Front Matter (C)
Cover (C)
Version Date (i)
Contents (v)
Tips for the Student (xxxi)
Miscellaneous Tips for the Student (xxxv)
Use of Graphing Calculators (xxxvii)
Apps You Can Use on Your TI Calculator (xxxix)
Preface/Rant (xl)
How and Why JC Maths Fails Our Students (xlii)
Work Work Work Work Work Work (xliv)
Taking Tests Seriously (xlv)
Genuine Understanding (xlvii)
Why Even Type 1 Pragmatists Should Read This Textbook (xlix)
PISA 2018 Maths Results (li)
0. A Few Basics (1)
1. Just To Be Clear (3)
2. PSLE Review: Division (4)
2.1. Long Division (6)
2.2. The Cardinal Sin of Dividing By Zero (8)
2.3. Is One Divided by Zero Infinity? (9)
3. Logic (10)
3.1. True, False, and Indeterminate Statements (11)
3.2. The Conjunction AND and the Disjunction OR (12)
3.3. The Negation NOT (13)
3.4. Equivalence ⟺ (14)
3.5. De Morgan's Laws: Negating the Conjunction and Disjunction (15)
3.6. The Implication ? ⟹ ? (17)
3.7. The Converse ? ⟹ ? (19)
3.8. Affirming the Consequent (or The Fallacy of the Converse) (21)
3.9. The Negation NOT-(? ⟹ ?) (22)
3.10. The Contrapositive NOT-? ⟹ NOT-? (25)
3.11. (? ⟹ ? AND ? ⟹ ?) ⟺ (? ⟺ ?) (27)
3.12. Other Ways to Express ? ⟹ ? (Optional) (28)
3.13. The Four Categorical Propositions and Their Negations (29)
3.14. Chapter Summary (34)
4. Sets (35)
4.1. The Elements of a Set Can Be Pretty Much Anything (36)
4.2. In ∈ and Not In ∉ (38)
4.3. The Order of the Elements Doesn't Matter (39)
4.4. n(?) Is the Number of Elements in the Set ? (40)
4.5. The Ellipsis ``…'' Means Continue in the Obvious Fashion (40)
4.6. Repeated Elements Don't Count (41)
4.7. ℝ Is the Set of Real Numbers (42)
4.8. ℤ Is the Set of Integers (43)
4.9. ℚ Is the Set of Rational Numbers (44)
4.10. A Taxonomy of Numbers (45)
4.11. More Notation: ⁺, ⁻, and ₀ (46)
4.12. The Empty Set ∅ (47)
4.13. Intervals (48)
4.14. Subset Of ⊆ (50)
4.15. Proper Subset Of ⊂ (51)
4.16. Union ∪ (53)
4.17. Intersection ∩ (54)
4.18. Set Minus ∖ (56)
4.19. The Universal Set ℰ (57)
4.20. The Set Complement ?' (58)
4.21. De Morgan's Laws (60)
4.22. Set-Builder Notation (or Set Comprehension) (62)
4.23. Chapter Summary (67)
5. O-Level Review (68)
5.1. Some Mathematical Vocabulary (68)
5.2. The Absolute Value or Modulus Function (69)
5.3. The Factorial ?! (71)
5.4. Exponents (72)
5.5. The Square Root Refers to the Positive Square Root (78)
5.6. Rationalising the Denominator with a Surd (81)
5.7. Logarithms (83)
5.8. Polynomials (85)
I. Functions and Graphs (88)
6. Graphs (90)
6.1. Ordered Pairs (90)
6.2. The Cartesian Plane (94)
6.3. A Graph is Any Set of Points (96)
6.4. The Graph of An Equation (98)
6.5. Graphing with the TI84 (100)
6.6. The Graph of An Equation with Constraints (101)
6.7. Intercepts and Roots (108)
7. Lines (111)
7.1. Horizontal, Vertical, and Oblique Lines (112)
7.2. Another Equation of the Line (113)
7.3. The Gradient of a Line (115)
7.4. Yet Another Equation of the Line (118)
7.5. Perpendicular Lines (120)
7.6. Lines vs Line Segments vs Rays (121)
8. Distance (122)
8.1. The Distance between Two Points on the Real Number Line (122)
8.2. The Distance between Two Points on the Cartesian Plane (123)
8.3. Closeness (or Nearness) (125)
9. Circles (126)
9.1. Graphing Circles on the TI84 (129)
9.2. The Number π (optional) (131)
10. Tangent Lines, Gradients, and Stationary Points (133)
10.1. Tangent Lines (133)
10.2. The Gradient of a Graph at a Point (134)
10.3. Stationary Points (135)
11. Maximum, Minimum, and Turning Points (136)
11.1. Maximum Points (136)
11.2. Minimum Points (140)
11.3. Extrema (144)
11.4. Turning Points (145)
12. Reflection and Symmetry (148)
12.1. The Reflection of a Point in a Point (148)
12.2. The Reflection of a Point in a Line (150)
12.3. Lines of Symmetry (152)
13. Solutions and Solution Sets (155)
14. O-Level Review: The Quadratic Equation ? = ??² + ?? + ? (163)
15. Functions (173)
15.1. What Functions Aren't (184)
15.2. Notation for Functions (187)
15.3. Warning: ? and ?(?) Refer to Different Things (194)
15.4. Nice Functions (195)
15.5. Graphs of Functions (197)
15.6. Piecewise Functions (201)
15.7. Range (206)
16. An Introduction to Continuity (211)
17. When a Function Is Increasing or Decreasing (215)
18. Arithmetic Combinations of Functions (219)
19. Domain Restriction (223)
20. Composite Functions (225)
21. One-to-One Functions (233)
21.1. Two More Characterisations of One-to-One (236)
21.2. A Strictly Increasing/Decreasing Function Is One-to-One (239)
22. Inverse Functions (240)
22.1. Only a One-to-One Function Can Have an Inverse (244)
22.2. Formal Definition of the Inverse (246)
22.3. A Method for Showing a Function Is One-to-One and Finding Its Inverse (247)
22.4. Inverse Cancellation Laws (251)
22.5. The Graphs of ? and ?⁻¹ Are Reflections in the Line ? = ? (252)
22.6. When Do the Graphs of ? and ?⁻¹ Intersect? (259)
22.7. Self-Inverses (263)
22.8. If ? Is Strictly Increasing/Decreasing, Then So Is ?⁻¹ (264)
22.9. A Function Is the Inverse of Its Inverse (265)
22.10. Domain Restriction to Create a One-to-One Function (267)
23. Asymptotes and Limit Notation (270)
23.1. Oblique Asymptotes (278)
24. Transformations (280)
24.1. ? = ?(?) + ? (280)
24.2. ? = ?(? + ?) (282)
24.3. ? = ??(?) (285)
24.4. ? = ?(??) (287)
24.5. Combinations of the Above (289)
24.6. ? = |?(?)| (293)
24.7. ? = ?(|?|) (295)
24.8. 1 / ? (297)
24.9. When a Function Is Symmetric in Vertical or Horizontal Lines (301)
25. ln, exp, and e (302)
25.1. The Exponential Function exp (306)
25.2. Euler's Number (308)
26. Trigonometry: Arcs (of a Circle) (310)
27. Trigonometry: Angles (313)
27.1. Angles, Informally Defined Again; Also, the Radian (315)
27.2. Names of Angles (318)
28. Trigonometry: Sine and Cosine (320)
28.1. The Values of Sine and Cosine at 0 and 1 (321)
28.2. Three Values of Sine and Cosine (323)
28.3. Confusing Notation (326)
28.4. Some Formulae Involving Sine and Cosine (327)
28.5. The Area of a Triangle; the Laws of Sines and Cosines (331)
28.6. The Unit-Circle Definitions (333)
28.7. Graphs of Sine and Cosine (343)
28.8. Graphs of Sine and Cosine: Some Observations (348)
29. Trigonometry: Tangent (351)
29.1. Some Formulae Involving Tangent (352)
29.2. Graph of Tangent (355)
29.3. Graph of Tangent: Some Observations (357)
30. Trigonometry: Summary and a Few New Things (358)
30.1. The Signs of Sine, Cosine, and Tangent (359)
30.2. Half-Angle Formulae (361)
31. Trigonometry: Three More Functions (363)
32. Trigonometry: Three Inverse Trigonometric Functions (366)
32.1. Confusing Notation (370)
32.2. Inverse Cancellation Laws, Revisited (371)
32.3. More Compositions (375)
32.4. Addition Formulae for Arcsine and Arccosine (378)
32.5. Addition Formulae for Arctangent (380)
32.7. Even More Compositions (optional) (384)
33. Elementary Functions (385)
34. Factorising Polynomials (391)
34.1. Long Division of Polynomials (391)
34.2. Factorising Polynomials (395)
34.3. Compare Coefficients (396)
34.5. The Remainder Theorem (400)
34.6. The Factor Theorem (401)
34.7. The Intermediate Value Theorem (IVT) (405)
34.8. Factorising a Quartic Polynomial (408)
34.9. Two Warnings (410)
35. Solving Systems of Equations (411)
35.1. Solving Systems of Equations with the TI84 (420)
36. Partial Fractions Decomposition (423)
36.1. Non-Repeated Linear Factors (424)
36.2. Repeated Linear Factors (425)
37. Solving Inequalities (427)
37.1. Multiplying an Inequality by an Unknown Constant (427)
37.2. ??² + ?? + ? > 0 (429)
37.3. (?? + ?) / (?? + ?) > 0 (436)
37.4. The Cardinal Sin of Dividing by Zero, Revisited (442)
37.5. Inequalities Involving the Absolute Value Function (443)
37.6. (??² + ?? + ?) / (??² + ?? + ?) > 0 (450)
37.7. Solving Inequalities by Graphical Methods (453)
38. Extraneous Solutions (456)
38.1. Squaring (456)
38.2. Multiplying by Zero (458)
38.3. Removing Logs (459)
39. O-Level Review: The Derivative (461)
39.2. The Derivative as the Gradient (463)
39.3. The Derivative as Rate of Change (466)
39.4. Differentiability vs Continuity (469)
39.5. Rules of Differentiation (473)
39.6. The Chain Rule (476)
39.7. Increasing and Decreasing (479)
39.8. Stationary and Turning Points (480)
40. Conic Sections (490)
40.1. The Ellipse ?² + ?² = 1 (The Unit Circle) (491)
40.2. The Ellipse ?² / ?² + ?² / ?²  = 1 (492)
40.3. The Hyperbola ? = 1 / ? (494)
40.4. The Hyperbola ?² − ?² = 1 (495)
40.5. The Hyperbola ?² / ?² − ?² / ?² = 1 (496)
40.6. The Hyperbola ?² / ?² − ?² / ?² = 1 (497)
40.7. The Hyperbola ? = (?? + ?) / (?? + ?) (498)
40.8. The Hyperbola ? = (??² + ?? + ?) / (?? + ?) (504)
40.9. Eliminating the Parameter ? (509)
41. Simple Parametric Equations (512)
II. Sequences and Series (517)
42. Sequences (519)
42.1. Sequences Are Functions (521)
42.2. Notation (523)
42.3. Arithmetic Combinations of Sequences (524)
43. Series (525)
43.1. Convergent and Divergent Series (526)
44. Summation Notation ∑ (529)
44.1. Summation Notation for Infinite Series (531)
45. Arithmetic Sequences and Series (534)
45.1. Finite Arithmetic Series (535)
45.2. Infinite Arithmetic Series (536)
46. Geometric Sequences and Series (537)
46.1. Finite Geometric Sequences and Series (538)
46.2. Infinite Geometric Sequences and Series (540)
47. Rules of Summation Notation (542)
48. The Method of Differences (544)
III. Vectors (549)
49. Introduction to Vectors (551)
49.1. The Magnitude or Length of a Vector (553)
49.2. When Are Two Vectors Identical? (554)
49.3. A Vector and a Point Are Different Things (555)
49.4. Two More Ways to Denote a Vector (556)
49.5. Position Vectors (557)
49.6. The Zero Vector (557)
49.7. Displacement Vectors (558)
49.8. Sum and Difference of Points and Vectors (559)
49.9. Sum, Additive Inverse, and Difference of Vectors (561)
49.10. Scalar Multiplication of a Vector (564)
49.11. When Do Two Vectors Point in the Same Direction? (565)
49.12. When Are Two Vectors Parallel? (566)
49.13. Unit Vectors (567)
49.14. The Standard Basis Vectors (569)
49.15. Any Vector Is A Linear Combination of Two Other Vectors (570)
49.16. The Ratio Theorem (572)
50. Lines (574)
50.1. Direction Vector (574)
50.2. Cartesian to Vector Equations (578)
50.3. Pedantic Points to Test/Reinforce Your Understanding (583)
50.4. Vector to Cartesian Equations (584)
51. The Scalar Product (587)
51.1. A Vector's Scalar Product with Itself (589)
52. The Angle Between Two Vectors (590)
52.1. Pythagoras' Theorem and Triangle Inequality (597)
52.2. Direction Cosines (598)
53. The Angle Between Two Lines (601)
54. Vectors vs Scalars (607)
55. The Projection and Rejection Vectors (610)
56. Collinearity (614)
57. The Vector Product (616)
57.1. The Angle between Two Vectors Using the Vector Product (619)
57.2. The Length of the Rejection Vector (620)
58. The Foot of the Perpendicular From a Point to a Line (621)
58.1. The Distance Between a Point and a Line (623)
59. Three-Dimensional (3D) Space (628)
59.1. Graphs (in 3D) (630)
60. Vectors (in 3D) (632)
60.1. The Magnitude or Length of a Vector (634)
60.2. Sums and Differences of Points and Vectors (635)
60.3. Sum, Additive Inverse, and Difference of Vectors (638)
60.4. Scalar Multiplication and When Two Vectors Are Parallel (642)
60.5. Unit Vectors (644)
60.6. The Standard Basis Vectors (645)
60.7. The Ratio Theorem (646)
61. The Scalar Product (in 3D) (647)
61.1. The Angle between Two Vectors (649)
61.2. Direction Cosines (652)
62. The Projection and Rejection Vectors (in 3D) (653)
63. Lines (in 3D) (657)
63.1. Vector to Cartesian Equations (660)
63.2. Cartesian to Vector Equations (665)
63.3. Parallel and Perpendicular Lines (668)
63.4. Intersecting Lines (670)
63.5. Skew Lines (672)
63.6. The Angle Between Two Lines (674)
63.7. Collinearity (in 3D) (678)
64. The Vector Product (in 3D) (681)
64.1. The Right-Hand Rule (685)
64.2. The Length of the Vector Product (686)
64.3. The Length of the Rejection Vector (687)
65. The Distance Between a Point and a Line (in 3D) (688)
66. Planes: Introduction (693)
66.1. The Analogy Between a Plane, a Line, and a Point (697)
67. Planes: Formally Defined in Vector Form (699)
67.1. The Normal Vector (703)
68. Planes in Cartesian Form (709)
68.1. Finding Points on a Plane (711)
68.2. Finding Vectors on a Plane (712)
69. Planes in Parametric Form (714)
69.1. Planes in Parametric Form (717)
69.2. Parametric to Vector or Cartesian Form (721)
70. Four Ways to Uniquely Determine a Plane (723)
71. The Angle between a Line and a Plane (727)
71.1. When a Line and a Plane Are Parallel, Perp., or Intersect (729)
72. The Angle between Two Planes (733)
72.1. When Two Planes Are Parallel, Perp., or Intersect (734)
73. Point-Plane Foot of the Perpendicular and Distance (739)
73.1. Formula Method (742)
74. Coplanarity (746)
74.1. Coplanarity of Lines (749)
IV. Complex Numbers (752)
75. Complex Numbers: Introduction (754)
75.1. The Real and Imaginary Parts of Complex Numbers (757)
75.2. Complex Numbers in Ordered Pair Notation (758)
76. Some Arithmetic of Complex Numbers (759)
76.2. Multiplication (760)
76.3. Conjugation (762)
76.4. Division (764)
77. Solving Polynomial Equations (765)
77.1. The Fundamental Theorem of Algebra (766)
77.2. The Complex Conjugate Root Theorem (768)
78. The Argand Diagram (770)
79. Complex Numbers in Polar Form (771)
79.1. The Argument: An Informal Introduction (772)
79.2. The Argument: Formally Defined (773)
79.3. Complex Numbers in Polar Form (775)
80. Complex Numbers in Exponential Form (776)
80.1. Complex Numbers in Exponential Form (777)
80.2. Some Useful Formulae for Sine and Cosine (778)
81. More Arithmetic of Complex Numbers (779)
81.1. The Reciprocal (783)
81.2. Division (784)
V. Calculus (787)
82. Limits (789)
82.1. Limits, Informally Defined (790)
82.2. Examples Where The Limit Does Not Exist (795)
82.3. Rules for Limits (804)
83. Continuity, Revisited (806)
83.1. Functions with a Single Discontinuity (809)
83.2. An Example of a Function That Is Discontinuous Everywhere (817)
83.3. Functions That Seem Discontinuous But Aren't (818)
83.4. Every Elementary Function Is Continuous (821)
83.5. Continuity Allows Us to ``Move'' Limits (822)
83.6. Every Elementary Function is Continuous: Proof (Optional) (825)
83.7. Continuity at Isolated Points (optional) (830)
84. The Derivative, Revisited (832)
84.1. Differentiable ⟺ Approximately Linear (838)
84.2. Continuity vs Differentiability (840)
84.3. The Derivative as a Function (841)
84.4. ``Most'' Elementary Functions Are Differentiable (845)
85. Differentiation Notation (847)
85.1. Using d /d ? as the Differentiation Operator (851)
85.2. Using d /d ? as Shorthand (854)
85.3. d ?/d ? Is Not a Fraction; Nonetheless ... (856)
86. Rules of Differentiation, Revisited (860)
86.1. Proving Some Basic Rules of Differentiation (optional) (861)
86.2. Proving the Product Rule (optional) (866)
86.3. Proving the Quotient Rule (optional) (868)
86.4. Proving the Chain Rule (optional) (870)
87. Some Techniques of Differentiation (871)
87.1. Implicit Differentiation (871)
87.2. The Inverse Function Theorem (IFT) (880)
87.3. Term-by-Term Differentiation (optional) (882)
88. Parametric Equations (884)
88.1. Parametric Differentiation (888)
89. The Second and Higher Derivatives (890)
89.1. Higher Derivatives (896)
89.2. Smooth (or ``Infinitely Differentiable'') Functions (900)
89.3. Confusing Notation (902)
90. The Increasing/Decreasing Test (906)
91. Determining the Nature of a Stationary Point (910)
91.1. The First Derivative Test for Extrema (FDTE) (914)
91.2. A Situation Where a Stationary Point Is Also a Strict Global Extremum (920)
91.3. Fermat's Theorem on Extrema (921)
91.4. The Second Derivative Test for Extrema (SDTE) (923)
92. Concavity (930)
92.1. The First Derivative Test for Concavity (FDTC) (934)
92.2. The Second Derivative Test for Concavity (SDTC) (936)
92.3. The Graphical Test for Concavity (GTC) (940)
92.4. A Linear Function Is One That's Both Concave and Convex (944)
93. Inflexion Points (945)
93.1. The First Derivative Test for Inflexion Points (FDTI) (947)
93.2. The Second Derivative Test for Inflexion Points (SDTI) (949)
93.3. The Tangent Line Test (TLT) (950)
93.4. Non-Stationary Points of Inflexion (optional) (952)
94. A Summary of Chapters 90, 91, 92, and 93 (953)
95. More Techniques of Differentiation (954)
95.1. Relating the Graph of ?′ to That of ? (954)
95.2. Equations of Tangents and Normals (955)
95.3. Connected Rates of Change Problems (957)
96. More Fun With Your TI84 (960)
96.1. Locating Maximum and Minimum Points Using Your TI84 (961)
96.2. Find the Derivative at a Point Using Your TI84 (963)
97. Power Series (967)
97.1. A Power Series Is Simply an ``Infinite Polynomial'' (967)
97.2. A Power Series Can Converge or Diverge (970)
97.3. A Power Series and Its Interval of Convergence (972)
97.4. (Some) Functions Can be Represented by a Power Series (974)
97.5. Term-by-Term Differentiation of a Power Series (975)
97.6. Chapter Summary (978)
98. The Maclaurin Series (979)
98.1. (Some) Functions Can Be Rep'd by Their Maclaurin Series (984)
98.2. The Five Standard Series and Their Intervals of Convergence (989)
98.3. Sine and Cosine, Formally Defined (991)
98.4. Maclaurin Polynomials as Approximations (993)
98.5. Creating New Series Using Substitution (996)
98.6. Creating New Series Using Multiplication (1000)
98.7. Repeated Differentiation to Find a Maclaurin Series (1006)
98.8. Repeated Implicit Differentiation to Find a Maclaurin Series (1007)
99. Antidifferentiation (1010)
99.1. Antiderivatives Are Not Unique ... (1011)
99.2. ... But An Antiderivative Is Unique Up to a COI (1011)
99.3. Shorthand: The Antidifferentiation Symbol ∫ (1013)
99.4. Rules of Antidifferentiation (1017)
99.5. Every Continuous Function Has an Antiderivative (1020)
99.6. Be Careful with This Common Practice (1021)
99.7. The ``Punctuation Mark'' dx (optional) (1022)
100. The Definite Integral (1025)
100.1. Notation for Integration (1028)
100.2. The Definite Integral, Informally Defined (1029)
100.3. An Important Warning (1031)
100.4. A Sketch of How We Can Find the Area under a Curve (1032)
100.5. If a Function Is Continuous, Then Its Definite Integral Exists (1036)
100.6. Rules of Integration (1037)
100.7. Term-by-Term Integration (optional) (1040)
101. The Fundamental Theorems of Calculus (1042)
101.1. Definite Integral Functions (1042)
101.2. The First Fundamental Theorem of Calculus (FTC1) (1045)
101.3. The Second Fundamental Theorem of Calculus (FTC2) (1049)
101.4. Why Integration and Antidifferentiation Use the Same Notation (1051)
101.5. Some New Notation and More Examples and Exercises (1053)
102. More Techniques of Antidifferentiation (1054)
102.1. Factorisation (1054)
102.2. Partial Fractions: Finding ∫1 / (??² + ?? + ?) d? where ?² − 4?? > 0 (1055)
102.3. Building a Divisor of the Denominator (1057)
102.4. More Rules of Antidifferentiation (1059)
102.5. Completing the Square: ∫1 / (??² + ?? + ?) d? where ?² − 4??

