of time is a factor) is to skip the proof of part (i) of Theorem G43 and concentrate on the proof of part (ii), which is where the main geometric ideas behind the proof reside. End of Pedagogical Comments.

20 See

the deﬁnition of alternate interior angles of a transversal in Section 5.2 of [Wu2020a].

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

261

Ceva’s theorem Our next theorem reveals an unexpected unifying idea behind the various phenomena of concurrency. It provides a uniform perspective on why the altitudes, the medians, and the angle bisectors must be concurrent. This is Ceva’s theorem, ﬁrst proved by Giovanni Ceva in 1678 (about the time that Newton and Leibniz discovered calculus). For its statement, recall that a segment AB consists of all the points N between A and B plus the endpoints A and B, i.e., all the points N so that A ∗ N ∗ B, together with A and B. Theorem G44 (Ceva’s theorem). Let L, M , and N be three points on the sides BC, AC, and AB, respectively. Then |AN | |BL| |CM | · · =1 |N B| |LC| |M A|

(6.20)

if and only if AL, BM , and CN are concurrent.

A

D

E

N

M P

B

L

C

Pedagogical Comments. Ceva’s theorem deserves to be known to all students, because it gives a totally new perspective on when three lines are concurrent (see, e.g., Exercise 9 on page 267). On the other hand, it would be pedagogically ill-advised to use it to give the only proofs of the concurrency of the altitude, the medians, and the angle bisectors—as it is done sometimes—because the standard proofs (pp. 255, 256, and 257) are more intuitive and are also more revealing of certain special properties of each of these lines. End of Pedagogical Comments. Proof. Let L, M , and N be three points on the sides BC, AC, and AB, respectively. We ﬁrst prove equation (6.20) under the assumption that AL, BM , CN meet at P . From B and C, let the lines parallel to LAL meet LCN and LBM at D and E, respectively. (Note that if, for example, the line through C parallel to LAL does not meet LBM , then this line would be parallel to both LAL and LBM , and we would have LAL LBM (by Lemma 4.3 of [Wu2020a]; see page 357 of the present volume); this would then contradict the fact that LAL and LBM meet at P .) Because AP DB, considerations of alternate interior angles and the AA criterion of similarity imply that DBN ∼ P AN so that |AP | |AN | = . |BN | |BD|

262

6. BASIC THEOREMS OF PLANE GEOMETRY

Similarly, from the fact that P L EC and the corollary to Theorem G11 on page 235, we have |BP | |BL| = . |LC| |P E| Finally, from AP EC, we have AP M ∼ CEM and therefore |CE| |CM | = . |AM | |AP | Altogether, we have |AN | |BL| |CM | · · |BN | |LC| |M A|

=

|AP | |BP | |CE| · · |BD| |P E| |AP |

|BP | |CE| · . |P E| |DB| But the last term is equal to 1 because DBP ∼ CEP , so that |BP | |DB| = . |EP | |CE| =

This proves equation (6.20). Acivity. Show that the preceding reasoning is enough to prove equation (6.20) for any L, M , and N that lie on the lines LBC , LAC , and LAB ; i.e., they do not need to lie in the segments BC, AC, and AB, respectively. Next we prove the converse; namely, if L, M , and N are points on the sides BC, AC and AB of ABC, respectively, so that equation (6.20) holds, i.e., |AN | |BL| |CM | · · = 1, |N B| |LC| |M A| then LAL , LBM , and LCN are concurrent. Suppose the segments AL, BM , and CN are not concurrent. Let BM and CN meet at a point P . Because ABC (as a region in the plane)—being the intersection of the three angles—is convex, P is a point in ABC.

A

N

B

P

L0 L

M

C

Consequently, if we let the ray RAP intersect LBC at a point L0 ,21 then we know L0 lies in the segment BC. By assumption, L also lies in the segment BC and since we are assuming the lines AL, BM , and CN are not concurrent, L0 = L. We will need the following lemma. 21 Mathematical Aside: This is entirely obvious geometrically, but formally we have to use the fact that P ∈ ∠BAC and invoke the crossbar axiom (page 217) at this point.

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

263

Lemma 6.3. Let BC be a segment and let Q, Q0 be points in the segment so that |BQ| |BQ0 | = . |QC| |Q0 C| Then Q = Q0 . B

Q0

Q

C

Proof. Applying the cross-multiplication algorithm to the hypothesis, we get |BQ| · |Q0 C| = |BQ0 | · |QC|. Now suppose Q = Q0 . We may assume without loss of generality that B ∗ Q0 ∗ Q, as shown above. Then |BQ| > |BQ0 | and |Q0 C| > |QC|, so that |BQ| · |Q0 C| > |BQ0 | · |QC|. The contradiction proves that Q = Q0 . There is another way to prove the lemma which may be of interest. From the hypothesis, we get |BQ0 | |BQ| =1+ . 1+ |QC| |Q0 C| The left side is equal to |QC| + |BQ| |BC| = . |QC| |QC| Similarly, the right side is equal to |BC|/|Q0 C|. Since the two sides are equal, we conclude that |QC| = |QC0 |. Since both Q and Q0 lie in BC, they are points in the ray RCB that are equidistant from C. By the uniqueness statement in assumption (L5)(ii) on page 215, we get Q = Q0 . The proof of Lemma 6.3 is complete. To return to the proof of the converse of Ceva’s theorem, since L0 = L, we may assume B ∗ L0 ∗ L. By the ﬁrst half of the theorem, the concurrency of AL0 , BM , and CN implies |AN | |BL0 | |CM | · · = 1. |N B| |L0 C| |M A| Comparing with equation (6.20) , we have |BL| |BL0 | = . |LC| |L0 C| Lemma 6.3 shows that L = L0 , a contradiction. Hence the lines LAL , LBM , and LCN are concurrent after all. The proof of Ceva’s theorem is complete. Remark. Ceva’s theorem can be generalized to the case where the points L, M , and N are allowed to lie on the lines LBC , LAC , and LAB rather than just in the segments BC, AC, and AB, respectively. But in that case, equation (6.20) no longer holds only for concurrent lines: the lines LBM , LAL , and LCN can be parallel (see Exercise 13 on page 267).

264

6. BASIC THEOREMS OF PLANE GEOMETRY

A M N A L

B B

C

L

C

P

N

M The complete proof of the general theorem is, however, a bit long. Since we have no critical need of the general case, we will give its proof in a ﬁle on the author’s homepage, https://math.berkeley.edu/~wu/. We now discuss how Ceva’s theorem implies the concurrency of the medians, altitudes, and angle bisectors of a triangle. The case of medians is straightforward because if A , B , and C are the midpoints of BC, AC, and AB, then obviously |AC | |BA | |CB | = = = 1, |C B| |A C| |B A| so that

|AC | |BA | |CB | · · = 1. |C B| |A C| |B A| Ceva’s theorem then implies that AA , BB , CC are concurrent. (But note that this proof does not give any information about the location of the centroid on each median, which is part of Theorem G41, page 256.) As for the altitudes, let the three altitudes of ABC be AD, BE, and CF . We will consider only the case where D, E, F lie in the sides BC, AC, and AB, respectively; the case of an obtuse triangle where two of these points lie outside the three sides can be treated in exactly the same way, although it will need the generalized version of Ceva’s theorem mentioned in the preceding remark.

A E F

B

D

C

(Note that in this picture, the three altitudes are intentionally drawn so that they appear not to be concurrent.) Now AF C ∼ AEB because these right triangles have angle A in common. Hence |AC| |AF | = . |AE| |AB|

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

265

Similarly, from ABD ∼ CBF , we have |BA| |BD| = |BF | |BC| and from ADC ∼ BEC, we have |CB| |CE| = . |CD| |CA| By rearranging the terms, we may rewrite |AF | |BD| |CE| · · |F B| |DC| |EA| as

|AF | |BD| |CE| · · . |AE| |BF | |CD| Therefore, using the preceding three equalities, we have |AC| |BA| |CB| |AF | |BD| |CE| · · = · · = 1. |F B| |DC| |EA| |AB| |BC| |CA|

By Ceva’s theorem, AD, BE, and CF are concurrent. The case of angle bisectors is more interesting. If the angle bisector of ∠A meets the opposite side BC at L, what do we know about |BL|/|LC|?

A

B

L

C

The answer is given in the following theorem. Theorem G45. Assume ABC. If the angle bisector of ∠A meets BC at L, then |AB| |BL| = . |LC| |AC| Postponing the proof for the moment, we ﬁrst use the theorem to show how the concurrence of the three angle bisectors of a triangle follows from Ceva’s theorem. With notation as above, let the angle bisectors from B and C meet AC and AB at M and N , respectively. Then applying Theorem G45 three times in a row, we get |AN | |BL| |CM | |AC| |AB| |BC| · · = · · = 1. |N B| |LC| |M A| |BC| |AC| |AB| So by Ceva’s theorem, the angle bisectors are concurrent. We now turn to the proof of Theorem G45. This proof is a good example of how, sometimes, a theorem can become very easy when it is looked at “the right way”. The proportion in the theorem is unlike anything we are familiar with, but we can try to convert it to something that looks like the proportions in FTS (page

266

6. BASIC THEOREMS OF PLANE GEOMETRY

219). Thus we extend CA along the ray RCA to a point D so that |AB| = |AD| (see the picture below). Then the proportion in the theorem becomes |DA| |BL| = , |LC| |AC| which is exactly what one ﬁnds in the FTS (see the corollary to Theorem G10 on page 231).

D A

B

L

C

This observation then suggests an approach to Theorem G45 via Theorem G11 (FTS*) on page 219. Proof of Theorem G45. Extend segment CA along the ray RCA to a point D so that D ∗ A ∗ C and |AB| = |AD|. By Theorem G26, |∠ADB| = |∠ABD|. By the exterior angle theorem (page 243), |∠BAC| = |∠ADB| + |∠ABD| = 2|∠ABD|. Since AL bisects ∠BAC, we see that |∠BAC| = 2|∠BAL|. It follows that |∠ABD| = |∠BAL|, and therefore AL DB because of equal alternate interior angles (Theorem G19 on page 219). By the corollary of Theorem G11 on page 235, we have |DA| |BL| = . |LC| |AC| Since |DA| = |AB| by construction, we have proved the desired proportion of Theorem G45. At the end of the next section, we will give a second proof of Theorem G45. Exercises 6.7. (1) (a) Prove that the angle bisector of a (convex) angle is the set of all the points in the angle that are equidistant from the sides of the angle. (b) Give the details of the proof of Theorem G42 on page 257. (2) (a) In ABC, prove that the angle bisector of ∠A and the angle bisectors of (either of) the exterior angles at B and C are concurrent. (This point of concurrency is an excenter of the triangle; see page 258. There are three excenters.) (b) Prove that the excenter in part (a) is equidistant from (the lines containing) the three sides of ABC. (3) (a) Prove the special case of Theorem G43 (Euler line) for an isosceles triangle directly without using Theorem G43. (b) What is the Euler segment (see page 258) of a right triangle? (You may refer to Exercise 6 on page 228.)

6.7. SOME BASIC PROPERTIES OF A TRIANGLE

267

(4) Prove that a triangle is equilateral ⇐⇒ its incenter coincides with its orthocenter. (5) Prove that a triangle is equilateral ⇐⇒ its circumcenter coincides with its orthocenter. (6) Let I be the incenter of ABC. Express |∠BIC| in terms of |∠A|. (7) (The following is just Exercise 9 on page 237. We are asking for a second proof.) Let D and E be points on AB and AC of a triangle ABC so that |AD|/|AB| = |AE|/|AC|. Prove that BE and CD intersect on the median from vertex A. (8) Let be a line passing through the centroid of ABC, with (let us say) A in one of its half-planes and B and C in the other. Prove that the distance from A to is equal to the sum of the distances from B and C to . (9) In ABC, let P be a point inside the triangle and let the lines LAP , LBP , and LCP meet BC, AC, and AB at D, E, and F , respectively. Let D , E , and F be points on BC, AC, and AB, respectively, so that |BD| = |D C|, |CE| = |E A|, and |AF | = |F B|. Prove that the three lines LAD , LBE , and LCF are also concurrent. (10) (a) Prove that the sum of (the lengths of) the medians of a triangle is greater than 34 of the perimeter of the triangle. (b) Can you think of a triangle in which the sum of its medians is “almost equal” to 34 of its perimeter? (Compare Exercise 9 on page 253.) (11) (a) Prove that if two medians of a triangle are equal, the triangle is isosceles; i.e., if B is the midpoint of AC and C is the midpoint of AB and if |BB | = |CC |, then |AB| = |AC|. (b) In ABC, let BB and CC be the medians from B and C. Prove that AC is longer than AB if and only if BB is shorter than CC . (Note: Naturally, part (b) implies part (a), but part (a) is easier and can be proved directly.) (12) The median of a triangle from vertex A of ABC is the angle bisector of ∠A if and only if |AB| = |AC|. (13) Given ABC, suppose parallel lines passing through A, B, and C intersect the lines LBC , LAC , and LAB at L, M , and N , respectively (see picture below). Prove that |AN | |BL| |CM | · · = 1. |N B| |LC| |M A|

N M A

B

L

C

268

6. BASIC THEOREMS OF PLANE GEOMETRY

(14) Prove Menelaus’s Theorem:22 The six segments determined by a transversal on the sides of a triangle have the property that the product of three nonconsecutive segments is equal to the product of the remaining three. Also formulate the converse statement carefully and prove it.

A

A

N

B

M

C

L

M B

C

L

N

(Hint: As in the case of Ceva’s theorem, it is perhaps easier to state the condition about the segments in the following way by referring to the above ﬁgures: |AN | |BL| |CM | · · = 1. |N B| |LC| |M A| Try to imitate the idea in the proof of Ceva’s theorem.) (15) Prove the converse of Theorem G45. More precisely, if L is a point on side BC of ABC so that |BL|/|LC| = |AB|/|AC|, then AL bisects ∠A. (16) Let the angle bisector of (either of) the exterior angle at A of ABC intersect the line LBC at M . Prove that |CM |/M B| = |CA|/|AB|. (b) With notation as in part (a), let the angle bisector of ∠A intersect BC at L. Prove that |BL| |CM | · = 1. |LC| |M B| (Note that B ∗ L ∗ C. The product on the left is called the cross ratio of the four points B, C, L, M . This concept plays a major role in projective geometry; see, for example, [Bix, Section 19].) (17) In ABC, let E and F be points on AB and AC so that |AE| = |AF |. Let AA be the median from A, and let AA intersect EF at P . Prove that |EP |/|P F | = |AC|/|AB|. (Hint: Consider the segment BK (K is on AC) so that BK EF , and let AA intersect BK at Q. What is |BQ|/|QK|?) (18) (a) Let A be the midpoint of side BC of ABC, and let M be the midpoint of the median AA . Let the ray RBM meet AC at E. Prove that |EC| = 2|AE| using Menelaus’s theorem (Exercise 14 above). (b) Find a second proof of part (a) without using Menelaus’s theorem.

22 Menelaus was an astronomer who spent his youth in Alexandria (the city where Euclid lived) around 100 AD. He was a contemporary of Ptolemy, another Alexandrian and the greatest astronomer of antiquity.

6.8. BASIC PROPERTIES OF THE CIRCLE

269

(19) (a) Let AA be the median of ABC so that A is the midpoint of BC. |AM | Let M be a point on AA so that |M A | = r for some positive number r. |AE| Let the ray RBM meet AC at a point E. Compute |EC| using Menelaus’s theorem (Exercise 14). (b) Find a second way to do part (a) without using Menelaus’s theorem. (20) Use coordinates to prove that the three medians of a triangle meet at a point, as follows. We may assume that the vertices of the triangle are A = (a, a ) (a > 0), B = (0, 0), and C = (c, 0) (c > 0); i.e., A is above the x-axis, B is the origin, and C lies on the positive x-axis. Let the midpoints of AB, AC, and BC be D, E, and F , respectively, and let BE and CD meet at a point G. Prove that A, G, and F are collinear by computing the coordinates of G and F .

6.8. Basic properties of the circle This section begins by providing proofs for the most mundane properties of a circle (its symmetry with respect to a diameter, the convexity of the corresponding disk, and the fact that a line and circle can meet at exactly two points or one point or no point). These properties are usually taken for granted, but as we shall see, they are not entirely trivial, especially the convexity of the disk. Next, we prove the elementary but spectacular theorem that all angles subtended by a given arc on a circle are equal. This theorem has many consequences, including necessary and suﬃcient conditions under which four points are concyclic, i.e., lie on a circle (recalling that any three noncollinear points lie on a unique circle, by Theorem G30 on page 226). When two lines passing through a given point P outside a circle C intersect C at four points, these four concyclic points turn out to be of special interest for the geometry of C and P . We will explore this situation a bit, and also the situation where one of the lines becomes tangent to C. Elementary properties of a circle (p. 269) Angles on a circle subtending the same arc (p. 276) Four concyclic points (p. 282) Tangents from a point to a circle, redux (p. 289) New proof of Theorem G45 (p. 293) Elementary properties of a circle Recall that the circle C of radius r and center O is the set of all points P in the plane so that |OP | = r (it will always be understood that r ≥ 0). Still with C, we also deﬁned the closed disk D of a given circle C to be all the points Q so that |OQ| ≤ r. This deﬁnition has to be understood in the context of what was pointed out in Section 4.1 of [Wu2020a], namely, the fact that in school mathematics, the word “circle” can mean both a closed disk and its circular boundary. In this section, we cannot aﬀord to have this confusion and therefore must draw a distinction between a “circle” and the “closed disk” enclosed by the circle.

270

6. BASIC THEOREMS OF PLANE GEOMETRY

From the deﬁnition of rotation (see page 355), a circle has maximum rotational symmetry, in the sense that if is any rotation around the center of a given circle C, then maps C onto itself; i.e., (C) = C. The next property about the intrinsic symmetry of a circle may be just as intuitive, but we had better prove it. Theorem G46. A circle is symmetric with respect to any line passing through its center; i.e., the reﬂection Λ across any line passing through the center of a circle C maps C onto itself: Λ(C) = C. Proof. As usual we have to prove two things: (i) Λ(C) ⊂ C and (ii) C ⊂ Λ(C). Let O be the center of C and be a line containing O, as shown: Pq '$ q C O q P = Λ(P ) &% To prove (i), we have to show that if P is a point on C, then Λ(P ) is also a point on C. Let us denote Λ(P ) by P . Since Λ is a congruence and Λ(O) = O, |OP | = |Λ(OP )| = |OP |. Therefore P and P lie on the same circle centered at O; i.e., Λ(P )(= P ) lies on C. Next, we prove (ii). If P is on C, we must prove P = Λ(Q) for some Q on C. With P = Λ(P ) as above, the fact that Λ ◦ Λ is the identity transformation implies that P = Λ(P ). We have just seen that both P and P lie on C, so letting Q = P gets the job done. The proof is complete. The converse of this theorem also holds: if a circle has bilateral symmetry with respect to a line L (i.e., the reﬂection of L maps the circle to itself), then L passes through the center of the circle (see Exercise 1 on page 294). Our next goal is the following theorem, which is usually taken for granted without proof. You will see that its proof requires the corollary of Theorem G33 on page 246 (the larger angle in a triangle faces the longer side) and is therefore not entirely trivial. This theorem should not be taken for granted at all. Theorem G47. A closed disk D is convex. Proof. Let the center of C be O and let its radius be r, and let A, B ∈ D. We have to show that the segment AB lies in D. If O, A, B are collinear, there is nothing to prove. Assume therefore that the three points are not collinear. There are two possibilities. Case I. One of ∠OBA and ∠OAB has degree ≥ 90. Case II. Both angles are acute. First consider Case I. We may let |∠OBA| ≥ 90◦ . To show that the segment AB lies inside D, let C ∈ AB, C = A, B, and we need to prove that |OC| ≤ r.

6.8. BASIC PROPERTIES OF THE CIRCLE

271

O

A

B C

By assumption, |OA| ≤ r, so if we can prove that |OC| < |OA|, we would be done. But by the exterior angle theorem (Corollary 1 of Theorem G32 on page 243), |∠OCA| > |∠OBA| ≥ 90. Thus ∠OCA is obtuse, and as there is at most one obtuse angle in a triangle (Theorem G32 on page 238), we see that in OCA, |∠OCA| > |∠OAC|. Therefore the corollary of Theorem G33 (page 246) implies that |OA| > |OC|, as desired. Now consider Case II, where both ∠OAB and ∠OBA are acute. Again, if C ∈ AB, we have to prove that |OC| ≤ r. We may as well assume C = A, B.

O

A

C

B

We observe that one of the angles ∠OCB and ∠OCA has to be ≥ 90◦ because the sum of their degrees is 180◦ . For deﬁniteness, let us say |∠OCB| ≥ 90◦ and the reasoning above shows that ∠OCB is the angle in OCB with the largest degree. Therefore |OB| > |OC| by the corollary to Theorem G33, page 246. Since |OB| ≤ r, we have |OC| < r. The proof of Theorem G47 is complete. Next, we prove another rather obvious property of the circle, which again depends on the corollary of Theorem G33 (page 246). Theorem G48. A circle and a line meet at no more than 2 points. Proof. Let the given circle be C with center O and radius r, and let the given line be L. Suppose there are at least three distinct points A, B, C in C ∩ L. Since these three points lie on the line L, we may assume without loss of generality that A ∗ B ∗ C. If O ∈ L, then the fact that |OA| = |OC| = r implies O is the midpoint of the segment AC. Since B lies in AC and |OB| = r, either B = A or B = C,

272

6. BASIC THEOREMS OF PLANE GEOMETRY

contradicting that A, B, and C are distinct. So we may assume O is not on L, as shown:

q A

Oq [email protected] D @ D @ D @ D @ Dq @q B C

L

The exterior angle theorem (page 243) implies that |∠OBA| > |∠OCA|. Since all three points A, B, C are also on C, we have |OA| = |OB| = |OC|. In particular, OAC is isosceles so that |∠OCA| = |∠OAC| or, equivalently, |∠OCA| = |∠OAB|. Thus in OAB, we have |∠OBA| > |∠OAB|. However, OAB is also isosceles, so |∠OBA| = |∠OAB|. Contradiction, and the theorem is proved. Notice that the theorem does not guarantee that there is a line that meets a given circle at 0, 1, or 2 points. All it says is that these are the only possibilities. Now if a circle is given, it is clear that there is a line meeting it at 0 points, i.e., not meeting it at all (exercise). It is also easy to see that there is a line meeting it at exactly 2 points: take two points on the given circle and the line joining them will have the requisite property because Theorem G48 says the line and the circle cannot meet at another point. Showing that there is a line meeting the circle at exactly 1 point, however, takes a bit of work. This will be the content of the next theorem. A line that meets a given circle C at exactly one point is called a tangent line of C at that point. We are going to construct a tangent line to a given circle at a pre-assigned point of the circle, and the next theorem tells us how. Recall the standard terminology: the segment joining the center of a circle to a point P on the circle is called the radius of the circle at P .23 Theorem G49. Let P be a point on a circle C. A line containing P is a tangent line to C at P ⇐⇒ it is perpendicular to the radius of C at P . If we assume this theorem for a minute, then because through a point on a line there is only one line perpendicular to the given line, we see that the tangent line to a circle at a given point is unique. This allows us to speak of the tangent line at a point of a circle. The proof of Theorem G49 can be given as another straightforward application of the corollary to Theorem G33 (page 246), but for a change of pace, we will invoke the Pythagorean theorem (page 220) instead. Proof. First assume a line L is tangent to the circle C at a point P . If the center of C is O, we have to prove that L ⊥ OP . Suppose not; let the perpendicular from O to L meet L at Q, where Q = P . 23 Notice the abuse of language: up to this point, the “radius” of a circle has been used to refer to the length of a segment joining the center to a point on the circle, but now the same “radius” is used to mean such a segment.

6.8. BASIC PROPERTIES OF THE CIRCLE

273

O

Q

L

P

Let Λ be the reﬂection across LOQ . Because L ⊥ LOQ , Λ(L) = L by the deﬁnition of the reﬂection Λ. Since LOQ passes through the center O of C, also Λ(C) = C (Theorem G46 on page 270). Therefore, since P lies in L ∩ C (the intersection of L and C), the point P = Λ(P ) being in Λ(L) ∩ Λ(C) (= L ∩ C) also lies in L ∩ C. Now, since P lies in a half-plane of LOQ , P lies in the other half-plane of LOQ and in particular, P = P . Thus L ∩ C includes two distinct points. Contradiction. Conversely, if a radius of C is perpendicular to a line L at P ∈ C, take any point Q ∈ L so that Q = P . By the Pythagorean theorem, the hypotenuse |OQ| is greater than the radius |OP | of C and therefore Q ∈ C so that C and L intersect only at P . Hence L is tangent to C, and Theorem G49 is proved. We now take a closer look at circles by investigating the arcs on a circle. Given a circle C, let P , Q be distinct points on C. Then the segment P Q is called a chord of C. If P Q contains the center C of C, it is called a diameter of C.24 The following is a straightforward consequence of the convexity of the closed disk (Theorem G47, page 270), and we leave its proof as an exercise (Exercise 9 on page 295). Lemma 6.4. If P , Q are distinct points on a circle C, then the closed disk D satisﬁes P Q = LP Q ∩ D. Without doubt, your ﬁrst reaction to Lemma 6.4 is, “Why bother?” This can only be because you are already biased by the preconception that the circle is “round”. The following picture shows you how P Q can fail to be LP Q ∩ D when D (the shaded ﬁgure) is not “round”:

P

Q

24 Be forewarned that “diameter” is usually used also to mean the length of a chord passing through the center. Cf. the preceding footnote about the abuse of “radius”.

274

6. BASIC THEOREMS OF PLANE GEOMETRY

Therefore to prove Lemma 6.4, you will have to learn to make explicit use of Theorem G47. Let P Q be a chord of a circle C. Let H+ and H− be the half-planes of the line LP Q and let H+ = H+ ∪ LP Q and H− = H− ∪ LP Q be the corresponding closed half-planes. Then each of the subsets (H+ ∩ C)

and (H− ∩ C)

of C is called an arc of C determined by P Q. In symbols, P Q. The points P and

Q are called the endpoints of P Q. As in the case of angles, the symbol P Q has a built-in ambiguity because it could refer to either of the two sets, H+ ∩ C or H− ∩ C. In general, it is usually obvious in context which arc is meant, or if there is danger of confusion, there are ways to get around it. For example, if A is a point in one

of these arcs, then the notation P AQ would clearly indicate the upper arc in the following picture:

A

Q

P

If P Q is a diameter, then both arcs are congruent to each other because the reﬂection across LP Q maps one arc onto the other (Theorem G46, page 270); these arcs are called semicircles of C. The following characterizes diameters among the chords of a circle. Theorem G50. Let P Q be a chord on a circle C and let A ∈ C be distinct from P and Q. Then |∠P AQ| = 90◦ ⇐⇒ P Q is a diameter.