##### Citation preview

H2 Mathematics Textbook CHOO YAN MIN

Revision in progress.1 Latest changes: Currently revising Part V (Calculus).

1

This textbook was first completed in Jul 2016. For 1.5 years afterwards, only minor changes were made. But starting Feb 2018, I’ll be completely rewriting this textbook. In particular, I’ll be working to (a) slow down the pace of this textbook; and (b) explain things more clearly and simply. Less importantly, I’ll also be removing the old (9740) material that is no longer on the current (9758) syllabus and making trivial formatting changes (to make the book beautifuller). As with everything I do, please let me know if you spot any errors or have any feedback. Thank you.

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, Errors? Feedback? Email me! , With your help, I plan to keep improving this textbook. Please refer to the latest version and also mention the relevant page number(s) (bottom left).

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You are free to: • Share — copy and redistribute the material in any medium or format • Adapt — remix, transform, and build upon the material The licensor cannot revoke these freedoms as long as you follow the license terms.

Under the following terms: • Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. • NonCommercial — You may not use the material for commercial purposes. • ShareAlike — If you remix, transform, or build upon the material, you must distribute your contributions under the same license as the original. • No additional restrictions — You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits.

Notices: You do not have to comply with the license for elements of the material in the public domain or where your use is permitted by an applicable exception or limitation. No warranties are given. The license may not give you all of the permissions necessary for your intended use. For example, other rights such as publicity, privacy, or moral rights may limit how you use the material. Author: Choo, Yan Min. Title: H2 Mathematics Textbook. ISBN: 978-981-11-0383-4 (e-book).

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The first thing to understand is that mathematics is an art. — Paul Lockhart (2009).

A mathematician, like a painter or a poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas. ... Beauty is the first test: there is no permanent place in the world for ugly mathematics. — G.H. Hardy (1940).

Le savant n’étudie pas la nature parce que cela est utile; il l’étudie parce qu’il y prend plaisir et il y prend plaisir parce qu’elle est belle. Si la nature n’était pas belle, elle ne vaudrait pas la peine d’être connue, la vie ne vaudrait pas la peine d’être vécue. The scientist does not study nature because it is useful to do so. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful. If nature were not beautiful it would not be worth knowing, and life would not be worth living. — Henri Poincaré (1908, 1914).

[W]hoever does not love and admire mathematics for its own internal splendours, knows nothing whatever about it. — Michael Polanyi (1959).

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Contents Cover

C

Version Date

i

iii

Contents

v

xxix

Tips for the Student

xxxi

Miscellaneous Tips for the Student

xxxv

Use of Graphing Calculators

xxxvii

Apps You Can Use on Your TI Calculator . . . . . . . . . . . . . . . . . . . . . . x . xxix

Preface/Rant

xl

How and Why JC Maths Fails Our Students . . . . . . . . Work Work Work Work Work Work . . . . . . . . . . . . . Taking Tests Seriously . . . . . . . . . . . . . . . . . . . . . Genuine Understanding . . . . . . . . . . . . . . . . . . . . Why Even Type 1 Pragmatists Should Read This Textbook PISA 2018 Maths Results . . . . . . . . . . . . . . . . . . .

Part 0.

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A Few Basics

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Just To Be Clear

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PSLE Review: Division

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2.1 2.2 2.3

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Long Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Cardinal Sin of Dividing By Zero . . . . . . . . . . . . . . . . . . . . Is One Divided by Zero Infinity? . . . . . . . . . . . . . . . . . . . . . . .

Logic 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

v, Contents

6 8 9 10

True, False, and Indeterminate Statements . . . . . . . . . . . . The Conjunction AND and the Disjunction OR . . . . . . . . . The Negation NOT . . . . . . . . . . . . . . . . . . . . . . . . . Equivalence ⇐⇒ . . . . . . . . . . . . . . . . . . . . . . . . . . De Morgan’s Laws: Negating the Conjunction and Disjunction The Implication P Ô⇒ Q . . . . . . . . . . . . . . . . . . . . . The Converse Q Ô⇒ P . . . . . . . . . . . . . . . . . . . . . . Affirming the Consequent (or The Fallacy of the Converse) . . The Negation NOT- (P Ô⇒ Q) . . . . . . . . . . . . . . . . . The Contrapositive NOT-Q Ô⇒ NOT-P . . . . . . . . . . . .

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3.11 3.12 3.13 3.14

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Sets 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23

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(P Ô⇒ Q AND Q Ô⇒ P ) ⇐⇒ (P ⇐⇒ Q) . . . . . Other Ways to Express P Ô⇒ Q (Optional) . . . . . . The Four Categorical Propositions and Their Negations Chapter Summary . . . . . . . . . . . . . . . . . . . . .

35 The Elements of a Set Can Be Pretty Much Anything . . In ∈ and Not In ∉ . . . . . . . . . . . . . . . . . . . . . . . The Order of the Elements Doesn’t Matter . . . . . . . . n(S) Is the Number of Elements in the Set S . . . . . . . The Ellipsis “. . . ” Means Continue in the Obvious Fashion Repeated Elements Don’t Count . . . . . . . . . . . . . . R Is the Set of Real Numbers . . . . . . . . . . . . . . . . Z Is the Set of Integers . . . . . . . . . . . . . . . . . . . . Q Is the Set of Rational Numbers . . . . . . . . . . . . . . A Taxonomy of Numbers . . . . . . . . . . . . . . . . . . More Notation: + , − , and 0 . . . . . . . . . . . . . . . . . . The Empty Set ∅ . . . . . . . . . . . . . . . . . . . . . . . Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subset Of ⊆ . . . . . . . . . . . . . . . . . . . . . . . . . . Proper Subset Of ⊂ . . . . . . . . . . . . . . . . . . . . . . Union ∪ . . . . . . . . . . . . . . . . . . . . . . . . . . . . Intersection ∩ . . . . . . . . . . . . . . . . . . . . . . . . . Set Minus ∖ . . . . . . . . . . . . . . . . . . . . . . . . . . The Universal Set E . . . . . . . . . . . . . . . . . . . . . The Set Complement A′ . . . . . . . . . . . . . . . . . . . De Morgan’s Laws . . . . . . . . . . . . . . . . . . . . . . Set-Builder Notation (or Set Comprehension) . . . . . . . Chapter Summary . . . . . . . . . . . . . . . . . . . . . .

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O-Level Review 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

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27 28 29 34

Some Mathematical Vocabulary . . . . . . . . . . . . The Absolute Value or Modulus Function . . . . . . The Factorial n! . . . . . . . . . . . . . . . . . . . . Exponents . . . . . . . . . . . . . . . . . . . . . . . . The Square Root Refers to the Positive Square Root Rationalising the Denominator with a Surd . . . . . Logarithms . . . . . . . . . . . . . . . . . . . . . . . Polynomials . . . . . . . . . . . . . . . . . . . . . . .

36 38 39 40 40 41 42 43 44 45 46 47 48 50 51 53 54 56 57 58 60 62 67 68

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Horizontal, Vertical, and Oblique Lines . Another Equation of the Line . . . . . . The Gradient of a Line . . . . . . . . . . Yet Another Equation of the Line . . . . Perpendicular Lines . . . . . . . . . . . Lines vs Line Segments vs Rays . . . . .

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The Distance between Two Points on the Real Number Line . . . . . . . . 122 The Distance between Two Points on the Cartesian Plane . . . . . . . . . 123 Closeness (or Nearness) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

126 Graphing Circles on the TI84 . . . . . . . . . . . . . . . . . . . . . . . . . 129 The Number π (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

133

Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 The Gradient of a Graph at a Point . . . . . . . . . . . . . . . . . . . . . 134 Stationary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Maximum Points Minimum Points Extrema . . . . . Turning Points .

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Reflection and Symmetry 12.1 12.2 12.3

112 113 115 118 120 121 122

Maximum, Minimum, and Turning Points 11.1 11.2 11.3 11.4

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Tangent Lines, Gradients, and Stationary Points 10.1 10.2 10.3

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Circles 9.1 9.2

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Ordered Pairs . . . . . . . . . . . . . . . . . . The Cartesian Plane . . . . . . . . . . . . . . A Graph is Any Set of Points . . . . . . . . . The Graph of An Equation . . . . . . . . . . Graphing with the TI84 . . . . . . . . . . . . The Graph of An Equation with Constraints Intercepts and Roots . . . . . . . . . . . . . .

Distance 8.1 8.2 8.3

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90

Lines 7.1 7.2 7.3 7.4 7.5 7.6

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Graphs 6.1 6.2 6.3 6.4 6.5 6.6 6.7

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Functions and Graphs

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136 140 144 145 148

The Reflection of a Point in a Point . . . . . . . . . . . . . . . . . . . . . 148 The Reflection of a Point in a Line . . . . . . . . . . . . . . . . . . . . . . 150 Lines of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

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Solutions and Solution Sets

155

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O-Level Review: The Quadratic Equation y = ax2 + bx + c

163

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Functions 15.1 15.2 15.3 15.4 15.5 15.6 15.7

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What Functions Aren’t . . . . . . . . . . . . . . Notation for Functions . . . . . . . . . . . . . . Warning: f and f (x) Refer to Different Things Nice Functions . . . . . . . . . . . . . . . . . . Graphs of Functions . . . . . . . . . . . . . . . Piecewise Functions . . . . . . . . . . . . . . . Range . . . . . . . . . . . . . . . . . . . . . . .

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184 187 194 195 197 201 206

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An Introduction to Continuity

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When a Function Is Increasing or Decreasing

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Arithmetic Combinations of Functions

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Domain Restriction

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Composite Functions

225

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One-to-One Functions

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21.1 21.2

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Inverse Functions 22.1 22.2 22.3 22.4 22.5 22.6 22.7 22.8 22.9 22.10

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Only a One-to-One Function Can Have an Inverse . . . . . . Formal Definition of the Inverse . . . . . . . . . . . . . . . . . A Method for Showing a Function Is One-to-One and Finding Inverse Cancellation Laws . . . . . . . . . . . . . . . . . . . . The Graphs of f and f −1 Are Reflections in the Line y = x . . When Do the Graphs of f and f −1 Intersect? . . . . . . . . . Self-Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . If f Is Strictly Increasing/Decreasing, Then So Is f −1 . . . . . A Function Is the Inverse of Its Inverse . . . . . . . . . . . . . Domain Restriction to Create a One-to-One Function . . . .

. . . . . . . 244 . . . . . . . 246 Its Inverse 247 . . . . . . . 251 . . . . . . . 252 . . . . . . . 259 . . . . . . . 263 . . . . . . . 264 . . . . . . . 265 . . . . . . . 267 270

Oblique Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

Transformations 24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8 24.9

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Asymptotes and Limit Notation 23.1

24

Two More Characterisations of One-to-One . . . . . . . . . . . . . . . . . 236 A Strictly Increasing/Decreasing Function Is One-to-One . . . . . . . . . 239

y = f (x) + a . . . . . . . . . . . . . . . . . y = f (x + a) . . . . . . . . . . . . . . . . . y = af (x) . . . . . . . . . . . . . . . . . . y = f (ax) . . . . . . . . . . . . . . . . . . Combinations of the Above . . . . . . . . y = ∣f (x)∣ . . . . . . . . . . . . . . . . . . y = f (∣x∣) . . . . . . . . . . . . . . . . . . 1/f . . . . . . . . . . . . . . . . . . . . . . When a Function Is Symmetric in Vertical

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280 282 285 287 289 293 295 297 301 302

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25.1 25.2

The Exponential Function exp . . . . . . . . . . . . . . . . . . . . . . . . 306 Euler’s Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

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Trigonometry: Arcs (of a Circle)

310

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Trigonometry: Angles

313

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Trigonometry: Sine and Cosine 28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8

29

320

The Values of Sine and Cosine at 0 and 1 . . . . . . . Three Values of Sine and Cosine . . . . . . . . . . . . Confusing Notation . . . . . . . . . . . . . . . . . . . . Some Formulae Involving Sine and Cosine . . . . . . . The Area of a Triangle; the Laws of Sines and Cosines The Unit-Circle Definitions . . . . . . . . . . . . . . . Graphs of Sine and Cosine . . . . . . . . . . . . . . . . Graphs of Sine and Cosine: Some Observations . . . .

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Trigonometry: Tangent 29.1 29.2 29.3

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Angles, Informally Defined Again; Also, the Radian . . . . . . . . . . . . . 315 Names of Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318

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Some Formulae Involving Tangent . . . . . . . . . . . . . . . . . . . . . . 352 Graph of Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 Graph of Tangent: Some Observations . . . . . . . . . . . . . . . . . . . . 357

Trigonometry: Summary and a Few New Things 30.1 30.2

321 323 326 327 331 333 343 348

358

The Signs of Sine, Cosine, and Tangent . . . . . . . . . . . . . . . . . . . 359 Half-Angle Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361

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Trigonometry: Three More Functions

363

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Trigonometry: Three Inverse Trigonometric Functions

366

32.1 32.2 32.3 32.4 32.5 32.6 32.7

Confusing Notation . . . . . . . . . . . . . . . Inverse Cancellation Laws, Revisited . . . . . More Compositions . . . . . . . . . . . . . . . Addition Formulae for Arcsine and Arccosine Addition Formulae for Arctangent . . . . . . Harmonic Addition . . . . . . . . . . . . . . . Even More Compositions (optional) . . . . .

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370 371 375 378 380 382 384

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Elementary Functions

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Factorising Polynomials

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Long Division of Polynomials . . Factorising Polynomials . . . . . Compare Coefficients . . . . . . . The Quadratic Formula . . . . . The Remainder Theorem . . . . The Factor Theorem . . . . . . . The Intermediate Value Theorem

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391 395 396 399 400 401 405

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Solving Systems of Equations 35.1

36

427

Multiplying an Inequality by an Unknown Constant ax2 + bx + c > 0 . . . . . . . . . . . . . . . . . . . . . (ax + b)/(cx + d) > 0 . . . . . . . . . . . . . . . . . . The Cardinal Sin of Dividing by Zero, Revisited . . Inequalities Involving the Absolute Value Function . (ax2 + bx + c) / (dx2 + ex + f ) > 0 . . . . . . . . . . . Solving Inequalities by Graphical Methods . . . . . .

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Squaring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 Multiplying by Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 Removing Logs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 The Gradient . . . . . . . . . . . . The Derivative as the Gradient . . The Derivative as Rate of Change Differentiability vs Continuity . . . Rules of Differentiation . . . . . . . The Chain Rule . . . . . . . . . . . Increasing and Decreasing . . . . . Stationary and Turning Points . .

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The Ellipse x2 + y 2 = 1 (The Unit Circle) . The Ellipse x2 /a2 + y 2 /b2 = 1 . . . . . . . . The Hyperbola y = 1/x . . . . . . . . . . . The Hyperbola x2 − y 2 = 1 . . . . . . . . . The Hyperbola x2 /a2 − y 2 /b2 = 1 . . . . . . The Hyperbola y 2 /b2 − x2 /a2 = 1 . . . . . . The Hyperbola y = (bx + c)/(dx + e) . . . The Hyperbola y = (ax2 + bx + c) /(dx + e) Eliminating the Parameter t . . . . . . . .

Simple Parametric Equations

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427 429 436 442 443 450 453 456

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Conic Sections 40.1 40.2 40.3 40.4 40.5 40.6 40.7 40.8 40.9

41

Non-Repeated Linear Factors . . . . . . . . . . . . . . . . . . . . . . . . . 424 Repeated Linear Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 Non-Repeated Quadratic Factors . . . . . . . . . . . . . . . . . . . . . . . 426

O-Level Review: The Derivative 39.1 39.2 39.3 39.4 39.5 39.6 39.7 39.8

40

423

Extraneous Solutions 38.1 38.2 38.3

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Solving Systems of Equations with the TI84 . . . . . . . . . . . . . . . . . 420

Solving Inequalities 37.1 37.2 37.3 37.4 37.5 37.6 37.7

38

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Partial Fractions Decomposition 36.1 36.2 36.3

37

Factorising a Quartic Polynomial . . . . . . . . . . . . . . . . . . . . . . . 408 Two Warnings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

461 463 466 469 473 476 479 480 490

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Sequences 42.1 42.2 42.3

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Convergent and Divergent Series . . . . . . . . . . . . . . . . . . . . . . . 526

534

Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 Infinite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 536

Geometric Sequences and Series 46.1 46.2

529

Summation Notation for Infinite Series . . . . . . . . . . . . . . . . . . . . 531

Arithmetic Sequences and Series 45.1 45.2

519

525

Summation Notation ∑ 44.1

517

Sequences Are Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523 Arithmetic Combinations of Sequences . . . . . . . . . . . . . . . . . . . . 524

Series 43.1

44

Sequences and Series

537

Finite Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . . 538 Infinite Geometric Sequences and Series . . . . . . . . . . . . . . . . . . . 540

47

Rules of Summation Notation

542

48

The Method of Differences

544

xi, Contents

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Part III. 49

The Magnitude or Length of a Vector . . . . . . . . . . . . When Are Two Vectors Identical? . . . . . . . . . . . . . . . A Vector and a Point Are Different Things . . . . . . . . . Two More Ways to Denote a Vector . . . . . . . . . . . . . Position Vectors . . . . . . . . . . . . . . . . . . . . . . . . The Zero Vector . . . . . . . . . . . . . . . . . . . . . . . . Displacement Vectors . . . . . . . . . . . . . . . . . . . . . Sum and Difference of Points and Vectors . . . . . . . . . . Sum, Additive Inverse, and Difference of Vectors . . . . . . Scalar Multiplication of a Vector . . . . . . . . . . . . . . . When Do Two Vectors Point in the Same Direction? . . . . When Are Two Vectors Parallel? . . . . . . . . . . . . . . . Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . The Standard Basis Vectors . . . . . . . . . . . . . . . . . . Any Vector Is A Linear Combination of Two Other Vectors The Ratio Theorem . . . . . . . . . . . . . . . . . . . . . .

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Direction Vector . . . . . . . . . . . . . . . . . . . . . Cartesian to Vector Equations . . . . . . . . . . . . . . Pedantic Points to Test/Reinforce Your Understanding Vector to Cartesian Equations . . . . . . . . . . . . . .

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574 578 583 584 587

A Vector’s Scalar Product with Itself . . . . . . . . . . . . . . . . . . . . . 589

The Angle Between Two Vectors 52.1 52.2

553 554 555 556 557 557 558 559 561 564 565 566 567 569 570 572 574

The Scalar Product 51.1

52

551

Lines 50.1 50.2 50.3 50.4

51

549

Introduction to Vectors 49.1 49.2 49.3 49.4 49.5 49.6 49.7 49.8 49.9 49.10 49.11 49.12 49.13 49.14 49.15 49.16

50

Vectors

590

Pythagoras’ Theorem and Triangle Inequality . . . . . . . . . . . . . . . . 597 Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598

53

The Angle Between Two Lines

601

54

Vectors vs Scalars

607

55

The Projection and Rejection Vectors

610

56

Collinearity

614

57

The Vector Product

616

57.1 57.2

58

The Foot of the Perpendicular From a Point to a Line 58.1

59

The Angle between Two Vectors Using the Vector Product . . . . . . . . 619 The Length of the Rejection Vector . . . . . . . . . . . . . . . . . . . . . . 620 The Distance Between a Point and a Line . . . . . . . . . . . . . . . . . . 623

Three-Dimensional (3D) Space

xii, Contents

621 628

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59.1

60

Vectors (in 3D) 60.1 60.2 60.3 60.4 60.5 60.6 60.7

61

Graphs (in 3D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630

632

The Magnitude or Length of a Vector . . . . . . Sums and Differences of Points and Vectors . . . Sum, Additive Inverse, and Difference of Vectors Scalar Multiplication and When Two Vectors Are Unit Vectors . . . . . . . . . . . . . . . . . . . . The Standard Basis Vectors . . . . . . . . . . . . The Ratio Theorem . . . . . . . . . . . . . . . .

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The Scalar Product (in 3D) 61.1 61.2

634 635 638 642 644 645 646 647

The Angle between Two Vectors . . . . . . . . . . . . . . . . . . . . . . . 649 Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652

62

The Projection and Rejection Vectors (in 3D)

653

63

Lines (in 3D)

657

63.1 63.2 63.3 63.4 63.5 63.6 63.7

64

Vector to Cartesian Equations . . Cartesian to Vector Equations . . Parallel and Perpendicular Lines Intersecting Lines . . . . . . . . . Skew Lines . . . . . . . . . . . . The Angle Between Two Lines . Collinearity (in 3D) . . . . . . .