A M P

O

Q

Proof. We ﬁrst prove that if P Q is a diameter of C, then ∠P AQ is a right angle. As usual, let O be the center of C. Then O ∈ P Q by hypothesis and, because OA, OP , and OQ are all radii, |OA| = |OP | = |OQ|. From the isosceles triangles OP A and OQA, we get |∠OP A| = |∠OAP |,

|∠OAQ| = |OQA|.

6.8. BASIC PROPERTIES OF THE CIRCLE

275

On the other hand, the angle sum of P AQ is 180◦ , so |∠AP O| + (|∠P AO| + |∠OAQ|) + |∠OQA| = 180◦ , which implies

2(|∠P AO| + |∠OAQ|) = 180◦ . Since O ∈ ∠P AQ (because O ∈ P Q), we see that ∠P AO and ∠OAQ are adjacent angles with respect to ∠P AQ (see page 215 for the deﬁnition of adjacent angles). By (L6)(iv) (page 216), we have |∠P AO| + |∠OAQ| = |∠P AQ|. Thus |∠P AQ| = 90◦ . Conversely, suppose P Q is a chord on C so that ∠P AQ is a right angle. We must prove that P Q is a diameter of C. Let O be the midpoint of P Q and let M be the midpoint of AP . By Theorem G15 (page 219), OM AQ, and (as QA ⊥ AP ) consequently also OM ⊥ AP . Thus OM is the perpendicular bisector of AP and therefore O is equidistant from P and A (Theorem G27 on page 222). In like manner, O also lies on the perpendicular bisector of AQ and is equidistant from A and Q. Thus the midpoint O of P Q is in fact the circumcenter of AP Q (part (ii) of Theorem G30 on page 226). In particular P Q is a diameter of C. This completes the proof. As a ﬁrst application of Theorem G50, we prove the following deﬁnitive result about tangent lines to a circle. For its statement, we follow the time-honored tradition in geometry of using ambiguous terminology in order to achieve brevity: if a tangent line from a point P in the exterior of a circle C intersects C at the point B, then the tangent from P to C will refer to the segment P B. Theorem G51. From a point P in the exterior of a given circle C, there are exactly two segments from P tangent to C. Moreover, these two tangents from P to C have the same length. Proof. Let the center of C be O. We ﬁrst prove that there is at least one tangent line to C from P . Let the circle having OP as diameter intersects C at some point B. By Theorem G50, OB ⊥ P B. By Theorem G49 (page 272), LP B is tangent to C at B.

B E P

O

B Next, we prove that there are at least two tangent lines from P to C. We know that C is symmetric with respect to LOP (Theorem G46, page 270). So the reﬂection of B across LOP is a point B lying on C, and also OB ⊥ P B because OB ⊥ P B and reﬂection preserves angles by (L7) (page 217). Thus LP B is a second tangent from P to C (Theorem G49 again). We also observe that because reﬂection also preserves length (again by (L7)), |P B| = |P B |.

276

6. BASIC THEOREMS OF PLANE GEOMETRY

Finally we prove that there are no more than two tangent lines from P to C. Suppose E is another point on C so that LP E is a third tangent line from P to C. Then by Theorem G49 again, OE ⊥ P E. There is no loss of generality by assuming that E is in the same half-plane of LOP as B. Then we have |OB| = |OE| = radius of C, and of course |OP | = |OP |. Hence the right triangles OP B and OP E are congruent (Theorem G25 (HL) on page 220), and this implies |∠P OB| = |∠P OE|. Since E and B lie in the same half-plane of LOP , the rays ROE and ROB coincide by part (ii) of (L6) (page 216). Since |OB| = |OE|, we conclude that B = E on account of (L5)(ii) (page 215). This proves that there are no more than two tangents from P to C. The proof is complete. Mathematical Aside: In the preceding proof, we asserted that the circle with OP as diameter must intersect C, because, intuitively, a circle containing a point O inside a circle C and a point P in the exterior of C must intersect C. While this is intuitively obvious, its validity can only be aﬃrmed by invoking the completeness property (sometimes called the continuity property) of real numbers (see, e.g., Exercise 12.3 on page 116 of [Hartshorne]).25 Because we do not want to bring such considerations into the discussion of plane geometry at this point, the only way out is to make an explicit assumption to this eﬀect. Consider this done but, formally, we will stop short of an explicit declaration because we prefer to let such subtle issues about “continuity” slide and concentrate instead on the geometry. See Section 8.3, especially the discussion after the parallel postulate on page 343, for a discussion of related issues. Angles on a circle subtending the same arc Theorem G50 on page 274 makes the study of chords very easy in case the chord is a diameter. In a sense, diameters among chords are the counterparts of the horizontal and vertical lines among arbitrary lines: they are usually the annoying exceptions to general theorems but are easy to handle separately. As in the case of lines, we shall henceforth ignore diameters in the consideration of chords. If a chord P Q is not a diameter, then one of the half-planes of LP Q contains the center C of C, and the other does not. Then the arc (see page 274 for the deﬁnition) in the half-plane of the line LP Q that contains the center is called the major arc determined by the chord P Q, and the other is called the minor arc of P Q. (We note that these concepts are often not precisely deﬁned in TSM.) We will refer to

these arcs as opposite arcs. Let P Q denote one of these arcs; for deﬁniteness let us say it is the minor arc in the following picture on the left, and let A be a point on the opposite (major) arc which is distinct from either P or Q. Then ∠P AQ is

said to be an angle subtended by arc P Q, or by the chord P Q, on the circle 25 There is a discussion of the completeness of the real numbers in Section 2.1 of [Wu2020c]. To see how something that is so intuitively obvious can break down, imagine, for example, a “plane” consisting of ordered pairs of rational numbers; i.e., a “point” is an ordered pair (x, y), where x and y are rational numbers. The “unit circle” in this plane is then the collection of all the ordered pairs (a, b) where a and b are rational numbers satisfying a2 + b2 = 1. Now the segment joining the interior point (0, 0) of the “unit circle” to the exterior √ point √ (1, 1) will have no point of intersection with this “unit circle” because the ordered pair ( 2/2, 2/2) is not in this plane but this is the point that would have been the point of intersection.

6.8. BASIC PROPERTIES OF THE CIRCLE

277

C. The angle ∠P AQ is also called the inscribed angle intercepting the arc

P Q or intercepting the chord P Q.

A

O O Q

P

Q

P

A

There is a related concept. If O is the center of C, then ∠P OQ is called the

central angle subtended by the arc P Q, or by the chord P Q. For conceptual

clarity, it is well to note that in case the arc P Q refers to the major arc, then the angle subtended by this arc as well as the central angle it subtends are the indicated angles in the above picture on the right. In this case, the central angle subtended

by P Q is the nonconvex angle determined by ROP and ROQ ; in particular, it has > 180 degrees (see nonconvex angles on page 354). The following theorem—a generalization of Theorem G50—is among the theorems in geometry that are at once surprising and elementary. Theorem G52. Fix an arc on a circle C. Then all the angles subtended by this arc on the circle C are equal. More precisely, they are all equal to half of the central angle subtended by the arc.

Proof. Fix an arc P Q on a circle C with center O, and let ∠P AQ be an angle

subtended by P Q. Then we want to prove that |∠P AQ| =

1 2 |∠P OQ|

for any A

in the opposite arc of P Q, A = P, Q. Let us take up the case where P AQ is the

major arc; the case of P AQ being a minor arc is left as an exercise (Exercise 10 on page 295).

If P AQ is the major arc, then by deﬁnition, A and the center O are in the same half-plane of the line LP Q . There are three possibilities: (a) O lies on a side of ∠P AQ (see ﬁgure on the left), (b) O lies in ∠P AQ (see ﬁgure in the middle), and (c) O lies outside ∠P AQ (see ﬁgure on the right). A A

A

O

O Q

P

O

Q

B P

P

Q B

278

6. BASIC THEOREMS OF PLANE GEOMETRY

Consider case (a). For deﬁniteness, let us say O lies on AQ. So AOQ is a diameter. Then by the exterior angle theorem (page 243), the degree of the central

angle ∠P OQ subtended by P Q is equal to |∠P AO|+|OP A|, which is twice |∠P AO| because |OA| = |OP | = radius of C and therefore Theorem G26 (page 221) applies. Hence |∠P OQ| = 2|∠P AO| = 2|∠P AQ|, as claimed. Next we tackle case (b). Thus O lies in ∠P AQ, and so does the ray RAO . Let RAO meet the circle C at a point B, and let B also lie in ∠P AQ. Since ∠P AB and ∠BAQ are adjacent angles with respect to ∠P AQ, we get (by (L6)(iv) on page 216) that (6.21)

|∠P AB| + |∠BAQ| = |∠P AQ|.

Since AOB is a diameter, by case (a), we have (6.22)

|∠P OB| = 2|∠P AB|,

|∠BOQ| = 2|∠BAQ|.

Now B also lies in ∠P OQ, so we also have (6.23)

|∠P OB| + |∠BOQ| = |∠P OQ|.

Putting (6.23) and (6.22) together, we get |∠P OQ| = 2(|∠P AB| + |∠BAQ|). By (6.21), we get |∠P OQ| = 2|∠P AQ|, as desired. Finally, consider case (c), where O does not lie in ∠P AQ. Let RAO meet the circle at B as usual; then B does not lie in ∠P AQ. Either Q lies in ∠P AB (as shown in the above ﬁgure on the right) or P lies in ∠BAQ, as shown here:

A

O B P

Q

Without loss of generality, we may assume that P lies in ∠BAQ. Thus ∠BAP and ∠P AQ are adjacent angles with respect to ∠BAQ, and therefore |∠BAP | + |∠P AQ| = |∠BAQ|. Equivalently, (6.24)

|∠P AQ| = |∠BAQ| − |∠BAP |.

Similarly, since P lies in ∠BOQ, (6.25)

|∠P OQ| = |∠BOQ| − |∠BOP |.

But by case (a), we know that |∠BOQ| = 2|∠BAQ|,

|∠BOP | = 2|∠BAP |.

6.8. BASIC PROPERTIES OF THE CIRCLE

279

Therefore, (6.25) implies that |∠P OQ| = 2(|∠BAQ| − |∠BAP |). But by (6.24), the right side is equal to 2|∠P AQ|. We conclude that |∠P OQ| = 2|∠P AQ|. The proof of Theorem G52 is complete. Pedagogical Comments. The preceding proof is very instructive, not only because it involves substantive geometric ideas, but also because it provides another good illustration of how a complicated proof can be broken down into several simpler steps (three, in this case). Compare the comment on the proof of SSS on page 223. This reasoning deserves to be learned by all school students. As is often the case, however, the straightforward argument—for valid pedagogical reasons—hides some tedious details. We will now ﬁll in these details after the fact. To begin with, although it is pictorially obvious that, in the proof of case (b), B lies in ∠P OQ, strictly speaking we have to give a proof that such is the case. To this end, let the ray RAB intersect the segment P Q at M (this is guaranteed by the crossbar axiom); i.e., we have P ∗ M ∗ Q. See the left picture below.

A

O

A

O

Q

B

M P

P

N Q

B We claim that we also have O ∗ M ∗ B. Indeed, since B is a point in the minor

arc P BQ, the center O of the circle and B are on opposite sides of LP Q , by the deﬁnition of a minor arc. Therefore OB must intersect the line LP Q at some point M . But we already know that RAB intersects the segment P Q at M , and if M = M , it would imply that the lines LOB and LP Q intersect at two distinct points and must therefore coincide, which is impossible. Thus M = M and OB intersects LP Q at M . Since P ∗ M ∗ Q, we see that the segment OB intersects the segment P Q at M . We can now show B lies in the convex angle, ∠P OQ (see page 352 for the deﬁnition of convex angle). We begin by showing that P and M lie on the same side of LOQ . This is so because we have P ∗ M ∗ Q. If P M contains a point Q of LOQ , then the lines LP M and LOQ intersect at two distinct points Q and Q and must therefore coincide, a contradiction. So P M does not intersect LOQ , and by (L4)(ii) (page 215), P and M lie on the same side of LOQ . Likewise, because O ∗ M ∗ B, M and B lie on the same side of LOQ. It follows that P , M , and B all lie on the same side of LOQ. In particular, P and B lie on the same side of LOQ. Since B ∗ M ∗ O and Q ∗ M ∗ P , the same reasoning shows that B and Q lie on the same side of LOP . Together, we have B lying in the convex angle ∠P OQ.

280

6. BASIC THEOREMS OF PLANE GEOMETRY

Next, in the proof of case (c), we claimed that either P lies in ∠BAQ or Q lies in ∠P AB. Let us also prove this pictorially obvious fact. If the former holds, then we are done. If not, i.e., suppose P does not lie in ∠BAQ, then we have to prove that Q lies in ∠P AB, as shown in the right ﬁgure above. To this end, we will prove (i) P and Q are on the same side of LAB and (ii) B and Q are on the same side of LAP . To prove (i), we claim that the segments AB and P Q are disjoint; i.e., (6.26)

AB ∩ P Q = ∅ (the empty set that contains no elements).

If not, suppose AB and P Q intersect at a point C. Since C ∈ P Q, we have C ∈ ∠P AQ. It then follows easily from the deﬁnition of a ray (page 354) and the deﬁnition of a convex angle (page 352) that the ray RAC lies in ∠P AQ. Since C ∈ AB, AB ⊂ RAC (this is straightforward to check). Since O ∈ AB, we have O ∈ RAC and therefore O ∈ ∠P AQ. This contradicts the basic hypothesis in case (c) that O lies outside ∠P AQ, and (6.26) is proved. Next, we go a step further: we claim that, in fact, P Q and the line LAB are disjoint. If not, let us say P Q intersects LAB at a point D, and we will deduce a contradiction. If D lies in the segment AB, this would imply that P Q and AB have a point D in common, contradicting (6.26). Thus D ∈ AB. Now, according to Lemma 6.4 on page 273, AB is in fact the intersection of the closed disk of C with LAB . So D ∈ AB means D is not in the intersection of the closed disk of C with LAB . Given that D is a point of LAB , the only way for this to happen is that D does not lie in the closed disk of C. However, by the convexity of the closed disk of C (Theorem G47 on page 270), the segment P Q lies in the closed disk. Since D ∈ P Q, D must lie in the closed disk of C. This is a contradiction and therefore P Q and LAB are disjoint. By assumption (L4)(ii) on page 216, this shows that P and Q lie on the same side of LAB , thereby proving (i). To prove (ii), we claim that P and B lie on opposite sides of the line LAQ . To see this, recall that we are assuming P does not lie in ∠QAB. Now P being in ∠QAB means both of the following hold: (a) P and Q lie on the same side of LAB and (b) B and P lie on the same side of LAQ . Hence for P not to lie in ∠QAB, one of (a) and (b) must fail. In view of (i) above, (a) is true. So the only way for P not to lie in ∠QAB is for (b) to fail, which means either one of P and B lies on LAQ or P and B lie on opposite sides of LAQ . The former being obviously impossible, we conclude that P and B lie on opposite sides of LAQ ; i.e., the claim is true. It follows that the segment P B intersects LAQ at a point N , so that P ∗ N ∗ B. We claim that we also have A ∗ N ∗ Q. To this end, we ﬁrst show N ∈ AQ. By the convexity of the closed disk of C (Theorem G47), P B lies in this closed disk. Since N ∈ P B, N lie in this closed disk as well. However, by Lemma 6.4, the chord AQ is the intersection of LAQ and this closed disk. Therefore, N being in this intersection, N ∈ AQ. But N = A would mean P = A and N = Q would mean P = Q, and neither is possible. Hence we have A ∗ N ∗ Q. Finally, the fact that P ∗ N ∗ B shows, by the usual arguments, that B and N lie on the same side of LAP , and the fact that A ∗ N ∗ Q shows, likewise, that N and Q lie on the same side of LAP . Consequently, B, N , and Q all lie on the same side of LAP , thereby proving (ii). It follows that Q lies in ∠P AB. The other assertion in the proof of case (c) to the eﬀect that P lies in ∠BOQ can be proved in a similar manner. We have now ﬁlled in all the gaps in the preceding proof of Theorem G52.

6.8. BASIC PROPERTIES OF THE CIRCLE

281

The geometric reasoning in the proof of Theorem G52 on pp. 277–279 is truly basic. Students will have their hands full learning this reasoning and developing their geometric intuition in the process. It is best not to get them sidetracked by the abstract and subtle arguments in these Pedagogical Comments about things that they consider to be obvious. Let us face it: the things these abstract arguments try to prove are too obvious, pictorially. While we want teachers to know these arguments because teachers have to answer questions from inquisitive students, we have to be realistic and recognize that these arguments do not belong in a typical school classroom. From a pedagogical standpoint, getting students to learn the proof of Theorem G52 on pp. 277–279 has to be the primary focus; one would consider attending to the subtle details of the preceding arguments only if time and interest allow. It may be of some comfort for school students to know that not all professional mathematicians consider these formal abstract arguments to be easy or instructive, but they just learn to do them out of professional necessity. End of Pedagogical Comments. The theorem leads to a generalization that complements Theorem G50.

Corollary. Suppose P Q and P Q are congruent arcs on circles. Then they subtend equal angles on their respective circles; they also subtend equal central angles.

Proof of corollary. Let P Q lie on circle C with center O, let P Q lie on

circle C with center O , and let φ be the congruence that maps P Q to P Q . There are two possibilities: φ(P ) = P and φ(Q) = Q , or φ(P ) = Q and φ(Q) = P . The following argument works the same way for either situation, so let us assume φ(P ) = P and φ(Q) = Q for deﬁniteness. We claim that φ(C) = C . Indeed, a congruence maps a circle to a circle (of

the same radius), so φ(C) is a circle. Let A be a point on P Q and let A = φ(A).

Then A lies on P Q and φ(P AQ) = P A Q , so that φ(C) is a circumcircle of P A Q (= φ(P AQ)). But C is a circumcircle of P A Q by hypothesis, so the uniqueness of the circumcircle of a triangle (Theorem G30 on page 226) implies that φ(C) = C . In particular, φ(O) = O .

ʹ ʹ

ʹ ʹ

We now have φ(P ) = P , φ(Q) = Q , and φ(O) = O . Therefore φ(∠P OQ) = ∠P O Q , and since congruences preserve degrees of angles, |∠P OQ)| = |∠P O Q |. This proves the part of the corollary about the central angles. The assertion about the angles subtended by the arcs now follows from Theorem G52. The proof is complete.

282

6. BASIC THEOREMS OF PLANE GEOMETRY

Four concyclic points In applications, we often have to decide whether a collection of points is concyclic, which means lying on the same circle. Recall that any three noncollinear points lie on a unique circle, by Theorem G30 on page 226. So we may look at a geometric ﬁgure of four points as already having a circle passing through three of these points and asking if the fourth point lies on this circle. In this connection, the following theorem is basic. Theorem G53. Let four points A, B, C, D be given. (i) If A and C lie on the same side of the line LBD , then the four points are concyclic ⇐⇒ |∠BAD| = |∠BCD| (see left ﬁgure below). (ii) If A and C lie on opposite sides of the line LBD , then the four points are concyclic ⇐⇒ |∠BAD| + |∠BCD| = 180 (see right ﬁgure below).

A

A

C

D B

B

D C

Proof. (i) If A, B, C, D are concyclic, then the hypothesis on A and C implies that they lie on the same (major or minor) arc determined by B and D. The preceding theorem, Theorem G52, then shows that |∠BAD| = |∠BCD|. Conversely, suppose A and C lie on the same side of the line LBD and |∠BAD| = |∠BCD|. Then we have to prove that A, B, C, D are concyclic. Suppose not, and we shall deduce a contradiction.

C A

E

E

A C

B

D

B

D

Let C be the circle passing through A, B, and D (Theorem G30, page 226), and suppose C does not lie on C. There are two possibilities, as depicted by the pictures above: either C lies in the exterior of C or C lies inside C.26 The proofs for both situations are essentially the same, so we will take the case on the left; i.e., C is in the exterior of C. Then BC intersects C at a point E, so that B ∗ E ∗ C. 26 For

the deﬁnition of in the exterior of (respectively, inside) a circle, see p. 353.

6.8. BASIC PROPERTIES OF THE CIRCLE

283

Clearly A and E also lie on the same side of LBD and Theorem G52 implies that |∠BAD| = |∠BED|. Since B ∗ E ∗ C, ∠BED is an exterior angle of DEC (see page 243) and therefore bigger than its opposite interior angle ∠BCD (Corollary 1 on page 243). Together, we have |∠BAD| > |∠BCD|, and this contradicts the hypothesis that they are equal. Thus part (i) is proved. (ii) First suppose A, B, C, D are concyclic and A and C lie on opposite sides of LBD , and we will show |∠BAD| + |∠BCD| = 180◦ . There is no loss of generality

to let the arc BAD be the minor arc and to let the arc BCD be the major arc, as shown:

A D

O

B

C By Theorem G52, 1 (|nonconvex ∠BOD| + |convex ∠BOD|) 2 1 1 · |full angle at O| = (360◦ ) = 180◦ . = 2 2 Next we prove the converse. Suppose A and C are on opposite sides of LBD and |∠BAD| + |∠BCD| = 180◦ . We will show that C lies on the circle C passing through A, B, D. Suppose not; then again there are two cases: C lies in the exterior of C and C lies inside C. The proofs of both cases being entirely similar, we will only prove one of the two cases; let us say the latter. So let C be inside C, as shown in the right ﬁgure below. |∠BAD| + |∠BCD| =

A

A

D

D

B

B E

C

E

C

Let the ray RBC meet C at E, so that B ∗ C ∗ E. Since A and C lie on opposite sides of LBD , clearly so do A and E. The four points A, B, E, D being concyclic, the preceding result says |∠BAD| + |∠BED| = 180◦ .

284

6. BASIC THEOREMS OF PLANE GEOMETRY

Now C is between B and E, so ∠BCD is an exterior angle of the opposite interior angle ∠CED in CED. Therefore |∠BCD| > |∠BED| (Corollary 1 on page 243). Thus 180◦ = |∠BAD| + |∠BED| < |∠BAD| + |∠BCD| = 180◦ , where the last equality comes from the hypothesis. This contradiction then completes the proof of Theorem G53. We will agree to say that a quadrilateral is cyclic if its vertices are concyclic. Then part (i) of Theorem G53 can be interpreted as saying that for a quadrilateral ABCD whose vertices A and D lie on the same side of line LBC to be cyclic, it is necessary and suﬃcient that ∠BAC equals ∠BDC. Therefore Theorem G53 presents two diﬀerent characterizations of a cyclic quadrilateral: the ﬁrst in terms of the angles at two adjacent vertices formed by the rays issuing from each vertex to the two remaining vertices, and the second in terms of the sum of the angles at opposite vertices. Pedagogical Comments. The preceding proof is normally what you would present in a school classroom, but as usual, you should know the full story. We will break up this story into two parts. (1) A minor problem arose twice in the proof of Theorem G53: how did we know that, in proving the converses of part (i) and part (ii), the ray RBC will always intersect the circle C at a point E? This again is an issue related to the nature of the real numbers R and can be proved using the assumption that a circle containing a point inside and a point outside another circle must intersect that circle. See the Mathematical Aside on page 276 (see also Proposition 11.6 on page 108 of [Hartshorne]). Just as we noted in that Mathematical Aside, we will purposely ignore such subtle points about R and will choose to concentrate on the geometric content instead. (2) Part (ii) of Theorem G53 is usually presented in TSM as: “A quadrilateral is cyclic if and only if the sum of its opposite angles is 180 degrees.” The problematic nature of this statement can be illustrated by the following quadrilateral ABCD which is obtained from the cyclic quadrilateral ABC D by reﬂecting C to C across the line LBD .

ʹ Consider then the statement (6.27)

|∠BAD| + |∠BCD| = 180◦ =⇒ ABCD is a cyclic quadrilateral.

Because we automatically take the convex part of an angle, the hypothesis in (6.27), namely, |∠BAD| + |∠BCD| = 180◦ , is still correct because ∠BCD is equal to ∠BC D so long as we are taking only the convex part of an angle into account.

6.8. BASIC PROPERTIES OF THE CIRCLE

285

Such being the case, (6.27) is a false statement and therewith the theorem that the sum of opposite angles of a quadrilateral being 180◦ guarantees concyclicity is also false. A common remedy for this situation is to deﬁne the interior angle at a vertex of a quadrilateral, so that the interior angle of ABCD in the preceding picture at the vertex C will be the nonconvex angle. Then, with “angle” of a quadrilateral understood to be “interior angle”, the validity of the statement that “a quadrilateral is cyclic if and only if the sum of its opposite angles is 180 degrees” will be restored. However, it is far from simple to deﬁne the concept of an “interior angle” of a polygon (compare the discussion at the end of Section 4.2 in [Wu2020a]), and we will not attempt to do so here. End of Pedagogical Comments. There is an equivalent formulation of Theorem G53 that is often used. Assume a quadrilateral ABCD so that A and C lie on opposite sides of LBD . The exterior angles of DBA at A and the exterior angles of DBC at C are therefore well-deﬁned (see page 243). Then the exterior angles of ABCD at A are by deﬁnition the exterior angles of DBA at A. See the picture below. The same is true for the other exterior angles of ABCD at C. As in the case of triangles, we usually refer to the exterior angle of ABCD at A (respectively, at C) because the two exterior angles are equal.

C

B D

A

E

F By deﬁnition, ∠C is the opposite interior angle of the exterior angle of ABCD at the vertex A, and similarly for vertex C. Corollary. Assume a quadrilateral ABCD so that A and C lie on opposite sides of LBD . Then the exterior angle at A is equal to the opposite interior angle at C ⇐⇒ ABCD is a cyclic quadrilateral. The proof is suﬃciently simple to be left as an exercise (Exercise 12 on page 295). The following companion criterion for points to be concyclic is equally useful, but since its proof is no more than a repeated application of Theorem G52, it will also be left as an exercise (the same Exercise 12). Theorem G54. Let points B, D be given and let A1 , A2 , . . . , An be points in the same half-plane of LBD . Then B, D, A1 , A2 , . . . , An are concyclic ⇐⇒ the angles ∠BAi D are equal for all i. We now explore some of the vast ramiﬁcations of Theorem G52. Take a point P and let a circle C be given in the plane. Let a line passing through P intersect C at two points A and D (Theorem G48 on page 271). Without looking at any pictures, we may ask if the product |P A| · |P D| is always the same

286

6. BASIC THEOREMS OF PLANE GEOMETRY

independent of the line. (Naturally, you are not likely to ask this question until after much experimentation that suggests that the product seems not to depend on the line used.) Now there are two cases: P is inside C and P is in the exterior of C, as shown:

D P

D

.