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The Vector Product (in 3D) 64.1 64.2 64.3

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660 665 668 670 672 674 678 681

The Right-Hand Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685 The Length of the Vector Product . . . . . . . . . . . . . . . . . . . . . . 686 The Length of the Rejection Vector . . . . . . . . . . . . . . . . . . . . . . 687

65

The Distance Between a Point and a Line (in 3D)

688

66

Planes: Introduction

693

66.1

67

Planes: Formally Defined in Vector Form 67.1

68

70

709

714

Planes in Parametric Form . . . . . . . . . . . . . . . . . . . . . . . . . . 717 Parametric to Vector or Cartesian Form . . . . . . . . . . . . . . . . . . . 721

Four Ways to Uniquely Determine a Plane

xiii, Contents

699

Finding Points on a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 711 Finding Vectors on a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 712

Planes in Parametric Form 69.1 69.2

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The Normal Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703

Planes in Cartesian Form 68.1 68.2

69

The Analogy Between a Plane, a Line, and a Point

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71

The Angle between a Line and a Plane 71.1

72

When Two Planes Are Parallel, Perp., or Intersect . . . . . . . . . . . . . 734 Formula Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742

Complex Numbers

752

Complex Numbers: Introduction 75.1 75.2

754

The Real and Imaginary Parts of Complex Numbers . . . . . . . . . . . . 757 Complex Numbers in Ordered Pair Notation . . . . . . . . . . . . . . . . . 758

Some Arithmetic of Complex Numbers 76.1 76.2 76.3 76.4

77

746

Coplanarity of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 749

Part IV.

76

739

Coplanarity 74.1

75

733

Point-Plane Foot of the Perpendicular and Distance 73.1

74

When a Line and a Plane Are Parallel, Perp., or Intersect . . . . . . . . . 729

The Angle between Two Planes 72.1

73

727

Addition and Subtraction Multiplication . . . . . . . Conjugation . . . . . . . . Division . . . . . . . . . .

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759 . . . .

Solving Polynomial Equations 77.1 77.2

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759 760 762 764 765

The Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . 766 The Complex Conjugate Root Theorem . . . . . . . . . . . . . . . . . . . 768

78

The Argand Diagram

770

79

Complex Numbers in Polar Form

771

79.1 79.2 79.3

80

Complex Numbers in Exponential Form 80.1 80.2

81

The Argument: An Informal Introduction . . . . . . . . . . . . . . . . . . 772 The Argument: Formally Defined . . . . . . . . . . . . . . . . . . . . . . . 773 Complex Numbers in Polar Form . . . . . . . . . . . . . . . . . . . . . . . 775 Complex Numbers in Exponential Form . . . . . . . . . . . . . . . . . . . 777 Some Useful Formulae for Sine and Cosine . . . . . . . . . . . . . . . . . . 778

More Arithmetic of Complex Numbers 81.1 81.2

xiv, Contents

776

779

The Reciprocal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 783 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784

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Part V. 82

Proving Proving Proving Proving

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838 840 841 845 847

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861 866 868 870 871

Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871 The Inverse Function Theorem (IFT) . . . . . . . . . . . . . . . . . . . . . 880 Term-by-Term Differentiation (optional) . . . . . . . . . . . . . . . . . . . 882

884

Parametric Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 888

890

Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896 Smooth (or “Infinitely Differentiable”) Functions . . . . . . . . . . . . . . 900 Confusing Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 902

The Increasing/Decreasing Test

xv, Contents

809 817 818 821 822 825 830 832

Some Basic Rules of Differentiation (optional) the Product Rule (optional) . . . . . . . . . . the Quotient Rule (optional) . . . . . . . . . . the Chain Rule (optional) . . . . . . . . . . .

The Second and Higher Derivatives 89.1 89.2 89.3

90

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Using d/dx as the Differentiation Operator . . . . . . . . . . . . . . . . . 851 Using d/dx as Shorthand . . . . . . . . . . . . . . . . . . . . . . . . . . . 854 df /dx Is Not a Fraction; Nonetheless ... . . . . . . . . . . . . . . . . . . . 856

Parametric Equations 88.1

89

Differentiable ⇐⇒ Approximately Linear . . . . Continuity vs Differentiability . . . . . . . . . . . The Derivative as a Function . . . . . . . . . . . “Most” Elementary Functions Are Differentiable

Some Techniques of Differentiation 87.1 87.2 87.3

88

Functions with a Single Discontinuity . . . . . . . . . . . . . An Example of a Function That Is Discontinuous Everywhere Functions That Seem Discontinuous But Aren’t . . . . . . . . Every Elementary Function Is Continuous . . . . . . . . . . . Continuity Allows Us to “Move” Limits . . . . . . . . . . . . Every Elementary Function is Continuous: Proof (Optional) . Continuity at Isolated Points (optional) . . . . . . . . . . . .

Rules of Differentiation, Revisited 86.1 86.2 86.3 86.4

87

806

Differentiation Notation 85.1 85.2 85.3

86

Limits, Informally Defined . . . . . . . . . . . . . . . . . . . . . . . . . . . 790 Examples Where The Limit Does Not Exist . . . . . . . . . . . . . . . . . 795 Rules for Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804

The Derivative, Revisited 84.1 84.2 84.3 84.4

85

789

Continuity, Revisited 83.1 83.2 83.3 83.4 83.5 83.6 83.7

84

787

Limits 82.1 82.2 82.3

83

Calculus

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91

Determining the Nature of a Stationary Point 91.1 91.2 91.3 91.4

92

The First Derivative Test for Extrema (FDTE) . . . . . . . . . . . . . . A Situation Where a Stationary Point Is Also a Strict Global Extremum Fermat’s Theorem on Extrema . . . . . . . . . . . . . . . . . . . . . . . The Second Derivative Test for Extrema (SDTE) . . . . . . . . . . . . .

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Concavity 92.1 92.2 92.3 92.4

93

910

930

The First Derivative Test for Concavity (FDTC) . . . . . . The Second Derivative Test for Concavity (SDTC) . . . . . The Graphical Test for Concavity (GTC) . . . . . . . . . . A Linear Function Is One That’s Both Concave and Convex

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Inflexion Points 93.1 93.2 93.3 93.4

914 920 921 923 934 936 940 944 945

The First Derivative Test for Inflexion Points (FDTI) . The Second Derivative Test for Inflexion Points (SDTI) The Tangent Line Test (TLT) . . . . . . . . . . . . . . . Non-Stationary Points of Inflexion (optional) . . . . . .

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947 949 950 952

94

A Summary of Chapters 90, 91, 92, and 93

953

95

More Techniques of Differentiation

954

95.1 95.2 95.3

96

More Fun With Your TI84 96.1 96.2

97

Locating Maximum and Minimum Points Using Your TI84 . . . . . . . . 961 Find the Derivative at a Point Using Your TI84 . . . . . . . . . . . . . . . 963

967

A Power Series Is Simply an “Infinite Polynomial” . . . A Power Series Can Converge or Diverge . . . . . . . . . A Power Series and Its Interval of Convergence . . . . . (Some) Functions Can be Represented by a Power Series Term-by-Term Differentiation of a Power Series . . . . . Chapter Summary . . . . . . . . . . . . . . . . . . . . .

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The Maclaurin Series 98.1 98.2 98.3 98.4 98.5 98.6 98.7 98.8

99

960

Power Series 97.1 97.2 97.3 97.4 97.5 97.6

98

Relating the Graph of f ′ to That of f . . . . . . . . . . . . . . . . . . . . 954 Equations of Tangents and Normals . . . . . . . . . . . . . . . . . . . . . 955 Connected Rates of Change Problems . . . . . . . . . . . . . . . . . . . . 957

xvi, Contents

967 970 972 974 975 978 979

(Some) Functions Can Be Rep’d by Their Maclaurin Series . The Five Standard Series and Their Intervals of Convergence Sine and Cosine, Formally Defined . . . . . . . . . . . . . . . Maclaurin Polynomials as Approximations . . . . . . . . . . . Creating New Series Using Substitution . . . . . . . . . . . . Creating New Series Using Multiplication . . . . . . . . . . . Repeated Differentiation to Find a Maclaurin Series . . . . . Repeated Implicit Differentiation to Find a Maclaurin Series .

Antidifferentiation

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99.1 99.2 99.3 99.4 99.5 99.6 99.7

Antiderivatives Are Not Unique ... . . . . . . . . . ... But An Antiderivative Is Unique Up to a COI . Shorthand: The Antidifferentiation Symbol ∫ . . . Rules of Antidifferentiation . . . . . . . . . . . . . Every Continuous Function Has an Antiderivative Be Careful with This Common Practice . . . . . . The “Punctuation Mark” dx (optional) . . . . . .

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100 The Definite Integral 100.1 100.2 100.3 100.4 100.5 100.6 100.7

1025

Notation for Integration . . . . . . . . . . . . . . . . . . . . . The Definite Integral, Informally Defined . . . . . . . . . . . An Important Warning . . . . . . . . . . . . . . . . . . . . . . A Sketch of How We Can Find the Area under a Curve . . . If a Function Is Continuous, Then Its Definite Integral Exists Rules of Integration . . . . . . . . . . . . . . . . . . . . . . . Term-by-Term Integration (optional) . . . . . . . . . . . . . .

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101 The Fundamental Theorems of Calculus 101.1 101.2 101.3 101.4 101.5

102.6 102.7 102.8

Definite Integral Functions . . . . . . . . . . . . . . . . . . . . . . The First Fundamental Theorem of Calculus (FTC1) . . . . . . . The Second Fundamental Theorem of Calculus (FTC2) . . . . . Why Integration and Antidifferentiation Use the Same Notation Some New Notation and More Examples and Exercises . . . . . .

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Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial Fractions: Finding ∫ 1/ (ax2 + bx + c) dx where b2 − 4ac > 0 Building a Divisor of the Denominator . . . . . . . . . . . . . . . . More Rules of Antidifferentiation . . . . . . . . . . . . . . . . . . . Completing the Square: ∫ 1/ (ax2 + bx + c) dx where b2 − 4ac < 0 . √ 2 ∫ 1/ ax + bx + c dx in the Special Case where a < 0 . . . . . . . . Using Trigonometric Identities . . . . . . . . . . . . . . . . . . . . Integration by Parts (IBP) . . . . . . . . . . . . . . . . . . . . . . .

103.5 103.6 103.7 103.8 103.9

The Substitution Rule Is Skipping Steps . . . . . ′ ∫ f exp f dx = exp f + C ′ ∫ f /f dx = ln ∣f ∣ + C . .

. 1042 . 1045 . 1049 . 1051 . 1053 1054

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103 The Substitution Rule 103.1 103.2 103.3 103.4

. 1028 . 1029 . 1031 . 1032 . 1036 . 1037 . 1040 1042

102 More Techniques of Antidifferentiation 102.1 102.2 102.3 102.4 102.5

. 1011 . 1011 . 1013 . 1017 . 1020 . 1021 . 1022

. 1054 . 1055 . 1057 . 1059 . 1061 . 1064 . 1067 . 1069 1073

the Inverse of the Chain Rule . . . . . . . . . . . 1079

. . . . . . . . . . . . . . . . . . . . . n n+1 ′ / (n + 1) + C . ∫ (f ) ⋅ f dx = (f ) Building a Derivative . . . . . . . . .

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104 Term-by-Term Antidifferentiation and Integration of a Power Series 1095 104.1 xvii, Contents

Formal Results (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . 1099 www.EconsPhDTutor.com

105 More Definite Integrals 105.1 105.2 105.3 105.4 105.5 105.6 105.7 105.8

1100

Area between a Curve and Lines Parallel to Axes Area between a Curve and a Line . . . . . . . . . Area between Two Curves . . . . . . . . . . . . . Area below the x-Axis . . . . . . . . . . . . . . . Area under a Curve Defined Parametrically . . . Volume of Rotation about the x-axis . . . . . . . Volume of Rotation about the y-axis . . . . . . . Finding Definite Integrals Using Your TI84 . . .

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106 Introduction to Differential Equations 106.1 106.2 106.3 106.4 106.5

dy/dx = f (x) . . dy/dx = f (y) . . d2 y/dx2 = f (x) . Word Problems . Yet Another Way

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107.1 107.2

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1116

. . . . to Define Sine and Cosine (optional) .

107 Revisiting ln, logb , and exp (optional)

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. 1101 . 1102 . 1103 . 1104 . 1105 . 1107 . 1113 . 1115

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Revisiting Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1131 Revisiting the Exponential Function . . . . . . . . . . . . . . . . . . . . . 1133

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Part VI.

Probability and Statistics

1135

108 How to Count: Four Principles 108.1 108.2 108.3 108.4

How How How How

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Addition Principle . . . . . . Multiplication Principle . . . Inclusion-Exclusion Principle Complements Principle . . .

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109 How to Count: Permutations 109.1 109.2 109.3 109.4

Permutations with Repeated Elements Circular Permutations . . . . . . . . . Partial Permutations . . . . . . . . . . Permutations with Restrictions . . . .

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110 How to Count: Combinations 110.1 110.2 110.3

Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1164 The Combination as Binomial Coefficient . . . . . . . . . . . . . . . . . . 1165 The Number of Subsets of a Set is 2n . . . . . . . . . . . . . . . . . . . . . 1167 Mathematical Modelling . . . . . . . . . The Experiment as a Model of Scenarios The Kolmogorov Axioms . . . . . . . . . Implications of the Kolmogorov Axioms

1168 . . . . . . . . . . . Involving Chance . . . . . . . . . . . . . . . . . . . . . . .

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112 Probability: Conditional Probability 112.1 112.2

The Conditional Probability Fallacy (CPF) . . . . . . . . . . . . . . . . . 1180 Two-Boys Problem (Fun, Optional) . . . . . . . . . . . . . . . . . . . . . . 1184

1186

Warning: Not Everything is Independent . . . . . . . . . . . . . . . . . . 1189 Probability: Independence of Multiple Events . . . . . . . . . . . . . . . . 1191

114 Fun Probability Puzzles 114.1 114.2

1192

The Monty Hall Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 1192 The Birthday Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1195

115 Random Variables: Introduction 115.1 115.2 115.3 115.4

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113 Probability: Independence 113.1 113.2

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111 Probability: Introduction 111.1 111.2 111.3 111.4

. 1138 . 1141 . 1144 . 1146

A Random Variable vs. Its Observed Values . . . . X = k Denotes the Event {s ∈ S ∶ X(s) = k} . . . . The Probability Distribution of a Random Variable Random Variables Are Simply Functions . . . . . .

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116 Random Variables: Independence

1204

117 Random Variables: Expectation

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117.2

The Expectation Operator is Linear . . . . . . . . . . . . . . . . . . . . . 1211

118 Random Variables: Variance 118.1 118.2 118.3 118.4

The Variance of a Constant R.V. is 0 . . . Standard Deviation . . . . . . . . . . . . . The Variance Operator is Not Linear . . . The Definition of the Variance (Optional)

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119 The Coin-Flips Problem (Fun, Optional)

1223

120 The Bernoulli Trial and the Bernoulli Distribution

1224

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Mean and Variance of the Bernoulli Random Variable . . . . . . . . . . . 1226

121 The Binomial Distribution 121.1 121.2

1227

Probability Distribution of the Binomial R.V. . . . . . . . . . . . . . . . . 1229 The Mean and Variance of the Binomial Random Variable . . . . . . . . . 1230

122 The Continuous Uniform Distribution 122.1 122.2 122.3 122.4

The Continuous Uniform Distribution . . . . The Cumulative Distribution Function (CDF) Important Digression: P (X ≤ k) = P (X < k) . The Probability Density Function (PDF) . .

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123 The Normal Distribution 123.1 123.2 123.3

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The Normal Distribution, in General . . . . . . . . . . . . . . . . . . . . . 1245 Sum of Independent Normal Random Variables . . . . . . . . . . . . . . . 1254 The Central Limit Theorem and The Normal Approximation . . . . . . . 1257

124 The CLT is Amazing (Optional) 124.1 124.2 124.3 124.4

The Normal Distribution in Nature . . . . . . . . Illustrating the Central Limit Theorem (CLT) . . Why Are So Many Things Normally Distributed? Don’t Assume That Everything is Normal . . . .

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125 Statistics: Introduction (Optional) 125.1 125.2

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Probability vs. Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1276 Objectivists vs Subjectivists . . . . . . . . . . . . . . . . . . . . . . . . . . 1277

126 Sampling 126.1 126.2 126.3 126.4 126.5 126.6 126.7 126.8 126.9

. 1233 . 1235 . 1236 . 1237

1279

Population . . . . . . . . . . . . . . . . . . . . . . . . . . . . Population Mean and Population Variance . . . . . . . . . . . Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Distribution of a Population . . . . . . . . . . . . . . . . . . . A Random Sample . . . . . . . . . . . . . . . . . . . . . . . . Sample Mean and Sample Variance . . . . . . . . . . . . . . . Sample Mean and Sample Variance are Unbiased Estimators The Sample Mean is a Random Variable . . . . . . . . . . . The Distribution of the Sample Mean . . . . . . . . . . . . .

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126.10

Non-Random Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1296

127 Null Hypothesis Significance Testing (NHST) 127.1 127.2 127.3 127.4 127.5 127.6 127.7 127.8

1297

One-Tailed vs Two-Tailed Tests . . . . . . . . . . . . . . . . . . . . . . . . 1301 The Abuse of NHST (Optional) . . . . . . . . . . . . . . . . . . . . . . . . 1304 Common Misinterpretations of the Margin of Error (Optional) . . . . . . 1305 Critical Region and Critical Value . . . . . . . . . . . . . . . . . . . . . . 1308 Testing of a Pop. Mean (Small Sample, Normal Distribution, σ 2 Known) 1310 Testing of a Pop. Mean (Large Sample, Any Distribution, σ 2 Known) . . 1312 Testing of a Pop. Mean (Large Sample, Any Distribution, σ 2 Unknown) . 1314 Formulation of Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . 1316

128 Correlation and Linear Regression 128.1 128.2 128.3 128.4 128.5 128.6 128.7 128.8 128.9

xxi, Contents

1317

Bivariate Data and Scatter Diagrams . . . . . . . . . . Product Moment Correlation Coefficient (PMCC) . . Correlation Does Not Imply Causation (Optional) . . Linear Regression . . . . . . . . . . . . . . . . . . . . . Ordinary Least Squares (OLS) . . . . . . . . . . . . . TI84 to Calculate the PMCC and the OLS Estimates Interpolation and Extrapolation . . . . . . . . . . . . . Transformations to Achieve Linearity . . . . . . . . . . The Higher the PMCC, the Better the Model? . . . .

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Part VII.

Ten-Year Series

1348

129 Past-Year Questions for Part I. Functions and Graphs

1352

130 Past-Year Questions for Part II. Sequences and Series

1367

131 Past-Year Questions for Part III. Vectors

1379

132 Past-Year Questions for Part IV. Complex Numbers

1389

133 Past-Year Questions for Part V. Calculus

1396

134 Past-Year Questions for Part VI. Prob. and Stats.

1430

135 All Past-Year Questions, Listed and Categorised

1475

135.1 135.2 135.3 135.4 135.5 135.6 135.7 135.8 135.9 135.10 135.11 135.12 135.13 135.14 135.15 135.16

2019 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2008 2007 2006

(9758) (9758) (9758) (9740) (9740) (9740) (9740) (9740) (9740) (9740) (9740) (9740) (9740) (9233) (9233) (9233)

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136 H1 Maths Questions (2016–19) 136.1 136.2 136.3 136.4 136.5 136.6 136.7 136.8

xxii, Contents

H1 H1 H1 H1 H1 H1 H1 H1

Maths Maths Maths Maths Maths Maths Maths Maths

2019 2019 2018 2018 2017 2017 2016 2016

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Part VIII.

Appendices

1523

137 Appendices for Part 0. A Few Basics 137.1 137.2 137.3

1524

Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1524 Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1524 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1526

138 Appendices for Part I. Functions and Graphs 138.1 138.2 138.3 138.4 138.5 138.6 138.7 138.8 138.9 138.10 138.11 138.12 138.13 138.14 138.15 138.16 138.17 138.18 138.19 138.20

1530

Ordered Pairs and Ordered n-tuples . . . . . . . . . . . . . . The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Maximum and Minimum Points . . . . . . . . . . . . . . . . . Distance between a Line and a Point . . . . . . . . . . . . . . Reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Quadratic Equation . . . . . . . . . . . . . . . . . . . . . Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . When a Function Is Increasing or Decreasing at a Point . . . One-to-One Functions . . . . . . . . . . . . . . . . . . . . . . Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . Standard Definitions of the Identity and Inclusion Functions . Left, Right, and “Full” Inverses . . . . . . . . . . . . . . . . . When Do f and f −1 Intersect? . . . . . . . . . . . . . . . . . Transformations . . . . . . . . . . . . . . . . . . . . . . . . . Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . Factorising Polynomials . . . . . . . . . . . . . . . . . . . . . Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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139 Appendices for Part II. Sequences and Series 139.1

1582

Convergence and Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . 1583

140 Appendices for Part III. Vectors 140.1 140.2 140.3 140.4 140.5 140.6 140.7 140.8 140.9 140.10 140.11 140.12 140.13 140.14 140.15 xxiii, Contents

. 1530 . 1532 . 1533 . 1536 . 1538 . 1539 . 1541 . 1543 . 1547 . 1547 . 1549 . 1551 . 1554 . 1555 . 1558 . 1559 . 1563 . 1575 . 1576 . 1581

Some General Definitions . . . . . . . . . . . Some Basic Results . . . . . . . . . . . . . . . Scalar Product . . . . . . . . . . . . . . . . . Angles . . . . . . . . . . . . . . . . . . . . . . The Relationship Between Two Lines . . . . . Lines in 3D Space . . . . . . . . . . . . . . . Projection Vectors . . . . . . . . . . . . . . . The Vector Product . . . . . . . . . . . . . . Planes in General . . . . . . . . . . . . . . . . Planes in Three-Dimensional Space . . . . . . Four Ways to Uniquely Determine a Plane . . The Relationship Between a Line and a Plane The Relationship Between Two Planes . . . . Distances . . . . . . . . . . . . . . . . . . . . Point-Plane Distance: Calculus Method . . .

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140.16 140.17

The Relationship Between Two Lines in 3D Space . . . . . . . . . . . . . 1620 A Necessary and Sufficient Condition for Skew Lines . . . . . . . . . . . . 1621

141 Appendices for Part IV. Complex Numbers

1622

142 Appendices for Part V. Calculus

1628

142.1 142.2 142.3 142.4 142.5 142.6 142.7 142.8 142.9 142.10 142.11 142.12 142.13 142.14 142.15 142.16 142.17 142.18 142.19 142.20 142.21 142.22 142.23 142.24 142.25

A Few Useful Results and Terms . . . . . . . . . . . . . . Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Infinite Limits and Vertical Asymptotes . . . . . . . . . . Limits at Infinity and Horizontal and Oblique Asymptotes Rules for Limits . . . . . . . . . . . . . . . . . . . . . . . . Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . The Derivative . . . . . . . . . . . . . . . . . . . . . . . . “Most” Elementary Functions are Differentiable . . . . . . Leibniz Got the Product Rule Wrong . . . . . . . . . . . . Proving the Chain Rule . . . . . . . . . . . . . . . . . . . Fermat’s Theorem on Extrema . . . . . . . . . . . . . . . Rolle’s Theorem and the Mean Value Theorem . . . . . . The Increasing/Decreasing Test . . . . . . . . . . . . . . . The First and Second Derivative Tests . . . . . . . . . . . Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . Inflexion Points . . . . . . . . . . . . . . . . . . . . . . . . Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . Power Series and Maclaurin series . . . . . . . . . . . . . Antidifferentiation . . . . . . . . . . . . . . . . . . . . . . The Definite Integral . . . . . . . . . . . . . . . . . . . . . Proving the First Fundamental Theorem of Calculus . . . 2 ∫ 1/ (ax + bx + c) dx . . . . . . . . . . . . . . . . . . . . √ 2 ∫ 1/ ax + bx + c dx . . . . . . . . . . . . . . . . . . . . . The Substitution Rule . . . . . . . . . . . . . . . . . . . . Revisiting Logarithms and Exponentiation . . . . . . . . .

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143 Appendices for Part VI. Probability and Statistics 143.1 143.2 143.3 143.4 143.5 143.6 143.7 143.8 143.9 143.10

xxiv, Contents

How to Count . . . . . . . . . . . . . . . . . . Circular Permutations . . . . . . . . . . . . . Probability . . . . . . . . . . . . . . . . . . . Random Variables . . . . . . . . . . . . . . . The Normal Distribution . . . . . . . . . . . . Sampling . . . . . . . . . . . . . . . . . . . . Null Hypothesis Significance Testing . . . . . Calculating the Margin of Error . . . . . . . . Correlation and Linear Regression . . . . . . Deriving a Linear Model from the Barometric

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Part IX.

1720

144 Part 0 Answers (A Few Basics) 144.1 144.2 144.3 144.4 144.5

Ch. Ch. Ch. Ch. Ch.

1 2 3 4 5

(Just To Be Clear) . . . (PSLE Review: Division) (Logic) . . . . . . . . . . (Sets) . . . . . . . . . . . (O-Level Review) . . . .

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145 Part I Answers (Functions and Graphs) 145.1 145.2 145.3 145.4 145.5 145.6 145.7 145.8 145.9 145.10 145.11 145.12 145.13 145.14 145.15 145.16 145.17 145.18 145.19 145.20 145.21 145.22 145.23 145.24 145.25 145.26 145.27

Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch.

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146 Part II Answers (Sequences and Series) 146.1 146.2 146.3 146.4 146.5 146.6 146.7

Ch. Ch. Ch. Ch. Ch. Ch. Ch.

42 43 44 45 46 47 48

(Sequences) . . . . . . . . . . . . . (Series) . . . . . . . . . . . . . . . (Summation Notation Σ) . . . . . (Arithmetic Sequences and Series) (Geometric Sequences and Series) (Rules of Summation Notation) . (Method of Differences) . . . . . .

147 Part III Answers (Vectors) xxv, Contents

. 1721 . 1721 . 1722 . 1727 . 1731 . 1734 . 1737 . 1738 . 1738 . 1739 . 1742 . 1742 . 1743 . 1744 . 1748 . 1748 . 1749 . 1750 . 1753 . 1754 . 1759 . 1760 . 1765 . 1766 . 1770 . 1775 . 1778 . 1781 . 1783 . 1791 . 1792 . 1800 1806

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. 1806 . 1806 . 1807 . 1808 . 1808 . 1809 . 1810 1813

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147.1 147.2 147.3 147.4 147.5 147.6 147.7 147.8 147.9 147.10 147.11 147.12 147.13 147.14 147.15 147.16 147.17 147.18 147.19 147.20 147.21 147.22 147.23 147.24 147.25

Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch.

49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 71 72 73 74

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148 Part IV Answers (Complex Numbers) 148.1 148.2 148.3 148.4 148.5 148.6 148.7

Ch. Ch. Ch. Ch. Ch. Ch. Ch.

75 76 77 78 79 80 81

1854

(Complex Numbers: Introduction) . . . . (Some Arithmetic of Complex Numbers) (Solving Polynomial Equations) . . . . . (The Argand Diagram) . . . . . . . . . . (Complex Numbers in Polar Form) . . . . (Complex Numbers in Exponential Form) (More Arithmetic of Complex Numbers)

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Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch.

82 83 84 85 85 87 89 90 91 92 93 94 95

. 1813 . 1816 . 1817 . 1818 . 1820 . 1820 . 1820 . 1821 . 1822 . 1823 . 1827 . 1827 . 1829 . 1830 . 1831 . 1834 . 1837 . 1840 . 1841 . 1842 . 1843 . 1846 . 1847 . 1849 . 1852 . 1854 . 1854 . 1856 . 1857 . 1858 . 1858 . 1859 1862

(Limits) . . . . . . . . . . . . . . . . . . . . . . . (Continuity, Revisited) . . . . . . . . . . . . . . (The Derivative, Revisited) . . . . . . . . . . . . (Differentiation Notation) . . . . . . . . . . . . . (Rules of Differentiation, Revisited) . . . . . . . (Some Techniques of Differentiation) . . . . . . . (The Second and Higher Derivatives) . . . . . . (The Increasing/Decreasing Test) . . . . . . . . (Determining the Nature of a Stationary Point) (Concavity) . . . . . . . . . . . . . . . . . . . . . (Inflexion Points) . . . . . . . . . . . . . . . . . (A Summary of Chapters 90, 91, 92, and 93) . . (More Techniques of Differentiation) . . . . . . .

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. 1862 . 1863 . 1864 . 1867 . 1868 . 1870 . 1872 . 1877 . 1879 . 1883 . 1884 . 1885 . 1886

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149.14 149.15 149.16 149.17 149.18 149.19 149.20 149.21 149.22 149.23 149.24

Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch.

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150 Part VI Answers (Probability and Statistics) 150.1 150.2 150.3 150.4 150.5 150.6 150.7 150.8 150.9 150.10 150.11 150.12 150.13 150.14 150.15 150.16 150.17 150.18

Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch.

108 109 110 111 112 113 115 116 117 118 119 120 121 122 123 126 127 128

1922

(How to Count: Four Principles) . . . (How to Count: Permutations) . . . . (How to Count: Combinations) . . . . (Probability: Introduction) . . . . . . (Conditional Probability) . . . . . . . (Probability: Independence) . . . . . . (Random Variables: Introduction) . . (Random Variables: Independence) . . (Random Variables: Expectation) . . . (Random Variables: Variance) . . . . (The Coin-Flips Problem) . . . . . . . (Bernoulli Trial and Distribution) . . . (Binomial Distribution) . . . . . . . . (Continuous Uniform Distribution) . . (Normal Distribution) . . . . . . . . . (Sampling) . . . . . . . . . . . . . . . (Null Hypothesis Significance Testing) (Correlation and Linear Regression) .

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151 Part VII Answers (2006–19 A-Level Exams) 151.1 151.2 151.3 151.4 151.5 151.6

Ch. Ch. Ch. Ch. Ch. Ch.

129 130 131 132 133 134

(Functions and Graphs) . . (Sequences and Series) . . . (Vectors) . . . . . . . . . . (Complex Numbers) . . . . (Calculus) . . . . . . . . . . (Probability and Statistics)

. 1888 . 1889 . 1890 . 1899 . 1901 . 1903 . 1904 . 1910 . 1914 . 1918 . 1921 . 1922 . 1925 . 1927 . 1930 . 1934 . 1935 . 1936 . 1940 . 1940 . 1942 . 1942 . 1942 . 1943 . 1944 . 1945 . 1951 . 1954 . 1958 1962

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. 1962 . 1996 . 2016 . 2030 . 2053 . 2110

Index

2141

Abbreviations Used in This Textbook

2141

Singlish Used in This Textbook

2145

Notation Used in the Main Text

2146

Notation Used (Appendices)

2147

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2148

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• Maths, Math, and Matzz. This book uses British English. And so the word mathematics shall be abbreviated as maths (and not math). By the way, Singaporeans used to pronounce maths as matzz (similar to how they pronounce clothes as klotes). But at some point between 2005 and 2015, perhaps after watching too many American TV and movies, Singaporeans decided they’d switch to the American math. Perhaps in 50 years, we’ll all be Amos Yees trying to speak annoying pseudo American English. But in the meantime, I’ll continue to say matzz. This is my way of promoting and preserving Singlish (and also sticking it to the ghost of LKY). • Hyperlinks This book has clickable hyperlinks that bring you to some other location in this book, a web-page, or email link. After clicking on any such hyperlink, to go back to where the hyperlink was, try clicking “Back” or using the “Previous View” feature available on many PDF readers (on desktop Adobe Acrobat, “Alt-Left Arrow” works for me).

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Tips for the Student • Read maths slowly. Reading maths is not like reading Harry Potter. Most of Harry Potter is fluff. You can skip 100 pages of Harry Potter without missing a beat.6 In contrast, there is little fluff in maths. Skip or misapprehend one line and it will cost you. So, go slowly. Dwell upon and carefully consider every sentence in this textbook. Make sure you completely understand what each statement says and why it is true. Reading maths is very different from reading most other subject matter. If you don’t quite understand some material, you might be tempted to move forward anyway. Don’t. In maths, later material usually builds on earlier material. So, if you simply move forward, then yea sure, this may save you some time and frustration in the short run, but it will almost always cost you far more in the long run. Better then to stop right there. Keep working on it until you “get” it. Help is all around— ask a friend or a teacher. Feel free to even email me! (I’m always interested to know what the common points of confusion are and how I can better clear them up.) • Examples and exercises are your best friends. A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one. — Paul Halmos (1983). Carefully work through all the examples and exercises. Merely moving your eyeballs is not the same as working. Working means having pencil and paper by your side and going through each example/exercise word-by-word, line-by-line.

For example, I might say something like “x2 − y 2 = 0. Thus, (x − y) (x + y) = 0.” If it’s not obvious to you why the first sentence implies the second, stop right there and work on it until you understand why. And again, if you can’t figure it out yourself, don’t be shy about asking someone for help. Don’t just let your eyeballs fly over these sentences and pretend that your brain “gets” it. Such self-deceit will only cost you in the long run. The Chinese believe in “eating bitterness” and work for work’s sake. That is not my view. The exercises in this textbook are not to make you suffer or somehow strengthen your moral fibre. Instead, as with learning to ride a bike or swim, practice makes perfect. The best way to learn and master any material is by practising, doing, and occasionally failing. You may struggle initially and get a few bruises, but eventually, you’ll get good. I strongly advise that you do every one of the exercises in this textbook. They will help you learn. They will also serve as a check that you’ve actually “got it”. And if you haven’t, then well, as mentioned, either keep working until you get it or go out and get help. The seemingly easy way out of pretending you’ve got it and flipping to the next page is actually the hard way out, because it will only cost you more grief in the long run.

6

Spoiler: Voldemort dies, Harry Potter lives happily ever after.

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• Confused? Good!

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• Learning by teaching. [Richard] Feynman was once asked by a Caltech faculty member to explain why spin 1/2 particles obey Fermi-Dirac statistics. He gauged his audience perfectly and said, “I’ll prepare a freshman lecture on it.” But a few days later he returned and said, “You know, I couldn’t do it. I couldn’t reduce it to the freshman level. That means we really don’t understand it.” — David L. Goodstein (1989). There is a view in some philosophical circles that anything that can be understood by people who have not studied philosophy is not profound enough to be worth saying. To the contrary, I suspect that whatever cannot be said clearly is probably not being thought clearly either. — Peter Singer (2016). I agree: If you can’t explain something simply, you don’t understand it well enough.7 This saying is useful for the instructor. But it also yields the learner the following useful corollary (in mathematics, a corollary is a statement that follows readily from another): An excellent test of whether you actually understand something is to see if you can explain it simply to yourself or to someone else. This learning technique may be dubbed learning by teaching and is a perfectly good one. As the Latin proverb goes, Docendo discimus—by teaching, we learn. One way to implement this technique8 is for you and your friends to get together explain concepts to each other aloud. For this technique to work, you and your friends must challenge each other and be demanding of each other. Do not be content until you’ve actually given each other the full and correct explanation of the concept at hand.

This quote or some similar variant is often misattributed to Einstein. But as Einstein himself once said, “73% of Einstein quotes are misattributed.” 8 Another way is to hand over the classroom to students. But this is probably too adventurous a move in Singapore, where the closest thing is probably the officially sanctioned and of course graded project work presentation. 7

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• Lessons from the science of learning. Unfortunately, the science of learning is still very much in its infancy. But do check out this short review of the literature: “How We Learn: What Works, What Doesn’t” (2013). According to this review, the two most effective study/learning techniques are: 1. Self-testing. 2. Distributed practice. (Sometimes also called spaced practice.) The next three are: 3. Elaborative interrogation. 4. Self-explanation. 5. Interleaved practice. Two commonly used but ineffective techniques are (a) highlighting; and (b) rereading. Again, be warned that the science of learning is very much in its infancy, so you should take with a large dose of salt any advice (including mine about learning by teaching). Nonetheless, there’s probably no harm trying out different techniques and seeing what works for you.

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Wolfram Alpha is somewhat more advanced (but also slower). Enter sin x for example and you’ll get graphs, the derivative, the indefinite integral, the Maclaurin series, and a bunch of other stuff you neither know nor care about. In this textbook, you’ll sometimes see (usually at the end of an example or an exercise answer) a clickable Wolfram Alpha logo that will bring you to the relevant computation on Wolfram Alpha. Symbolab is a much less powerful alternative to Wolfram Alpha. However, it’s perfectly good for simple algebra and somewhat quicker, so you may sometimes prefer using it.

Derivative Calculator • With Steps!

https://www.derivative-calculator.net/

OK

The Derivative Calculator and the Integral Calculator are probably unbeatable for the specific purposes of differentiation and integration. Both give step-by-step solutions for anything you want to differentiate or integrate. As with Wolfram Alpha, you’ll sometimes see clickable logos and that’ll bring you to the relevant computations. (Note that unfortunately, after clicking on these logos, you’ll also have to either click “Go!” or hit Enter .)10 Also check the Integral Calculator! Calculadora de Derivadas en español Ableitungsrechner auf Deutsch

Calculate derivatives online — with steps and graphing!

I also made this Collection of Spreadsheets (click “Make a copy”). These are for doing tedious and repetitive calculations you’ll often encounter in H2 Maths (with vectors, complex numbers, etc.).11 Some kiasu JCs, like HCI, even require that you got at least a B3 for both Maths & Additional Maths. The site author informed me that not having a direct link was deliberate and defensive. 11 As with anything I do, I welcome any feedback on these spreadsheets. (Perhaps in the future I will make a more attractive version.)

9

10

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• Other online resources. There are way too many websites that try to cover primary, secondary, and lower-level undergraduate maths. Unfortunately, some of them are awful and get things wrong. Three websites I like (though are probably a bit advanced for JC students) are: 1. The Stack Exchange (SE)

family of Q&A websites.12

Just for maths alone, there are three SE sites!13 (a) At , you can ask questions and often get them answered fairly promptly. Note though that this site is mostly frequented by fairly advanced users of maths (including many mathematicians), so they can be pretty impatient and quick to downvote questions they perceive to be “stupid”. Nonetheless, if you make an effort to write down a carefully crafted question and show also that you’ve made some effort to look for an answer (either on your own or online), they can be very helpful. (b) Mathematics Educators beta focuses on pedagogy (teaching and learning). Unfortunately, . Nonetheless, many of the discussions there are it’s not as active as filled with insights—insights that I’ve tried to incorporate into this textbook. (c)

is for research-level mathematics and is thus way beyond anything that’s of use to us.

In this textbook, you’ll occasionally see sent to related SE discussions.

and

logos. Click/touch them and you’ll be

2. ProofWiki gives succinct and rigorous definitions and proofs. Unfortunately it is very incomplete. 3. Mathworld.Wolfram is also great, but at times excessively encyclopaedic, at the cost of clarity and brevity. And of course, you can find countless free maths textbooks online (some less legal than others). One wonderful and (mostly) legal resource is the Internet Archive, which has many old books—including very many that were scanned by Google but which are no longer available on Google Books. Two totally illegal14 resources are: Library Genesis for books and Sci-Hub for articles.15 And of course, an old reliable is BitTorrent. (I have, of course, never used any of these illegal resources, but I hear they are great.) The flagship SE site is where you can ask any programming question and (often) see it answered amazingly quickly. For computing in general, there are also many other SE sites. SE sites are like Yahoo! Answers or Quora, but less stupid. The worst SE site is probably Politics beta , but even there, the average question or answer is probably better than that on Yahoo! Answers or Quora. 13 There are also many other wonderful SE sites that you should explore. Unfortunately, despite my magnificent contributions, Economics beta is not exactly thriving. It seems that economists, unlike programmers or mathematicians, have learnt all too well that contributing to the public good is folly. 14 Well, depending on which jurisdiction you live in. Of course, in Singapore, unless told otherwise, you should assume that everything is illegal. 15 Note though that these sites are constantly playing whac-a-mole with the fascist authorities and so the URLs often change. If the links given here aren’t correct, the first thing you should do is to let me know so I can correct these broken links. Then simply google to find the current working URLs. For Sci-Hub, the page usually lists the latest up-and-running URLs.

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Use of Graphing Calculators You are required to know how to use a graphing calculator.16 This textbook will give only a few examples involving graphing calculators. There is no better way of learning to use it than to play around with it yourself. By the time you sit down for your A-Level exams, you should have had plenty of practice with it. You can also use any of the seven calculators in the following list:17 LIST OF APPROVED GRAPHING CALCULATORS1 The following graphing calculator models are approved for use ONLY in subjects examined at H1, H2 and H3 Levels of the A-Level curriculum. Note: All graphing calculators must be reset prior to any examination. S/N

Calculator Brand

Calculator Model

Operating System

Approved Period1

1

CASIO

FX-9860GIIs

OS 2.04 OS 2.09

2014 – 2022

TI-84 Plus

OS 2.43 OS 2.53MP OS 2.55MP

2006 – 2024

2 3 4

TEXAS INSTRUMENTS

TI-84 Plus Silver Edition TI-84 Plus Pocket SE

2006 – 2024 2012 – 2021

5

TI-84 Plus C Silver Edition

OS 4.0 OS 4.2

2015 – 2023

6

TI-84 Plus CE

OS 5.0.1 OS 5.1.5 OS 5.2.0 OS 5.2.2 OS 5.3.0 OS 5.3.1 OS 5.4.0

2017 – 2024

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√ 3 or π.”

In Spider-Man: Homecoming (2017), Spider-Man uses a TI calculator to bust out of the “most secure facility on the Eastern Seaboard”—the Damage Control Deep Storage Vault (also known as the Department of Damage Control Vault).20

20

This scene takes place at about 57:30–58:00 of the movie (YouTube clip). They removed the brand name and also the model name, so I can’t quite tell what model it is. One website claims it’s a TI-86, while another claims it’s an exact match to the TI-83 PLUS. Zooming in, it looks like Spider-Man is doing some sort of combinatorics on his notepad. I can’t tell what exactly’s on the calculator screen. Many of the buttons on the calculator seem to be permanently depressed, which is weird. Slowing down, we see that he first punches in 5 8 ( , then a moment later 5 7 (or maybe 2 6 )—so it’s probably all just rubbish that he’s punching in. The latter keystrokes do get him out of the place though. (That’s all for my brilliant movie analysis of the week.)