A

.P A It turns out that the answer is always aﬃrmative in each case, and we can even give the exact value of the product in both cases. Theorem G55. (i) Let P be a point inside a circle C with center O and radius r, and let a line passing through P intersect C at A and D. Then |P A|·|P D| = r 2 −|OP |2 . (ii) Let P be a point in the exterior of a circle C with center O and radius r, and let a line passing through P intersect C at A and D. Then |P A| · |P D| = |OP |2 − r 2 . Proof. (i) Let P be inside C, and let the diameter passing through P intersect the circle at B and E.

D E P

.O A

B

By Theorem G52 (page 277), |∠E| = |∠A| and |∠D| = |∠B|. Therefore P DE ∼ P BA by the AA criterion for similarity (page 220). Therefore, by Theorem G20 (page 220), |P A| |P B| = . |P E| |P D| By the cross-multiplication algorithm, we have |P A| · |P D| = |P B| · |P E|. Now because BE is a diameter of C, |P B| = |BO| + |OP | = r + |OP |, |P E| = |OE| − |OP | = r − |OP |. Thus, |P A| · |P D| = |P B| · |P E| = (r + |OP |)(r − |OP |) = r 2 − |OP |2 and (i) is proved.

6.8. BASIC PROPERTIES OF THE CIRCLE

287

(ii) Now suppose P is in the exterior of a circle C with center O and radius r, and let a line passing through P meet C at A and D. Let the line joining P and O meet C at E and B, as shown:

D A

B

.

O

E

P

We have |∠B| = |∠D| because of Theorem G52, so that P ED ∼ P AB because of the AA criterion for similarity (the triangles share ∠P ). By Theorem G20 again, |P B| |P A| = . |P E| |P D| By the cross-multiplication algorithm, we have |P A| · |P D| = |P B| · |P E|. Now because BE is a diameter of C, |P B| = |OP | + |OB| = |OP | + r, |P E| = |OP | − |OE| = |OP | − r. Thus, |P A| · |P D| = |P B| · |P E| = (|OP | + r)(|OP | − r) = |OP |2 − r 2 and (ii) is proved and therewith Theorem G55. Each of the two parts in Theorem G55 has a converse. Let us take up part (i) ﬁrst. Let P be a point inside a circle with center O and radius r. If two chords BP E and AP D pass through P , then we know that |P A| · |P D| = |P B| · |P E| because they are both equal to r 2 − |OP |2 .

E D P A B We now prove the converse. Theorem G56. Let two lines intersect at P and let A, D be points on one line separated by P (in the sense that A and D lie on opposite sides of the other line) and let B, E be points on another line also separated by P . If |P A| · |P D| = |P B| · |P E|, then the four points A, B, D, E are concyclic.

288

6. BASIC THEOREMS OF PLANE GEOMETRY

Proof. Join B and D. Then the hypothesis that P separates A and D means the segment P A does not meet the line LBD (if it does, then the lines LP A and LBD meet at two points: D and a point in the segment P A, contradicting (L1) on page 214). Thus P and A lie in the same half-plane of LBD . For similar reasons, P and E also lie in the same half-plane of LBD . We therefore conclude that A and E lie on the same side of line LBD . Now ∠AP B and ∠EP D are equal because they are opposite angles. Moreover, we have |P B| |P A| = , |P E| |P D| which is an immediate consequence of |P A| · |P D| = |P B| · |P E| by the crossmultiplication algorithm. Therefore P AB ∼ P ED on account of the SAS theorem for similarity (page 220). By Theorem G20 (page 220), |∠P AB| = |∠P ED|. Since A and E lie on the same side of line LBD , it follows from part (i) of Theorem G53 (page 282) that A, B, D, E are concyclic, and the theorem is proved. Sometimes Theorem G56 is stated in terms of a quadrilateral. Assume a quadrilateral ABCD. Suppose we know that the diagonals AC, BD, as segments, intersect at a point P . Then Theorem G56 and part (i) of Theorem G55 imply the following statement: Let ABCD be a quadrilateral whose diagonals AC, BD intersect at a point P . Then ABCD is a cyclic quadrilateral if and only if |P A| · |P C| = |P B| · |P D|. We next take up the converse of Theorem G55 in the case where the point P is in the exterior of a circle C with center O and radius r. Let two lines through P intersect C at A, D and E, B, respectively. Without loss of generality, we may assume that A is between D and P and that E is between B and P .

D A O. B

E

P

Then |P A| · |P D| = |P E| · |P B| because by Theorem G55(ii) both are equal to |OP |2 − r 2 . The converse then states: Theorem G57. Suppose two lines intersect at a point P . Let A be between D and P on one line, and let E be between B and P on another. If |P A|·|P D| = |P E|·|P B|, then the four points A, D, B, E are concyclic. Proof. From |P A| · |P D| = |P E| · |P B| and the cross-multiplication algorithm, we get |P B| |P A| = . |P E| |P D|

6.8. BASIC PROPERTIES OF THE CIRCLE

289

The triangles P AB and P ED also share an angle, namely, ∠P . Therefore triangles P AB and P ED are similar because of the SAS theorem for similarity (Theorem G21 on page 220). By Theorem G20 (page 220), we have |∠D| = |∠B|. From the hypothesis that D ∗ A ∗ P and B ∗ E ∗ P , we see that D and P lie in opposite half-planes of LAE and, likewise, B and P lie in opposite half-planes of LAE , by assumption (L4)(ii) (page 215). It follows that D and B lie in the same half-plane of LAE . Since |∠D| = |∠B|, Theorem G52 implies that A, E, B, D are concyclic. The proof is complete. Tangents from a point to a circle, redux Theorem G57 is not the end of the story, however, because we can imagine the geometric conﬁguration in the preceding picture to evolve in the following way. Let C be the circle containing the four points A, D, B, E. Now hold the line P AD ﬁxed, but allow the line P EB to rotate counterclockwise around P until it becomes tangent to the circle C and so that B and E merge to become one point; i.e., B = E, as suggested by the picture below.

D

A

O. B

E P

B=E Intuitively, we would expect that |P A| · |P C| = |P E| · |P B| becomes |P A| · |P C| = |P B|2 at the end. Our next goal is to give a simple, direct proof of this assertion. We ﬁrst need a preliminary result which is of interest in its own right. Theorem G58. Let C be a circle tangent to LAP at P . Let P Q be a chord of C and let B be a point on C not lying in the convex angle ∠QP A. Then |∠AP Q| = |∠P BQ|.

B O.

P

Q Q B

A

O.

P

A

Proof. [See the Pedagogical Comments after the proof.] If the chord P Q is a diameter, the theorem is an immediate consequence of Theorems G49 and G50 (pp. 272 and 274, respectively). From now on, we will assume that the center O of C

290

6. BASIC THEOREMS OF PLANE GEOMETRY

does not lie on P Q. Then there are two cases: (i) the center O of C does not lie in (the convex) ∠QP A (see left ﬁgure above) or (ii) O lies in (the convex) ∠QP A (see right ﬁgure above).

First consider case (i). Let P Q denote the minor arc determined by P Q in this proof. We claim that

(6.28)

P Q= {C ∩ ∠QP A without P and Q}.

To see this, we begin with an observation. Recall that for a point to lie in ∠QP A, both of the following must hold: (a) Q and the point lie on the same side of LP A and (b) A and the point lie on the same side of LP Q . Since O is not in ∠QP A, one of (a) and (b) does not hold. But O and Q always lie on the same side of the tangent LP A (see Exercise 2 on page 294); thus the only possibility is that (b) does not hold; i.e., O and A do not lie on the same side of LP Q . Since neither O nor A can lie on LP Q , we conclude that O and A must lie on opposite sides of LP Q (as in left ﬁgure above).

Now suppose D is a point of P Q; note that D = P, Q by the deﬁnition of P Q. By the deﬁnition of minor arc (see page 276), O and D lie on opposite sides of LP Q . But we already know that O and A lie on opposite sides of LP Q , so (6.29)

A and D lie on the same side of LP Q .

Since a circle lies in a closed half-plane of a tangent line (see Exercise 2 on page 294), D and Q—being points of C distinct from P —lie on the same side of LP A . Together with (6.29), we see that D ∈ ∠QP A. We have therefore proved that the left side of (6.28) is contained in the right side. Conversely, suppose D is a point of C in ∠QP A, but D = P, Q. Then D and A are on the same side of LP Q . Since O and A lie on opposite sides of LP Q , we conclude that D and O lie on opposite

sides of LP Q . Recall that D is a point of C, so D is on the minor arc P Q (see the deﬁnition of minor arc on page 276), thereby proving that the right side of (6.28) is contained in the left side. The proof of (6.28) is complete. We can now prove the theorem for case (i). Thus B is a point of C not lying in ∠QP A, and we will prove |∠AP Q| = |∠P BQ|. In view of (6.28), this is equivalent to proving that for any B in the major arc of P Q, |∠AP Q| = |∠P BQ|. Thus B and the center O of C lie on the same side of LP Q . By Theorem G52 (page 277), |∠P BQ| does not depend on the location of B in the major arc. Consider therefore the diameter P B of C, as shown:

B

O Q P

A

6.8. BASIC PROPERTIES OF THE CIRCLE

291

Now B does lie in the major arc because, since P ∗ O ∗ B, the segment OB contains no point of LP Q (as otherwise LBP and LP Q would have two distinct points in common) and therefore O and B lie on the same side of LP Q . In other words, B is in the major arc. By Theorem G50 (page 274), ∠P QB is a right angle in P BQ. Therefore, |∠P BQ| + |∠BP Q| = 90◦ .

(6.30)

Since A and O lie on opposite sides of LP Q while O and B lie on the same side of LP Q , we see that A and B lie on opposite sides of LP Q . It is therefore clear that Q lies in the convex angle ∠AP B (see the deﬁnition of convex angle on page 352). Thus ∠AP Q and ∠QP B are adjacent angles with respect to ∠AP B, and (L6)(iv) (page 216) implies that |∠AP Q| + |∠QP B| = |∠AP B|. But by Theorem G49 (page 272), ∠AP B is also a right angle so that |∠AP Q| + |∠BP Q| = 90◦ . In view of (6.30), we get |∠AP Q| = |∠P BQ|. The theorem is proved for case (i). Consider now case (ii), where O lies in ∠QP A but the given point B on C does not lie in ∠QP A. Let E be a point of LP A so that E ∗ P ∗ A, and let U be a point in the major arc of P Q. Applying (6.28) to ∠QP E, we see that the part of C in ∠QP E is the minor arc of P Q plus P and Q. Therefore U , being in the major arc of P Q, does not belong to ∠EP Q and therefore by case (i), |∠EP Q| = |∠QU P |.

(6.31)

U

Q

B

E

O.

P

A

To return to B, since B does not lie in ∠QP A, it lies in ∠QP E and is therefore in the minor arc of P Q. Since U is in the major arc of P Q, clearly B and U lie on opposite sides of LP Q . Hence, by Theorem G53(ii) (page 282), we have (6.32)

|∠P BQ| = 180◦ − |∠P U Q|.

We can now ﬁnish the proof of the theorem. Since E ∗ P ∗ A, |∠AP Q| = 180◦ − |∠EP Q|. By equation (6.31), we have |∠AP Q| = 180◦ − |∠P U Q|. Comparing with equation (6.32), we ﬁnally get |∠AP Q| = |∠P BQ|. The proof of Theorem G58 is complete.

292

6. BASIC THEOREMS OF PLANE GEOMETRY

Pedagogical Comments. The preceding proof appears to be so long only because we have given detailed proofs about certain points being on the same side of LP Q or on opposite sides of LP Q . Such tedious arguments will not be welcome in a school classroom. Our suggestion, which is by now a standard one, is the following: explicitly cut out most of those arguments in a high school setting by relying on pictures. For example, can there be any doubt in students’ minds that if the angle between a chord and a tangent is acute, then the angle intercepts the minor arc of the chord? But this is all that (6.28) tries to say. Along this line, please read the comments in Section 4.7 of [Wu2020a]. End of Pedagogical Comments. The following is the theorem we are after. Theorem G59. Let P be a point in the exterior of a circle C and let line P B be tangent to C at B. If another line through P intersects C at A and D, then |P B|2 = |P A| · |P D|. Conversely, if a line through a point P in the exterior of a circle C meets C at two points A and D and if a point B ∈ C satisﬁes |P B|2 = |P A|·|P D|, then the segment P B is tangent to the circle C at B. Proof. Let P B be tangent to C at B and let line LP A intersect C at D. We may assume A is between D and P .

B

P D

A

Since D ∗ A ∗ P , the segment P D intersects LAB at A and, therefore, D and P lie in opposite half-planes of LAB (see assumption (L4) on page 215). We can now apply Theorem G58 to get |∠P BA| = |∠P DB|. Since P BD and P AB also share ∠P , we have P BD ∼ P AB because of the AA criterion for similarity (page 220). Thus by Theorem G20 (page 220), |P D| |P B| = , |P A| |P B| which is equivalent to |P B|2 = |P A| · |P D|. Conversely, suppose |P B|2 = |P A| · |P D| where LP B and LP A are lines which intersect the circle C at B and at A, D, respectively. We have to prove that LP B is tangent to C.

6.8. BASIC PROPERTIES OF THE CIRCLE

293

B

P

O A D

We know by part (ii) of Theorem G55 (page 286) that |P A| · |P D| = |OP |2 − r 2 , where r is the radius of C. Since |OB| = r, we have |P B|2 = |OP |2 − r 2 = |OP |2 − |OB|2 , which is equivalent to |OP |2 = |OB|2 + |P B|2 . The converse of the Pythagorean theorem (Theorem G24, page 220) now implies that OB ⊥ P B. By Theorem G49 (page 272), P B is tangent to C. The proof of Theorem G59 is complete. New proof of Theorem G45 We end this section by giving a new proof of Theorem G45 on page 265 in the last section about angle bisectors in a triangle. We need a simple lemma which is of independent interest: it has a bearing on the paradox of Rouse Ball that “every triangle is isosceles” (see Exercise 20 on page 296). Lemma 6.5. Suppose in ABC, AB = AC. Then the perpendicular bisector of BC meets the angle bisector of ∠A at the midpoint M of the arc on the circumcircle

of ABC which is the opposite arc of the arc BAC.

A

B

L

Aʹ

C

M Proof. Observe that A does not lie on the perpendicular bisector of BC, because if it did, AB = AC by Theorem G27 on 222, which would contradict the hypothesis that AB = AC. Let A be the midpoint of BC; then the perpendicular bisector of BC passes through A . Let this perpendicular bisector meet the circumcircle at M . The line LAM is now distinct from the perpendicular bisector LA M of BC. We claim that AM is the angle bisector of ∠A. This is because the line A M passes through the center of the circle so that the reﬂection across A M

294

6. BASIC THEOREMS OF PLANE GEOMETRY

is a congruence that maps BM to M C (Theorem G46 on page 270), so that by the corollary of Theorem G52 (page 277), ∠BAM and ∠M AC are equal because they subtend congruent arcs. This proves Lemma 6.5. Proof of Theorem G45. With notation as in the proof of Lemma 6.5, let the angle bisector of ∠A meet BC at L and meet the circumcircle of ABC at M ,

the midpoint of the opposite arc of the arc BAC (by Lemma 6.5). First observe that if AB = AC, then by Theorem G26 on page 221, the angle bisector of ∠A is the perpendicular bisector of BC, and Theorem G45 becomes trivial. We may henceforth assume that AB = AC, so that Lemma 6.5 becomes applicable. We claim that ABL ∼ AM C. This is because |∠BAL| = |∠M AC| and because

|∠ABL| = |∠AM C| (both subtend AC). Thus |AM | |AB| = . |BL| |M C|

(6.33)

Similarly, by considering CAL ∼ M AB, we get |AC| |AM | = . |CL| |M B|

(6.34)

But A M is the perpendicular bisector of BC, so |M C| = |M B| (Theorem G27, page 222). Therefore the right-hand sides of both equation (6.33) and (6.34) are equal. It follows that |AC| |AB| = , |BL| |CL| which, by the cross-multiplication algorithm, is equivalent to |AB| |BL| = . |AC| |CL| Theorem G45 is proved. Exercises 6.8. (1) Assume a circle C and a line L. If the reﬂection across L maps C onto itself, prove that L passes through the center of C. (2) (i) Prove that, given a circle C, there is a line that does not intersect C. (ii) Prove that a circle lies in a closed half-plane of any of its tangent lines, as does the closed disk of the circle. (3) Given any triangle, prove that there is one and only one circle, inside the triangle, so that it is tangent to all three sides of the triangle, meaning that the lines containing the sides are tangent lines of this circle. This is called the incircle of the triangle. (4) Given ABC, let its incircle (see Exercise 3) be tangent to BC at D, AC at E, and AB at F . Prove that AD, BE, and CF are concurrent. This is the Gergonne point of ABC. (5) (i) Prove that the incircle (see Exercise 3) of the right triangle with side lengths 3, 4, and 5 is the unit circle. (ii) Prove that if the legs of a right triangle have lengths a and b and if the length of the hypotenuse is c, then the radius of the incircle of the right triangle is 12 (a + b − c).27 27 I

got this exercise from Yuki Hotta, who took this course at Berkeley in 2008.

6.8. BASIC PROPERTIES OF THE CIRCLE

295

(6) Let ABC and A B C be two similar triangles inscribed in the same circle (see page 353 for the deﬁnition). Prove that ABC ∼ = A B C . (7) Let two circles intersect at M and N . (a) Prove that the line joining the centers of the circles is perpendicular to LM N . (b) Let a line be tangent to both circles. Prove that the circles have equal radii if and only if line LM N is perpendicular to line . (8) Let a line L be tangent to a circle C at A, and let a line L parallel to L intersect C at two points B and C. (i) Prove that |AB| = |AC|. (ii) Let D be a point on C so that A and D are in opposite half-planes of LBC . Let AD and BC intersect at K. Prove that |BD| |BK| = . |CD| |CK| (9) Give a detailed proof of Lemma 6.4 on page 273.

(10) (a) Prove Theorem G52 on page 277 for the case of P AQ being the minor arc. (b) Prove the corollary of Theorem G52 on page 281. (11) Prove the converse to the corollary of Theorem G52: if the angles subtended by two arcs on a circle are equal, then the arcs are congruent. (12) Prove Theorem G54 on page 285. (13) Prove that two distinct circles can intersect at no more than two points. (14) (a) Prove that in a cyclic quadrilateral, each vertex lies in the (convex) angle of the opposite vertex; i.e., if quadrilateral ABCD is cyclic, then A lies in the (convex) angle ∠BCD. (b) Give a counterexample to part (a) if ABCD is not cyclic. (15) Use Theorem G52 to give another proof of the concurrency of the altitudes of a triangle ABC by showing that if the two altitudes BE and CF meet at a point H, then the line AH is perpendicular to BC. (16) If AD, BE, and CF are the altitudes of an acute ABC, the triangle DEF is called the orthic triangle of ABC. Prove that the orthocenter of ABC is the incenter of its orthic triangle. (The orthic triangle is distinguished by the fact that it has the smallest perimeter (the sum of the lengths of the sides) among all triangles with vertices on the sides of ABC. See pp. 75–77 of [Kazarinoﬀ].) (17) Prove Ptolemy’s theorem: a quadrilateral is cyclic ⇐⇒ the product of the (lengths of the) diagonals is equal to the sum of the products of the (length of the) opposite sides.28 (Hint: The main diﬃculty is to ﬁnd a geometric conﬁguration in which the product of the diagonals arises naturally, and one way to do it is to construct a segment AE so that |∠EAB| = |∠CAD| and E ∈ LBC , as shown:

28 This theorem is included in the great astronomer’s magnum opus Almagest (circa 150 AD). This is an inﬂuential theorem in the development of trigonometry as it is essentially the sine addition formula. See [Maor] or [Katz] for further details.

296

6. BASIC THEOREMS OF PLANE GEOMETRY

A D

E

C

B

Now prove AEB ∼ ACD and AEC ∼ ABD, and go on from there.) (18) Let the altitude from the vertex A of triangle ABC meet LBC at D and let O be the circumcenter of ABC. Prove that |∠BAD| = |OAC|. (There are three cases to consider. Case 1: ∠B is 90 degrees. Case 2: ∠B is acute. Case 3: ∠B is obtuse.) Remark: This theorem has vast implications in the geometry of the triangle (see pp. 267–284 of the book [Altshiller-Court]). (19) We say two circles C and C are tangent to each other if they intersect at exactly one point. (a) Prove that two circles are tangent to each other if and only if the line joining the centers passes through a point of intersection of the circles (which will then be the point of tangency). (b) Let B be a point of intersection of two distinct circles. Prove that the circles are tangent to each other if and only if they have a common tangent line at B. (20) The Rouse Ball paradox (discovered in 1892). Here is a purported proof that every triangle ABC is isosceles. Let the angle bisector of ∠A meet the perpendicular bisector of BC at K, and let perpendiculars KF , KE be dropped from K to both AB and AC, as shown:

A

F B

.K D

E C

We know |KB| = |KC| (Theorem G24 on page 220). We also know |KF | = |KE| (Exercise 3 on page 294). The two right triangles KF B and KEC therefore have a pair of equal legs and equal hypotenuses and must be congruent (Theorem G25 on page 220). Hence |F B| = |EC|. But the two right triangles KF A and KEA also have two pairs of equal sides: |KF | = |KE| and |KA| = |KA|. Thus also KF A ∼ = KEA, and

6.9. POWER OF A POINT WITH RESPECT TO A CIRCLE

297

we also have |AF | = |AE|. Consequently, |AB| = |AF | + |F B| = |AE| + |EC| = |AC| and ABC is isosceles. Prove that (i) if the (line containing the) angle bisector of ∠A coincides with the perpendicular bisector of BC, then ABC is isosceles and (ii) if the (line containing the) angle bisector of ∠A is distinct from the perpendicular bisector of BC, then this purported proof must be wrong. (Hint: F and E cannot both lie in the segments AB and AC, respectively, nor can they both lie outside the segments AB and AC, respectively.) (21) Let be the rotation of β degrees around a ﬁxed point O, where 0 < β < 180◦ . Let P be a point distinct from O and let L be a line passing through P ; let Q be a point on L and let |∠OP Q| = θ ◦ for some number θ, 0 < θ < 90. Let P = (P ) and L = (L). Then: (a) Prove that L and L must intersect at some point V . (b) Prove that |∠P V P | = β ◦ if V and Q are on the same half-line of L with respect to P , and |∠P V P | = 180◦ −β ◦ if V and Q are on opposite half-lines of L with respect to P .

6.9. Power of a point with respect to a circle Theorem G55 in the last section (page 286) gives rise to a new concept—the power of a point with respect to a circle—which has many interesting applications. For lack of space, we can only mention the one given in Exercise 4 on page 299. We single out this application because it is about the concurrence of three lines—a leitmotif in Section 6.7. For further reading on the power of a point with respect to a circle, we recommend [Altshiller-Court, pp. 190–227]. Let P be a point not lying on a given circle C with center O and radius r. Let a line passing through P intersect C at two points A and C. According to Theorem G55, the product |P A| · |P C| is a constant independent of : in case P is in the exterior of C (left picture below), the constant is |OP |2 − r 2 and in case P is inside C (right picture below), the constant is r 2 − |OP |2 . We deﬁne the power of P with respect to C to be |P A| · |P C|.

C

P

A

C

r r O

A

O

P It is worth pointing out that for a point in the exterior of C the power of P with respect to C is the square of the length of the tangent from P to C (Theorem G59, page 292).

298

6. BASIC THEOREMS OF PLANE GEOMETRY

The reason we are interested in the power of a point with respect to a circle is because of the following theorem. Theorem G60. Assume two circles with distinct centers O1 and O2 , respectively. Then the set of all the points whose powers with respect to the two circles are equal is a line, called theradical axis of the circles. Moreover, the radical axis is perpendicular to the line joining their centers. Because the proof is somewhat involved, it will be posted on the author’s homepage, https://math.berkeley.edu/~wu/. What we will do here is point out the main idea of the proof and give a rather striking application of the theorem. The heart of the proof is the following lemma: Lemma 6.6. Assume two points A and B and a ﬁxed constant β. The set of points P so that |P A|2 − |P B|2 = β is a line L perpendicular to LAB . The intuition behind the lemma is easy to convey. By interchanging A and B if necessary, we may assume β > 0. If β = |AB|2 , then it is easy to believe—because of the Pythagorean theorem—that L is just the line ⊥ LAB and passing through B. L P A

B

If β < |AB| , then we can also believe that L is the line ⊥ LAB and passing through the point C in AB so that |AC|2 − |BC|2 = β (see picture below). This is because the Pythagorean theorem implies that if P ∈ L, then 2

|P A|2 − |P B|2

= (|P C|2 + |CA|2 ) − (|P C|2 + |CB|2 ) = |AC|2 − |BC|2 = β. L

P A A A A A A

A C B Finally, if β > |AB|2 , then it is also plausible that L is the line ⊥ LAB and passing through the point C in LAB so that A ∗ B ∗ C and so that |AC|2 − |BC|2 = β. This is because the Pythagorean theorem implies that if P ∈ L, then we have the same

6.9. POWER OF A POINT WITH RESPECT TO A CIRCLE

299

calculation: |P A|2 − |P B|2

= (|P C|2 + |CA|2 ) − (|P C|2 + |CB|2 ) = |AC|2 − |BC|2 = β. L

P A

B

C

In any case, if we assume Lemma 6.6, then we can prove Theorem G60. Proof of Theorem G60. As in the theorem, the centers of the circles in question are O1 and O2 , where O1 = O2 . Let the radii of the circles be r1 and r2 , respectively. For a point P , its powers with respect to these circles being equal means that |O1 P |2 − r12 = |O2 P |2 − r22 . Since this equality holds if and only if |O1 P |2 − |O2 P |2 = r12 − r22 , where r12 − r22 is a constant, Lemma 6.6 implies that the set of all such P is a line perpendicular to the line joining O1 and O2 . This completes the proof. The concept of the radical axis of two circles may seem a little elusive, but in special cases it is very explicit. If the two circles intersect at two points A and B, then their radical axis is the line LAB , and if the two circles are tangent to each other, then their radical axis is their common tangent line. Since the proofs of these facts become routine once we have Theorem G60, they will be left as exercises (see Exercises 1 and 2 below). However, they lead to the beautiful and striking fact that if three circles have the property that any two of them meet at two points, then the three common chords are concurrent (Exercises 3 and 4 below).

Exercises 6.9. (1) Let two circles intersect at two points A and B. Prove that their radical axis is LAB . (2) We say two circles C and C are tangent to each other if they intersect at exactly one point. Prove that if two circles are tangent to each other, then their common tangent line is their radical axis. (3) Let A, B, and C be three mutually tangent circles; i.e., any two of them are tangent to each other. Prove that the three common tangent lines meet at a point K. If the centers of A, B, and C are A, B, and C, respectively, how is K related to ABC? (4) Given three circles such that any two of them meet at exactly two points. Prove that the three common chords are concurrent.

300

6. BASIC THEOREMS OF PLANE GEOMETRY

6.10. Two interesting theorems about the circle One of the problems facing a course in geometry is that there does not seem to be any meaningful reward for learning how to prove geometric theorems. Although few students of geometry can point to any truly memorable facts they have learned in their coursework, the subject is chock-full of them. In the last section of this chapter, we will present two that are accessible at this point. For further reading, we can recommend Chapter 10 of [Altshiller-Court]. The Miquel points of a triangle (p. 300) The nine-point circle (p. 301) The Miquel points of a triangle We will prove a simple theorem that was ﬁrst published by A. Miquel in 1838 (Journal de Math´ematiques Pures et Appliqu´es 1 (1838), 485–487), but the theorem was probably known before that. It has a surprising consequence that is equally intriguing (see Exercise 3 on page 304). Theorem G61. Let D, E, F be three points on the lines AB, BC, AC, respectively, of ABC, distinct from A, B, C. Then the circles passing through the vertices of ADF , BDE, and CEF , respectively, have a point in common, the Miquel point determined by D, E, F .

A D F M B

E

C

Proof. The proof is a simple consequence of Theorems G52 and G53, depending on where the points D, E, F happen to be. In the picture above, D, E, and F all lie in the segments AB, BC, and CA, respectively. Let the circles DAF and DBE intersect at M , as shown. It suﬃces to prove that the quadrilateral EM F C

6.10. TWO INTERESTING THEOREMS ABOUT THE CIRCLE

301

is cyclic. But |∠BEM | = |∠M DA| (corollary of Theorem G53, page 285) and |∠M DA| = |∠M F C| (same corollary). Thus |∠BEM | = |∠M F C|, and EM F C is cyclic (corollary of Theorem G53 again). However, if ∠A is obtuse, for example, M could be in the exterior of ABC.

A

F

D E

B

C

M In this case, we will prove that quadrilateral M EF C is cyclic by proving |∠M EC| = |∠M F C| (Theorem G53, page 282). To this end, it suﬃces to prove that |∠M EB| = |M F A|. But |∠M F A| = |∠M DB| (corollary of Theorem G53, page 285), and |∠M DB| = |∠M EB| (Theorem G52, page 277). So we are done again. If D ∗ B ∗ A or B ∗ A ∗ D, again the picture will be diﬀerent but the idea behind the proof above does not change. Similarly for the positions of E and F on LBC and LCA , respectively. The details of the other cases are left as an exercise (Exercise 1 on page 304). The nine-point circle The next theorem is one of the most famous theorems not only in Euclidean geometry but in all of elementary mathematics: the nine-point circle.29 The way we present it here, the theorem lacks romance because its proof is almost clinical. Historically, the circle was discovered in stages, and then it was observed that from the point of view of dilations, this circle is nothing but the circumcircle shrunk to half the size relative to the orthocenter. For conceptual clarity, however, we will present the proof in its ﬁnal form without any historical detours. Recall that the centroid G of a triangle ABC is the point of intersection of the three medians (Theorem G41 on page 256), the orthocenter H is a point of intersection of the three altitudes (Theorem G40, page 255), and the circumcenter O is the point of intersection of the three perpendicular bisectors (Theorem G30 on page 226). In addition, the circumcircle of ABC is the unique circle centered at O and passing through its vertices A, B, and C (Theorem G30, page 226). Recall also from Theorem G43 (page 258) that if G = O, then G = O = H and ABC is equilateral. In this case, the following theorem becomes rather trivial (see Exercise 29 See

page 304 for an additional reason why this circle is so famous.

302

6. BASIC THEOREMS OF PLANE GEOMETRY

6 on page 304). For the rest of this section, we will tacitly assume that G = O and the Euler segment (see page 258) is a segment (and not a point); see Theorem G43. Theorem G62 (The nine-point circle). Let H be the orthocenter of a given triangle ABC and let D be the dilation centered at H with scale factor 12 . If C is the circumcircle of ABC and if N denotes the circle D(C), then the center of N is the midpoint of the Euler segment of ABC and N passes through the following nine points: (α) the three feet of the altitudes (i.e., the points where each altitude meets the line containing the opposite side), (β) the three midpoints A , B , C of the sides of ABC, and (γ) the midpoints of the segments AH, BH, and CH. Of course this N is called the nine-point circle of ABC. See the picture below.

A

Cʹ

Bʹ H Aʹ

B

C

We begin the proof with a lemma. Lemma 6.7. Let AD be an altitude of ABC, and let the ray RAD meet the circumcircle of ABC at D . Let H be the orthocenter of ABC as usual. Then |HD| = |DD |.

A E

H B

D

C

Dʹ Proof of the Lemma. We claim that |∠BHD| = |∠ACB|. To prove this, let RBH intersect LAC at a point E. Then in the right triangle BEC, |∠ACB| + |∠HBD| = 90◦ . But also in the right triangle BHD, |∠BHD| + |∠HBD| = 90◦ .

6.10. TWO INTERESTING THEOREMS ABOUT THE CIRCLE

303

Thus |∠BHD| = |∠ACB| as each is equal to 90◦ − |∠HBD|. On the other hand,

|∠ACB| = |∠AD B| because they both subtend the same arc AB. Hence, together, we get |∠BHD| = |∠AD B|, and triangle BD H is isosceles. But by hypothesis, BD ⊥ HD , and so by Theorem G26(b), BD is the perpendicular bisector of the base HD ; i.e., |HD| = |DD |. The lemma is proved. Proof of Theorem G62. Let H be the orthocenter of ABC and let D be the dilation with center H and with scale factor 12 . Let O be the center of the circumcircle C of ABC. We are going to show that the circle N = D(C) passes through all nine points described in (α), (β), and (γ) of the theorem.

Recall that, by deﬁnition, if Q is a point in the plane, D(Q) is just the midpoint of the segment HQ. It follows that the circle N consists of all the points Q so that if a ray issuing from H intersects C at a point Q, then Q is the midpoint of HQ. It follows immediately from this deﬁnition of N that all three points described in (γ), namely, the midpoints of AH, BH, and CH, are on N . For example, the midpoint P of AH in the above picture is on N . Moreover, if we extend the ray from A to H and let it intersect BC (or the line containing BC) at D and C at D , then Lemma 6.7 above implies that D(D ) = D. Similarly for the vertices B and C, so that the three points described in (α) are on N . It remains to prove that the three points described in (β) of the theorem lie on N . Let the ray RAO intersect C at X, as shown, and let the ray RHX intersect BC at a point A . We are going to prove that |HA | = |A X|, so that D(X) = A ; this will show that the point A lies on N . We will then show that A is the midpoint of BC, so that the midpoint of BC is on N .30 Since the same argument can be made for the midpoints of AC and AB, we will then have shown that the three points described in (α) lie on N . Let us ﬁrst show that |HA | = |A X|. Join D to X; then ∠AD X is a right angle, by Theorem G50. Since both D X and BC are perpendicular to the altitude AD, DA D X. But D is the midpoint of HD (Lemma 6.7, page 302). By FTS (page 219), A is also the midpoint of HX, as desired. Next, we show A is also the midpoint of BC. Indeed, consider triangle AHX. Then OA is the segment joining 30 This

argument is classical, but it was ﬁrst suggested to me by Steven Deng.

304

6. BASIC THEOREMS OF PLANE GEOMETRY

the midpoints of AX and HX, and therefore it is parallel to AH (Theorem G15, page 219). But AH ⊥ BC, so also OA ⊥ BC. By Lemma 6.2 on page 258, OA is the perpendicular bisector of BC and, in particular, A is the midpoint of BC. It remains to prove that the center N of N (which is D(C)) lies in the Euler segment. Now N is the image under D of the center O of C, so it follows from the deﬁnition of D that N is the midpoint of HO. Since OH is the Euler segment of ABC (Theorem G43, page 258), the proof of Theorem G62 is complete. There are many interesting facts about the nine-point circle, but none more famous than Feuerbach’s theorem, discovered only in 1822, about twenty-one centuries after Euclid. To state the theorem, recall that there is a unique circle inside any triangle tangent to all three sides (see Exercise 3 on page 294), called the incircle of the triangle. The Feuerbach theorem states that the nine-point circle and the incircle are tangent to each other in the sense of page 296; i.e., these two circles intersect at only one point.31 Exercises 6.10. (1) In the proof of Theorem G59, we left many remaining cases open. It may be too tedious to try to prove all of them as they are quite similar, so prove at least the following two cases: (1) D ∗ B ∗ A, while B ∗ C ∗ E and A ∗ F ∗ C. (2) B ∗ A ∗ D, while B ∗ C ∗ E and A ∗ F ∗ C. (2) Given ABC, let D, E, F be points on the lines BC, AC, and AB, respectively, distinct from A, B, C. Prove that the Miquel point determined by D, E, F lies on the circumcircle of ABC ⇐⇒ D, E, F are collinear. (3) Use the preceding exercise to prove that the perpendicular lines from a point on the circumcircle of a triangle to (the lines containing) the three sides of the triangle intersect the latter at three points that are collinear. (The line containing these collinear points is called the Simson line of the triangle.) (4) (Converse of Simson line) Given a triangle, suppose the perpendicular lines from a point P to (the lines containing) the three sides of the triangle intersect the latter at three collinear points; then P lies on the circumcircle of the triangle. (5) Describe the nine points of the nine-point circle of a right triangle, and give a direct proof that these points are concyclic. (6) Prove Theorem G62 (the nine-point circle) directly for the case of an equilateral triangle. (Observe that in this case, every altitude is a median so that the nine-point circle becomes the six-point circle; it therefore suﬃces to prove by a computation that all six points are equidistant from the centroid.) (7) (a) If the nine-point circle of ABC is tangent to side AB, what can you say about ABC? (b) When does the incircle of a triangle coincide with its nine-point circle?

31 See, e.g., page 105 of [Altshiller-Court]; see also page 117 of [Coxeter-Greitzer]. The Feuerbach theorem also gives the tangency of the nine-point circle with the three excircles associated with the excenters (see page 258 for the deﬁnition of excenter).

CHAPTER 7

Ruler and Compass Constructions The problem of what plane geometric ﬁgures could be constructed using ruler and compass started with the Greeks no later than the ﬁfth century BC and, like most other things done before the third century BC in Greek geometry, it was codiﬁed by Euclid in his Elements ([Euclid-I] and [Euclid-II]) around 300 BC. Euclid’s work gave the logical context in which to formulate the three unsolved classical construction problems with ruler and compass: (i) (angle trisection) given an angle of x degrees, to construct an angle of 13 x degrees, (ii) (doubling the cube) to construct a segment AB from a given segment CD so that the cube with side AB has volume equal to twice that of the cube with side CD, and (iii) (squaring the circle) given a circle of radius r, to construct a segment so that the square with the segment as a side has the same area as the circle. By the time of Euclid, the Greeks had already been struggling with these problems for at least a century and a half. Without any success, of course. Concomitant with these three well-known problems there is a fourth which is no less important, both historically and mathematically: (iv) (constructing regular polygons) to construct a regular n-gon for any integer n ≥ 3. For n = 3 or 4 (an equilateral triangle or a square) and therefore for all n = 2m 3 and n = 2m 4 where m is a positive integer, this problem is easily solved (see, e.g., (h) on page 313). In Section 7.2, we will show how to construct a regular pentagon, thus solving the problem for n = 2m 5 for all positive integers m; this construction is a high point of Euclid’s treatise. We would like to dispel a possible misconception about ruler-compass construction problems before proceeding further. Let us illustrate with the angle-trisection problem (i) above. Assume an angle ∠AOB of x◦ . Assumption (L6)(ii) (page 216) implies that there is an angle ∠COA and an angle ∠DOA of ( 13 x)◦ and ( 32 x)◦ , respectively, so that C and D lie in the same half-plane of the line LOA as B. It follows that ROC and ROD together trisect ∠AOB and, therefore, there is no question that we are assured of the existence of rays issuing from O that trisect a given angle.

B D C

x˚ A

O 305

306

7. RULER AND COMPASS CONSTRUCTIONS

The same comment can be made about the other three unsolved construction problems. Questions about whether a geometric object can be constructed by ruler and compass are therefore not about whether the object exists but about whether or not we can draw it after a ﬁnite number of steps using only a ruler and a compass. In other words, such a construction is strictly a game and nothing more. It is what may be called recreational mathematics. One can easily see why the Greeks allowed the drawing of lines using a ruler, but the use of a compass is equally understandable if we realize that, to the Greeks in the BC era, circles typiﬁed perfection. The fascination that these construction problems exerted on mathematicians after Euclid had more of a mathematical basis, however. Consider the angle-trisection problem. Given an angle, it is elementary to draw its angle bisector using ruler and compass (see (e) on page 309). So why can’t we trisect it too? Similarly, given a segment of length k, √ one can easily construct—again by ruler and compass alone—a segment that of length 2k (see () on page 316 and (k) on page 315). So it is tantalizing √ one cannot also construct by similar methods a segment of length 3 2k, which is all it takes to double the cube (see the comments on page 326). The reason we take up these construction problems is not because their solutions tell a good story, but because they have inspired a great deal of important mathematics all through the ages. We already mentioned (page 114) how they apparently led Menaechmus to the discovery of conic sections, and Descartes (1596– 1650) clearly had the solutions of these construction problems in mind when he discovered analytic geometry (see pp. 312–314 of [Kline]). In Section 7.3 below, we will describe, albeit only superﬁcially, the mathematics that led to their ultimate solutions in the nineteenth century. In case you wonder why it took so long (some twenty-two centuries) for them to be solved, simply note that the solutions came from algebra, number theory, and analysis1 but not from geometry itself. In the ﬁrst section, we will give the most basic ruler and compass constructions together with their justiﬁcations. 7.1. The basic constructions This section gives twelve of the most basic constructions (pp. 307–316) with ruler and compass, together with proofs that each construction does exactly what it claims to do. Some of these constructions are well known, such as the construction of the bisector of an angle, but others may be less so, such as the construction of a segment of length ab when √ segments of lengths a and b are given or the construction of a segment of length c when a segment of length c is given. We begin by specifying how the ruler and compass will be used in general: (1) Given any two points A and B, the ruler can be used to draw the line LAB . (2) Given a point O and a length r (r > 0), the compass can be used to draw a circle with center O and radius r. Note that a ruler has markings in inches or centimeters. However, (1) says explicitly that a ruler is used strictly for drawing the line joining two points, so that the markings are never used. For this reason, a more accurate description 1 “Analysis” is the ﬁeld of mathematics that, among other things, provides the theoretical underpinning for calculus.

7.1. THE BASIC CONSTRUCTIONS

307

of the constructions would be construction with straightedge and compass, which emphasizes the irrelevance of the markings. In practice, the inaccurate terminology of “construction with ruler and compass” is, by far, the more popular one. (a) Reproduce a line segment on a ray with a speciﬁed endpoint. Suppose a line segment AB is given; a line L is also given, together with a point C on L and a ray determined by C on L. The problem is to construct a line segment on this ray which has C as an endpoint and which has the same length as AB.

L D A

B C

The construction: (1) Open the compass to the length of AB. With C as center and |AB| as radius, draw a circle which intersects the given ray at D. (2) CD is the desired segment; i.e., |CD| = |AB|. Remarks. (a) A general remark that applies not only to this particular construction but to all such constructions is that the statement “draw a circle” does not really call for the drawing of a complete circle, only the drawing of an arc on the circle that gets the job done. In this particular construction, for instance, we only need the part of the circle that intersects the line LAB at B and the part of the circle that intersects the line LCD at D. (b) The fact that |CD| = |AB| is obvious in this case. Note that the circle in step (1) intersects L at two points, and the choice of which of the two points to be designated as D depends on which half-line of C on L we want the desired segment to lie in. (b) Reproduce an angle with one side speciﬁed. Suppose we are given an angle ∠ABC and a ray to be denoted by REF . The problem is to construct a ray RED so that ∠DEF equals ∠ABC.

N

D

A Q

B

M

C E

P

F

308

7. RULER AND COMPASS CONSTRUCTIONS

The construction: (1) Draw a circle (any radius) with B as center; then draw another circle with E as center with the same radius. (2) Let the ﬁrst circle intersect the sides of ∠ABC at N and M , as shown, and let the second circle (to be called circle E) intersect the ray REF at P. (3) With P as center and with |M N | as radius, draw a circle and let one of the points of intersection of this circle with circle E be Q. (4) Let RED be the ray that contains E and Q. Then ∠DEF is the soughtafter angle. Remarks. The reason |∠DEF | = |∠ABC| is that, by construction, N BM ∼ = QEP because of SSS, so the corresponding angles ∠N BM and ∠QEP are equal, which is equivalent to |∠DEF | = |∠ABC|. Note that the circle in step (3) meets the circle with center E and radius |EP | at another point Q and Q, Q lie on opposite half-planes of LEF . If RED is the ray that contains Q , then also |∠D EF | = |∠ABC|. The choice of which of the two angles ∠DEF and ∠D EF we want as our solution depends on which closed half-plane of LEF we want the angle to lie in. (c) Construct a line perpendicular to a given line L from a given point. Let the given point be P . There are two cases to consider: P lies on L, and P does not lie on L. It will be seen that the following construction takes care of both cases at the same time.

P

A

B

L

Q

The construction: (1) Draw a circle with P as center so that it intersects L at two points A and B. (2) Draw two circles with the same (suﬃciently large) radius but with two diﬀerent centers A and B so that they intersect; let one of the points of intersection be Q, and make sure that Q = P . (3) The line LP Q is the line we seek. Remark. The reason why LP Q ⊥ L is that LP Q is the perpendicular bisector of AB on account of Theorem G27 which characterizes the perpendicular bisector of a segment as all the points equidistant from the endpoints: by step (1), P is equidistant from A and B, and by step (2), Q is also equidistant from A and B.

7.1. THE BASIC CONSTRUCTIONS

309

(d) Construct the perpendicular bisector of a given line segment. Let the segment be AB. We have to construct a line which is perpendicular to AB and which passes through the midpoint of AB. Incidentally, this construction also yields a method that locates the midpoint of a given segment.

The construction: (1) Draw two circles with the same (suﬃciently large) radius but with two diﬀerent centers A and B so that they intersect at two distinct points P and Q. (2) The line containing P and Q is the line we seek. (By implication, the point of intersection M of P Q and AB is the midpoint of AB.) Remark. The reason why LP Q is the perpendicular bisector of AB has already been given in construction (c).

(e) Construct the angle bisector of a given angle. Let the angle be ∠ABC. The problem is to construct a ray from B that bisects this angle. (Note the similarity of this construction with construction (c).)

The construction: (1) Draw a circle with center at B (with any radius); let it intersect the rays RBA and RBC at P and Q, respectively.

310

7. RULER AND COMPASS CONSTRUCTIONS

(2) Draw two circles with the same (suﬃciently large) radius but with two diﬀerent centers P and Q so that they intersect at two points. Let one of them, say M , be in the angle ∠ABC (if both points are in ∠ABC, either one would do). (3) The ray RBM is the angle bisector of ∠ABC. Remark. The reason RBM is the angle bisector of ∠ABC is that, by construction, P BM ∼ = QBM because of SSS. Therefore the corresponding angles ∠P BM and ∠QBM are equal, which is equivalent to |∠ABM | = |∠CBM |.

(f) Construct a line parallel to a given line through a given point. Let the line be L and let the point be P ; P does not lie on L. We have to construct a line passing through P and parallel to L.

The construction: (1) Take a point Q on L and join P to Q. (2) Let A be a point on the ray RQP so that P is between A and Q. Let R be some point on L. On the ray RP A , construct an angle ∠AP B so that it is equal to ∠AQR and so that B and R are on the same half-plane of LP Q . (See construction (b) above.) (3) The line LP B is the line parallel to L and passing through P .

Remarks. (i) This construction assumes that you are already completely ﬂuent in reproducing an angle (construction (b)). (ii) The reason LP B L is by virtue of Theorem G19 about corresponding angles; the fact that ∠AP B and ∠AQR are corresponding angles follows from step (2) which guarantees that B and R are on the same half-plane of LP Q (see the deﬁnition of alternative interior angles on page 351). (iii) It is not easy to do this construction accurately, mainly because it is not easy to do construction (b) accurately. In practice, a better alternative is to use a plastic triangle to construct parallel lines, as follows.

7.1. THE BASIC CONSTRUCTIONS

311

There are two kinds of plastic triangles on the market, the 90-45-45 triangle and the 90-60-30 triangle: @ @ @ @ @@ @ @ @ @ @

HH HH HH HH HH HH HH HHHH H

To draw a line through a given point P parallel to a given line L which does not contain P , either triangle can be used, but we will use the 90-45-45 triangle for illustration. (1) Place a ruler along the line L. (2) Holding the ruler in place with one hand, put the triangle ﬂush against the ruler as shown in the left picture below. (3) Now hold the triangle ﬁrmly in place and put the ruler ﬂush against the vertical side of the triangle, as shown in the right picture below.

L

@ @ r @ @ @ @ P @ @ @ @ @

L

@ @ r @ @ @ @ P @ @ @ @ @

(4) Hold the ruler ﬁrmly in place and slide the triangle along the ruler until the horizontal side passes though the point P , as shown in the picture on the left below. (5) Gently remove the ruler and draw the parallel line through P , or place the ruler ﬂush against the horizontal side of the triangle before drawing the line, as shown in the following picture on the right.

@ @ @ @ @@ @ @ @ @ q @ P

L

@ @ @ @ @@ @ @ @ @ q @ P

L

312

7. RULER AND COMPASS CONSTRUCTIONS

(g) Divide a given line segment into any number of equal segments. Let segment AB be given. We show how to trisect AB. The construction can obviously be generalized to equal division into any number of parts.

The construction: (1) Let RAK be any ray issuing from A which is diﬀerent from RAB . Let AC be any segment on RAK . (2) Reproduce AC successively on AK so that AC is equal to CD and is equal to DE (see construction (a)). (3) Join EB. From D and C, construct lines parallel to LEB (see construction (f)), and let these lines intersect AB at G and F , respectively. (4) AF , F G, and GB have the same length. Remarks. (i) The fact that |AF | = |F G| = |GB| is because of Theorem G11 (FTS*) on page 219, as follows. Consider ADG. Then this theorem implies that |AC| 1 |AF | = = , |AG| |AD| 2 so that F is the midpoint of AG. Thus |AF | = |F G|. Now consider the parallel lines LCF and LEB in ABE. The same theorem implies that |AF | |AC| 1 = = , |AB| |AE| 3 so that |AB| = 3|AF |. Also |AG| = |AF | + |F G| = 2|AF |. Therefore, |GB| = |AB| − |AG| = 3|AF | − 2|AF | = |AF |. Hence |AF | = |F G| = |GB|. (ii) In terms of drawing, step (3) above can be done more accurately using a plastic triangle. See remark (ii) at the end of construction (f).

7.1. THE BASIC CONSTRUCTIONS

313

(h) Construct an equilateral triangle on a given side. Let a segment AB be given. We have to construct an equilateral triangle with AB as one of its sides. (This is the ﬁrst proposition in [Euclid-I].) C

A

B

D

The construction: (1) Draw two circles with radius |AB| with two diﬀerent centers at A and B. (2) Let C and D be the points of intersection of the two circles. Then ABC or ABD is the sought-after triangle. Remarks. Again, the fact that |AB| = |BC| = |AC| is obvious from the construction. The vertex C could lie on either of the half-planes of LAB that one can specify ahead of time. (i) Construct a regular hexagon inscribed in a circle. Recall that a polygon is inscribed in a circle if all its vertices lie on that circle (page 324). Given a circle with center O, we will refer to it as circle O. The problem is to locate six points on circle O so that they form the vertices of a regular hexagon.

D

E

O

F

A

C

B

314

7. RULER AND COMPASS CONSTRUCTIONS

The construction: (1) Take a point A on circle O whose radius will be denoted by r. With A as center and with r as radius, draw a circle which intersects circle O at B and F . (2) With B as center and r as radius, draw a circle which intersects circle O at an additional point C. (3) Repeat the drawing of circles with center C and then D, as shown, so that we obtain two more points D and E. (4) Connect the successive points A, B, . . . , F and A to get the desired hexagon. Remark. The proof that ABCDEF is a regular hexagon is left as an exercise (Exercise 1 on page 317; but compare Exercise 1 on page 324.) (j) Draw tangents to a circle from a point in the exterior of the circle. Let P be a point in the exterior of circle O. The problem is to construct a line passing through P and tangent to circle O.

A

O

.

M

P

The construction: (1) Join P to O to obtain segment OP . (2) Locate the midpoint M of OP (see construction (d)). (3) With M as center and M P as radius, draw a circle that intersects circle O at two points. (4) Let A be one of the two points of intersection in step (3). Then the line LP A is tangent to circle O. Remark. The reason why LP A is tangent to circle O is because of Theorems G49 and G50 (pp. 272 and 274, respectively). Indeed, by construction, OP is a diameter of the circle of step (3) so that ∠P AO is a right angle (Theorem G50) and therefore LP A is tangent to circle O (Theorem G49). For the next two constructions, we adopt special terminology. Essentially we want to equate a number with the length of a segment. We will say that a number r is given when we mean that a segment of length r is given. We also ask for the

7.1. THE BASIC CONSTRUCTIONS

315

construction of a number r when we mean the construction of a segment of length r. Throughout the discussion, the implicit assumption is that a segment of length 1 is given. In other words, we assume that the number 1 is given.

(k) Construct the sum, diﬀerence, product, and quotient of two given positive numbers. Let two segments of length r and s be given, both r and s being assumed to be positive. We have to construct a segment with length equal to r +s, r −s (assuming r > s), rs, and rs , respectively. The constructions of segments of lengths r±s follow routinely from construction (a) and from the deﬁnitions of sums and products of numbers in Chapter 1 of [Wu2020a] (keep in mind FASM). Now we construct a segment of length rs.

The construction: (1) On a ray RAC , let B and C be chosen so that |AB| = 1 and |AC| = r. (See construction (a).) A @ @ @ @s 1 @ @ @ B @E @ @ C F (2) On another ray RAE , let |AE| = s and let the line passing through C and parallel to LBE intersect RAE at F (see construction (g)). (3) Then |AF | = rs. Remark. The reason for the assertion in step (3), |AF | = rs, is that by Theorem G11 (FTS*) on page 219,

|AE| |AB| = , |AC| |AF | which then becomes 1/r = s/|AF | (steps (1) and (2)), so that |AF | = rs, as desired.

316

7. RULER AND COMPASS CONSTRUCTIONS

We next construct a segment of length

r s

using the same idea.