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Apps You Can Use on Your TI Calculator On 2020-04-03, I received this information from SEAB: The following applications for Texas Instruments graphing calculators are approved for use in the GCE A-Level national examinations: 1. 2. 3. 4. 5.

Finance Conics Inequalz PlySmlt2 Transfrm

Please ensure that your graphing calculator is installed with the approved operating system. The Singapore Reset (pressing “On” key while holding the “2” and “8” keys) will reset the calculator without deleting these applications. On my (emulated) TI-84 Plus Silver Edition, the Finance app was already pre-installed. So I simply had to google, download, install the other four apps.21 After doing so, here’s what my “APPLICATIONS” screen looks like after executing these two steps: 1. Press ON to turn on your calculator. 2. Press APPS to bring up the “APPLICATIONS” screen. Step 1.

Step 2.

(Screenshots are of the end of each Step.)

21

Webpages to download these four apps for the TI-83 PLUS or TI-84 PLUS: Conics, Inequalz (seems like we’re supposed to use the “International” version), PlySmlt2, and Transfrm (links retrieved on 2020-0404)—ZIP file containing these four apps and the accompanying guidebooks (all written in 2001 or 2002!). (Unfortunately, the guidebook they have for PlySmlt2 seems to be for an old version of the same app.) Note that if you have the TI-84 PLUS CE, you may have to look for different webpages from which to download the appropriate apps.

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Preface/Rant When you’re very structured almost like a religion ... Uniforms, uniforms, uniforms ... everybody is the same. Look at structured societies like Singapore where bad behaviour isn’t tolerated. You are extremely punished. Where are the creative people? Where are the great artists? Where are the great musicians? Where are the great singers? Where are the great writers? Where are the athletes? All the creative elements seem to disappear. — Steve Wozniak (2011).22 That ghost, authoritarianism, sees education as a way to instill in all students the same knowledge and skills deemed valuable by the authority. — Zhao Yong (2014). The most dangerous man, to any government, is the man who is able to think things out for himself. — H.L. Mencken (1919). Divide students into two extremes: • Type 1 students are happy to learn absolutely nothing, so long as they get an A. • Type 2 would rather learn a lot, even if this means getting a C. The good Singaporean is taught that pragmatism is the highest virtue (obedience is second). She is thus also trained to be a Type 1 student (and indeed a Type 1 human being). If you’re a Type 1 student, then this textbook may not be the best use of your time,23 though you may still find the exercises and Ten Year Series (TYS) questions useful. (But do read Why Even Type 1 Pragmatists Should Read This Textbook below.) Of course, any careful student of this textbook will be rewarded with an A. But getting an A is not the goal of this textbook. Instead, the goal is to impart genuine understanding. Contrast this with the goal of the Singapore education system: Create a docile labour force that generates GDP growth. At first glance, these two goals do not seem to be in conflict. After all, we’d expect a student who genuinely understands her H2 Maths to also contribute to GDP growth. 2011 BBC interview. Also quoted in Zhao (2012, p. 103). Unfortunately, the audio at the given BBC link is broken. (Over the years, I’ve reported this broken audio to the BBC at least thrice, but it’s never been fixed. I have also not been able to find this audio anywhere else.) 23 The efficient Type 1 student may be interested in these resources: (a) The H2 Mathematics CheatSheet, all the formulae you’ll ever need on two sides of an A4 sheet of paper (this was written in 2016 and I hope to update it “soon”); (b) The H1 Mathematics Textbook, which is written more simply and covers a subset of the H2 syllabus; (c) The H2 Maths Exercise Book (coming “soon”), which teaches you how to mindlessly apply formulae and give the “correct” answer to every exam question; (d) My totally awesome tuition classes!

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the US as “stupid and boring”, “formulaic”, and “mindless” “pseudo-mathematics”.30 The same may be said of maths education in Singapore. As currently taught and tested for, JC maths is a mere and mindless collection of isolated recipes. Little or no understanding is required. Where do these mysterious recipes come from? Why do they “work”? No matter. Instead, all that’s required is for the student to mug 31 and reproduce these recipes during exams. Imagine a maths education system in which students are simply required to mug a milliondigit number. Those who manage to correctly write out the first 10 000 digits get an A; those who manage only 5 000 get a B; etc. Those who manage 100 000 digits get government scholarships, with the n President’s Scholarships going to the top n muggers.32 This imaginary system sounds absurd, but isn’t really all that different from the existing system. Indeed, it’d probably be an improvement. If we replaced the existing maths “education” system with the simple goal of mugging a million-digit number, then • Students would understand no less of maths than they do now. • Students would find maths no more “stupid and boring” than they do now. • We’d have a selection device that’s no worse than the present one. (Under the present system, students are selected and sorted based on their ability to mug and regurgitate isolated recipes that are no less random or arbitrary than a million-digit number.) • Resources currently dedicated to JC maths education would be freed for other uses. “[M]emory works far better when you learn networks of facts rather than facts in isolation,” writes the British mathematician Timothy Gowers (2012) in a critique of A-Level maths. I agree. When we learn a network of facts and seek to understand why something works, we are far more likely to remember what we’ve learnt and make use of what we’ve learnt. Moreover, further learning also becomes easier. It becomes easier to acquire and assimilate even more facts, knowledge, and wisdom. In contrast, when we mug isolated facts and recipes that we do not understand, we aren’t likely to remember them mere months after our A-Level exam. When in a few years, we actually have to make use of the material that we were supposed to have learnt in JC, we find ourselves having to learn everything from scratch. What then was the point? We may as well have spent those two years mugging a random million-digit number.

By the way, Lockhart explains why this is so and what maths really is far more eloquently and clearly than I ever could. I strongly recommend that every student and instructor of maths read A Mathematician’s Lament. There are two versions—a 2002 25-page PDF that circulated online and a 2009 book version. 31 To mug is to study hard and, especially, to engage in rote learning or memorisation. I consider mug to be Singlish and so italicise it. It seems though that the phrase mug up may have originated in Britain. To my knowledge, this phrase isn’t in current usage in Britain. I have however come across South Asians using mug up in this sense. (In efficient Singlish though, the preposition up is simply dropped.) 32 Not coincidentally, this is pretty much how the Chinese imperial examinations worked for many centuries. 30

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Work Work Work Work Work Work Pre-tertiary maths education in Singapore is no more “stupid and boring” than in the US (or elsewhere). The difference is that by the time the typical student in the US (or elsewhere) completes high school, she will only have squandered a very small portion of her life on such “mindless” and “formulaic” “pseudo-maths”. The same cannot be said for the typical Singaporean student. By the time she turns 18, she will have—just for the single subject of maths alone—clocked many thousands of hours attending school and tuition classes; doing homework, practice exam questions, assessment books, and TYS; taking common tests, promos, prelims, mid-year exams, and end-of-year exams; ad infinitum, ad nauseam. The Confucian East Asian countries33 perform splendidly on international tests. For example, in the 2018 Programme for International Student Assessment (PISA) Maths test, they swept the top seven spots (see p. li).34 “What,” the Western educator enquires, “is the magic here?”35 But there is no magic. To me, the explanation for why East Asian students do so well on these tests is obvious: They’re forced to work their butts off. While the American teen is “wasting” her time on typical, “useless” teenager-ly pursuits, the Singaporean teen is seated obediently in front of his desk, doing yet another soulcrushing TYS question. And once in a while, children as young as ten commit suicide due to poor exam results.36 Kids the world over commit suicide for a variety of reasons, but only in East Asia do they regularly do so because of exams and schoolwork. In South Korea, legislators have even passed (ill-enforced) laws barring hagwons (private cram schools) from operating past 10 p.m. To the American teen, it is mind-blowing that (a) anyone would be in school past 10 p.m.; or that (b) this practice would grow so common that legislators saw fit to take action. But to the East Asian, this isn’t at all strange. To me, the fact that East Asian students bust their butts is obviously the single most important explanation for why they do so well on international tests. Yet strangely, in the countless papers and books that I’ve come across seeking to explain why some countries do better than others, this explanation is rarely ever considered.37 China, Hong Kong, Japan, Korea, Macao, Singapore, Taiwan, and Viet Nam. Viet Nam’s results have been omitted from the headline charts in the 2018 edition. I’m not sure why exactly and haven’t looked into this matter. But we have for example this somewhat-cryptic statement: “the statistical uniqueness of Viet Nam’s response data implies that performance in Viet Nam cannot be validly reported on the same PISA scale as performance in other countries” (“PISA 2018 Results, Volume I, What Students Know and Can Do”, 2019, p. 190, PDF). 35 In the US, “Singapore Math” has acquired something of a mythical status. As with weight loss, Americans are constantly on the lookout for some magic, painless solution to their mediocre education systems. 36 In 2001, ten-year-old Lysher Loh jumped to her death from her fifth floor apartment. She “had been disappointed with her mid-year examination results and had found the workload heavy.” She had also “told her maid Lorna Flores two weeks before her death she did not want to be reincarnated as a human being because she never wanted to have to do homework again.” In 2016, an 11-year-old “killed himself over his exam results by jumping from his bedroom window in the 17th-storey flat”. More suicides here. 37 This explanation is neglected in part because the relevant statistics are difficult and expensive to collect. But a bigger reason for such neglect is probably that writers on education (be it in the West or in East Asia) are simply not quite aware of how much harder students in East Asia are made to work.

33

34

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Taking Tests Seriously A closely related explanation is that East Asians are trained from young to take every test seriously. In contrast, US kids couldn’t care less about some inconsequential PISA test.38 And yes, the PISA test is truly inconsequential. Each individual student’s results are never revealed (not to the student herself, her teachers, her parents, her school, or anyone else). Students are simply forced to take a two-hour test39 whose results they’ll never know and which will have absolutely no consequence on their lives. Under such circumstances, I suspect the truly intelligent and educated kid would simply click through the test as quickly as possible. In contrast, the dull-witted, obedient, and well-trained student would actually take the test seriously. This explanation has received some academic attention. For example, in “Measuring Success in Education: The Role of Effort on the Test Itself” (2019), experimenters went to two high schools in the US40 and four in Shanghai.41 Students at these schools42 were made to do a 25-minute, 25-question PISA-like maths test. At each school, students were divided into a control group and a treatment group. Students in the control group got nothing. Each student in the treatment group was given US\$25 or ¥90 upfront, with US\$1 or ¥3.60 later taken away for each wrong answer.43 In the US, the treatment groups did significantly better than the control groups. In contrast, in Shanghai, the treatment groups did no better: 25

US

Shanghai

Score (out of 25)

20

15

21.3

10

19.3 17.0 14.2 11.4

5

19.9 17.5 17.6

20.7 20.3

21.6

22.4

18.3

13.3

8.3 9.1 6.1 6.4

0

School 1 School 1 School 1 School 2 School 2 low regular honors regular honors Control

School 1 School 2 School 3 School 4

Treatment

East Asians care about exams as much as Americans care about sports, and vice versa. The standard PISA test is two hours. More time may be spent on other questionnaires and assessments. 40 “[A] high-performing private boarding school and a large public school with both low- and averageperforming students” (p. 295). 41 “[O]ne below-average-performing school, one school with performance that is just above average, and two schools with performance that is well above average” (p. 295). 42 “All students present on the day of testing took part in the experiment” (p. 295). 43 Using The Economist’s Big Mac Index, the authors reckoned that ¥90 was the purchasing-power equivalent of US\$25 (see their p. 295 and n. 11).

38

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This suggests that robotic Shanghai students have been trained and conditioned to always try their hardest, whether or not there’s any financial incentive. In contrast, US students may not be trying particularly hard when the stakes are low (or zero as is the case with PISA), but will try a little harder when there’s some financial incentive. Note also that Importantly, students learn about the incentive just before taking the test, so any impact on performance can only operate through increased effort on the test itself rather than through, for example, better preparation or more studying. Hence, any remaining performance gap between US and Shanghai students could very well be eliminated if US students were incentivised (through carrot and stick) to prepare and work half as hard as Shanghai students.44 In two separate studies, the authors arrive at similar conclusions: • “A country can rise up to 15 places in rankings if its students took the exam seriously.”45 • “our measures of student effort explain between 32 and 38 percent of the variation in test scores across countries.”46 Altogether then, there is very little that Western educators can learn from East Asia. The only lesson is this: If you want your students to do well on PISA-like tests, then • Train them from young to take every test seriously; and • Force them to work their butts off. This may mean destroyed childhoods and adolescences, plus the occasional academic suicide. But that, surely, is a small price to pay for topping the PISA charts.

44

Two more factors that may explain East Asians’ stellar performance (more pure speculation on my part):

1. In the US, it would take a very unusual teacher or student to devote even a second preparing for PISA. In contrast, at many East Asian schools, many teachers and students will likely have spent at least some time preparing for it. 2. In the peculiar case of China, there will likely also be a “patriotism effect”: Students, teachers, and administrators may be conditioned and exhorted to think, “It is our patriotic duty to do our best in order to show the rest of the world that China is great and glorious! China is no longer the Sick Man of Asia!” (Cue Chinese heroes defeating Somali pirates to the tune of March of the Volunteers.) Chinese teachers and students will therefore try harder than even other East Asians. 45 46

“Taking PISA Seriously: How Accurate Are Low Stakes Exams?” (2019). “When Students Don’t Care: Reexamining International Differences in Achievement and Student Effort” (2019).

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Genuine Understanding Again, the goal of this textbook is to impart genuine understanding. I suspect the sincere pursuit of this goal will do more to promote GDP growth than any of Gahmen’s current educational policies. But quite aside from any such instrumental value, I believe that a genuine understanding of maths (and indeed any other material) is intrinsically valuable. (GDP growth is lovely, but despite what Gahmen would have you believe, it is not all that makes life worthwhile.) And heck, learning can even be that three-letter F word banished from the Singapore education system—and apparently also from playgrounds and HDB void decks:

These were actual signs posted in Singapore. They became viral and were removed in June 2013 (New Paper story) and in March 2016 (Straits Times story), respectively.

Now, what do I mean by “imparting genuine understanding”? Personal anecdote: As a JC student, I remember being deeply mystified by why the scalar product had such a simple algebraic definition and yet could at the same time also tell us about the cosine of the angle between the two vectors. I never figured it out. But this didn’t matter, because this was simply “yet another formula” that we learnt for the sole purpose of answering exam questions.47 I remember being confused about the difference between the sample mean, the mean of the sample mean, the variance of the sample mean, and the sample variance. But this confusion didn’t matter, because once again, all we needed to do to get an A was to mindlessly apply formulae and algorithms. Monkey see, monkey do. 47

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This textbook is thus partly in response to my unhappy and unsatisfactory experience as a cog in the Singapore educational system. In other words, this is the textbook I wish I had had when I was a JC student. In the Singapore JC approach, teachers work through one example after another, demonstrating to students that something “works”. The student is then expected to faithfully replicate such “workings” on the exams. If the student acquires any understanding of what she’s doing, then that’s a nice but not terribly important bonus—what’s far more important is that the student produces the “correct” answers on the exams. Such perverse priorities are reversed in this textbook. Our foremost goal will always be to explain and understand why something is true. To then also be able to ace the exams is a natural but unimportant side effect. Almost all results are proven. I try to supply the intuition for each result in the simplest possible terms. Many proofs are relegated to the appendices, but where a proof is especially simple and beautiful, I encourage the student to savour it by leaving it in the main text (often as a guided exercise). In the rare instances where proofs are entirely omitted from this book—usually because they are too advanced—I make sure to clearly state so, lest the student be confused over where a result comes from or whether it’s supposed to be obvious. This textbook follows the Singapore A-Level syllabus.48 And so, a good deal of mindless formulae is unavoidable. Even so, I try in this textbook to give the student a tiny glimpse of what maths really is: “the art of explanation”.49 I try to plant a thoughtcrime in the student’s mind: Maths is not merely another pain to be endured, but can at times be a joy. And so for example, this textbook explains • A bit of intuition behind differentiation, integration, and the Fundamental Theorems of Calculus. (To get an A, no understanding of these is necessary. Instead, one need merely know how to “do” differentiation and integration problems.) • A bit of intuition behind why the Maclaurin series “works”. (To get an A, it suffices to mindlessly write out various Maclaurin series without having the slightest clue what they are or where they come from.) • Why the Central Limit Theorem (CLT) is so amazing. (To get an A, one need merely treat the CLT as yet another mysterious mathematical “trick” whose sole purpose is to solve exam questions. It isn’t necessary to appreciate why it is so amazing, where it might possibly come, or what relevance it has to everyday life.) • Why it is terribly wrong to believe that “a high correlation coefficient means a good model”. (Yet this is exactly your A-Level examiners seem to believe—see Ch. 128.9.)

A quixotic, longer-term goal of mine is to entirely change that too. As Lockhart says, “[T]hrow the stupid curriculum and textbooks out the window!” 49 Lockhart (2002, 2009).

48

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Why Even Type 1 Pragmatists Should Read This Textbook Two reasons: 1. The A-Level exams now include more curveball or out-of-syllabus questions. Previously, the A-Level exam questions were always perfectly predictable. If you had no problem doing past-year exam questions, then you’d have no problem getting an A. But starting in 2017 (coinciding with the new and supposedly reduced 9758 syllabus), curveball questions now carry a weight of perhaps 10–20%. For example, in 2017, out of absolutely nowhere, students were suddenly asked to use something called D’Alembert’s ratio test and to explain whether a series converges (see Exercise 574). I can find no official, publicly available statement announcing this change (much less explaining it). I have heard only that JC maths teachers were informed by MOE of this change ahead of time. My guess is that this is the MOE’s highly creative method of creating creative students. In my humble opinion, this change is cow manure. Challenging students with curveball questions is a perfectly good idea. What isn’t a good idea is to pose such questions on the high-stakes A-Level exam, thereby exerting additional pressure on the already pressurecooked Singapore student.50 But I will confess that selfishly, I welcome this change because it increases the value of this textbook. The student who carefully studies this textbook will be rewarded with a true and deep understanding of all the H2 Maths material and hence be fully prepared to bat away any curveball. Take for example N2017/I/6. This unfamiliar problem will likely have come as a shock to the Singapore monkey drilled a thousand times over to “do” computational problems involving 3D geometry, without ever understanding what he was doing. In contrast, any student who bothered to read Part III of this textbook even once will always have understood what she’d been “doing” and will thus have breezed through this problem.

50

Of course, the ancient East Asian education system has conditioned students, teachers, and administrators to believe that anything that isn’t on the exam is worthless. And so, it is difficult to persuade them to learn or teach any material that isn’t also on the exams. Difficult yes, impossible no. Overcoming this ancient mindset will require years of hard work, cultural change, and courage. Simply adding curveball questions to the A-Level exams certainly does not help and indeed merely serves to reinforce that mindset: “Oh you have to learn this too, not because it’s interesting or intrinsically valuable, but because it might show up on your exam.”

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I conclude this Preface/Rant by expressing my hope that even if you the instructor or student do not use this textbook as your primary instructional or learning material, you will still find it perfectly useful as an authoritative and reliable reference. P.S. This textbook is far from perfect. To steal a certain neighbourhood school’s motto, the best is yet to be. I hope to keep improving this textbook, but I can only do so with your help. So if you have any feedback or spot any errors, please feel free to email me. (As you can tell, I am pretty merciless about criticising others. So please don’t be shy about pointing out the many foolish mistakes that are surely still lurking in this textbook.)

And of course, I would’ve saved even more time by skipping the Singapore education system, but alas, that wasn’t an option. 52 Here I intend absolutely no knock or diss on my JC Maths teachers. They and indeed most of my Singapore teachers were generally pretty good (especially when compared to the US). They did the best they could—within the stultifying confines of the Singapore educational system. My critique here applies to that system. As the hip-hop cliché goes, I don’t hate the player(s); I hate the game.

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PISA 2018 Maths Results B-S-J-Z Singapore Macao Hong Kong Taiwan Japan Korea Estonia Netherlands Poland Switzerland Canada Denmark Slovenia Belgium Finland Sweden UK Norway Ireland Germany Czechia Austria Latvia France Iceland

591 569 558 551 531 527 526 523 519 516 515 512 509 509 508 507 502 502 501 500 500 499 499 496 495 495

NZ Portugal Australia Russia Italy Slovakia Spain Luxembourg Hungary Lithuania US Belarus Malta Croatia Israel Turkey Ukraine Greece Cyprus Serbia Malaysia Albania Bulgaria UAE Romania Montenegro

494 492 491 488 487 486 483 483 481 481 478 472 472 464 463 454 453 451 451 448 440 437 436 435 430 430

Brunei Kazakhstan Moldova Baku Thailand Uruguay Chile Qatar Mexico Bosnia & Herz. Costa Rica Jordan Peru Georgia N. Macedonia Lebanon Colombia Brazil Argentina Indonesia Saudi Arabia Morocco Kosovo Panama Philippines Dominican R.

430 423 421 420 419 418 417 414 409 406 402 400 400 398 394 393 391 384 379 379 373 368 366 353 353 325

Notes: “B-S-J-Z” = “Beijing-Shanghai-Jiangsu-Zhejiang”; Baku is in Azerbaijan; I’ve abbreviated or otherwise slightly changed some country names.53 Source: “PISA 2018 Results, Volume I, What Students Know and Can Do” (2019, pp. 17–18, PDF).

53

And yes, I consider Macao, Hong Kong, and Taiwan countries that are separate from China. I hope the PRC gahmen and people will forgive me for this egregious sin (please don’t kidnap me and force me to make a TV confession).

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Part 0. A Few Basics

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The glory of [maths] is its complete irrelevance to our lives. That’s why it’s so fun! — Paul Lockhart (2002, 2009).

I have never done anything ‘useful’. No discovery of mine has made, or is likely to make, directly or indirectly, for good or ill, the least difference to the amenity of the world. — G.H. Hardy (1940).

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1.

Just To Be Clear

In this textbook, we’ll stick to these standard conventions: • Greater than means “strictly greater than” (>). So I won’t bother saying “strictly”, unless it’s something I want to emphasise. • Similarly, less than means “strictly less than” ( 0) and non-positive means “less than or equal to zero” (≤ 0). • Similarly, negative means “less than zero” (< 0) and non-negative means “greater than or equal to zero” (≥ 0). • Zero is neither positive nor negative. Instead, it is both non-negative and nonpositive.54 Names of some punctuation marks: Left Parenthesis ( Bracket [ Brace {

Right Parenthesis ) Bracket ] Brace }

A pair of Parentheses () Brackets [] Braces {}

Remark 1. Some writers refer to (), [], and {} as round, square, and curly brackets— we’ll avoid these terms. Instead, as stated above, we’ll strictly refer to (), [], and {} as parentheses, brackets, and braces.55

• The symbol ∵ and ∴ stand for because and therefore.

• The punctuation mark “ means ditto or “the same as above/before”: Example 1. Tokyo is the capital of Japan. Beijing “ China. Lima “ Peru.

• The multiplication symbol ⋅ is (sometimes) preferred to × because there is (sometimes) the slight risk of confusing × with the letter x.

Note though that in France, positif and négatif mean ≥ 0 and ≤ 0, so that 0 is both positif and négatif. On this, see e.g. Wiktionary. 55 There is actually another pair of brackets ⟨⟩ called angle brackets. If we’re using angle brackets, then we’ll want to be careful to distinguish them from [] by referring to the latter as square brackets. Happily, we won’t be using angle brackets at all in this textbook. And so, we’ll simply call [] brackets.

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2.

PSLE Review: Division 9 1 =2 . 4 4

Example 2. Consider 9 ÷ 4: We call

• 9 the dividend • 4 the divisor

• q = 2 the quotient—q is the largest integer such that 4q ≤ 9 • r = 1 the remainder—r is defined so that 9 = 4q + r or equivalently, Example 3. Consider 17 ÷ 3:

r = 9 − 4q = 9 − 4 × 2 = 1. 17 2 =5 . 3 3

We call

• 17 the dividend • 3 the divisor

• q = 5 the quotient—q is the largest integer such that 3q ≤ 17 • r = 2 the remainder—r is defined so that 17 = 3q + r or equivalently, r = 17 − 3q = 17 − 3 × 5 = 2.

The above two examples used an algorithm56 called Euclidean division: Definition 1. (Euclidean division) Given integers x and d that are both positive,57 let q be the largest integer that satisfies dq ≤ x. Also, let r be the integer that satisfies x = dq + r.

Given the above equation, we call x the dividend, d the divisor, q the quotient, and r the remainder. And if r = 0, then we say that x divided by d leaves no remainder and that d is a factor of (or divides ) x.

It is possible to prove that thus defined, the quotient and remainder are unique.58 56 57

Algorithm is just a fancy synonym for method. For simplicity, the above definition considers only the case where both x and d are positive. But we can easily cover the other three cases too:

1. If both x and d are negative, then let q be the largest integer that satisfies dq ≥ x.

2. If x is negative while d is positive, then let q be the smallest integer that satisfies dq ≥ x.

3. If x is positive while d is negative, then let q be the smallest integer that satisfies dq ≤ x. 58

This is proven in Theorem 51 (Appendices).

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Example 4. Find 95 ÷ 4 using Euclidean division:

1. Find the largest integer q such that 4q ≤ 95: q = 23.

2. Now compute the remainder: r = 95 − 4q = 95 − 4 × 23 = 3. 95 3 3. Conclude: = 23 . 4 4

Example 5. Find 18 ÷ 6 using Euclidean division:

1. Find the largest integer q such that 6q ≤ 18: q = 3. 2. Now compute the remainder: r = 18 − 6q = 18 − 3 × 3 = 0. 18 3. Conclude: = 3. 6 Since r = 0, we say that 18 ÷ 6 leaves no remainder and that 6 is a factor of or divides 18.

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2.1.

Long Division

Remember long division from primary school? It’s simply Euclidean division repeated at each decimal place: Example 6. Find 95 ÷ 4 using long division. Seven steps:

1. First, draw two lines, one horizontal and one vertical.59 Then write the dividend 95 “inside” and the divisor 4 on the “left”. Long division is repeated Euclidean division—specifically, we do Euclidean division at each decimal place (ones, tens, hundreds, etc., but starting from the biggest place): 2. Starting at the tens place, we ask, “What is the largest integer such that 40q1 ≤ 95?” Answer: q1 = 2. So, we write 2 above the horizontal line, at the tens place. 3. Compute 40q1 = 40 × 2 = 80 and write “80” below our dividend 95. Also, draw a horizontal line below “80”. 4. Compute the remainder 95 − 80 = 15 and write “15” below the line just drawn. Step 1.

Step 2.

Step 3.

Step 4.

Step 6.

Step 7.

4 95

2 4 95

2 4 95 80

2 4 95 80 15

23 4 95 80 15 12

23 4 95 80 15 12 3

5. Now move to the ones place. Ask, “What is the largest integer such that 4q0 ≤ 15?” Answer: q0 = 3. So, write 3 above the top horizontal line, at the ones place. 6. Compute 4q0 = 4 × 3 = 12 and write “12” below our remainder “15”. Also, draw a horizontal line below “12”. 7. Compute the remainder 15 − 12 = 3 and write “3” below the line just drawn.

We’re done! The quotient is 23 and the remainder is 3: 95 3 = 23 . 4 4

We could actually keep going, repeating the same procedure for subsequent decimal places: the tenths, hundredths, etc. If we did this, we’d find that 95 = 23.75. 4

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Example 7. Find 87 ÷ 7 using long division:

12 7 87 70 17 14 3

Thus,

87 3 = 12 . 7 7

Example 8. Find 912 ÷ 17 using long division:

53 17 912 850 62 51 11

Thus,

912 11 = 53 . 17 17

Exercise 1. Find 8 057 ÷ 39 using long division. Identify the dividend, divisor, quotient, and remainder. (Answer on p. 1721.)

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2.2.

The Cardinal Sin of Dividing By Zero

Dividing by zero is a common mistake. It’s easy to avoid this mistake when the divisor is obviously a big fat zero. It gets harder when the divisor is an unknown constant or variable that may or may not be zero. Example 9. Solve x (x − 1) = (2x − 2) (x − 1). (That is, find the values of x for which the equation is true. We call such values of x the solutions or roots to the equation.) Here’s a wrong answer: “Divide both sides by x − 1 to get x = 2x − 2. So x = 2.” 7 Here’s a correct answer that considers two possible cases:

Case 1. If x − 1 = 0, then the equation is true. So, x = 1 is a possible solution. 1 M Pwe L can E Edivide Q_U A both T 1 osides N. S. by x − 1 to get Io 3x = 2x − 2. Case Art. 2. 24. If x − 1 ≠ 0,sthen The Proof. So, x = 2 is another possible solution. Conclusion. The two possible solutions are x = 1 and x = 2. The original equation, x= 6; therefore 2x=12; Moral of the story. Dividing by zero may cause us4.5 to lose perfectly valid solutions. So, 4.5 therefore 2 x + 3 = 15 ; therefore 2x+3 15 3 : again, 4x=24; always make sure the divisor is non-zero. If you’re not = sure whether it equals zero, then –37––37– 45 break up yourtherefore analysis4x— into 5two cases, as was done in the above example: = 19 ; therefore Iz-z-z-3; therefore 2x+3

#;=Iº;

-

-

Case 1. The quantity equals zero (and see what happens in this case). ——%

4 x – c’is non-zero (in which case go ahead and divide). Case 2. TheT quantity 4. 5 Example 12. Exercise 2. What’s the following seven-step 128 wrong 2 I with 6 6482–864“proof” that 1 = 0?

3 x—4 T 5x-6° therefore 128 ==== ; therefore 640 x

1. Let x and y be positive numbers such that x = y.

2. Square both x2 == y 28x—864; . thatsides: is, -768 therefore +864–768 =83, that is, 8x= 96 ; and x = 12. 3. Rearrange: x2 − y 2 = 0. The Proof.

—768 = 648 x–864; therefore – 768 = 648 x – 640 x —864,

4. Factorise: (x − y) (x + y) = 0. 128 216 The original equation, 5. Divide both sides by x − y to get3x–4Tsz-6 x + y = 0.

x = 12 ; therefore 3 x

6. Since x = y, y = x into the above get 2x 0. 128 —=128 : again =plug 36; therefore 3 x–4 = 32;equation thereforeto======4; again, 7. Divide both sides by 2x to get 1 = 0. 216 216(Answer on p. 1721.) - -

5 x = 60 ; therefore 5x-6 = 54; therefore :=E===4; 60

I 28 error(s) 216 Exercise 3. Identify any below.

therefore

3x–4

-

5 x–6'

Example 13.

#=#. divide both –tº– 3. – 2

-

35

numerators by x, and you will have

; therefore 42 = 35 x-70 ; therefore 42 x– 126 =

x -

3: -

3.5 x–70; therefore 42 x–35 x–126=–70, that is, 7 x — 126 =–70; therefore 7x = 126–70, that is, 7x = 56; and x = 8.

The Proof. 60

This was taken from an 18th-century textbook Elements of Algebra, 1740, p. 103), written by one 42* (The35% The original ; x = Lucasian 8 ; therefore x–2=6; Nicholas Saunderson (1682–1739). Saunderson the fourth Professor of Mathematics (at ginal equation, z_2 wasx–3 the University of Cambridge)—this is a fancy title that’s been held by Isaac Newton 42 ×(1642–1727) and Stephen Hawking (1942–2018). -

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2.3.

Is One Divided by Zero Infinity?

By the way, let’s take this opportunity to clear up a related and popular misconception. You’ve probably heard someone saying that 1 = ∞. 0

7

1 ≠ ∞. 0

3

Any non-zero number divided by zero is undefined.61 Undefined is the mathematician’s way of saying You haven’t told me what you’re talking about. So what you’re saying is meaningless. And so, the following five expressions are all undefined: 1 , 0

61

50 , 0

,

−17.1 , 0

∞ , 0

−∞ . 0

The special case is 0÷0, which is indeterminate. This means that 0÷0 is sometimes undefined, but can sometimes be defined under certain circumstances. We will not however encounter such circumstances in H2 Maths. So, we’ll say no more about 0 ÷ 0.

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3.

Logic

Big surprise—you’ve secretly been using logic your whole life. Logic isn’t explicitly on your H2 Maths syllabus.62 But spending an hour or two on logic pays huge dividends—you’ll learn to reason better, both in maths and in everyday life. This chapter is thus a brief and gentle introduction to logic. Here we merely present some of the most basic but also some of the most useful results from logic. (If you truly can’t be bothered, please at least check out the one-page summary of this chapter on p. 34.) First, try this appetiser.63 Example 10. (Wason Four-Card Puzzle) In a special deck of cards, each card has a letter on one side and a number on the other. You are shown these four cards:

A

Z

1

8

Betsy the Bimbotic Blonde now comes along and makes this claim: “If a card has a vowel on one side, then it has an even number on the other side.” You suspect that Betsy is wrong. To prove that she’s wrong, which of the above four cards should you turn over? (The goal is to turn over as few cards as possible.)64 The above puzzle baffles most who are untrained in logical thinking. Right now, that probably includes you. But by p. 24 of this textbook, you’ll have had some training in logic and thus be able to solve this puzzle easily.

If it were up to me, the H2 Maths syllabus would devote at least a little time to logic. Instead, that time is spent on learning to mindlessly compute such objects as the volume of the revolution of a curve around the y-axis. Which is a doggie trick that (1) students will forget two weeks after the final A-Level exam; and (2) is completely useless unless you’re planning to be an engineer or physicist, in which case it is still completely useless since down the road, you’ll be learning it again (and probably more properly). 63 The wording here is a slightly modified version of Wason (1966, pp. 145–146). 64 Answer: A and 1. We’ll explain this on p. 24.

62

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3.1.

True, False, and Indeterminate Statements

Example 11. Let A, B, C, and D be these statements: A: “Germany is in Europe.” B: “Germany is in Asia.” C: “1 + 1 = 2.” D: “1 + 1 = 3.”

3 7 3 7

(True) (False) (True) (False)

Then statements A and C are true, while statements B and D are false. As shorthand, we’ll use the green checkmark 3 to indicate that a statement is true; and a red crossmark 7 to indicate that it’s false. Example 12. Let M , N , and O be these statements: M : “x > 0.” N : “x > 1.” O: “x is a positive number.” Note that the truth values of M , N , and O depend on the value of x. That is, whether each statement is true or false depends on the value of x. So, without being given further information, each of these three statements can neither be said to be true nor said to be false. Instead, we say that each is indeterminate. But if we’re told that

• x = 5, then all three statements are true.

• x = 0.5, then statements M and O are true, while N is false. • x = −1, then all three statements are false.

Remark 2. In this textbook, we’ll leave the terms statement, true, false, and indeterminate undefined.

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3.2.

The Conjunction AND and the Disjunction OR

Example 13. As before, let A, B, C, and D be these statements: A: “Germany is in Europe.” B: “Germany is in Asia.” C: “1 + 1 = 2.” D: “1 + 1 = 3.”

3 7 3 7

Using the logical connective AND (called the conjunction), we can form these statements (which we also call conjunctions): A AND B: “Germany is in Europe AND Germany is in Asia.” 7 A AND C: “Germany is in Europe AND 1 + 1 = 2.” 3 B AND D: “Germany is in Asia AND 1 + 1 = 3.” 7

• The conjunction A AND B is false because B is false.

• The conjunction A AND C is true because both A and C are true. • The conjunction B AND D is false because B is false. (Indeed, D is also false.) Using the logical connective OR (called the disjunction) we can form these statements (which we also call disjunctions): A OR B: “Germany is in Europe OR Germany is in Asia.” 3 A OR C: “Germany is in Europe OR 1 + 1 = 2.” 3 B OR D: “Germany is in Asia OR 1 + 1 = 3.” 7

• The disjunction A OR B is true because A is true. • The disjunction A OR C is true because A is true. (Indeed, C is also true.) • The disjunction B OR D is false because both B and D are false. Definition 2. Let P and Q be statements. The conjunction of P and Q is the statement denoted P AND Q and that is • True if both P and Q are true; and • False if at least one of P or Q is false. The disjunction of P and Q is the statement denoted P OR Q and that is • True if at least one of P or Q is true; and • False if both P and Q are false. Exercise 4. Continue with the above example. Explain if each of the following statements is true or false. (Answer on p. 1722.) (a) B AND C. (d) B OR C. 12, Contents

(b) A AND D. (e) A OR D.

(c) C AND D. (f) C OR D. www.EconsPhDTutor.com

3.3.

The Negation NOT

Informally, a statement’s negation is the “opposite” or contradictory statement. Formally, Definition 3. Let P be a statement. The negation of P is the statement denoted NOT-P and that is true if P is false

and

false if P is true.

Example 14. As before, let A, B, C, and D be these statements: A: “Germany is in Europe.” B: “Germany is in Asia.” C: “1 + 1 = 2.” D: “1 + 1 = 3.”

3 7 3 7

The negations of statements A, B, C, and D are simply these statements: NOT-A: “Germany is not in Europe.” NOT-B: “Germany is not in Asia.” NOT-C: “1 + 1 ≠ 2.” NOT-D: “1 + 1 ≠ 3.”

7 3 7 3

• A and C are true; thus, their negations NOT-A and NOT-C must be false. • B and D are false; thus, their negations NOT-B and NOT-D must be true. The negation of the negation simply brings us back to the original statement: NOT-NOT-A: “Germany is in Europe.” NOT-NOT-B: “Germany is in Asia.” NOT-NOT-C: “1 + 1 = 2.” NOT-NOT-D: “1 + 1 = 3.”

3 7 3 7

Exercise 5. Let E: “It’s raining”, F : “The grass is wet”, G: “I’m sleeping”, and H: “My eyes are shut”. Write down NOT-E, NOT-F , NOT-G, and NOT-H.

Remark 3. We’ve just learnt about the three basic logical connectives: AND, OR, and NOT. Using these three connectives, we can build ever more complex statements.

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3.4.

Equivalence ⇐⇒

Definition 4. If two statements P and Q always have the same truth value, then we say that they are equivalent and write P ⇐⇒ Q.

Remark 4. The symbol ⇐⇒ stands for is equivalent to or if and only if.65 Example 15. Let M , N , and O be these statements:

M : “x > 0.” N : “x > 1.” O: “x is a positive number.”

Observe that if M is true, then O is also true. And if M is false, then O is also false. Since M and O always have the same truth value, we say that M and O are equivalent and write M ⇐⇒ O. In contrast, if x = 0.5, then M is true while N is false. Since M and N do not always have the same truth value, we say that M and N are not equivalent and write M ⇐⇒ / N.

So, if we can find even one instance where two statements P and Q do not have the same truth value, then P ⇐⇒ / Q. If we can’t, then P ⇐⇒ Q.

Example 16. Let α, β, and γ be these statements: α: “x = 3.” β: “x + 2 = 5.” γ: “x2 = 9.”

Observe that if α is true, then β is true. And if α is false, then β is also false. Since α and β always have the same truth value, we say that α and β are equivalent and write α ⇐⇒ β.

Exercise 6. Explain if each pair of statements is equivalent. (a) N : “x > 1” (b) α: “x = 3”

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and and

O: “x is a positive number” γ: “x2 = 9”.

If and only if is sometimes shortened to iff.

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3.5.

De Morgan’s Laws: Negating the Conjunction and Disjunction

The negation of P AND Q is denoted NOT- (P AND Q). Remark 5. Note the use of parentheses.

In arithmetic, 2 + 3 × 7 = 23 is different from (2 + 3) × 7 = 35.

Likewise, in logic, NOT- (P AND Q) is different from NOT-P AND Q.

In both arithmetic and logic, we sometimes add parentheses to be clear about the order of operations. Indeed, just to be extra clear, we sometimes add parentheses even when strictly speaking, they aren’t necessary. Fact 1. Suppose P and Q are statements. Then NOT- (P AND Q)

Proof. See p. 1524 (Appendices).

⇐⇒

(NOT-P OR NOT-Q).

Example 17. As before, let A: “Germany is in Europe”; B: “Germany is in Asia”; C: “1 + 1 = 2”, and D: “1 + 1 = 3”. Consider these three statements:

1. A AND B: “Germany is in Europe AND Germany is in Asia.” 7 2. C AND D: “1 + 1 = 2 AND 1 + 1 = 3.” 7 3. A AND C: “Germany is in Europe AND 1 + 1 = 2.” 3

Let’s now consider the negation of each of the above statements: 1. NOT- (A AND B) is true. There are two ways to see this:

• Since A AND B is false, its negation NOT- (A AND B) must be true. • By Fact 1, NOT- (A AND B) is equivalent to (NOT-A OR NOT-B): “Germany is not in Europe OR Germany is not in Asia”. Which is true because NOT-B: “Germany is not in Asia” is true.

2. NOT- (C AND D) is true. Two ways to see this:

• Since C AND D is false, its negation NOT- (C AND D) must be true. • By Fact 1, NOT- (C AND D) is equivalent to (NOT-C OR NOT-D): “1 + 1 ≠ 2 OR 1 + 1 ≠ 3”. Which is true because NOT-D: “1 + 1 ≠ 3” is true.

3. NOT- (A AND C) is false. Two ways to see this:

• Since A AND C is true, its negation NOT- (A AND C) must be false. • By Fact 1, NOT- (A AND C) is equivalent to (NOT-A OR NOT-C): “Germany is not in Europe OR 1 + 1 ≠ 2”. Which is false because both NOT-A: “Germany is not in Europe” and NOT-C: “1 + 1 ≠ 2” are false.

Exercise 7. Continue with the above example. Is the negation of each statement true? (a) B AND C. 15, Contents

(b) A AND D.

(c) B AND D. (Answer on p. 1722.) www.EconsPhDTutor.com

The negation of P OR Q is denoted NOT- (P OR Q). Fact 2. Suppose P and Q are statements. Then NOT- (P OR Q)

Proof. See p. 1525 (Appendices).

⇐⇒

(NOT-P AND NOT-Q).

Remark 6. Together, Facts 1 and 2 are called De Morgan’s Laws. Example 18. As before, let A: “Germany is in Europe”; B: “Germany is in Asia”; C: “1 + 1 = 2”, and D: “1 + 1 = 3”. Consider these three statements:

1. A OR B: “Germany is in Europe OR Germany is in Asia.” 3 2. C OR D: “1 + 1 = 2 OR 1 + 1 = 3.” 3 3. A OR C: “Germany is in Europe OR 1 + 1 = 2.” 3

Let’s now consider the negation of each of the above statements: 1. NOT- (A OR B) is false. There are two ways to see this:

• Since A OR B is true, its negation NOT- (A OR B) must be false. • By Fact 2, NOT- (A OR B) is equivalent to (NOT-A AND NOT-B): “Germany is not in Europe AND Germany is not in Asia”. Which is false because NOT-A: “Germany is not in Europe” is false.