The construction: (1) On a ray RAC , let B and C be chosen so that |AB| = s and |AC| = 1. (See construction (a).) A @ @ @ @r s @ @ @ B @E @ @ C F (2) On another ray RAE , let |AE| = r and let the line passing through C and parallel to LBE intersect RAE at F (see construction (g)). (3) Then |AF | = rs . Remark. The reason for the assertion in step (3), |AF | = rs , is that by Theorem G11 (FTS*) on page 219, |AC| |AF | = , |AB| |AE| which then becomes 1/s = |AF |/r (steps (1) and (2)), so that |AF | = rs , as desired. () Construct the square root of a positive number. Given a segment of length r, we have to construct a segment of length

√ r.

The construction: (1) Construct a segment AB of length 1 + r (see construction (k)).

C

A

1

r

D

B

(2) Draw a semicircle with AB as diameter (see construction (d) for locating the center of this circle). (3) Let D be the point in AB so that |AD| = 1 and |DB| = r. Construct the line perpendicular to AB at D, and let it meet the semicircle at C (see construction (c)). √ (4) Then |CD| = r. √ Remarks. The proof that |CD| = r is left as an exercise (Exercise 1 on page 317). Constructions (k) and () will be further discussed in Section 7.3.

7.1. THE BASIC CONSTRUCTIONS

317

Exercises 7.1. In each of the following problems, “construction” means “construction with ruler and compass”. Moreover, it is understood that every construction has to be accompanied by a proof (explanation) of why the construction is correct. (1) (1) Explain the construction of a regular hexagon in construction (i). (2) Explain the construction of a square root in construction (). (3) In the picture of construction (), suppose the length of AC is s rather than 1. What is the length of CD in terms of r and s? (2) Give another explanation of construction (g) by making use of Theorem G15* on page 358 (proved in Section 5.1 of [Wu2020a]). (3) Given a triangle, construct its circumcircle. (4) (i) Construct a square with one side being a given segment. (ii) Construct a regular hexagon with one side being a given segment. (5) Given a point P on a circle with center O, construct the tangent to the circle at P . (6) Given an angle ∠ABC which is less than 180◦ , let P ∈ ∠ABC. Construct a line passing through P so that if intersects RBA and RBC at M and N , respectively, then |BM | = |BN |. (7) Given a segment AB and two segments of lengths r and s, construct a |AC| = rs . point C in AB so that |CB| √ (8) Given positive numbers r and s, construct rs and rs . (9) Let L1 and L2 be parallel lines, and let P be a point not on either line. Let s be a positive number exceeding the distance between L1 and L2 (see page 353 for the deﬁnition of the distance between parallel lines). Construct a line passing through P , so that the segment intercepted on by L1 and L2 is of length s. (10) Let be a line passing through the center O of a circle. Construct a point P on but in the exterior of the circle so that the length of the tangent from P to the circle is equal to the radius of the circle. (11) Assume right triangle ABC so that ∠B is the right angle. On BC, construct a point E so that |BE| is equal to the distance from E to the hypotenuse AC. (12) Given a point P , a line not containing P , and a point Q on , construct a circle passing through P and tangent to at Q. (13) Let P be a point inside a circle with radius r. (a) Let D be the collection of all the chords of circle O which pass through P . Determine the maximum E and minimum e of the lengths of the chords in D. (b) Let d be a number so that e ≤ d ≤ E. Construct a chord in D of length exactly d. (14) Assume triangle ABC with points M and N on RAB and RAC , respectively. Construct D and E on RAB and RAC , respectively, so that LDE LM N and so that area(ABC) = area(ADE). (15) Let three positive numbers a, m, and d be given. Construct ABC, so that a = |BC|, m = |∠BAC|, and so that if AD is the altitude from A to the opposite side BC, then d = |AD|. (16) Given positive numbers b, c, and m so that b + c > 2m, construct ABC so that |AB| = c, |AC| = b, and so that the median from A has length m.

318

7. RULER AND COMPASS CONSTRUCTIONS

7.2. The regular pentagon We will present a ruler and compass construction of a regular pentagon. The constructibility of a regular pentagon is one of the high points of Greek geometry. The method we use—diﬀerent from the one given in [Euclid-II, Book IV, Proposition 11]—is a mixture of geometry and algebra; it will be seen that the quadratic formula plays a crucial role. Euclid’s original construction is outlined in Exercise 5 on page 324. A criterion of constructibility (p. 318) A special isosceles triangle (p. 319) The two constructions (p. 322) A criterion of constructibility We ﬁrst recall that a polygon is a regular polygon if all its sides are of the same length, all its angles (at the vertices) have the same degree, and it is inscribed in a circle; i.e., all its vertices lie on a circle. Because we will be focussed on the ruler and compass construction of a regular pentagon, let us ﬁrst clear up a conceptual issue: do we know that there exist regular pentagons or, in fact, regular n-gons for any positive integer n, regardless of how they are constructed? The answer is aﬃrmative and the easy proof is part of Exercise 1 on page 324. Therefore the only issue of interest here is whether a regular pentagon can be constructed by ruler and compass. We will prove a series of lemmas that lead up to the construction itself. Here is the fundamental observation. Lemma 7.1. A regular pentagon is constructible (by the use of ruler and compass) if and only if an angle of 72 degrees is constructible.

A

E

B O C

D

Proof. First, assuming that a regular pentagon ABCDE is constructible, we will show that the angle of 72 degrees is constructible. By the deﬁnition of a regular polygon, ABCDE is inscribed in a circle. Because the chords AB, BC, . . . , EA are equal, SSS implies that the triangles OAB, OBC, etc., are congruent, and therefore central angles ∠AOB, . . . , ∠EOA are all equal. Consequently, 360◦ = 72◦ . (7.1) |∠AOB| = 5 Since ABCDE is constructible by assumption and since angle bisectors are constructible (construction (e) on page 309), we conclude that ∠AOB is constructible. We have therefore constructed an angle of 72 degrees.

7.2. THE REGULAR PENTAGON

319

Conversely, suppose an angle of 72 degrees is constructible. In a given circle, we construct four central angles of 72 degrees (construction (b) on page 307): ∠AOB,

∠BOC,

∠COD,

∠DOE,

where A, B, . . . , E are points on the circle, B is in the counterclockwise direction of A, C is in the counterclockwise direction of B, etc. We claim that also |∠EOA| = 72◦ . This is because |∠EOA| = 360 − |∠AOB| − |∠BOC| − |∠COD| − |∠DOE| = 72◦ . It follows that the rotation of 72 degrees around O would map A to B, B to C, . . . , E to A and therefore map AB to BC, BC to CD, . . . , EA to AB and ∠A to ∠B, ∠B to ∠C, . . . , ∠E to ∠A. Since a rotation is a congruence, all the sides of ABCDE are equal and all its angles are also equal. Since ABCDE is inscribed in this circle, it is a regular pentagon. The lemma is completely proved. A special isosceles triangle Lemma 7.1 says that to construct a regular pentagon, it suﬃces to construct a 72◦ angle. We do not know how to construct such an angle directly; for example, we do not know how to divide (with ruler and compass) the full angle of 360◦ into ﬁve angles of the same degree. What we are going to do is to construct, instead, segments with certain lengths so that when an isosceles triangle has sides with these lengths, its base angles will be proved to have 72 degrees. Now, such a triangle does not come out of nowhere because there is one hidden in a regular pentagon ABCDE, namely, the isosceles triangle ACD, as shown:

A

E

B O

D

C

Acivity. Use congruent triangles to prove directly that |AC| = |AD|. The fact that |∠ACD| = |∠ADC| = 72◦ is because, by equation (7.1), the central angles ∠AOD and ∠AOC are 144 degrees. By Theorem G52 on page 277, we have |∠ACD| = |∠ADC| = 72◦ . Observe that this implies |AC| = |AD|, by Theorem G29 on page 224. Our ability to construct such a triangle as ACD will depend, as we mentioned above, on how much we can ﬁnd out about the lengths of |AD| and |CD|. To this end, let us state clearly what our assumptions are: ACD is a triangle so that |AC| = |AD|, and so that |∠ACD| = |∠ADC| = 72◦ . Let |CD| = s.

320

7. RULER AND COMPASS CONSTRUCTIONS

Our goal is to compute the length of AD in terms of s. Since the angle sum of ACD is 180 degrees, we know that ∠CAD is 36 degrees. It follows that if the angle bisector of ∠ACD meets AD at F , then F CD ∼ CAD because |∠DCF | = 36◦ = |∠DAC| and the two triangles share the angle at vertex D (AA criterion for similarity).

A

F C

s

D

Therefore, |AD| |CD| = . |CD| |F D|

(7.2)

But CAD is isosceles; therefore so is the similar triangle F CD. In particular, |CD| = |CF |. Furthermore, both ∠ACF and ∠CAF are 36◦ and therefore |CF | = |AF | (Theorem G29 on page 224). Therefore, |F D| = |AD| − |AF | = |AD| − |CD| = |AD| − s. Thus equation (7.2) becomes s |AD| = . s |AD| − s Writing |AD| = x, we have

x s

=

s x−s ,

or x2 − sx − s2 = 0.

By the quadratic formula, we obtain √ √ 1 1 x = (s ± 5s) = (1 ± 5)s. 2 2 Since x must be positive, we discard the negative solution to get √ √ 1 1 (7.3) |AD| = (1 + 5)s = (1 + 5) · |CD|. 2 2 √ 1 Incidentally, the number 2 (1 + 5) is known as the golden ratio.2 Thus what equation (7.3) says is that the ratio of |AD| to |CD| is the golden ratio. In summary, we have proven the following lemma. 2 Google golden ratio, mathematics to see how often this number pops up in various unexpected situations, including the Fibonacci numbers. This is another example of the underlying connections in mathematics that reveal themselves only through an ample supply of technical details but not through abstruse discussions of conceptual understanding.

7.2. THE REGULAR PENTAGON

321

Lemma 7.2. Let s be any positive number, and let A0 C0 D0 be the isosceles triangle with base angles√of 72◦ and a base C0 D0 of length s. Then the two equal sides have length 12 (1 + 5)s. Lemma 7.2 now makes it obvious how to construct an angle of 72◦ . Lemma 7.3. For any positive number √ s, let ACD be an isosceles triangle with sides of lengths |AC| = |AD| = 12 (1 + 5)s and |CD| = s. Then |∠C| = |∠D| = 72◦ . Proof. Compare ACD of this lemma with the triangle A0 C0 D0 of Lemma 7.2: they must be congruent because of SSS. Therefore |∠C| = |∠D| = 72◦ . This proves the lemma. However, some would object to this proof as being too tricky, so we oﬀer a second one that is more straightforward. With triangle ACD as given in Lemma 7.3, then |AD| > s. Let F be the point on AD so that |AF | = s. Then we claim that ACD ∼ CF D.

A

s F C

s

D

To this end, observe that

√ √ 5−1 1+ 5 s−s= s. |F D| = |AD| − |AF | = 2 2

This implies |AD| |CD| = |CD| |F D|

(7.4) because now it is equivalent to 1 2 (1

√ + 5)s = s

1 2(

s √ , 5 − 1)s

which, by the cross-multiplication algorithm, is in turn equivalent to √ 1 1 √ (1 + 5) ( 5 − 1) = 1. 2 2 But the last equality is simple to verify. Consequently, equation (7.4)) holds. It follows that the two triangles ACD and CF D have an angle in common, ∠D, and also have a pair of proportional sides (by (7.4)). The SAS criterion for similarity (page 220) therefore implies that ACD ∼ CF D, as claimed.

322

7. RULER AND COMPASS CONSTRUCTIONS

Since |AC| = |AD| by hypothesis, the similarity implies that |CF | = |CD| = s. Since also |AF | = s, triangle ACF is isosceles and |∠A| = |∠ACF |. Thus |∠D|

= |∠CF D| = |∠A| + |∠ACF | = 2|∠A|.

(|CF | = |CD| = s) (exterior angle theorem)

Thus |∠ACD| = |∠D| = 2|∠A|. Therefore the fact that |∠ACD| + |∠D| + |∠A| = 180◦ becomes 5|∠A| = 180◦ and we have |∠A| = 36◦ . Standard arguments now show that |∠ACD| = |∠D| = 72◦ . The lemma is again proved. The following lemma provides the ﬁnishing touch to the construction of a regular pentagon: it says that the triangle of Lemma 7.3 is constructible. Lemma 7.4. Given a segment CD, one can construct with ruler and compass an isosceles triangle ACD so that its base is CD, AC and AD are the equal sides, and ∠ACD is 72◦ . √ Proof. A segment of length 5 is constructible, e.g., as the hypotenuse of the following right triangle, by virtue of the Pythagorean theorem: HH HH 1 HH HH 2 (Of course, we can simply use construction () on √ page 316, but that would be an overkill.) Therefore a segment of length 12 (1 + 5) is constructible. Now let the √ length of the given segment CD be s. By construction (k) on page 315, 12 (1 + 5)s √ is also constructible. Hence the circles with radius 12 (1 + 5)s and with centers C and D will intersect at a point A.3 This ACD now satisﬁes all the requirements of Lemma 7.3. By that lemma, ∠ACD is 72◦ . This completes the proof. To construct the desired triangle, it suﬃces to draw a segment CD and draw √ two circles with radius 12 (1 + 5) and let A be one of the points of intersection of the circles. Then triangle ACD has the required properties. This proves Lemma 7.4. The two constructions There are actually two kinds of constructions of the regular pentagon. Construction I. Given a circle O, construct a regular pentagon inscribed in O. By Lemma 7.4, we can construct an angle of 72 degrees. By construction (b) on page 307, we can reproduce this angle as ∠AOB in circle O. Similarly, we can reproduce this angle as ∠BOC in the circle so that C is in the counterclockwise direction of B. Repeat, and we get ∠COD and ∠DOE in the circle so that D is 3 For

the reason that these circles must intersect, see the Mathematical Aside on page 276.

7.2. THE REGULAR PENTAGON

323

in the counterclockwise direction of C and E is in the counterclockwise direction of D. Since the full angle at O has 360 degrees, it follows that ∠EOA also has 72 degrees because |∠EOA| = 360◦ − |∠AOB| − |∠BOC| − |∠COD| − |∠DOE| = 72◦ .

A

B

E O

D

C

We claim that ABCDE is a regular pentagon. Since ABCDE is already inscribed in a circle, it suﬃces to prove that it has congruent sides and congruent angles. To this end, let be the rotation of 72◦ around O. Then maps A to B, B to C, C to D, D to E, and also E to A. In particular, maps AB to BC, BC to CD, CD to DE, and DE to EA, ∠ABC to ∠BCD, ∠BCD to ∠CDE, ∠CDE to ∠DEA, and ∠DEA to ∠EAB. Since is a congruence, we have |AB| = |BC| = |CD| = |DE| = |EA|, |∠ABC| = |∠BCD| = |∠CDE| = |∠DEA| = |∠EAB|. This shows that ABCDE has congruent sides and congruent angles. By an earlier remark, it is a regular pentagon. Euclid’s construction of the regular pentagon is outlined in Exercise 5 below. Construction II. Given a segment CD, construct a regular pentagon with CD as one side. By Lemma 7.4 on page 322, we can construct a triangle ACD so that |∠ACD| = |∠ADC| = 72◦ . Since we can construction the circumcircle of ACD (see Exercise 3 on page 317), let its center be O, and let it meet the angle bisectors of ∠ACD and ∠ADC at E and D, respectively, as shown below. We claim that ABCDE is a regular pentagon. A

E

B O

C

D

324

7. RULER AND COMPASS CONSTRUCTIONS

Since CE is the angle bisector of the 72◦ angle, |∠DCE| = 36◦ . By Theorem G52 on page 277, the central angle ∠DOE has 72◦ . In the same way, we show that |∠EOA| = |∠AOB| = |∠BOC| = 72◦ . And of course, ∠COD is also 72◦ . Now the same reasoning as in Construction I shows that ABCDE is a regular pentagon. Exercises 7.2. (1) If C is a circle with center O, let A1 , A2 be two points on C so that |∠A1 OA2 | = 360 n degrees for a positive integer n. Let be the rotation of 360 degrees around the center O so that (A1 ) = A2 . Now let A3 = (A2 ), n A4 = (A3 ), . . . , An = (An−1 ). Prove that A1 = (An ) and that A1 A2 · · · An is a regular n-gon. (2) Prove that all regular n-gons for a ﬁxed n are similar to each other. (3) (a) Prove that a pentagon inscribed in a circle so that all its angles are equal is a regular pentagon. (b) Is it true that a quadrilateral inscribed in a circle so that all its angles are equal is a square? (c) Can you generalize part (a)? (4) (a) Prove that the length of one side of a regular pentagon inscribed in a circle of radius r is √ 1 (5 − 5). r· 2 (b) Use part (a) to prove that a regular pentagon is constructible. (5) (This exercise is, in essence, an elaboration of the preceding exercise.) Prove that the following construction of a regular pentagon in [Euclid-II, Book IV, Proposition 11] is correct:

C

F A

E

O

M

B

D Let AB and CD be perpendicular diameters of a circle with center O. Let M be the midpoint of OB, and let the circle with center M and radius |M C| intersect AB at E. Let F be a point of intersection of the circle with center C and radius |CE| and the circle O. Then CF is one side of a regular pentagon inscribed in circle O. (6) Let ACD be the triangle constructed in Lemma 7.4 on page 322 so that |AC| = |AD| and |CD| = 1. Let O be the center of the circumcircle of ACD, and let the perpendicular bisector from O to AC (respectively, AD) meet the circumcircle at B (respectively, E). Prove that ABCDE is a regular pentagon.

7.3. A SHORT HISTORY OF THE CONSTRUCTION PROBLEMS

325

7.3. A short history of the construction problems We give a historical survey of the mathematical developments that led to the complete solutions of the four construction problems. The following account of the solutions in the nineteenth century, more than two thousand years after the problems were ﬁrst posed, will not be completely self-contained because some advanced concepts will have to be mentioned. However, it should be possible to follow the ﬂow of ideas even if some of the technical details are not completely understood. We ﬁrst recall the classical (ruler-compass) construction problems: • Trisecting an angle: To trisect a given angle. • Doubling the cube: To construct a segment AB from a given segment CD so that the cube with side AB has volume equal to twice that of the cube with side CD. • Squaring the circle: Given a circle of radius r, to construct a segment so that the square with the segment as a side then has the same area as the circle. • Constructing regular polygons: To construct a regular n-gon for any integer n ≥ 3. The general framework for understanding the construction problems is the one set forth above construction (k) in Section 7.1: a number r will be identiﬁed with a segment of length r, and each of the four problems will be reduced to the construction of a segment of a certain length. Thus, in this terminology, the four problems become problems on whether certain numbers are constructible by ruler and compass. Because the number 1 is always assumed to be given (i.e., the unit segment is always a given), construction (k) of Section 7.1 implies that all positive rational numbers can be constructed. On the number line, this means the negative fractions can also be constructed because the circle centered at 0 with radius r (r > 0) intersects the negative number line at −r. So all negative fractions can be considered constructible, and we say that the rational numbers Q √ are constructible. Furthermore, construction () of Section 7.1 implies that if Q denotes all the numbers √ to Q, that can be obtained √ by applying the ﬁve operations +, −, ×, ÷, √ and then every number in Q can be constructed. A typical number in Q is √ 4 15 3 4 − √ . 11 8 + 12.7 √ √ (Of course 4 3 = 3. See Exercise 14 on page 153.) The following theorem says the converse is also true. Theorem √ 7.5. With the number 1 given, a number is constructible if and only if it belongs to Q. An intuitive argument for the validity of the “only if” part of the theorem is as follows. Let us set up coordinates in the plane. Then the equation of a circle is x2 + bxy + y 2 + cx + dy + e = 0 (see Section 2.3), and the equation of a line is Ax + by + C = 0, where b, c, d, e, A, B, C are constants whose nature we wish to discuss. Since these are the only geometric ﬁgures that can be drawn with a ruler and compass, the only points in the plane that can be constructed are those whose

326

7. RULER AND COMPASS CONSTRUCTIONS

coordinates are the intersections of constructible circles and lines. These coordinates are solutions (x0 , y0 ) of one of the following three types of simultaneous equations: Ax + By + C = 0, A x + B y + C = 0, Ax + By + c = 0, x2 + bxy + y 2 + cx + dy + e = 0, 2 x + bxy + y 2 + cx + dy + e = 0, x2 + b xy + y 2 + c x + d y + e = 0. (The symbols A, . . . , C , a, . . . , e denote constants.) Because of constructions (k) and () in Section 7.1, one√can show that the radius and center of a constructible circle have coordinates in Q and likewise for the slopes of constructible lines (use mathematical induction, for example). Thus each of the √ above equations may be assumed to have coeﬃcients A, . . . , C , a, . . . , e in Q. In view of the explicit solutions of simultaneous linear equations (see equation (6.42) in Section 6.7 of [Wu2020a]) and the quadratic formula on page 77, the solutions of the above si√ multaneous equations √ are also in Q. Therefore the coordinates of a constructible point are numbers in Q. Now a constructible segment is one between two con√ structible points. Let (x1 , y1 ) and (x2 , y2 ) be two such points, so that xi , yi ∈ Q for all i = 1, 2. Then the length of the segment joining these points is (x2 − x1 )2 + (y2 − y1 )2 , √ which is clearly in Q, as desired (see, e.g., pp. 21ﬀ. of [Hadlock] for a fuller discussion). Already one detects from the preceding reasoning a distinct pattern of moving from geometry into the domain of algebra in seeking solutions of these geometric construction problems. The idea of “transposing” geometric problems into an algebraic setting was in fact a main impetus behind the discovery of analytic geometry by Ren´e Descartes (1596–1650), as we mentioned in the introduction to this chapter on page 306. He wanted to solve the ﬁrst three construction problems through algebra and guessed correctly that the answers to all of them are no, but he could not supply the necessary proofs. (As is well known, Pierre Fermat (1601–1665) was a codiscoverer of analytic geometry, but there is no record to show that Fermat’s discovery was similarly inspired by these construction problems. He seemed to have been motivated mainly by problems in calculus, such as the drawing of the tangent to a given point of a curve.) Let us take up the four construction problems one at a time. The one on the doubling of the cube is conceptually the simplest, so we begin with that. We claimed √ 3 2. Let us ﬁrst that this problem is equivalent to the constructibility of the number √ prove that if every cube can be doubled, then 3 2 is constructible. Now if every cube can be doubled, then, in particular, the unit cube can be doubled. Thus one can construct a positive number s so that if a cube has one side of length s, then it has√volume that is twice the volume of the unit cube. Thus s3 = 2 and therefore √ √ 3 3 s = 2. Since s is constructible, 2 is constructible. Conversely, suppose 3 2 is constructible. Given a cube with one side of length a, we will construct a cube with

7.3. A SHORT HISTORY OF THE CONSTRUCTION PROBLEMS

327

volume equal to twice the volume of√the given cube, i.e., 2a3 . By construction √ (k) on page 315, √ we can construct s = 3 2 · a. Then the cube with side length 3 2 · a has volume ( 3 2 · a)3 = 2a3 , as desired. √ 3 2 is not constructible. If it We now give a heuristic argument as to √ why √ 3 were, this would be equivalent to saying that 2 ∈ Q (see Theorem 7.5 on page 325). However, it is intuitively implausible that the positive cube root of 2 can be expressed in terms of the square roots of a collection of rational numbers by the use of the four arithmetic operations on these numbers. This intuitive statement can be easily proved using elementary arguments in the theory of ﬁeld extensions (see, e.g., [Hadlock, page 25]). So no cube can be doubled (by ruler and compass). The ﬁrst person to write down the proofs of the impossibility of doubling the cube and trisecting some angles is the French mathematician Pierre Wantzel (1814– 1848); he did so in 1837. However, Wantzel’s name is not generally known, possibly because by then the work of Gauss (we will encounter him later) had made these proofs almost inevitable. The problem of squaring the circle is also conceptually simple. However, this is not to say that the problem is easy to solve! We ﬁrst prove that the problem is equivalent to the constructibility of π. If every circle can be squared, we will prove that π is constructible. The area of the unit circle being π, this means there is a constructible number s so that a square whose sides have length s has the same area as the unit circle; i.e., s2 = π. But s being constructible implies that s2 is also constructible (see construction (k) on page 315). Thus π is constructible. Conversely, if π is constructible, let a circle of radius r be given. We will show that this circle can be squared; i.e., there is a constructible number s so that s2 (= the area of the square √ with a side of length s) is equal to πr 2 . Now π being constructible √ implies that π is constructible (construction () on page 316).√r is given, so πr is constructible (see construction (k) on page 315). Let s = πr. Then s2 = πr 2 , as claimed. In summary, every circle can be squared if and only if π is a constructible number. √ According to Theorem 7.5 on page 325, π is constructible if and√ only if π ∈ Q. We will now√ present a heuristic argument that shows π cannot lie in Q.√Let us ﬁrst understand Q better. The key observation is that every number t in Q must be 4 This can be made believable by a root of a polynomial with integer √ coeﬃcients. √ √ showing how a number such as t = 4 8 − 13 5 in Q is a root of a polynomial of √ √ degree 8 with integer coeﬃcients. Rewrite the equation as t + 13 5 = 4 8. Square both sides to get 2 √ 5 √ t2 + t 5 + = 8. 3 9 Putting all the terms involving a square root to the right, we obtain 5 √ 2 √ t2 + = 8 − t 5. 9 3 Squaring both sides again, we have 4 √ 10 25 20 t4 + t 2 + = 8 − t 40 + t2 . 9 81 3 9 4 It is irrelevant in the present context, but let it be noted that the degree of this polynomial must be a power of 2.

328

7. RULER AND COMPASS CONSTRUCTIONS

Putting the term involving

√ 40 on the right, we arrive at 10 623 4 √ t4 − t2 − = − t 40. 9 81 3

Now squaring both sides gets rid of the last square root (i.e., collecting terms, we then have t8 −

√

40) for good. After

20 6 1146 4 39380 2 388129 t − t − t + = 0. 9 81 729 6561

If we now multiply both sides by 6561, we get ﬁnally 6561t8 − 1620t6 − 10314t4 − 39380t2 + 388129 = 0. √ √ √ This means that the number t = 4 8 − 13 5 in Q is a root of the polynomial equation (7.5) in x of degree 8 with integer coeﬃcients. √ A similar calculation can be made for any number in Q to arrive at the same result that this number must be a root of a polynomial equation in x with integer coeﬃcients. However, one can arrive at the same conclusion with ease if a little ﬁeld theory in abstract algebra is available (see, e.g., page 32 of [Hadlock]). We now return to the constructibility of π. We ﬁrst need a new concept. A number is said to be transcendental if it is not a root of any polynomial equation with integer coeﬃcients. It is far from obvious that transcendental numbers exist, but the French mathematician Joseph Liouville (1809–1882) proved in 1844 that they do ([Wiki-transcendental]). Using inﬁnite series, he constructed an inﬁnite number of them.5 However, proving that a given number such as π is transcendental turns out to be a task of an altogether higher level of diﬃculty. For a long time, mathematicians had suspected that π must be transcendental, but they had no idea how to prove it until the major breakthrough in 1873, when another French mathematician, Charles Hermite (1822–1901), proved that the number e is transcendental (e being the base of the natural logarithm; see page 164). In 1882, the German mathematician C. L. F. Lindemann (1852–1939) ﬁnally proved that π is also transcendental by extending Hermite’s method (the transcendence of π is far from easy to prove even in 2020; see pp. 47–55 of [Hadlock]). In particular, since √ Q consists of √ roots of polynomial equations with integer coeﬃcients, π cannot be a number in Q, and therefore by Theorem 7.5, π is not constructible. Thus no circle can be squared. The works of Hermite and Lindemann inaugurated the modern theory of transcendental numbers. (7.5)

The problem of trisecting an angle is more subtle than the preceding two problems in that some angles can be trisected, e.g., angles of 180◦ , 90◦ , 135◦ , 45◦ , and 22.5◦ (see Exercises 1 on page 330). The question therefore becomes whether there exist angles that cannot be trisected. We give a heuristic argument why the 60◦ angle cannot be trisected. This depends on the trigonometric identity cos 3x = 4 cos3 x − 3 cos x

for all x ∈ R

5 Mathematical Aside: It is not intuitive, but nevertheless true, that there are “more” transcendental numbers in R than rational numbers. This was the highly original work of the German mathematician Georg Cantor (1845–1938), to whom we owe the theory of sets and therefore the foundation of modern mathematics. See [Wiki-Cantor].

7.3. A SHORT HISTORY OF THE CONSTRUCTION PROBLEMS

329

(see Exercise 8 in Exercises 1.4 of [Wu2020c]). Letting x = 20, then cos 3x = 12 . Therefore we have 8 cos3 20 − 6 cos 20 − 1 = 0. Letting t = cos 20, then t is a root of the cubic equation 8t3 − 6t − 1 = 0. So suppose the 60-degree angle can be trisected; then a 20-degree angle would be constructible. Let ∠ABC be a 20-degree angle. A 1 B C On the ray RBA , we may assume that the point A is of distance 1 from B (construction (a) on page 307). From A drop a perpendicular to the other side of the angle and let the two lines meet at C (see construction (c) on page 308). Then |BC| = cos 20. Because |∠ABC| is constructible, so is the length |BC| and, there√ fore, the number t = cos 20 belongs to Q (see Theorem 7.5). Thus this t can be expressed in terms of a collection of rational numbers using the four arithmetic operations and the operation of taking square root. However, we know that t is a root of the cubic 8t3 − 6t − 1 = 0, and this cubic is irreducible over Q in the sense that it is not equal to a product of polynomials with integer coeﬃcients of degrees 1 and 2 (see page 65 of [Hadlock]). Consequently, t cannot be a root of a quadratic equation (or a linear equation) and, therefore, it is intuitively plausible that t cannot be expressible with rational numbers using only the square root operations and +, −, ×, and ÷. This would be the desired contradiction. Again the theory of ﬁelds easily provides a simple proof of this fact (see, e.g., pp. 27–30 of [Hadlock]) and we see that the 60-degree angle cannot be trisected by ruler and compass. Finally we come to the construction of regular polygons of n sides, n ≥ 3. We saw in Section 7.2 that a regular n-gon is constructible for n = 3, 4, 5, 6. These were already known to the Greeks. Clearly if a regular n-gon is constructible, then so is every regular polygon of 2k n sides (see Exercise 3 on page 330). Moreover, if m, n are relatively prime integers and both the regular m-gon and the regular n-gon are constructible, then so is the regular mn-gon (see Exercise 3 on page 330). Thus for n ≤ 17 the following regular n-gons are constructible: n = 3, 4, 5, 6, 8, 10, 12, 15, 16. The ﬁrst unknown cases are therefore n = 7, 9, 11, 13, 14, and 17. For over two thousand years after Euclid, no one had any idea how to approach the constructibility of any of them, and the fact that the regular 17-gon might be constructible—it is fair to say—was never suspected. This changed in 1796 when Gauss6 made the discovery of the constructibility of the 17-gon one month before his twentieth birthday. 6 Carl Friedrich Gauss (1777–1855) is usually mentioned in the company of Archimedes and Newton as one of the three greatest mathematicians of all time. He was extremely precocious, and had already done epoch-making work in number theory by the time he was twenty-one. He was the master of every ﬁeld of mathematics and physics in his lifetime.

330

7. RULER AND COMPASS CONSTRUCTIONS

This discovery was made in connection with his number-theoretic study of the n-th roots of unity, i.e., the roots of xn − 1 = 0. The reason this equation is relevant is that its n complex roots are exactly the vertices of a regular n-gon inscribed in the unit circle of the complex plane (see Section 1.6 in [Wu2020c]). To state Gauss’s theorem—together with some minor details ﬁlled in later by Pierre Wantzel—we k need a deﬁnition. A prime number of the form 22 + 1, where k is a whole number, is called a Fermat prime. For k = 0, 1, 2, 3, 4, the corresponding numbers are 3, 5, 17, 257, 65537, and these are all primes. It is now known that for k = 5, 6, . . . , k 16, the numbers 22 + 1 are composite (see [Wiki-Fermat]). It is not known if there are an inﬁnite number of Fermat primes or not. In any case, Gauss’s theorem is the following. Theorem 7.6 (Gauss’s theorem). (i) If p is a prime, then a regular p-gon is constructible if and only if p is a Fermat prime. (ii) For a whole number n, a regular n-gon is constructible if and only if n = 2k p1 p2 · · · p , where k is a whole number and the p’s are distinct Fermat primes. In particular, this shows that, among the numbers in the above list of 7, 9, 11, 13, 14, and 17, the regular n-gon for n equal to 7, 9, 11, 13, or 14 is not constructible, but the regular 17-gon, 257-gon, and 65537-gon are all constructible. The proof of this theorem depends on algebra and number theory (see pp. 97 and 106 of [Hadlock]), thereby reminding us once again that mathematics is coherent, in the sense that its diﬀerent parts are interrelated. The solution of a problem in one area of mathematics sometimes comes from some other areas. Exercises 7.3. (1) Prove that the angles of 180, 90, 135, 45, and 22.5 degrees can be trisected by ruler and compass. (2) (a) If a regular n-gon is constructible, then so is every regular polygon of 2k n sides, where K is any whole number. (b) If a regular n-gon is constructible, then so is every regular m-gon where m|n. (3) If m, n are relatively prime integers and both the regular m-gon and the regular n-gon are constructible, then so is the regular mn-gon. (Hint: We want to show that an angle of 360/(mn) degrees is constructible. One way to show this is to express 360/(mn) as α(360/m) + β(360/n) for some integers α and β.)

CHAPTER 8

Axiomatic Systems We have thus far tackled four major topics in school mathematics: fractions, negative numbers (rational numbers), algebra, and geometry.1 It is easy to recognize that the exposition of fractions and negative numbers, while following closely the order of presentation in school textbooks, deviates noticeably from TSM2 in terms of providing precise deﬁnitions for all the concepts and backing up each statement with precise reasoning. The rationale for this deviation has been explained in the overview at the beginning of Chapter 1 in [Wu2020a] as unequivocally a matter of necessity. On the other hand, the exposition of algebra and geometry departs from TSM in multiple substantial ways: the introduction of similar triangles before the concept of slope is deﬁned, the emphasis on the correct use of symbols and on the graphs of ax2 in discussing quadratic functions, the use of the basic isometries (translations, rotations, and reﬂections) and dilation as the starting point of the development of plane geometry, and especially the intentional neglect of axioms and the advocacy for soft-pedaling certain correct proofs (see, for example, the Pedagogical Comments following the proof of Theorem G14 in Section 5.1 of [Wu2020a] or those on pp. 242ﬀ. in the present volume about the proof of the angle sum theorem for a triangle). These were not whimsical decisions, and we would like to begin this chapter with an explanation for the decisions concerning geometry. At the moment, the teaching of geometry in schools appears to ﬂuctuate between two extremes: it is either a long series of hands-on activities that provide heuristic explanations for geometric theorems but no proofs, or a matter-of-fact listing of axioms from the beginning and an equally matter-of-fact litany of proofs of theorems with almost no pauses in between. For a general discussion of this unsatisfactory situation, together with its background, consult the book review [Wu2004]. The former kind of teaching, one that allows hands-on activities to replace proofs, seemed to have peaked nationwide in the two decades following 1990, if anecdotal evidence is to be believed. For the purpose of secondary mathematics education, whose goal is to educate our students about mathematics—more specifically, about mathematical reasoning—this kind of teaching must be rejected out of hand (except as the most basic kind of introductions for students who were never exposed to hands-on activities in middle school, as explained in Chapters 4 and 5 of [Wu2016a]). Future historians of education will have to address the question of how such an aberration could have been accepted as the norm in most schools 1 As mentioned in the preface, the CCSSM standards related to these four topics are mostly consistent with the exposition in these volumes. It remains to be seen if these standards can be successfully—and properly—implemented nationwide in the years to come. 2 See page xi for the deﬁnition of TSM.

331

332

8. AXIOMATIC SYSTEMS

in the last decade of the twentieth century. Our attention here will concentrate instead on the second kind of instruction, one that is characterized by the precipitous introduction of axioms and proofs in the geometry curriculum. There is a perception that the only meaningful way for students to learn about proofs in geometry is by introducing axioms at the beginning of a course on geometry and by making students prove every theorem on the basis of the axioms. Such a perception is the natural consequence of having modeled geometric instruction on Euclid’s original work3 for over two thousand years. During the twentieth century, this model rigidiﬁed into a dogma, and in year 2020, it is time to take a second look at this dogma from the perspective of school mathematics education. As we said, what was in the school mathematics curriculum in the recent past was TSM but not mathematics, so that almost all proofs (and perhaps all proofs) resided only in the high school course on geometry. If the course on geometry was the only place where reasoning and proofs could be found, then according to TSM, this was where the Euclidean ideal must be ruthlessly pursued and, therefore, every theorem in geometry must be proved no matter what. Since it is the common belief that, in Euclid, a small collection of axioms is suﬃcient to provide a solid foundation for proving every theorem, then the prevailing dogma dictates that every student must also learn to begin with axioms and learn to prove everything in order to acquire a modicum of reasoning. We will attack the fallacy of this dogma from two diﬀerent directions: ﬁrst, school mathematics education cannot achieve its goal of teaching students how to reason if proofs are provided only in high school geometry, and second, Euclid’s model of “proving every statement from axioms” has been known to be seriously ﬂawed for two centuries. What we have demonstrated in these three volumes (this volume, [Wu2020a] and [Wu2020c]), together with [Wu2011a], [Wu2016a], and [Wu2016b], is that in the school mathematics curriculum, every assertion in it4 can be proved in a way that a school student can understand, and most of these proofs deserve to be an integral part of the school mathematics curriculum. If we can provide gradeappropriate proofs for the major theorems in every school mathematics course—and these volumes have shown that this is possible—rather than just in the course on geometry, then the latter will no longer be subjected to the extra pressure of being the only source of reasoning and proofs. When that happens, one will be able to gain a more balanced view on the need for proving everything from axioms and come to appreciate how unrealistic it is to pursue the goal of proving every theorem in geometry. We are now more than a hundred years removed from Hilbert’s pioneering work on the foundations of geometry (see below) and we have a fairly robust understanding of the immense subtleties of an adequate set of axioms that would make possible the rigorous proofs of all the theorems in Euclid’s geometry. We now know, for example, that a complete proof of even the fundamental fact about the angle sum of a triangle being 180 degrees is not something that an average high school student could tolerate with any modicum of grace (see the Pedagogical Comments on pp. 242ﬀ. about the proof of Theorem G32). More is true. By insisting on proving 3 [Euclid-I]

and [Euclid-II]. a small number of exceptions, such as the fundamental theorem of algebra and the fundamental assumption of school mathematics (FASM). 4 With

8. AXIOMATIC SYSTEMS

333

every theorem ab initio, an inordinate amount of instructional time would have to be spent on the deduction of immediate consequences of the axioms. Two things should be known about this kind of deductive activities. First of all, it is wrong to assume that deductions from axioms are elementary and therefore easy for a beginner. Because such deductions are, as a rule, strictly formal (i.e., far removed from intuition and dictated solely by logical considerations), they are diﬃcult not only for beginners, but also for professional mathematicians, for the simple reason that this kind of reasoning cannot rely on geometric intuition for guidance. One can easily get a taste of such proofs by reading the ﬁrst two chapters of [Hartshorne] or the ﬁrst four chapters of [Greenberg]. A second fact is that the deduction of obvious consequences from axioms is boring even for the average college student and therefore deadly in a high school setting. To illustrate the last comment about the axiomatic treatment of school geometry, let us look at one of the best textbooks of this genre: the book [Moise-Downs] by E. Moise and F. Downs. Consider the following three theorems in [Moise-Downs]: Theorem 4-3. Any two right angles are congruent. Theorem 5-2. Every angle has exactly one bisector. Theorem 6-5. If M is a point between points A and C on a line L, then M and A are on the same side of any other line that contains C. I hope no one will try to argue that these are the kinds of geometric facts that will ﬁre up school students’ geometric imaginations. Let us also take note that Theorem 6-5 appears on p. 177 of [Moise-Downs]. Now, if students have to work through 177 pages to be convinced that what one sees at a glance in the following picture is true, who can blame them for feeling that geometry is not worth the trouble to learn it? Aq M Cq q L Such, alas, is the peril of having to “prove every theorem in geometry”. The unfortunate fact is that, even after such a valiant attempt at achieving rigor, [Moise-Downs] still falls considerably short of its goal. For example, the proof of the exterior angle theorem (Theorem 7-3) on page 189 of [Moise-Downs] is incomplete, for a reason that is well known in the post-Hilbert era; see, for example, page 36 of [Hartshorne]. Another example is the proof of the angle sum theorem of a triangle (Theorem 9-4) on page 242 of [Moise-Downs]; it is too simplistic as it misses the subtleties we mentioned after the proof of Theorem G32 (pp. 242ﬀ.) in Chapter 6. The only purpose of pointing out these mathematical and pedagogical missteps (among others) in the book [Moise-Downs] is to underscore the futility of trying to “prove every theorem” in a school course on geometry. We repeat: There are valid mathematical as well as pedagogical reasons for us to reject the naive belief that proving every theorem in a course on school geometry is a worthwhile educational goal.

334

8. AXIOMATIC SYSTEMS

It remains to point out that the work of developing a complicated subject like plane geometry by starting with a set of axioms is really not a job suitable for beginners. Historically, the organization of a subject in an axiomatic format has always been an afterthought: when a subject has reached maturity, the need will arise that there be a better organization to display its logical structure. The available evidence points to the fact that it was exactly under such circumstances that Euclid wrote his Elements ([Euclid-I] and [Euclid-II]), and the same is true of the axiomatization of calculus in the nineteenth century (which amounts to the axiomatization of the real numbers) and many other subjects. For example, groups, rings, homology, cohomology, etc. A very small number of mathematicians are known to have learned mathematics eﬃciently and productively by starting with axioms, but most others rely on ﬁrst acquiring prior experience with various natural examples. In terms of school mathematics education, the most important skill that students must acquire is how to move from a hypothesis to a conclusion by the use of logical reasoning. For two given statements A and B, if students can detect the underlying connections that allow them to go from A to B, then they have already made the most signiﬁcant ﬁrst steps towards achieving mathematical proﬁciency (in the sense of the National Research Council volume Adding It Up ([NRC])). For this purpose, it is not essential that the hypothesis or the conclusion be at the level of the axioms or that every theorem be proved. All that matters is whether one learns to move from A to B by the use of reasoning. There are in fact many illustrations of this philosophy in [Wu2020a] and the present volume, e.g., the Pedagogical Comments following the proof of Theorem G14 in Section 5.1 of [Wu2020a], where it is explicitly suggested that certain facts be assumed but not proved in the school classroom, the way we make use of the intermediate value theorem in Section 3.1 (p. 121), and the way we take the fundamental theorem of algebra for granted in Section 5.3 (p. 196). All this is not to say that students need not learn about axiomatic systems. They do, but for pedagogical reasons it is not to their advantage to do so at the beginning of a high school course on geometry. In fact, we are going to embark on a brief discussion of axioms in this chapter, and the reason we can aﬀord to do so is that we have already proved enough theorems in Chapters 4 and 5 of [Wu2020a] and Chapters 6 and 7 of the present volume to be somewhat familiar with the subject of plane geometry. Therefore we are now in a position to step back and contemplate how the subject might be more tightly reorganized from a mathematical perspective. What we are suggesting is that the concept of an axiomatic system can be more proﬁtably discussed at the end of a school geometry course rather than at the beginning. 8.1. The concept of an axiomatic system The intuitive idea of an axiomatic system is very simple. Suppose we want to explain a given assertion A. In so doing, let us say we have to make use of another assertion B. But then why is B true? So we explain B in terms of another assertion C. This means that if we accept the truth of C, we can explain A because C explains B and B explains A. But then the same question returns: why is C true? To answer that, we need to invoke another assertion D, so that now D explains A, and so on. One might ask again why D is true, whereupon the logical regression goes another

8.1. THE CONCEPT OF AN AXIOMATIC SYSTEM

335

step. It is clear that there is no end to this regression and a sensible way to handle this is to simply accept D, for example, on faith and say that D is our starting point. In that case, we say D is an axiom. If it happens that, for all the facts we care to explain, there is a collection of n axioms A1 , A2 , . . . , An that suﬃce to serve as our starting point for the explanation of all the things under consideration, then these A1 , A2 , . . . , An together with all the things they can explain are then said to form an axiomatic system. Of course the formal deﬁnition of an axiomatic system is a lot more precise, but precision is not our main concern here in an informal discussion. The ﬁrst person to present an axiomatic system that has survived for the scrutiny of subsequent generations is of course Euclid (circa 300 BC). That was a monumental achievement. However, like most pioneering eﬀorts, Euclid’s system has ﬂaws. He made use of only ﬁve axioms, but he seemed not to take seriously the fact that, once he declared them to be his axioms, then everything he tried to prove would have to be based on these ﬁve axioms and these ﬁve axioms alone. Quite the contrary, Euclid felt no inhibition about supplementing his axioms with his intuitive beliefs in providing proofs. For example, in the very ﬁrst proposition, he claimed as obvious that if two circles have the same radius r and the distance between their centers is also r, then they must intersect at two points (see the picture in construction (h) on page 313). Few of us would dispute this fact, but that is not the issue. The issue in question is that, regardless of whether this claim is true or not, his ﬁve axioms nowhere come close to guaranteeing this fact, whereas he left us with the impression that they do. See [Hartshorne, pp. 29–43] for a more thorough discussion of the many gaps in Euclid’s axiomatic system. We can point to another ﬂaw in Euclid’s system, one that is of an entirely diﬀerent nature. In stating his ﬁve axioms, Euclid used the terms “point”, “line”, “plane”, etc., and he added deﬁnitions of these terms. For example: A point is that which has no part. A line is breadthless length. A straight line is a line which lies evenly with the points on itself. (Note that Euclid’s “line” is what we call a curve, and his “straight line” corresponds to our line.) Such deﬁnitions raise further questions. Beyond the fact that the prose is perhaps even more abstruse than the terms it tries to elucidate (one wonders, for example, what “breadthless length” could mean), suppose Euclid were to explain it in terms of some other words; then one would pursue this line of inquiry and press for an explanation of those other words. We would then fall back on the cycle of inﬁnite logical regressions that had prompted the formulation of axioms in the ﬁrst place. Can the terms used in the axioms be deﬁned after all? These issues were raised and rigorously examined in the nineteenth century, especially after the discovery of non-Euclidean geometries (see Section 8.4 on pp. 345)ﬀ. Building on the work of many of his predecessors, especially Moritz Pasch (German, 1843–1930) and G. Peano (Italian, 1858–1932), David Hilbert5 clariﬁed 5 David Hilbert (German, 1862–1943) and Henri Poincar´ e (French, 1854–1912) dominated mathematics in the decades around 1900. They also share the distinction of having made contributions of lasting importance in every major ﬁeld of mathematics and in theoretical physics. In 1900, Hilbert proposed a list of 23 open problems, and the search for solutions of these Hilbert’s problems played a signiﬁcant role in the development of the mathematics of the twentieth century.

336

8. AXIOMATIC SYSTEMS

once and for all what an axiomatic system ought to be. In 1899, he published his short tract Grundlagen der Geometrie (Foundations of Geometry),6 which changed not only the face of Euclidean geometry but also the faces of mathematics and mathematical logic. The Grundlagen addresses many profound issues that lie at the foundation of mathematics, such as the meaning of an axiom being independent of other axioms in the same system, or whether an axiomatic system is consistent. It is in investigations into these questions that the Grundlagen has exerted its greatest inﬂuence on mathematics. This fact notwithstanding, the contribution of the Grundlagen that is directly relevant to the learning of plane geometry lies in its recognition that the terms used in a set of axioms (such as “point”, “line”, etc.) must remain undeﬁned and that their meaning can only be inferred from the axioms themselves. In other words, all we know about points, lines, etc., is completely contained in the axioms, no more and no less. Mathematical Aside: In year 2020, of course, the preceding recognition is commonplace. For example, in an undergraduate course on abstract algebra, the usual axioms of group theory do not describe what a group is or what its elements are. A group is blandly described as a set G, together with a binary operation G × G → G, so that the binary operation obeys the associative law, has a neutral element, and every element in G has an inverse relative to this neutral element. The same goes for the axioms of rings, ﬁelds, and vector spaces over a ﬁeld. As a consequence, diverse objects such as the nonzero real numbers under multiplication and the collection of all bijections of a set onto itself under composition are both groups, as is the collection of automorphisms of a vector space under composition. Back in 1899, however, these abstract considerations were a novelty and such a precise concept of an axiomatic system was a revelation. Another departure of Hilbert from Euclid is the conferring of “neutral” status on each axiom, in the following sense. Whereas Euclid’s axioms (in accordance with the view of the time, especially that of Aristotle) were supposed to be “self-evident truths”, in Hilbert’s system, an axiom is what it is, the starting point of logical deductions, and it matters not at all whether it is a “self-evident truth” or not. This fact liberates mathematical investigations from psychological barriers and eases the acceptance of various axioms such as those that lead to non-Euclidean geometries (see Section 8.4 on page 345). The fundamental idea of Hilbert’s work, to the eﬀect that the terminology used in the axioms (e.g., point, line, etc.) is irrelevant and the content of the axioms is everything, captures the essence of abstract mathematics. For example, we freely talk about the vector space of real-valued continuous functions deﬁned on some set without the slightest concern for the fact that such a collection of functions has nothing to do with our intuition of physical “space”. There are too many such examples in contemporary mathematics, and it is unlikely that this would be the case were it not for Hilbert’s work. 8.2. The role of axioms in school geometry We will discuss Hilbert’s axioms for Euclidean geometry in the next section. What is worth pointing out is that, precisely because of its abstract nature and because it is designed to give a mathematically complete description of the Euclidean 6 An

English translation, [Hilbert], can be downloaded for free.

8.2. THE ROLE OF AXIOMS IN SCHOOL GEOMETRY

337

plane, this set of axioms is unsuitable for the high school curriculum. Moreover, the usual axioms for high school geometry make use of real numbers, but the Hilbert axioms do not. The Hilbert axioms stand on their own and they assume no prior knowledge of any kind. It must also be said that, even if we simplify the Hilbert axioms by incorporating the real numbers into the mix (which can certainly be done), the proofs in the resulting system are not necessarily much simpler. For example, a complete proof of Theorem G14 (page 218) that is deemed acceptable in the Hilbert axiomatic system will still be subtle (see the Pedagogical Comments in Section 5.1 in [Wu2020a] after the proof of Theorem G14), and there are no two ways about it. Examples such as this should be enough to convince educators and textbook authors to give up any pretensions to a completely rigorous axiomatic treatment of geometry. Once we give up on the impossible, then we can contemplate a reasonable approach to geometric instruction. In this context, we can now explain the rationale behind the presentation of school geometry given in Chapters 4 and 5 of [Wu2020a] and Chapters 6 and 7 of the present volume. We begin by noting that our approach to geometry deﬁnitely assumes a knowledge of real numbers. Indeed, we assume throughout all three volumes ([Wu2020a], [Wu2020c], and this volume) the following: (L3) Every line can be made into a number line so that any two given points on the line are the 0 and 1, respectively, of the number line. We also assume (L1), (L2), (L4)–(L6) (see pp. 214–216). On this basis, we deﬁne the basic isometries (rotations, reﬂections, and translations) and further assume that they satisfy assumption (L7), to the eﬀect that they map lines to lines, rays to rays, and segments to segments; furthermore, they are length- and degree-preserving. We also observe that there are as many basic isometries as we want: (1) Given any point O in the plane and any number θ satisfying −360 ≤ θ ≤ 360, there is a rotation of θ degrees around O (Lemma 4.16 in Section 4.4 of [Wu2020a]). (Λ1) Given any line L, there is a reﬂection across L (Lemma 4.19 in Section 4.5 of [Wu2020a]). −− → (T 1) Given any vector AB, there is a unique translation along −−→ AB (Lemma 4.22 in Section 4.5 of [Wu2020a]). Clearly, we assume a lot before we embark on any proofs! This is consistent with our desire not to get bogged down in reasoning on the formal level. We should also mention that the last explicit assumption we make, the crossbar axiom (L8) (page 217), could be proved on the basis of (L1), (L3), and (L4). Again, our goal is to get the proofs of theorems started at a level where geometric intuition can play a role; redundancy is not a primary concern. In other words, our intent is not to derive every mathematical assertion strictly on the basis of the smallest possible collection of axioms, but rather to give students as many demonstrations as possible on how to go from statement A (the hypothesis) to statement B (the conclusion) via logical deductions in a geometric context. As explained above, we believe the latter

338

8. AXIOMATIC SYSTEMS

should be the focus of school geometry and—for that matter—school mathematics as a whole. Students should get to know the role that axioms play in mathematics and what an axiomatic system is, but they should not suﬀer pedagogically for doing this. The most persuasive argument in favor of the treatment of geometry that is being proposed here may be that its implementation requires no major changes in the sequencing of the existing school mathematics curriculum. The need for teaching geometry beginning with the elementary grades is now universally recognized and, along with this recognition, also the need for teaching the basic isometries (rotations, reﬂections, and translations). What has happened in TSM is that the basic isometries are dutifully introduced and advertised as important concepts, but little evidence is oﬀered to demonstrate their putative importance. The instruction on the basic isometries in middle school usually consists of nothing more than presenting some examples from nature or the arts—tessellations of the plane and elaborate mosaics—to illustrate the omnipresence of symmetries and engaging students in a few activities to move geometric ﬁgures around in the coordinate plane. Then at the end of the high school geometry course, there may be some discussion about realizing congruences using the basic isometries. Beyond that, there is little or no serious eﬀort to convince students that the basic isometries have any mathematical signiﬁcance. What Chapters 4 and 5 of [Wu2020a] and Chapters 6 and 7 of this volume try to accomplish is to leave unchanged the emphasis on the basic isometries but, instead of using them to engage in nonmathematical activities, to introduce the basic isometries as mathematical concepts. They explain why the basic isometries are truly basic to the study of geometry by revealing to students the meaning of congruence for the ﬁrst time. Students get to see in a hands-on way why certain segments or certain angles in a geometric ﬁgure (such as opposite sides or opposite angles in a parallelogram) are congruent by the use of these isometries. These chapters also show how the basic isometries underlie the all-important concept of congruence (see the deﬁnition in Section 4.6 of [Wu2020a]). This approach to school geometry lays bare the fact that the basic isometries are a truly basic part of the school mathematics curriculum. On a more practical level, how will Chapters 4 and 5 of [Wu2020a] and Chapters 6 and 7 of this volume improve the learning of school geometry? The reason geometry is a fearsome subject to teachers and students is their lack of understanding of what a proof is and how to approach the writing of a proof, because proving theorems before high school geometry is a rare (often nonexistent) activity in TSM. By contrast, these four chapters on geometry are embedded in an environment where proofs are taken for granted as a normal activity. Moreover, by making the basic isometries the foundation of plane geometry, the beginning of a high school geometry course is no longer exclusively devoted to the writing of formal deductions from axioms but is, for the most part, grounded in hands-on activities using transparencies. Because we do not insist on proving everything along the way—and there are many explicit pedagogical suggestions in those chapters on what to skip in a school classroom—students can get quickly to proofs of less obvious, and therefore more interesting, theorems. We hope these changes will ease students’ entry into geometry by removing some unnecessary roadblocks in the path of mathematics learning.