2. NOT- (C OR D) is false. Two ways to see this:

• Since C OR D is true, its negation NOT- (C OR D) must be false. • By Fact 2, NOT- (C OR D) is equivalent to (NOT-C AND NOT-D): “1 + 1 ≠ 2 AND 1 + 1 ≠ 3”. Which is false because NOT-C: “1 + 1 ≠ 2” is false.

3. NOT- (A OR C) is false. Te two ways to see this:

• Since A OR C is true, its negation NOT- (A OR C) must be false. • By Fact 2, NOT- (A OR C) is equivalent to (NOT-A AND NOT-C): “Germany is not in Europe AND 1 + 1 ≠ 2”. Which is false because NOT-A: “Germany is not in Europe” is false. Indeed, NOT-C: “1 + 1 ≠ 2” is also false.

Exercise 8. Continuing with the above example, three statements are given below. Is the negation of each true? (Answer on p. 1723.) (a) B OR C.

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(b) A OR D.

(c) B OR D.

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3.6.

The Implication P Ô⇒ Q

The statement “P Ô⇒ Q” is read aloud as “If P , then Q” or “P implies Q”. Example 19. Let E: “It’s raining” and F : “The grass is wet”.

Then the implication E Ô⇒ F is this statement:

“If it’s raining, then the grass is wet.”

Or equivalently,

“That it’s raining implies that the grass is wet.”

Which is true. Example 20. Let G: “I’m sleeping” and H: “My eyes are shut”. Then the implication G Ô⇒ H is this statement:

“If I’m sleeping, then my eyes are shut.”

Or equivalently,

“That I’m sleeping implies that my eyes are shut.”

Which is true.66 Example 21. Let M : “x > 0” and N : “x > 1”.

Then the implication M Ô⇒ N is this statement:

“If x > 0, then x > 1.”

“That x > 0 implies that x > 1.”

Or equivalently,

Which is false (counterexample: x = 0.5).

Example 22. Let α: “x = 3.” and γ: “x2 = 9.”

Then the implication γ Ô⇒ α is this statement: Or equivalently,

“If x2 = 9, then x = 3.”

“That x2 = 9 implies that x = 3.”

Which is false (counterexample: x = −3).

This all seems simple enough. However, you may find the formal definition of P Ô⇒ Q a little strange and unintuitive:

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For most healthy human beings.

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Definition 5. Given statements P and Q, the implication P Ô⇒ Q is defined as NOT-P OR Q.

That is, P Ô⇒ Q is the statement that is • True if P is false OR Q is true; and • False if P is true AND Q is false.

Given the implication P Ô⇒ Q, we call P the hypothesis, premise, or antecedent; and Q the conclusion or consequent.

What’s most confusing about the above definition is this: From a false hypothesis, any conclusion may be drawn! That is,

If P is false, then P Ô⇒ Q is always true!

Example 23. Consider these three statements:

1. “If goldfish can walk on land, then pigs can fly.” 2. “If there are 17 washing machines on Mars, then I am a billionaire.” √ 3. “If x > 0 and x < 0, then 2 = 77.” √ Pigs cannot fly, I am not a billionaire, and 2 ≠ 77. It may thus seem that all three statements must be false.

But strangely enough, all three are true! To see why, use the above Definition to rewrite the three statements as 1. “Goldfish cannot walk on land OR pigs can fly.” Which is true, because “Goldfish cannot walk on land” is true. 2. “There aren’t 17 washing machines on Mars OR I am a billionaire.” Which is true, because “There aren’t 17 washing machines on Mars” is true. √ 3. “NOT-(x > 0 AND x < 0) OR 2 = 77.” Which is true, because “NOT-(x > 0 AND x < 0)” is true—it is impossible that x is both more than AND less than 0.

The examples here illustrate that from a false hypothesis, any conclusion may be drawn! Exercise 9. Is each statement true?

(a) “If Tin Pei Ling (TPL) is a genius, then the Nazis won World War II (WW2).” (b) “If TPL is a genius, then the Allies won WW2.” (c) “If π is rational, then I am the king of the world.” (d) “If π is rational, then Lee Hsien Loong is Lee Kuan Yew’s son.”

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3.7.

The Converse Q Ô⇒ P

Informally, converse = “Flip”. Formally,

Definition 6. Given the implication P Ô⇒ Q, its converse is the statement Q Ô⇒ P . Example 24. Let E: “It’s raining” and F : “The grass is wet”. Then E Ô⇒ F : “If it’s raining, then the grass is wet.”

The converse of the implication E Ô⇒ F is this implication: Or equivalently,

F Ô⇒ E: “If the grass is wet, then it’s raining.” “That the grass is wet implies that it’s raining.”

Note that F Ô⇒ E is false. One way to prove that a statement is false is by supplying a counterexample. A counterexample that shows that F Ô⇒ E is false is any scenario where the grass is wet even though it isn’t raining. We can easily think of such counterexamples: 1. The rain just stopped. 2. Someone is watering the grass. 3. A dog is peeing on the grass. 4. You are peeing on the grass. Observe that in the above example, E Ô⇒ F is true but its converse F Ô⇒ E is false. This proves that an implication and its converse are not always equivalent. Let’s jot this down formally: Fact 3. Given two statements P and Q, it is not always true that (P Ô⇒ Q)

⇐⇒

Instead, by Definition 5, we have this result:

(Q Ô⇒ P ).

Fact 4. Suppose P and Q are statements. Then (Q Ô⇒ P )

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⇐⇒

(NOT-Q OR P ).

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Exercise 10. Let G: “I’m sleeping”; H: “My eyes are shut”; M : “x > 0”; N : “x > 1”; α: “x = 3”; and γ: “x2 = 9”. Explain whether each of the three statements below is true. Then write down its converse and explain whether this converse is true. (Answer on p. 1723.) (a) G Ô⇒ H.

(b) M Ô⇒ N .

(c) γ Ô⇒ α.

Exercise 11. In Exercise 9, we already explained whether each of the four statements below is true. Now, write down the converse of each statement. Then explain whether this converse is true. (Answer on p. 1723.) (a) “If Tin Pei Ling (TPL) is a genius, then the Nazis won World War II (WW2).” (b) “If TPL is a genius, then the Allies won WW2.” (c) “If π is rational, then I am the king of the world.” (d) “If π is rational, then Lee Hsien Loong is Lee Kuan Yew’s son.” Exercise 12. Let A: “Germany is in Europe”, B: “Germany is in Asia”, C: “1 + 1 = 2”, and D: “1 + 1 = 3”. Explain whether each of the four statements below is true. Then write down its converse and explain whether this converse is true. (Answer on p. 1724.) (a) A Ô⇒ B.

(b) A Ô⇒ C.

(c) A Ô⇒ D.

(d) C Ô⇒ D.

Exercise 13. Fill in each blank below with (i) must be true; (ii) must be false; or (iii) could be true or false. (Answer on p. 1724.) (a) If P is true, (b) If P is false, (c) If P Ô⇒ Q is true, (d) If P Ô⇒ Q is false,

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then then then then

P P Q Q

Ô⇒ Ô⇒ Ô⇒ Ô⇒

Q Q P P

_____ and Q Ô⇒ P _____. _____ and Q Ô⇒ P _____. _____. _____.

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3.8.

Affirming the Consequent (or The Fallacy of the Converse)

Affirming the consequent or the fallacy of the converse is a common error that people make in everyday life. Formally, it takes this form:

1. “P Ô⇒ Q.” 2. “Q.” 3. “Therefore, P .”

Example 25. 1. “If it’s raining, then the grass is wet.” 2. “The grass is wet.” 3. “Therefore, it’s raining.”67 That the grass is wet does not imply that it’s raining. The grass may be wet because (a) it just stopped raining; (b) someone is watering it; (c) a dog is peeing on it; (d) you are peeing on it; or a billion other reasons. Example 26. 1. “If John is undergoing chemotherapy, then John is bald.” 2. “John is bald.” 3. “Therefore, John is undergoing chemotherapy.” That John is bald does not imply he’s undergoing chemotherapy. John may be bald because (a) he’s suffering from male-pattern baldness; (b) he likes to keep his scalp cleanshaven; or a billion other reasons. Example 27. 1. “If Mary drinks a lot of Coke, then Mary is fat.” 2. “Mary is fat.” 3. “Therefore, Mary drinks a lot of Coke.” That Mary is fat does not imply she drinks a lot of Coke. Mary may be fat because (a) she drinks a lot of Pepsi; (b) eats a lot of junk food; (c) suffers from low metabolism; or a billion other reasons. When spelt out so explicitly, affirming the consequent or the fallacy of the converse seems rather silly. But unfortunately, people make this error all the time. Hopefully you’ll now be able to avoid it. Exercise 14. Is the following chain of reasoning valid? 1. “If Warren Buffett owns Google, then Warren Buffett is rich.” 2. “Warren Buffett is rich.” 3. “Therefore, Warren Buffett owns Google.”

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By the way, a chain of reasoning of this form (whether valid or invalid) is called a syllogism. A syllogism has two or more statements called premises, followed by a conclusion.

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3.9.

The Negation NOT- (P Ô⇒ Q)

Example 28. Let I: “x is German” and J: “x is European”. Then we can contruct this implication:

I Ô⇒ J: “If x is German, then x is European.”

Which of the following correctly negates I Ô⇒ J? In other words, which of the following statements is NOT- (I Ô⇒ J)?

(a) (b) (c) (d)

“If x is German, then x is not European.” “If x is not German, then x is European.” “Some x is German and not European.” “Some x is European and not German.”

This is tricky and you should take as long as you need to think about it, before reading the answer/explanation on the next page. The point of this exercise is to demonstrate to yourself that it isn’t obvious what the negation of an implication is. (Or if it’s obvious, it’ll demonstrate that you’re pretty smart.) (As I’ve repeatedly stressed, do not do the intellectually lazy thing of skipping ahead. Give it at least three minutes of honest effort before going to the next page.)

Or don’t.

Whatevs.

If you dun care, I oso dun care.

(Example continues on the next page ...)

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With the following fact, we’ll find it very easy to negate any implication: Fact 5. Suppose P and Q are statements. Then NOT- (P Ô⇒ Q)

⇐⇒

(P AND NOT-Q).

Proof. You are asked to prove this in the next Exercise. Exercise 15. Prove the above Fact.68

(... Example continued from the previous page.) Continue to let I: “x is German” and J: “x is European”.

By Fact 5, NOT- (I Ô⇒ J) is equivalent to

I AND NOT-J: “x is German AND x is not European”.

That is,

I AND NOT-J: “Some x is German AND not European.”

And so, the negation of I Ô⇒ J: “If x is German, then x is European” is (c) “Some x is German and not European.”

(In Ch. 3.13, we’ll study statements of the form “Some x is A” in greater detail.) Exercise 16. Let K: “x is donzer” and L: “x is kiki”. Consider this implication: K Ô⇒ L: “If x is donzer, then x is kiki.”

Which of the following statements is the negation of the above implication? That is, which of the following is NOT- (K Ô⇒ L)?

(a) “If x is kiki, then x is not donzer.” (b) “Some x is kiki and not donzer.” (c) “If x is donzer, then x is not kiki.” (d) “Some x is donzer and not kiki.”

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Hint: Look at the definition of P Ô⇒ Q (Definition 5). What is its negation?

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We can now reproduce and solve the Wason Four-Card Puzzle: Example 10. (Wason Four-Card Puzzle) In a special deck of cards, each card has a letter on one side and a number on the other. You are shown these four cards:

A

Z

1

8

Betsy the Bimbotic Blonde now comes along and makes this claim: “If a card has a vowel on one side, then it has an even number on the other side.” You suspect that Betsy is wrong. To prove that she’s wrong, which of the above four cards should you turn over? (The goal is to turn over as few cards as possible.) The answer is that we should turn over A and 1.

Explanation I. By Fact 5, the negation of P Ô⇒ Q is P AND NOT-Q. Thus, the negation of Betsy’s claim is “A card has a vowel on one side AND an odd number on the other side.”

So, we should turn over any vowels and odd numbers.

In case you weren’t convinced, here’s Explanation II, which doesn’t directly use Fact 5. We can call this the brute-force case-by-case method: 1. An odd number behind A would prove Betsy wrong. So, we should turn over A. 2. An odd number behind Z would not prove Betsy wrong. Nor would an even number. So, we needn’t turn over Z. 3. A vowel behind 1 would prove Betsy wrong. So, we should turn over 1. 4. A vowel behind 8 would not prove Betsy wrong. Nor would a consonant. So, we needn’t turn over 8.

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3.10.

The Contrapositive NOT-Q Ô⇒ NOT-P

Informally, contrapositive = “Flip and negate both”. Formally,

Definition 7. Let P and Q be statements. Given the implication P Ô⇒ Q, its contrapositive is the statement NOT-Q Ô⇒ NOT-P . Example 29. Let I: “x is German” and J: “x is European”.

Consider the implication I Ô⇒ J and its contrapositive NOT-J Ô⇒ NOT-I: I Ô⇒ J: “If x is German, then x is European.” NOT-J Ô⇒ NOT-I: “If x is not European, then x is not German.”

Both I Ô⇒ J and its contrapositive NOT-J Ô⇒ NOT-I are true.

Next, consider the converse J Ô⇒ I and its contrapositive NOT-I Ô⇒ NOT-J: J Ô⇒ I: “If x is European, then x is German.” NOT-I Ô⇒ NOT-J: “If x is not German, then x is not European.”

Both J Ô⇒ I and its contrapositive NOT-I Ô⇒ NOT-J are false.

Example 30. Let E: “It’s raining” and F : “The grass is wet”.

Consider the implication E Ô⇒ F and its contrapositive NOT-F Ô⇒ NOT-E: E Ô⇒ F : “If it’s raining, then the grass is wet.” NOT-F Ô⇒ NOT-E: “If the grass is not wet, then it’s not raining.”

Both E Ô⇒ F and its contrapositive NOT-F Ô⇒ NOT-E are true.

Next, consider the converse F Ô⇒ E and its contrapositive NOT-E Ô⇒ NOT-F : F Ô⇒ E: “If the grass is wet, then it’s raining.” NOT-E Ô⇒ NOT-F : “If it’s not raining, then the grass is not wet.”

Both E Ô⇒ F and its contrapositive NOT-F Ô⇒ NOT-E are false.

As the above examples suggest, every implication is equivalent to its contrapositive: Fact 6. Suppose P and Q are statements. Then (P Ô⇒ Q)

⇐⇒

(NOT-Q Ô⇒ NOT-P ).

Proof. By Definition 5, P Ô⇒ Q is NOT-P OR Q. And NOT-Q Ô⇒ NOT-P is Q OR NOT-P . Hence, P Ô⇒ Q and NOT-Q Ô⇒ NOT-P are equivalent.69 69

This proof implicitly assumes that the logical connective OR is commutative.

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Fact 6 is especially useful on those occasions when it’s hard to prove an implication but easy to prove its contrapositive:70 Example 31. It’s not obvious how we can prove this implication: If x4 − x3 + x2 ≠ 1, then x ≠ 1.

But proving its contrapositive is easy:

If x = 1, then x4 − x3 + x2 = 1.

Proof. Simply plug in x = 1 to verify that x4 − x3 + x2 = 1.

Example 32. It’s not obvious how we can prove this implication: If x2 is even, then x is even. But proving its contrapositive is easy: If x is odd, then x2 is odd. Proof. Let x = 2k + 1 where k is some integer. Then x2 = 4k 2 + 4k + 1, which is odd.

But don’t worry. In H2 Maths, you won’t be required to write any proofs; the above is just FYI and to illustrate why the contrapositive is useful. Exercise 17. The statement “If x is German, then x is European” is true. Which of the following statements is its contrapositive? Which are true? (Answer on p. 1725.) (a) “If x is European, (b) “If x is not German, (c) “If x is not German, (d) “If x is not European, (e) “If x is not European,

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These examples are from

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then then then then then

x x x x x

is is is is is

German.” not European.” European.” not German.” German.”

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3.11.

(P Ô⇒ Q AND Q Ô⇒ P ) ⇐⇒ (P ⇐⇒ Q)

Example 33. As before, let • I: “x is German.” • J: “x is European.” • M : “x > 0.”

• N : “x > 1.” • O: “x is a positive number.”

To show that two statements are equivalent, we can use Definition 4. So for example, • M ⇐⇒ O, because both statements always have the same truth value.

• I ⇐⇒ / J (counterexample: if x = Emmanuel Macron, then I is false while J is true). • M ⇐⇒ / N (counterexample: if x = 0.5, then M is true while N is false).

Alternatively, we can use this fact:

Fact 7. Suppose P and Q are statements. Then (P Ô⇒ Q AND Q Ô⇒ P )

Proof. See p. 1525 (Appendices).

⇐⇒

(P ⇐⇒ Q).

So, to show that P ⇐⇒ Q, we can show that both P Ô⇒ Q and Q Ô⇒ P are true. And to show that P ⇐⇒ / Q, we can show that either P Ô⇒ Q or Q Ô⇒ P is false.

Example 34. The implication M Ô⇒ O (“if x > 0, then x is a positive number”) is true. Also, the implication O Ô⇒ M (“if x is a positive number, then x > 0”) is true. So, by Fact 7, M ⇐⇒ O.

Example 35. The implication J Ô⇒ I (“if x is European, then x is German”) is false. So, by Fact 7, I ⇐⇒ / J.

Example 36. The implication N Ô⇒ M (“if x > 1, then x > 0”) is false. So, by Fact 7, M ⇐⇒ / N.

Exercise 18. Continue with the last example: Is N ⇐⇒ O true? (Answer on p. 1725.) Exercise 19. Consider the following three statements. Are any two equivalent? X: “John is a Singapore citizen.” Y : “John has a National Registration Identity Card (NRIC).” Z: “John has a pink NRIC.” 27, Contents

3.12.

Other Ways to Express P Ô⇒ Q (Optional)

Eskimos supposedly have 50 different words for snow, presumably because snow is so important and ubiquitous in their lives.71

Similarly, because the implication P Ô⇒ Q is so important and ubiquitous in both mathematics/logic and everyday life, we have many equivalent ways to express it: Example 37. Let E: “It is raining” and F : “The grass is wet”. Then all of the following statements are exactly equivalent: Maths/Logic E Ô⇒ F E Ô⇒ F If E, then F . If E, F . F if E. F when E. F follows from E. E is sufficient for F . F is necessary for E.

Everyday English That it’s raining implies that the grass is wet. It’s raining only if the grass is wet.72 If it’s raining, then the grass is wet. If it’s raining, the grass is wet. The grass is wet if it’s raining. The grass is wet when it’s raining. That the grass is wet follows from the fact that it’s raining. That it’s raining is sufficient for the grass to be wet. It is necessary that the grass is wet, for it to be raining.

But don’t worry. The above is just FYI. In the main text of this textbook, we’ll avoid using the terms “only if”, “sufficient”, and “necessary”. Instead, we’ll stick to using only these three (equivalent) statements: “P Ô⇒ Q”,

“P implies Q”,

and

“If P , then Q”.

Exercise 20. Let G: “I’m sleeping” and H: “My eyes are shut”. Construct the exact same table as we just did, but for G Ô⇒ H. (Answer on p. 1725)

71 72

See this Washington Post story: “There really are 50 Eskimo words for ‘snow’”. It’s far from obvious, but implies is logically equivalent to only if. And thus, the symbol Ô⇒ can be read aloud not only as implies, but also as only if.

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3.13.

The Four Categorical Propositions and Their Negations

In mathematics and logic, the word some means at least one. Here are the four categorical propositions:73 1. The universal affirmative (UA): 2. The universal negative (UN): 3. The particular affirmative (PA): 4. The particular negative (PN):

“All S are P ” (or “Every S is P ”). “No S is P ” (or “Every S is NOT − P ”). “Some S is P .” “Some S is NOT − P .”

In each, we call S the subject and P the predicate.

Example 38. We give six examples of each categorical proposition:

1. 2. 3. 4. 5. 6.

The Universal Affirmative (UA) “All S are P ” “All Koreans are Asian.” (Or: “Every Korean is Asian.”) “All Germans are European.” (Or: “Every German is European.”) “All animals are dogs.” (Or: “Every animal is a dog.”) “All mammals are bats.” (Or: “Every mammal is a bat.”) “All Koreans eat dogs.” (Or: “Every Korean eats dogs.”) “All Germans eat bats.” (Or: “Every German eats bats.”)

7. 8. 9. 10. 11. 12.

The Universal Negative (UN) “No S is P ” “No Korean is Asian.” (Or: “Every Korean is not Asian.”) “No German is European.” (Or: “Every German is not European.”) “No animal is a dog.” (Or: “Every animal is not a dog.”) “No mammal is a bat.” (Or: “Every mammal is not a bat.”) “No Korean eats dogs.” (Or: “Every Korean does not eat dogs.”) “No German eats bats.” (Or: “Every German does not eat bats.”)

The six subjects used are: “Korean”, “German”, “animal”, “mammal”, “Korean”, and “German”. The six predicates used are: “Asian”, “European”, “a dog”, “a bat”, “eats dogs” (or “a dog-eater”), and “eats bats” (or “a bat-eater”). (Example continues on the next page ...)

73

Also called the A, E, I, and O propositions.

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(... Example continued from the previous page.)

13. 14. 15. 16. 17. 18.