8.3. HILBERT’S AXIOMS

339

But more is true. Beyond the ﬁrst few theorems, e.g., Theorems G1–G8, G10, and G16, the proofs of most of the theorems in this volume are substantially the same as the standard ones using axioms. If anything, they are conceptually simpler. For example, in the proof of the “midpoints theorem”, Theorem G15 (see page 219), the usual proof of |CF | = |AD| and CF AD (where D and E are the midpoints of AB and AC, respectively, and |DE| = |EF |; see the picture below) uses congruent triangles, whereas it is simpler and more transparent to note (as we did in Section 5.1 of [Wu2020a]) that the rotation of 180 degrees around E maps CF to AD, so that |CF | = |AD| (because rotation is an isometry) and CF AD (because of Theorem G1 on page 217). A @ @ @ F D @E @ @ @ @C B At other times, there is no diﬀerence at all between the proofs in this book and the usual proofs using axioms, e.g., the proof of the Pythagorean theorem in Section 5.3 of [Wu2020a]. Therefore, the advantage of this approach is that it clariﬁes the foundational materials on the one hand and builds on teachers’ existing geometric knowledge on the other. We look forward to the day when school geometry will be taught in a more reasonable manner.7 We have oﬀered Chapters 4 and 5 of [Wu2020a] and Chapters 6 and 7 of this volume as one alternative. What can be claimed about this approach is that, given the available options, it is a relatively simple way to make mathematical sense of the school geometry curriculum as of 2020. A more reasonable school geometry curriculum is of course fundamental to improving school mathematics education; it transcends the CCSSM. 8.3. Hilbert’s axioms We have stressed all along that teaching school geometry from axioms is neither feasible nor desirable. Part of the reason is that developing a subject from axioms is an endeavor so sophisticated that one should not impose it on the average beginner if there is a more productive alternative. However, even school students should know something about axiomatic systems if for no other reason than the fact that the concept of an axiomatic system is an integral part of mathematics. A workable compromise is to give a presentation about axiomatic systems at the end of a geometry course, in much the same way that we are doing now, namely, discussing Hilbert’s axiomatic system for geometry after we have proved many geometric theorems. Thus, we are going to give a brief introduction to the Hilbert axioms from the standpoint of school mathematics. These axioms can be given in slightly different formulations. From the perspective of Chapter 4 in [Wu2020a], we choose to follow the collection in Greenberg’s book [Greenberg]; see also Chapter 2 in [Hartshorne]. 7 Such would be the case if CCSSM is competently implemented. See the ﬁrst footnote on page 331.

340

8. AXIOMATIC SYSTEMS

Strictly speaking, the following discussion has a gap, as it avoids any mention of the “continuity axiom” having to do with the “completeness” of the number line. Our axioms (L1)–(L8) include (L3), which explicitly allows the conversion of every line in the plane into a number line, so this “completeness” is built into our system. However, Hilbert’s axioms do not assume the possibility of this conversion. Therefore, when we want to show that plane geometry does take place in R2 , the “continuity axiom” will have to enter the scene. See page 343 below for more information. The Hilbert axioms are statements about what are called the undeﬁned terms; i.e., these are words that are not given any meaning a priori, and what they are or what properties they have can only be inferred from the subsequent axioms. Hilbert’s undeﬁned terms are the following: point, line, incident, between, congruent. Deﬁnitions will be given later which make use of these undeﬁned terms. There are four groups of axioms: the the the the

incidence axioms, betweenness axioms, congruence axioms, parallel postulate.

We will enumerate them ﬁrst and leave a general discussion of their signiﬁcance to pp. 343ﬀ.

The incidence axioms. Intuitively, a point being “incident” with a line or a line being “incident” with a point means that the point lies on the line. I-1. For every point P and for every point Q not equal to P , there exists a unique line incident with P and Q. I-2. For every line there exist at least two distinct points incident with the line. I-3. There exist three distinct points with the property that no line is incident with all three of them. We will denote the line incident with two distinct points A and B by LAB . This notation is consistent with the one we have been using, of course. We will also adopt the usual terminology of saying A and B are on a line if is incident with points A and B. It is also common to describe points which are incident with a single line as collinear points. Axiom I-3 can therefore be stated as asserting the existence of three noncollinear points in the plane.

The betweenness axioms. For collinear points A, B, and C, we denote “B is between A and C” by the symbol A ∗ B ∗ C. Intuitively, B being “between” A and C means exactly what it normally means. B-1. If A ∗ B ∗ C, then A, B, C are three distinct points all lying on the same line, and C ∗ B ∗ A. B-2. Given any two distinct points B and D, there exist points A, C, and E lying on line LBD such that A ∗ B ∗ D, B ∗ C ∗ D, and B ∗ D ∗ E.

8.3. HILBERT’S AXIOMS

Aq

B q

Cq

D q

341

Eq

B-3. If A, B, C are three distinct points lying on the same line, then one and only one of the points is between the other two. For the statement of the ﬁnal betweenness axiom, we introduce some deﬁnitions. Given distinct points A and C, the segment AC is the collection of A and C, together with all the points B so that A ∗ B ∗ C. We call A and C the endpoints of AC. Next, let be any line and let A and B be any two points not on . If A = B or if AB contains no point on , we say A and B lie on the same side of . Otherwise we say A and B lie on opposite sides of . B-4. For every line and for any three points A, B, C not on : (i) If A and B are on the same side of and B and C are on the same side of , then A and C are also on the same side of . (ii) If A and B are on opposite sides of and B and C are on opposite sides of , then A and C are on the same side of . It follows from this axiom that for a given line and a point Q not on , any point not on is either on the same side of as Q or on the opposite side of as Q; there is no other possibility. In this sense, separates the plane into two sides. The part of the plane on one side of is also called a half-plane of . Thus the two half-planes of , together with , form a disjoint union (see page 353) of the plane.8 Before we can state the next group of axioms, we have to prove a theorem, which is in eﬀect Lemma 4.5 (line separation) of [Wu2020a] (see page 357). Theorem 8.1. Given a line L and a point P on L. If Q is a point on L not equal to P , then any point A on L not equal to P satisﬁes one and only one of the following conditions: (i) AQ contains P . (ii) AQ does not contain P . Pq Qq L

Proof. By axiom I-3, there is at least one point E so that E is not incident with the given L. Let denote the line LP E . q E P Qq q L By axiom B-4, the points on L are in particular separated into those that are on the same side of as Q and those that are on the opposite side of as Q. Recalling the deﬁnition of “being on the same side”, we see that Theorem 8.1 has been proved. 8 It

is easy to see that B-4 is essentially the plane separation assumption (L4).

342

8. AXIOMATIC SYSTEMS

With L and Q as in Theorem 8.1, the ray RP Q consists of P and all the points A on L so that AQ does not contain P . If S is a point on L so that SQ contains P , then the ray RP S is sometimes called the opposite ray of RP Q . The two rays have only P in common, and P is called the vertex of either ray. We can now deﬁne the notion of an angle. Let RP A and RP B be distinct rays with a common vertex P but which are not opposite rays. We have the following two well-deﬁned half-planes: the half-plane of the line LP B containing A and the half-plane of LP A containing B. The intersection of these two half-planes, together with the two rays RP A and RP B , is called the angle determined by these rays and is denoted by ∠AP B (the dotted region below). Ar q q q hhhh hhhh P hhhhq q q q q hhqhhq hhh r hq q hhh q hh B Note that, for the moment, this is a restrictive deﬁnition of an angle as there is neither an “angle of 0 degrees” so to speak as we require RP A and RP B to be distinct rays, nor any “angle of 180 degrees” as the rays are required to be nonopposite rays, nor any “angle of degree > 180 degrees” because this deﬁnition ignores the “nonconvex” subset determined by these rays. We also introduce the deﬁnition of a triangle ABC as the collection of all the points in the segments AB, AC, and BC, where A, B, and C are three noncollinear points. With the notation as given, the above three segments are called the sides of ABC, and the angle ∠BAC is called the included angle of the sides AB and AC. B

A C We are now ready to state the next group of axioms.

The congruence axioms. These are axioms that describe the meaning of “being congruent to”, including the congruence between segments and the congruence between angles. We denote “is congruent to” by the symbol ∼ =. Notice that SAS is now an axiom (see C-6 below). C-1. If A and B are distinct points and A is any point, then for each ray with vertex A , there is a unique point B on this ray so that AB ∼ = A B . C-2. If AB ∼ = CD and AB ∼ = EF , then CD ∼ = EF . Moreover, every segment is congruent to itself. C-3. If A ∗ B ∗ C and A ∗ B ∗ C and if AB ∼ = A B and BC ∼ =AC. = B C , then AC ∼ C-4. Given any angle ∠AOB and any ray RO B with vertex O , there is a unique ray RO C on a given side of the line LO B so that ∠AOB ∼ = ∠A O B . C-5. If ∠AOB ∼ = ∠COD, then ∠A O B = ∠A O B and ∠AOB ∼ ∼ = ∠COD. Moreover, every angle is congruent to itself.

8.3. HILBERT’S AXIOMS

343

∼ A B C Assume two triangles ABC and A B C . By deﬁnition, ABC = if there is a congruence between corresponding sides and corresponding angles; explicitly, this means if AB ∼ = A B , AC ∼ = A C , and BC ∼ = B C , and if ∠ABC ∼ = ∠A B C , ∠ACB ∼ = ∠A C B , and ∠BAC ∼ = ∠B A C . C-6 (SAS). If two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then the two triangles are congruent. The last group of axioms consists of one axiom only, and it is the parallel postulate. Formally, two lines are deﬁned to be parallel if there is no point incident with both. In symbols, the lines and L being parallel is denoted by L. The parallel postulate. For any line and any point not on , there is at most one line passing through the point and parallel to . A collection of points which satisfy these four groups of axioms—together with another axiom which we have not stated, the continuity axiom of Dedekind—is called a Euclidean plane, and the collection of these four groups of axioms, together with all the theorems deduced from them, is called Euclidean geometry. As mentioned earlier, we have purposely left out any discussion of the Dedekind axiom because it is beyond the level of school mathematics. One can consult [Greenberg, pp. 129–138] or [Hartshorne, pp. 112–116] for the precise statement of the axiom. The Dedekind axiom would guarantee that what we call a Euclidean plane is in fact the unique coordinate plane R2 . Therefore, once we have Dedekind’s axiom, everything that we usually expect to be true in a plane becomes true for a Euclidean plane. For example, we will know that in a Euclidean plane, every segment has a length, every angle has a degree, and if a line contains an interior point and an exterior point of a circle, then it must intersect the circle (see the Pedagogical Comments on page 276, for example). In addition, if a circle containing an exterior point and an interior point of another circle C, then it must intersect C at two points. We refer the reader to [Hartshorne, Chapters 3 and 4] for more information about the Dedekind axiom. Not much needs to be said about the simple incidence axioms, except to note that axiom I-1 is exactly (L1) of Section 4.1 in Chapter 4 of [Wu2020a]. But the betweenness axioms are perhaps the most subtle among the four groups and, while all other axioms can be more or less traced back to Euclid, the explicit recognition of the need for these betweenness axioms escaped mathematicians for over two thousand years after Euclid until Pasch. Pasch’s work was stimulated by the creation of hyperbolic geometry around 1830 (see page 345). We have so far not brought up any discussion of the betweenness axioms because we assumed from the beginning that every line can be made into a number line (see (L3) on page 337). In eﬀect, once we assume that every line is a number line relative to the choice of a 0 and a 1 on the line, it is intuitively clear that a number B (i.e., a point on the line) is between the numbers A and C if either A < B < C or C < B < A.9 Thus, 9 There is actually something quite subtle here: how do we know that this concept of B being between A and C does not depend on the particular choice of 0 and 1 on the line? The fact that it does not requires a careful proof, e.g., see the discussion after (L3) in Section 4.1 of [Wu2020a]. We repeat that in a typical high school classroom, one may not wish to get entangled in a proof of the all too obvious theorem that “betweenness” is independent of the choice of 0 and 1.

344

8. AXIOMATIC SYSTEMS

once a line is converted to a number line, the fact that A ∗ B ∗ C automatically means one of two things: either A < B < C or C < B < A. But now, imagine that you have to give up the concept of comparing numbers and yet you still want to express the intuitive idea that a point B is “between” two other points A and C. How do we do it? Clearly we will have to somehow formalize the concept of betweenness so that it would behave the way we usually think of “B is between A and C”. This is the intent of axioms B-1 to B-4, and their exact (and nonintuitive) formulation was of course the result of many trials and errors, from Pasch to Hilbert. Note that we need B-3 to make sure that a line is really “straight” and doesn’t come back on itself like a circle, because the following conﬁguration of three points A, B, and C is ruled out by B-3: on a circle, all three possibilities A ∗ B ∗ C, B ∗ C ∗ A, and C ∗ A ∗ B simultaneously hold:

C

A B

As noted earlier, B-4 is essentially (L4). Also note that (L8) (crossbar axiom on page 217) is a consequence of (L4) (see [Greenberg, pp. 116–117]), but we did not pause to prove the implication in Chapter 4 of [Wu2020a] precisely because of the point we have been trying to make, namely, that if at all possible, we skip those logical deductions that are made purely on the formal level. One more comment may be relevant. One can easily convince oneself that the concept of betweenness is absolutely essential for geometric discussions. For example, axiom C-3 is false without the betweenness assumptions, as the following picture shows: Aq

Bq

Cq

Aq

Cq

Bq

As for the congruence axioms, what they manage to accomplish is to give meaning to “congruence” without mentioning basic isometries. The axioms C-2 and C-5 are of course the statement that ∼ = is an equivalence relation among segments and among angles. Axiom C-6 is Theorem G8 (see page 218). It is far from obvious that these six axioms are enough to prove everything we want to say about congruent segments, angles, or triangles. The fact that they do implies that if we know SAS, then we can prove the other congruence criteria (SSS, ASA, and HL for right triangles). Finally, we look at the parallel postulate. It stands alone. This axiom held the fascination of countless geometers after Euclid until about 1830, and the fascination centered on why such a complicated statement could be accepted as an axiom (recall: until Hilbert clariﬁed what an axiomatic system really is, an axiom had to be “self-evident truth”). The main mathematical activity associated with this fascination was to try to prove the parallel postulate on the basis of the other four

8.4. HYPERBOLIC GEOMETRY

345

axioms of Euclid. One can imagine the confusion surrounding this attempt because, the concept of an axiomatic system not being well understood, it became diﬃcult to tell a valid proof from a fake one. Indeed, many false proofs were recorded. However, in the early 1800s, three mathematicians independently proposed that the parallel postulate had to be an axiom, because assuming the opposite of the parallel postulate seemed also to lead to an entirely consistent body of theorems. The three were C. F. Gauss, who had a fairly complete overview of the whole situation no later than 1828, N. Lobachevsky (Russian, 1792–1856) in 1829, and J. Bolyai (Hungarian, 1802–1860) in 1831. Gauss never published his ﬁndings, but the other two did. (See Chapter 6 of [Greenberg].) We now turn to a brief discussion of this new geometry.

8.4. Hyperbolic geometry As we indicated in the last section, it would be to our advantage to separate the parallel postulate from the other Hilbert axioms. Thus consider the axiomatic system based on the three incidence axioms, the four betweenness axioms, the six congruence axioms. A plane whose points satisfy the collection of these thirteen axioms together with the aforementioned Dedekind axiom (page 343) is called a neutral plane, and the collection of all the theorems deduced from these axioms is called neutral geometry. Thus Theorem 8.1 of the preceding section is an example of a theorem in neutral geometry. A more sophisticated example of a theorem in neutral geometry is Theorem G19 (see page 219). With transversal and alternate interior angles deﬁned as on pp. 351 and 352, this theorem reads: Theorem G19. If the alternate interior angles of a transversal with respect to a pair of distinct lines are equal, then the lines are parallel. It is not straightforward to see that this is really a theorem in neutral geometry; i.e., it is not at all obvious that there is a proof of Theorem G19 which does not use the parallel postulate. In Exercise 2 on page 349, we give an outline of the argument that leads to the veriﬁcation of this fact and ask for details to be ﬁlled in. It is instructive to state without proof two basic theorems in neutral geometry due, respectively, to Saccheri-Legendre and Saccheri, because they set the stage for the understanding of our next topic, which is hyperbolic geometry. For this purpose, we need the real numbers R, especially the continuity axiom of Dedekind,10 so that we can assign a degree to every angle and a length to every segment (see [Hartshorne, Chapters 3 and 4]). On a practical level, however, all we care about 10 Mathematical Aside: The Dedekind axiom is the following. Suppose the points of a line are divided into two nonempty subsets S and T so that no point of S is between two points of T and no point of T is between two points of S. Then there is a unique point P on so that for any A ∈ S and B ∈ T , either A = P or B = P or the point P is between A and B.

346

8. AXIOMATIC SYSTEMS

at this point is to have available the concept of the degree of an angle as a real number. With this understood, we record the ﬁrst basic fact. Recall that we use the term the angle sum of a triangle (or any convex polygon) to mean the sum of the degrees of all the angles. The Saccheri-Legendre theorem. In a neutral plane, the angle sum of every triangle is ≤ 180 degrees. The proof is a bit long and will not be given here (see [Hartshorne, p. 320] or [Greenberg, p. 186]). G. Saccheri (1667–1773) was an Italian Jesuit priest and A.-M. Legendre (1752–1833) was a French mathematician who lived through the French Revolution. Legendre rediscovered Saccheri’s result because the latter was not known to the general public until 1899. The Saccheri-Legendre theorem puts in proper perspective our “common sense” notion that the angle sum of a triangle is always 180 degrees: it tells us that we cannot lay claim to this fact if we do not know the parallel postulate. The last remark about the angle sum of a triangle is ampliﬁed further in the next theorem in neutral geometry. It was essentially proved by Saccheri. Saccheri’s theorem. In a neutral plane, the following conditions are equivalent: (i) There is one triangle whose angle sum is 180 degrees. (ii) Every triangle has angle sum equal to 180 degrees. (iii) The parallel postulate holds. Again, the proof is somewhat long. See [Greenberg, p. 183] or [Hartshorne, p. 321]. From the perspective of neutral geometry, we may describe Euclidean geometry as the geometry of a neutral plane in which the parallel postulate holds. Therefore, what Saccheri discovered can be paraphrased as follows: a neutral plane becomes a Euclidean plane as soon as one triangle in it has an angle sum of 180 degrees. These two theorems together shed an entirely diﬀerent light on the seemingly innocuous statement that the angle sum of a triangle is 180 degrees. The moral is that it is not easy for any triangle in a neutral plane to have an angle sum of 180 degrees. Also observe that we now understand perfectly the role of the parallel postulate in the angle sum of a triangle only because we have axiomatized neutral geometry and Euclidean geometry. Now consider the negation (i.e., the opposite) of the parallel postulate: Hyperbolic axiom. For every line and for any point not on , there are at least two distinct lines passing through the point and parallel to . A neutral plane that satisﬁes the hyperbolic axiom is called a hyperbolic plane, and the collection of all the axioms of neutral geometry plus the hyperbolic axiom, together with all the theorems deduced therefrom, is called hyperbolic geometry.11 Thus we have neutral geometry = Euclidean geometry hyperbolic geometry. Hyperbolic geometry is one of the non-Euclidean geometries. While in the nineteenth century, people were incredulous that hyperbolic geometry could be no less “true” than Euclidean geometry, we now fully understand, after the fundamental 11 We note again that, for the ease of the ensuing discussion, we will also add the continuity axiom of Dedekind.

8.4. HYPERBOLIC GEOMETRY

347

work of Riemann,12 why this is the case. The reason cannot be given without using advanced mathematics, but suﬃce it to say that Euclidean geometry corresponds to “zero curvature”, hyperbolic geometry corresponds to “curvature = −1”, and therefore both geometries are very special cases of surfaces whose “curvature” can vary arbitrarily. For our purpose, these precise deﬁnitions of Euclidean geometry and hyperbolic geometry exhibit the most persuasive reason why the use of axioms is such a powerful way to organize a subject. In this case, we see that the diﬀerence between the two geometries lies only in the diﬀerent choice of a single axiom at the starting point. We should prove at least one theorem in hyperbolic geometry to give an idea of the diﬀerence between Euclidean and hyperbolic geometries. Notice that Saccheri’s theorem and the Saccheri-Legendre theorem together imply that, in a neutral plane, no triangle can have an angle sum of 180 degrees unless the neutral plane is already Euclidean. In particular, the angle sum of every triangle in a hyperbolic plane must be less than 180 degrees. We are now ready to state and prove the theorem in hyperbolic geometry alluded to above. Recall that two triangles in the Euclidean plane are similar if and only if the triangles have proportional sides (i.e., corresponding lengths are proportional) and equal angles (i.e., corresponding angles have the same degrees); see Theorem G20 on page 220. We do not yet have a deﬁnition of similarity in the hyperbolic plane (and for a good reason, as we shall see presently), but we can at least deﬁne similar triangles in the hyperbolic plane by using Theorem G20 as a takeoﬀ point. Thus we say two triangles in a hyperbolic plane are similar if they have equal corresponding angles and proportional sides. What we now prove is the following stunning fact. Theorem 8.2. In a hyperbolic plane, if two triangles are similar, then they are congruent. Observe that photography is possible only because we have similar ﬁgures in the space around us. So had we lived in hyperbolic 3-space, there would not have been any nifty digital cameras. For the proof of Theorem 8.2, we begin with an observation. Lemma 8.3. In a hyperbolic plane, the angle sum of a convex quadrilateral is < 360 degrees.13 Proof of lemma. Let the convex quadrilateral be ABCD. 12 Bernhard Riemann (1826–1866) was a German mathematician and a student of Gauss. In the works of Gauss and Riemann, mathematics reaches two of its highest peaks. Riemann is remembered by Riemannian geometry, the Riemannian curvature tensor, Riemann surfaces, the Riemann-Roch theorem, the Riemann zeta function, the Riemann hypothesis, the CauchyRiemann equations, the Riemann-Hilbert problem, and the Riemann mapping theorem. To college students, he is of course famous for the Riemann integral, but that is unfortunately not one of his great achievements. 13 The same statement holds for a quadrilateral even when it is not convex, but for that the angle sum would refer to the sum of the degrees of the interior angles, and the latter would have to be deﬁned with care.

348

8. AXIOMATIC SYSTEMS

C B ,,\ \ ,, \ , \ \D A,

By the convexity of ABCD, the diagonal AC separates the quadrilateral into two triangles, ABC and ACD. Recalling the previous remark (following the SaccheriLegendre theorem) that the angle sum of every triangle in a hyperbolic plane is < 180 degrees, we have angle sum of ABCD

=

1. Thus |AB| > |A B | and |AC| > |A C |. Let B0 and C0 be chosen on the sides AB and AC, so that |AB0 | = |A B | and |AC0 | = |A C |. A

@

@

@

@ B0

@C0 @C B

A

@ @

@

@

@ C B

Consider AB0 C0 and A B C . We have |∠A| = |∠A | because ABC ∼ A B C . Thus AB0 C0 ∼ = A B C (SAS). Therefore

|∠AB0 C0 | = |∠B |

and

|∠AC0 B0 | = |∠C |.

But by the assumption that ABC ∼ A B C , we also have |∠B| = |∠B | and |∠C| = |∠C |. Thus |∠AB0 C0 | = |∠B|

and

|∠AC0 B0 | = |∠C|.

Now consider the angle sum of the quadrilateral B0 BCC0 . It is convex, since its interior is the intersection of the convex triangular region ABC and the convex half-plane of the line passing through B0 C0 which is opposite to A. By Lemma 8.3, the angle sum of B0 BCC0 is < 360. However, this same angle sum can also be

8.4. HYPERBOLIC GEOMETRY

349

directly computed: |∠B| + |∠C| + |∠CC0 B0 | + |∠C0 B0 B| = |∠B| + |∠C| + (180 − |∠AC0 B0 |) + (180 − |∠AB0 C0 |) = |∠B| + |∠C| + (180 − |∠C|) + (180 − |∠B|) = 180 + 180 = 360. This contradiction shows that the two triangles ABC and A B C had better be congruent to begin with. The proof of Theorem 8.2 is complete. Exercises 8.4. (1) Suppose ∠ABC ∼ = ∠A B C . Suppose also that D and D are points so that D ∗ B ∗ C and D ∗ B ∗ C . Then prove that ∠ABD ∼ = ∠A B D . ∼ ∼ ∼ Hint: We may assume AB = A B , BC = B C , and DB = D B . Then ABC ∼ = A D C . Now prove = A B C , and consequently ADC ∼ ∼ ADB = A D B to ﬁnish the proof. (2) Give a new proof of Theorem G19 (page 345) by ﬁlling in the following outline of a proof. HH \ H \H Q1\HH \ HH HH \ rH P1 \ HH \ HH \r L2 H S Q2 H P2 L1 Let the distinct lines be L1 and L2 , and let the transversal be ; let the equal alternate interior angles be the ones indicated with a dot, as shown. Suppose L1 is not parallel to L2 so that they meet at a point Q2 . If P2 is on and L2 and if P1 is on and L1 , let the point Q1 be a point on L1 so that Q1 and Q2 are on the opposite sides of and so that P1 Q1 ∼ = P2 Q2 . Then P1 P2 Q2 ∼ = P2 P1 Q1 . Consequently, ∠P1 P2 Q1 ∼ = ∠P2 P1 Q2 . But if S is a point on L2 so that S and Q2 are on opposite rays of L2 with P2 as vertex, then ∠P1 P2 S ∼ = ∠P2 P1 Q2 . Therefore ∠P1 P2 Q1 ∼ = ∠P1 P2 S, and this implies the rays RP2 S and RP2 Q1 are equal, and therefore Q1 lies on L2 . Contradiction.

Appendix: Facts from [Wu2020a] There are three parts in this appendix: Part 1. Assumptions Part 2. Deﬁnitions Part 3. Theorems and lemmas (p. 356)

Part 1. Assumptions Fundamental Assumption of School Mathematics (FASM): The laws of operations for both addition and multiplication (associative, commutative, and distributive), the formulas (a)–(d) for rational quotients (page 356), and the basic facts about inequalities (A)–(E) for rational numbers (p. 356) continue to be valid when the rational numbers are replaced by real numbers. Geometric assumptions (L1)–(L8): See pp. 214ﬀ.

Part 2. Deﬁnitions Alternate interior angles: Let two distinct lines L1 , L2 be given. A transversal of L1 and L2 is any line that meets both lines at distinct points. Suppose meets L1 and L2 at P1 and P2 , respectively. Let Q1 , Q2 be points on L1 and L2 , respectively, so that they lie in opposite halfplanes of . Then ∠Q1 P1 P2 and ∠P1 P2 Q2 are said to be alternate interior angles of the transversal with respect to L1 and L2 . Angle: Given two rays ROA and ROB with a common vertex O, if A, O, B are not collinear, then the angle ∠AOB is understood to be the convex angle (see page 352) determined by these rays; the nonconvex angle ∠AOB will refer to the union of the two rays ROA and ROB with the complement of the convex angle ∠AOB. The rays ROA and ROB are the sides of ∠AOB. If A, O, B are collinear and A and B are in diﬀerent halflines of O, then either of the half-planes of LAB is called a straight angle determined by LAB (see page 355). If A, O, B are collinear and A and B are in the same half-line of O, then ROA = ROB ; the ray ROA is called the zero angle determined by ROA , and the whole plane is called the full angle determined by ROA . 351

352

APPENDIX: FACTS FROM [Wu2020a]

Average speed: For an object in motion, its average speed over the time interval from t1 to t2 , t1 < t2 , is distance traveled from t1 to t2 . t2 − t1 Basic isometries (of the plane): Rotations, reﬂections, and translations. Bilateral symmetry: A geometric ﬁgure is said to have bilateral symmetry with respect to a line L if the reﬂection across L maps the ﬁgure onto itself. Boundary: The boundary of a set (in the plane) is the collection of all the boundary points of the region. Boundary point: A point is a boundary point of a set S (in the plane) if every disk of positive radius—no matter how small—centered at the point, contains both a point in S and a point not in S. Bounded set: A set in the plane is bounded if it is contained in a circle (of ﬁnite radius). Closed half-plane: A line L separates the plane into two half-planes by (L4) on page 215. The union of L with one of its half-planes is called a closed half-plane of L. Closed set: It is a set that contains its boundary. Collinear: A collection of points is said to be collinear if they all lie on a line. Complex fraction: A complex fraction is a fraction obtained by a division A B of two fractions A and B (B = 0). We continue to call A and B the A numerator and denominator of B , respectively. Congruence: A congruence is a transformation of the plane that is the composition of a ﬁnite number of reﬂections, rotations, and translations. Congruent ﬁgures: A geometric ﬁgure S is congruent to another geometric ﬁgure S , in symbols, S ∼ = S , if there is a congruence ϕ so that ϕ(S) = S . Convex angle: Given two rays ROA and ROB with a common vertex O so that A, O, B are not collinear, the convex angle determined by these rays is the intersection of the closed half-plane of LOA containing B and the closed half-plane of LOB containing A. Convex polygon: A polygon whose associated polygonal region is convex. Convex set: A subset R in a plane is said to be convex if given any two points A, B in R, the segment AB lies completely in R. (Half-planes and closed half-planes are convex, and so are their intersections thereof.) Corresponding angles: A pair of angles relative to two lines intersected by a transversal are called corresponding angles if they are obtained by replacing one angle in a pair of alternate interior angles (relative to this transversal) by its opposite angle. Dilation: A transformation D of the plane is a dilation with center O and scale factor r (r > 0) if, for every point P in the plane, (1) either P = O and D(O) = O (2) or P = O and the point D(P ), to be denoted by P , is the point on the ray ROP so that |OP | = r|OP |.

APPENDIX: FACTS FROM [Wu2020a]

353

Disjoint sets: A collection of subsets in a given set is said to be disjoint if no two of them have any element in common. Disjoint union: A set S is said to be the disjoint union of its subsets T1 , . . . , Tn if the Tj ’s are disjoint and the union of T1 , . . . , Tn is S itself. Distance between parallel lines: Let L1 and L2 be parallel lines and let P1 be a point on L1 . If the line perpendicular to L1 at P1 intersects L2 at P2 , then the length of the segment P1 P2 is called the distance between L1 and L2 . Distance of a point from a line: The distance of a point Q from a line L is the length of the segment QQ , where Q is the point of intersection of L with the line passing through Q and perpendicular to L. Equal geometric ﬁgures: Two geometric ﬁgures S and S are equal, in symbols S = S , if (i) every point P of S is also a point in S and (ii) every point Q of S is also a point in S. Equality of sets: Two sets A and B are equal if they have the same collection of elements; i.e., every element of A is an element of B, and every element of B is also an element of A. Equation: An equation in a number x is a question asking whether two given expressions in x are equal. If the expressions are 3x4 − x3 + 2 and 0, then the corresponding equation is 3x4 − x3 + 2 = 0. If x = 1, then the answer to the question is no. Equilateral triangle: A triangle all whose sides are of the same length. Equivalence relation: A relation among elements in a set that is reﬂexive, symmetric, and transitive is called an equivalence relation. Exterior of a circle: If a circle C has center O and radius r, then the exterior of C consists of all the points whose distance from O exceeds r. Figure: A ﬁgure, or geometric ﬁgure, is just a subset of the plane. Occasionally, it also refers to a subset of 3-dimensonal space. Fractions in lowest terms: A fraction m n is said to be in lowest terms if the only whole number that divides both m and n is 1. Fraction multiplication: The multiplication of two fractions k × m n is by ] is partideﬁnition the length of the concatenation of k parts when [0, m n tioned into equal parts. k m Fraction subtraction: If k > m n , then the subtraction − n is by deﬁnition the length of the remaining segment when a segment of length m n is taken from one end of a segment of length k . Geometric ﬁgure: A geometric ﬁgure, or ﬁgure, is just a subset of the plane. Occasionally, it also refers to a subset of 3-dimensonal space. Graph of an equation (in two variables): It is the collection of all the points (x, y) in R2 so that the coordinates x and y of each (x, y) are solutions of the given equation in two variables. Inscribed in a circle: A polygon is said to be inscribed in a circle if all its vertices lie on that circle. Inside a circle: Let C be a circle with center O and radius r; a point is inside C if its distance from O is ≤ r.

354

APPENDIX: FACTS FROM [Wu2020a]

Isometry: A transformation F of the plane is called an isometry (of the plane) if F preserves distance; i.e., for all P , Q in the plane, dist(P, Q) = dist(P , Q ) where P = F (P ) and Q = F (Q). Linear equation in two variables: It is a question about whether two given expressions in two numbers x and y are equal; for such an equality, ax + by = c, where a, b, and c are constants, it is required in addition that at least one of a and b be nonzero. Linear polynomial: A polynomial of degree one. Multiplicative inverse: The multiplicative inverse of a number x is the number x−1 so that x · x−1 = 1. Nonconvex angle: Given two rays ROA and ROB with a common vertex, the nonconvex angle determined by these rays is the union of the closed half-plane of LOA not containing B and the closed half-plane of LOB not containing A. Parallelogram: A quadrilateral with four sides of the same length. Perpendicular bisector of a segment: It is the line perpendicular to the segment and passing through its midpoint. −− → Pointing in the same direction (for two vectors): Two vectors AB −−→ and P Q are said to be pointing in the same direction if (i) either LAB = LP Q or LAB LP Q and (ii) there is a line L0 , distinct from LAB and LP Q and parallel to neither, so that one of the closed half-planes of L0 contains both rays, RAB and RP Q . Polygon: Let n be any positive integer ≥ 3. An n-sided polygon (or more simply an n-gon) is by deﬁnition a geometric ﬁgure consisting of n distinct points A1 , A2 , . . . , An in the plane (the polygon’s vertices), together with the n segments A1 A2 , A2 A3 , . . . , An−1 An , An A1 (the polygon’s sides), so that (i) none of these segments intersect each other except at the endpoints as indicated, i.e., A1 A2 intersects A2 A3 at A2 , A2 A3 intersects A3 A4 at A3 , etc., but otherwise no other intersections are allowed, and (ii) consecutive segments A1 , A2 and A2 A3 , . . . , Ai−1 Ai and Ai Ai+1 , . . . , An−1 An and An A1 do not lie in a line. It is also standard to use the term “polygon” to denote the polygonal region enclosed by a polygon. Polygonal region: Given a polygon P, the polygonal region enclosed or bounded by P is the union of P and the set B in Theorem 4.13 on page 358. Polynomial: Let x be a number. A sum of multiples of nonnegative integer powers of x is called a polynomial in x (here “multiple” means multiplication by a real number which may or may not be a whole number). Quadrilateral: A 4-sided polygon. Rational quotient: A number that is the quotient (or division) of one rational number by another. For example, if x and y are rational numbers and y = 0, then xy is a rational quotient. Ray: A ray with vertex O is a subset of a line L passing through O consisting of O and one of the half-lines (page 214) of L determined by O. Rectangle: A quadrilateral all of whose angles are right angles.

APPENDIX: FACTS FROM [Wu2020a]

355

Reﬂection: Given a line L, the reﬂection across L (or with respect to L) is by deﬁnition the transformation ΛL of Π, so that: (1) If P ∈ L, then ΛL (P ) = P . (2) If P ∈ L, then ΛL (P ) is the point Q so that L is the perpendicular bisector of the segment P Q. Regular polygon: A polygon is a regular polygon if all its sides are of the same length, all its angles (at the vertices) have the same degree, and it is inscribed in a circle, i.e., all its vertices lie on a circle. Relatively prime: Two whole numbers are relatively prime if the only whole number that divides both is 1. Rhombus: A quadrilateral whose four sides have the same length. Right triangle: A triangle in which one angle is 90◦ . Rotation: Let O be a point in the plane Π and let a number θ be given so that −360 ≤ θ ≤ 360. Then the rotation of θ degrees around O (or sometimes we say with center O) is the transformation θ deﬁned as follows: θ (O) = O, and if P ∈ Π and P = O, let C be the circle of radius |OP | centered at O. Then: If θ ≥ 0, θ (P ) is the point Q on C so that Q is obtained from P by turning θ degrees in the counterclockwise direction along C (in other words, |∠QOP | = θ ◦ ). If θ < 0, θ (P ) is the point Q obtained from P by turning |θ| degrees in the clockwise direction along C (in other words, |∠P OQ| = |θ|◦ ). Similar ﬁgures: A geometric ﬁgure S is similar to another geometric ﬁgure S , in symbols, S ∼ S , if there is a similarity F so that F (S) = S . Similarity: A similarity is a transformation of the plane that is the composition of a ﬁnite number of congruences and dilations. Slope: For a nonvertical line L in a coordinate plane, let P = (p1 , p2 ) and Q = (q1 , q2 ) be any two points on L. Then the slope of L is by deﬁnition the diﬀerence quotient (p2 − q2 )/(p1 − q1 ). Solution of an equation: Given an equation in a number x, if x = b makes the equation true, then b is called a solution of the equation. For example, x = −1 is a solution of 5x3 − 4x2 − 7x + 2 = 0. Square: A rectangle all of whose sides are of the same length. Straight angle: Let A, O, B be three collinear points and suppose A∗O∗B. Then the straight angle ∠AOB refers to either closed half-plane of the line LAB , and O is the vertex of the straight angle. Symmetry with respect to a line: A geometric ﬁgure S is said to be symmetric with respect to a line L if the reﬂection with respect to L maps S to S. Transformation: A transformation of the plane is a rule that assigns to each point P of the plane a unique point F (P ) (read: “F of P ”) in the plane. −→ −− → Translation: Given a vector AB, the translation along AB is the transformation TAB of the plane so that, for a point P in the plane, TAB (P ) = −−→ Q, where Q is the endpoint of the vector P Q which points in the same −− → direction as AB and so that |P Q| = |AB|.

356

APPENDIX: FACTS FROM [Wu2020a]

Triangle: A 3-sided polygon, equivalently, a geometric ﬁgure consisting of three noncollinear points, together with the segments that join these points. − − → Vector: A vector AB for two points A and B in the plane is the segment AB, together with the designation that A is the starting point and B is the endpoint.

Part 3. Theorems and lemmas Basic facts about inequalities in Section 2.6 of [Wu2020a]: If x, y, z, . . . are rational numbers, then: (A) x < y ⇐⇒ −x > −y. (B) x < y ⇐⇒ x + z < y + z. (C) x < y ⇐⇒ y − x > 0. (D) If z > 0, then x < y ⇐⇒ xz < yz. (E) If z < 0, then x < y ⇐⇒ xz > yz. Cancellation law for rational quotients: If x, y, z ∈ Q and y, z = 0, then, zx x = . y zy Corollary 1 of Theorem 2.9 in [Wu2020a]: If x, y ∈ Q and xy = 0, then x = 0 or y = 0. Cross-multiplication algorithm: (i) For rational numbers x, y, z, and w, with y = 0 and w = 0: xy = wz if and only if xw = yz. (ii) For positive complex fractions x, y, z, and w, xy < wz if and only if xw < yz. Distance formula: Let (a, b) and (c, d) be two points in the coordinate plane. Then distance between (a, b) and (c, b) = (a − c)2 + (b − d)2 . Exercise 4 in Exercises 6.6 of [Wu2020a]: Let O be the origin of a coordinate system in R2 , and let r be a positive number. Prove that the transformation T of R2 which assigns to each point (a, b) the point (ra, rb) is exactly the dilation with center O and scale factor r. Exercise 7 in Exercises 4.1 of [Wu2020a]: (a) Prove that the intersection of a ﬁnite number of convex sets is convex. (b) Prove that the intersection of any number of convex sets is convex (i.e., the number of sets could be inﬁnite). (c) Prove that a closed half-plane is convex. Exercise 18 in Exercises 2.6 of [Wu2020a]: (a) Prove the following inequality of arithmetic and geometric means for two numbers: if s and t √ are nonnegative numbers, then st ≤ 12 (s + t), and equality holds if and only if s = t. (b) Prove that among all rectangles with a ﬁxed perimeter, the square has the biggest area. Exercise 19 in Exercises 2.6 of [Wu2020a]: If x and y are positive, then prove that (a) x2 = y 2 if and only if x = y and (b) x2 < y 2 if and only if x < y. Formulas for rational quotients in Section 2.5 of [Wu2020a]: Let x, y, z, w, . . . be rational numbers so that they are nonzero where appropriate in the following.

APPENDIX: FACTS FROM [Wu2020a]

357

(a) Cancellation law: xy = zx zy for any nonzero z. (b) Cross-multiplication algorithm: xy = wz if and only if xw = yz. (c) xy ± wz = xw±yz yw . x z xz (d) y × w = yw . Fundamental theorem of arithmetic: Every whole number n ≥ 2 is the product of a ﬁnite number of primes: n = p1 p2 · · · pk . Moreover, this collection of primes p1 , . . . , pk , counting the repetitions, is unique. Lemma 3.4 (Key lemma) of [Wu2020a]: Suppose , m, n are nonzero whole numbers and |mn. If and m are relatively prime, then |n. Lemma 4.3 of [Wu2020a]: If three lines L1 , L2 , and L3 satisfy L1 L2 and L2 L3 , then either L1 = L3 or L1 L3 . Lemma 4.4 of [Wu2020a]: Given three distinct collinear points A, B, C, then exactly one of the following three possibilities holds: A ∗ B ∗ C, B ∗ C ∗ A, or C ∗ A ∗ B. Lemma 4.5 (Line separation) of [Wu2020a]: A point O on a line L separates L into two nonempty subsets, L+ and L− , called the half-lines of O, and L+ and L− satisfy the following two properties: (i) The line L is the disjoint union of L+ , L− , and {O} (the set containing O alone), and the half-lines L+ and L− are convex. (ii) If two points A and B on L belong to diﬀerent half-lines, then the line segment AB contains O. Lemma 4.6 of [Wu2020a]: Given three distinct points O, A, and B on a line L. Then: (i) B lies on the opposite ray of ROA ⇐⇒ A ∗ O ∗ B. (ii) B lies on the ray ROA ⇐⇒ either O ∗ A ∗ B or O ∗ B ∗ A. (iii) The ray RAB is contained in the ray ROA ⇐⇒ O ∗ A ∗ B. Lemma 4.8 of [Wu2020a]: Let L and L be two distinct lines and let P1 , P2 , and P3 be three distinct points on L so that P1 ∗ P2 ∗ P3 . Let three mutually parallel lines passing through P1 , P2 , and P3 intersect L at Q1 , Q2 , and Q3 , respectively. Then Q1 ∗ Q2 ∗ Q3 . (It is possible that Pi = Qi for some i = 1, 2, or 3.) Lemma 4.16 of [Wu2020a]: Given any point O in the plane and any number θ satisfying −360 ≤ θ ≤ 360, there is a rotation of θ degrees around O. Lemma 4.19 of [Wu2020a]: Given any line L, there is a reﬂection across L. −−→ Lemma 4.22 of [Wu2020a]: Given any vector AB, there is a unique trans−− → lation along AB. Lemma 5.1 of [Wu2020a]: Opposite angles at a point are equal. Lemma 5.2 of [Wu2020a]: (i) The composition of a ﬁnite number of similarities is a similarity. (ii) Each similarity has an inverse transformation that is also a similarity. (iii) The similarity of two ﬁgures S ∼ S* is a symmetric relation. (iv) The similarity of two ﬁgures S ∼ S* is also a transitive relation. Lemma 6.13 of [Wu2020a]: The equation of a nonvertical line L is of the form y = mx + k, where m is the slope of L and k is the y-intercept of L.

358

APPENDIX: FACTS FROM [Wu2020a]

Lemma 6.14 of [Wu2020a]: Let the distinct points P1 = (x1 , y1 ) and P2 = (x2 , y2 ) lie on a nonvertical line L. Then the equation of L is y − y1 = m(x − x1 ), where y2 − y1 m= . x2 − x1 Lemma 6.15 of [Wu2020a]: Let two distinct points P = (p, p ) and Q = (q, q ) be given. (i) If p = q but p < q , then the segment P Q consists of all the points {(p, t)}, where p ≤ t ≤ q . (ii) If p < q, let P and Q lie on the line whose equation is y = mx+k for some constants M and k. Then P Q consists of all the points {(t, mt+k)}, where p ≤ t ≤ q. Lemma 6.16 of [Wu2020a]: (i) Let a nonvertical line L be the graph of y = ax + b, and let P be a point on L with coordinates (p, ap + b). Then the two half-lines of L determined by P are the following two subsets of L: P − = all the points (x, ax + b) so that x < p. P + = all the points (x, ax + b) so that p < x. (ii) For a vertical line L deﬁned by x = c, if P = (c, p) is a point of L, then the half-lines of L determined by P are the subsets P − = all the points (c, y) so that y < p. P + = all the points (c, y) so that p < y. Lemma 6.20 of [Wu2020a]: Let T be the translation along the vector −−→ BC, where B = (b1 , b2 ) and C = (c1 , c2 ). Then for all (x, y) in R2 , T (x, y) = (x + a1 , y + a2 ), where (a1 , a2 ) = (c1 − b1 , c2 − b2 ). Theorem G15*: Let ABC be given and let D be the midpoint of AB. Suppose a line parallel to BC passing through D intersects AC at E. Then E is the midpoint of AC and 2|DE| = |BC|. Theorem 1 in the appendix of Chapter 1 of [Wu2020a]: For any ﬁnite collection of numbers, the sums obtained by adding them up in any order are all equal. Theorem 2 in the appendix of Chapter 1 of [Wu2020a]: For any ﬁnite collection of numbers, the products obtained by multiplying them in any order are all equal. Theorem 1.7 [Wu2020a]: The area of a rectangle with sides of lengths k k m and m n is equal to × n . Theorem 2.9, Corollary 1 of [Wu2020a]: If x, y ∈ Q and xy = 0, then x = 0 or y = 0. Theorem 3.9 of [Wu2020a]: Let n be a whole number which is not a perfect square. If there is a positive number r so that r 2 = n, then r is irrational. Theorem 4.13 of [Wu2020a]: The complement of a polygon P consists of two nonempty planar regions B and E with the following properties: (i) B and E are both connected, B is bounded and E is unbounded, and P is their common boundary. Moreover, the three sets B, E, and P are disjoint and their union is the whole plane. (ii) A segment joining a point of B to a point of E must intersect the polygon P.

APPENDIX: FACTS FROM [Wu2020a]

359

In addition, suppose we have two nonempty planar regions B and E so that P is their common boundary and the three sets B, E, and P are disjoint. Then after a change of notation if necessary, B = B and E = E. Theorem 6.9 of [Wu2020a]: Two lines passing through the same point and having the same slope are identical. Theorem 6.17 of [Wu2020a]: Two distinct nonvertical lines have the same slope ⇐⇒ they are parallel. Theorem 6.18 of [Wu2020a]: Two distinct nonvertical lines are perpendicular ⇐⇒ the product of their slopes is −1. Theorem 6.21 of [Wu2020a]: A simultaneous system of linear equations has a solution (x0 , y0 ) ⇐⇒ the point (x0 , y0 ) lies in the intersection of the (linear) graphs of the equations in the linear system. Trichotomy law: Given two numbers x and y, one and only one of the following three possibilities holds: x = y or x < y or x > y.

Glossary of Symbols Those symbols that are standard in the mathematics literature or were introduced in [Wu2020a] usually will be listed without a page reference; those that are introduced in this volume are given a page reference. N : the whole numbers, 2 Q : the rational numbers, 2 R : the real numbers, 2 Z : the integers, 2 ⇐⇒ : is equivalent to, 194 =⇒ : implies · · · a for a number a and a positive integer n, 138 an : the product aaa n

n!

n: n factorial for a whole number n, 204 n! k : binomial coeﬃcient deﬁned by k!(n−k)! , 203 x · y : product of the numbers x and y |x| √ : absolute value of a number x √ α : if α is a positive (real) number, α denotes the √ unique positive square root of α, but √if α is a complex number, then α is a complex number that satisﬁes ( α)2 = α, 65 and 195 [a, b] : the segment from a to b on the number line or the closed interval from a to b for numbers a < b, 10 (a, b) : the open interval from a to b on the number line for numbers a < b (it could also mean the point (a, b) in the coordinate plane), 122 < : less than ≤ : less than or equal to > : greater than ≥ : greater than or equal to αx : the value of the exponential function with base α (α > 0) at the real number x, 139 e : the base of natural logarithm, 164 logα t : the logarithm of t (t > 0) with base α (α > 0), 164 R[X] : the polynomial ring over R, 177 R(X) : the ﬁeld of rational forms over R, 178 f (X)|g(X) : for polynomial forms f (X) and g(X) in R[X] (respectively, C[X]), this means there is an h(X) ∈ R[X] (respectively, h(X) ∈ C[X]) so that g(X) = h(X)f (X), 184 C : the complex numbers, 189 C[X] : the polynomial ring over C, 194 ∈ : belongs to (as in a ∈ A), 3 A ⊂ B : A is contained in B, 6 361

362

GLOSSARY OF SYMBOLS

∪ : union (of sets), 32 ∩ : intersection (of sets), 32 R2 : the coordinate plane, 2 (x, y) : coordinates of a point in the plane (it could also mean the open interval from the number x to the number y), 6 AB : the segment joining the two points A and B in the plane, 28 dist(A, B) : the distance between two points A and B in the plane, 214 |AB| : the length of segment AB for two points A and B in the plane, 28 −− → AB : the vector from the point A to the point B in the plane, 356 A ∗ C ∗ B : the point C is between points A and B, 214 LP Q : the line joining P to Q for two points P and Q in the plane, 34, 213 H+ , H− : half-planes of a line L, 31 ROP : the ray on LOP from O to P , 213 : is parallel to, 213 ⊥ : is perpendicular to, 227 ∠AOB : the angle with vertex O and sides ROA and ROB , 351 ABC : the triangle with vertices A, B, and C, 355 ∠A : the angle of a triangle or a polygon at a vertex A x◦ : x degree (of an angle), where x is a positive real number ∼ = : is congruent to, 218 ∼ : is similar to, 220

AB : one of two arcs joining the point A to the point B on a circle, 276

Bibliography [Altshiller-Court] N. Altshiller-Court, College geometry: An introduction to the modern geometry of the triangle and the circle, reprint of the second (1980) edition, Dover Publications, Inc., Mineola, NY, 2007. MR2351498 [Arbaugh et al.] F. Arbaugh, M. Smith, J. Boyle, and M. Steele, We Reason & We Prove for All Mathematics, Corwin, Thousand Oaks, CA, 2018. [Askey] R. Askey, Orthogonal polynomials and special functions, Society for Industrial and Applied Mathematics, Philadelphia, Pa., 1975. MR0481145 [Bashmakova-Smirnova] I. Bashmakova and G. Smirnova, Beginning & Evolution of Algebra, Mathematical Association of America, Washington DC, 2000. MR1735499 [Bix] R. Bix, Topics in geometry, Academic Press, Inc., Boston, MA, 1994. MR1244576 [Boyer-Merzbach] C. B. Boyer, A history of mathematics, 2nd ed., edited and with a preface by Uta C. Merzbach, John Wiley & Sons, Inc., New York, 1989. MR996888 [CCSSM] Common Core State Standards for Mathe