The Particular Affirmative (PA) “All S is P ” “Some Korean is Asian.” (Or: “At least one Korean is Asian.”) “Some German is European.” (Or: “At least one German is European.”) “Some animal is a dog.” (Or: “At least one animal is a dog.”) “Some mammal is a bat.” (Or: “At least one mammal is a bat.”) “Some Korean eats dogs.” (Or: “At least one Korean eats dogs.”) “Some German eats bats.” (Or: “At least one German eats bats.”)

19. 20. 21. 22. 23. 24.

The Particular Negative (PN) “Some S is NOT − P ” “Some Korean is not Asian.” (Or: “At least one Korean is not Asian.”) “Some German is not European.” (Or: “At least one German is not European.”) “Some animals is not a dog.” (Or: “At least one animal is not a dog.”) “Some mammal is not a bat.” (Or: “At least one mammal is not a bat.”) “Some Korean does not eat dogs.” (Or: “At least one Korean does not eat dogs.”) “Some German does not eat bats.” (Or: “At least one German does not eat bats.”)

Exercise 21. For each given pair of subject S and predicate P , write down the corresponding UA, UN, PA, and PN. (Answer on p. 1725.)

(a) (b) (c) (d)

S P Donzer Kiki Donzer Cancer Bachelor Married Bachelor Smoke

Exercise 22. Is each of the two statements below always true? If so, explain why. If not, supply a counterexample. (Answer on p. 1726.) (a) The UA and UN are negations of each other. (b) The PA and UN are negations of each other.

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In the last exercise, we proved that the UA and the UN are not generally negations of each other. Nor are the PA and the UN. So, what then is the correct negation of each categorical proposition? Example 39. Consider this UA: “All Koreans eat dogs.” To prove that this UA is wrong, we need merely show that there is at least one Korean who doesn’t eat dogs. Hence, the negation of this UA is a PN: “Some (i.e. at least one) Korean does not eat dogs”. Example 40. Consider this UN: “No Korean eats dogs.” To prove that this UN is wrong, we need merely show that there is at least one Korean who eats dogs. Hence, the negation of this UN is a PA: “Some (i.e. at least one) Korean eats dogs”. The above examples show that in general, the negation of the UA is the PN, while the negation of the UN is the PA: Statement UA: “All S are P .” UN: “No S is P .”

Negation PA: “Some S is NOT-P .” PN: “Some S is P .”

Example 41. The same examples as before, with their negations:

1. 2. 3. 4. 5. 6.

UA “All S are P ” “All Koreans are Asian.” “All Germans are European.” “All animals are dogs.” “All mammals are bats.” “All Koreans eat dogs.” “All Germans eat bats.”

Negation: PN “Some S is NOT-P ” “Some Korean is not Asian.” “Some German is not European.” “Some animal is not a dog.” “Some mammal is not a bat.” “Some Korean does not eat dogs.” “Some German does not eat bats.”

7. 8. 9. 10. 11. 12.

UN “No S is P ” “No Korean is Asian.” “No German is European.” “No animal is a dog.” “No mammal is a bat.” “No Korean eats dogs.” “No German eats bats.”

Negation: PA “Some S is P ” “Some Korean is Asian.” “Some German is European.” “Some animal is a dog.” “Some mammal is a bat.” “Some Korean eats dogs.” “Some German eats bats.”

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Exercise 23. Is each of the following statements a UA, UN, PA, or PN? Also, write down its negation. (Answer on p. 1726.)

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p)

Statement “All donzers are kiki.” “No donzer is kiki.” “Some donzer is kiki.” “Some donzer is not kiki.” “All bachelors are married.” “No bachelor is married.” “Some bachelor is married.” “Some bachelor is not married.” “All donzers cause cancer.” “No donzer causes cancer.” “Some donzer causes cancer.” “Some donzer does not cause cancer.” “All bachelors smoke.” “No bachelor smokes.” “Some bachelor smokes.” “Some bachelor does not smoke.”

Exercise 24. While trying to excuse the less-than-perfect play of a basketball player, a commentator remarks, (Answer on p. 1726.) “Everybody is not LeBron James.” (a) Rewrite this statement into the form of a categorical proposition. Identify the type of categorical proposition, the subject, and the predicate. (b) Write down this statement’s negation. (c) Is the commentator’s statement is true? If false, what should he have said instead?

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3.14.

Chapter Summary

• The conjunction P AND Q is true if both P and Q are true. • The disjunction P OR Q is true if either P or Q is true. • The negation NOT-P is the statement that’s true if P is false and false if P is true.

• P and Q are equivalent (written P ⇐⇒ Q) if both always have the same truth value. • De Morgan’s Laws: – NOT- (P AND Q) ⇐⇒ (NOT-P OR NOT-Q). – NOT- (P OR Q) ⇐⇒ (NOT-P AND NOT-Q).

• The implication P Ô⇒ Q – – – –

Is defined as NOT-P OR Q. Is equivalent to its contrapositive NOT-P Ô⇒ NOT-Q. Has the negation P AND NOT-Q. Is not generally equivalent to its converse Q Ô⇒ P .

∗ If P Ô⇒ Q is true, then Q Ô⇒ P could be true or false. ∗ If P Ô⇒ Q is false, then Q Ô⇒ P must be true.

• The four categorical propositions and their negations:

Negation Statement UA: “All S are P .” PN: “Some S is NOT − P .” UN: “No S is P .” PA: “Some S is P .” We call S the subject and P the predicate.

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4.

Sets

The set is the basic building block of mathematics. Informally, a set is a “container” or “box” whose contents we call its elements (or members). Example 42. Let A = {1, 3, 5} and B = {100, 200}.

1 3

100

5

The set A

200

The set B

The set A contains three elements—namely, the numbers 1, 3, and 5. Informally, it is a “box” containing the numbers 1, 3, and 5. The set B contains three elements—namely, the numbers 100 and 200. Informally, it is a “box” containing the numbers 100 and 200. Note that when we talk about a set, we refer to both the box and the things inside it. Mathematical punctuation:

• Braces {}—are used to denote the “container”.

• A comma means “and” and is used to separate the elements within a set. Exercise 26. Write down C, the set of the first 7 positive integers.(Answer on p. 1727.) Exercise 27. Write down D, the set of even prime numbers.

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4.1.

The Elements of a Set Can Be Pretty Much Anything

Note that for the A-Levels and hence also in this textbook, the elements of a set will almost always be numbers. However, in general, they needn’t be numbers; they can be pretty much any objects whatsoever:74 Example 43. Let V be the set of the four largest cities in the US. Then V = {New York City, Los Angeles, Chicago, Houston} .

Example 44. Let L be the set of suits in the game of bridge. Then L = {♠, ♡, ♢, ♣} .

Example 45. Let E = {3, π2 , The Clementi Mall, Love, the colour green}.

3

π2

The Clementi Mall

❤ ▮ The set E The set E contains exactly five elements: two numbers (3 and π2 ); a shopping centre (The Clementi Mall); an abstract concept called love (denoted in the figure above by a red heart); and even the colour green (denoted by a green rectangle). Example 46. Let S be the set of Singapore citizens. Then S contains about 3.5M elements,75 three of which are Lee Hsien Loong, Ho Ching, and Chee Soon Juan. Example 47. Let U be the set of United Nations (UN) member states. Then U contains exactly 193 elements,76 three of which are Afghanistan, Singapore, and Zimbabwe. Exercise 28. Write down X, the set of Singapore Prime Ministers (both past and present). (Answer on p. 1727.) Remark 7. We’ll leave the terms set, element, and object undefined.

Actually, there are some restrictions on what can go into a set, but these technicalities are beyond the scope of the A-Levels. 75 According to SingStat, the number of Singapore citizens in 2017 was about 3,439,200. 76 This UN webpage states that the most recent and 193rd state to join the UN was South Sudan in 2011.

74

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A set can even contain other sets: Example 48. Let F be the set that contains the sets A = {1, 3, 5} and B = {100, 200}.

That is, let

The set A

1

F = {A, B} = {{1, 3, 5} , {100, 200}}.

The set B

5 3

1

100

100

5 3

200

The set F

200

The set G

Now consider instead this set: G = {1, 3, 5, 100, 200}.

The sets F and G look very similar. So, is G the same set as F ? Nope. The set F contains exactly two elements, namely the sets A and B. In contrast, G contains exactly five elements, namely the numbers 1, 3, 5, 100, and 200. And so, F and G are not the same. You can think of F as a box that itself contains two boxes—namely, A and B, each of which contain some numbers. In contrast, G is a box that contains no boxes; instead, it simply contains five numbers. The following exercises continue with the above example: Exercise 29. Let H be the set whose elements are F and G.

(a) How many elements does H have? (b) Write down H without using the letters A, B, F , or G. Exercise 30. Let I be the set whose elements are A, B, and G.

(a) How many elements does I have? (b) Write down I without using the letters A, B, F , or G. (c) Compare the sets H (from the previous Exercise) and I. Are they the same?

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4.2. Mathematical punctuation:

In ∈ and Not In ∉

• ∈ means “is in” (or “is an element of”);

• ∉ means “is not in” (or “is not an element of”). Example 49. Let J = {1, 2, 3, 4, 5, 6, 7}. Then 1 ∈ J,

2 ∈ J,

3 ∈ J,

4 ∈ J,

5 ∈ J,

6 ∈ J,

7 ∈ J.

You can read these statements aloud as “1 is in J” (or “1 is an element of J”), “2 is in J” (or “2 is an element of J”), “3 is in J” (or “3 is an element of J”), etc. We can also write 1, 2, 3 ∈ J (read aloud as “1, 2, and 3 are in J” or “1, 2, and 3 are elements of J”). Also, 8 ∉ J, 9 ∉ J, 10 ∉ J, etc. (read aloud as “8 is not in J” or “8 is not an element of J”, “9 is not in J” or “9 is not an element of J”, “10 is not in J” or “10 is not an element of J”, etc.).

We can also write 8, 9, 10 ∉ J (read aloud as “8, 9, and 10 are not in J” or “8, 9, and 10 are not elements of J”). Example 50. Let K = {Cow, Chicken}. Then

• Cow ∈ K (“Cow is in K” or “Cow is an element of K”).

• Chicken ∈ K (“Chicken is in K” or “Chicken is an element of K”). • Cow, Chicken ∈ K (“Cow and Chicken are in K” or “Cow and Chicken are elements of K”). • Cat ∉ K (“Cat is not in K” or “Cat is not an element of K”).

Exercise 31. Fill in the blanks with either ∈ or ∉.

(a) Los Angeles ___ The set of the four largest cities in the US. (b) Tharman ___ The set of Singapore Prime Ministers (past and present).

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4.3.

The Order of the Elements Doesn’t Matter

Definition 8. Two sets A and B are equal if every element that is in A is also in B and every element that is in B is also in A. One implication of the above definition77 is that the order in which we write out the elements of a set does not matter: Example 51. Let A = {2, 4, 6} and B = {6, 2, 4}.

Then A = B because every element that is in A is also in B and every element that is in B is also in A.

2

6 4 6

=

The set A

2 4

The set B

In fact, {2, 4, 6} = {2, 6, 4} = {4, 2, 6} = {4, 6, 2} = {6, 2, 4} = {6, 4, 2}.

Example 52. Let C = {Cow, Chicken} and D = {Chicken, Cow}.

Then C = D because every element that is in C is also in D and every element that is in D is also in C.

🐄

🐔

=

The set C Exercise 32. Are the two given sets equal? (a) {1, 2, 3} and {3, 2, 1}.

77

🐔

🐄

The set D (Answer on p. 1727.) (b) {{1} , 2, 3} and {{3} , 2, 1}.

Actually, in set theory, this is not a definition, but an axiom (known as the Axiom of Extensionality). But here for simplicity, I’ll just call it a definition.

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4.4.

n(S) Is the Number of Elements in the Set S

Let S be a set. Then the number of elements in S 78 is denoted by

Example 53. n ({2, 4, 6}) = 3.

n (s) .

Example 54. n ({Cow, Chicken}) = 2.

Exercise 33. Let X be the set of Singapore Prime Ministers (past and present). Then what is n (X)? (Answer on p. 1727.) Remark 8. Note that most writers denote the number of elements in the set S by ∣S∣.79

But for some reason, your A-Level syllabus (p. 16) instead uses the notation n (S), so that’s what we’ll have to use too.

4.5.

The Ellipsis “. . . ” Means Continue in the Obvious Fashion

More mathematical punctuation—the ellipsis “. . . ” means “continue in the obvious fashion”. Example 55. L is the set of all odd positive integers smaller than 100. So in set notation, we can write L = {1, 3, 5, 7, 9, 11, . . . , 99}.

Example 56. M is the set of all negative integers greater than −100. So in set notation, we can write M = {−99, −98, −97, . . . , −2, −1}.

What is “obvious” to you may not be obvious to your reader. So only use the ellipsis when you’re confident it will be obvious to your reader! And as I did with the sets above, never be shy to write a few more of the set’s elements (doing so costs you nothing except maybe a few more seconds and some ink). Exercise 34. In the above examples, what are n (L) and n (M )? (Answer on p. 1727.)

Exercise 35. Let N be the set of even integers greater than 100 but smaller than 1, 000. Write down N in set notation. (Answer on p. 1727.)

78 79

Or the cardinality of S. Some also use card A. See ISO 80000-2:2009, Item No. 2-5.5.

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4.6.

Repeated Elements Don’t Count

Another implication of Definition 8 is that repeated elements don’t count (they’re simply ignored): Example 57. Let A = {2, 4, 6} and B = {2, 2, 4, 6}.

Then A = B because every element that is in A is also in B and every element that is in B is also in A.

2 4 6

=

The set A

2

2

4

6

The set B

In fact, {2, 4, 6} = {2, 2, 4, 6} = {4, 2, 6, 2, 2, 6, 4, 2}.

Moreover, n ({2, 4, 6}) = n ({2, 2, 4, 6}) = n ({4, 2, 6, 2, 2, 6, 4, 2}) = 3.

Example 58. {Cow, Chicken} = {Cow, Cow, Chicken} = {Chicken, Cow, Chicken}.

🐄

🐔

=

🐄 🐔 🐄

=

🐔

🐄

🐔

Moreover, n ({Cow, Chicken}) = n ({Cow, Cow, Chicken}) = n ({Chicken, Cow, Chicken}) = 2.

Exercise 36. Let W = {Apple, Apple, Apple, Banana, Banana, Apple}. Find n (W ). Also, rewrite W more simply. (Answer on p. 1727.)

Exercise 37. Let C be the set of even prime numbers. Find n (C).(Answer on p. 1727.)

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4.7.

R Is the Set of Real Numbers

Like most students entering college, mathematicians of the midnineteenth century thought they understood real numbers. In fact, the real number line turned out to be much subtler and more complicated than they imagined. — David M. Bressoud (2008). Few mathematical structures have undergone as many revisions or have been presented in as many guises as the real numbers. Every generation re-examines the reals in the light of its values and mathematical objectives. — Faltin, Metropolis, Ross, and Rota (1975). So far, we’ve encountered only finite sets, i.e. sets with finitely many elements. In this and the next two subchapters, we introduce several infinite sets, i.e. sets with infinitely many elements. First, we have the set of real numbers (or simply reals): Definition 9. The set of real numbers is denoted R. Example 59. 16, −1.87, π ≈ 3.14159, and

√ 2 ≈ 1.41421 are all real numbers.

Now, by the way, what exactly is a real number? This sounds like a “dumb” question, but is actually a profound one that was satisfactorily resolved only from the late 19th century. Indeed, this question is a little beyond the scope of the A Levels. And so for the A Levels, we’ll simply pretend—as we did in secondary school—that “everyone knows” what real numbers are (even though, as the quotes above suggest, they actually don’t). We shall not attempt to define or construct the real numbers.

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4.8.

Z Is the Set of Integers

Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk. God made the integers, the rest is the work of man. — Leopold Kronecker (1886).80 Next, Z is the set of integers: Definition 10. The set of integers, denoted Z, is Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . } .

Z is for Zahl, German for number.

Example 60. 16 is an integer, while −1.87, π, and

√ 2 are non-integers.

Note that with an infinite set, we cannot explicitly list out all its elements. And so, when writing out an infinite set, we’ll sometimes find it helpful to use the ellipsis. When writing out Z above, we used two ellipses. The first ellipsis says we continue “leftwards” in the “obvious” fashion, with −4, −5, −6, etc. The second says we continue “rightwards” in the “obvious” fashion, with 4, 5, 6, etc. Exercise 38. H is the set of all prime numbers. With the aid of an ellipsis, write down H in set notation. (Answer on p. 1727.) Definition 11. An even integer is any element of the set {⋅ ⋅ ⋅ − 6, −4, −2, 0, 2, 4, 6, . . . }.

An odd integer is any element of the set {⋅ ⋅ ⋅ − 5, −3, −1, 1, 3, 5, . . . }.

It’s not on your syllabus, but a natural number is simply any positive integer:81 Definition 12. The set of natural numbers, denoted N, is N = {1, 2, 3, . . . } .

According to Heinrich Weber (in his 1893 obituary for Kronecker), this quoted remark was made by Kronecker at an 1886 lecture to the Berliner Naturforscher-Versammlung. 81 There’s actually some debate as to whether N should include 0.

80

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4.9.

Q Is the Set of Rational Numbers

Definition 13. A rational number (or simply rational) is any real number that can be expressed as the ratio of two integers. Any other real number is called an irrational number (or simply irrational). Definition 14. The set of rational numbers is denoted Q. Q is for quotient.82 Example 61. 16 ∈ Q because we can express 16 as the ratio of two integers (e.g. 16/1).

−1.87 ∈ Q because we can express −1.87 as the ratio of two integers (e.g. −187/100).

√ √ Example 62. 2, π ∉ Q. In √ words: “ 2 and π are not elements of the set of rational numbers.” Or more simply: “ 2 and π are irrational.” √ (Note though that this is far from obvious. It takes a little work to prove that 2 is irrational and even more work to prove that π is irrational.)

As you probably already know, a number is rational if and only if it digits eventually recur:83 Example 63. 1/3 = 0.33333 ⋅ ⋅ ⋅ = 0.3 is rational and sure enough, it has the recurring digit 3. We will use the overbar to denote recurring digit(s).84 1/7

= 0.142857142857142857 ⋅ ⋅ ⋅ = 0.142857 is rational and sure enough, it has the recurring digits 142857.

Similarly, 16 = 16.000 ⋅ ⋅ ⋅ = 16.0 and 1.87 = 1.87000 ⋅ ⋅ ⋅ = 1.87 are rational and have the recurring digit 0. (Of course, when the recurring digit is 0, we usually don’t bother writing it.) √ Example 64. 2 ≈ 1.4142135623 . . . and π ≈ 3.1415926535 . . . are irrational. And sure enough, their digits never recur. (But again, this is far from obvious and takes some work to prove.)

Also quoziente in Italian, Quotient in German, and quotient in French. See Proposition 24 (Appendices). 84 ˙ But we shan’t. Some writers also use an overdot instead, like so: 1/3 = 0.33333 ⋅ ⋅ ⋅ = 0.3.

82

83

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4.10.

A Taxonomy of Numbers

Below is a taxonomy of the types of numbers you’ll encounter in this textbook. √ We’ll study complex and imaginary numbers only later on in Part IV (spoiler: i = −1 is an imaginary number and the imaginary unit). Real numbers are either rational or irrational. In turn, rational numbers are either integers or non-integers.

Complex numbers C

Reals R

Rationals Q

Integers Z

Imaginary numbers

Irrationals

Non-integers

Three miscellaneous remarks: 1. R, Z, and Q are infinite sets (i.e. they contain infinitely many elements).

We may thus write: n (R) = n (Z) = n (Q) = ∞. 2. When handwritten, R is an R with an extra vertical line on the left; Z is a Z with an extra diagonal line; and Q is a Q with an extra vertical line:

3. Infinity (∞) and negative infinity (−∞) are NOT numbers.85

As with real numbers, we shall not attempt to formally define what ∞ or −∞ is. Instead, we shall simply and informally say that ∞ is the “thing” that is greater than every real number; while −∞ is the “thing” that is smaller than every real number. In case that wasn’t clear, let me repeat:

INFINITY IS NOT A NUMBER. Remark 9. This textbook uses blackboard bold font and writes R, Z, and Q. Note though that some other writers instead use bold font and write R, Z, and Q. Your A-Level syllabus and exams use only the former, so that’s what we’ll do too. 85

Actually, some writers sometimes find it convenient to call ∞ and −∞ numbers—for example, some call ∞ and −∞ extended real numbers. But in this textbook, I’ll keep it simple and insist that neither ∞ nor −∞ is a number.

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4.11.

More Notation: + , − , and

0

To create a new set that contains only the positive elements of the old set, append a superscript plus sign (+ ) to the name of a set: 1. Z+ = {1, 2, 3, . . . } is the set of all positive integers.

2. Q+ is the set of all positive rational numbers. 3. R+ is the set of all positive real numbers.

To create a new set that contains only the negative elements of the old set, append a superscript minus sign (− ) to the name of a set: 1. Z− = {−1, −2, −3, . . . } is the set of all negative integers. 2. Q− is the set of all negative rational numbers.

3. R− is the set of all negative real numbers.

(As we’ll learn later, there is no such thing as a positive or negative complex number. Hence, there are no sets denoted C+ or C− .) To add the number 0 to a set, append a subscript zero (0 ) to its name:

1. Z+0 = {0, 1, 2, 3, . . . } is the set of all non-negative integers.

2. Z−0 = {0, −1, −2, −3, . . . } is the set of all non-positive integers. 3. Q+0 is the set of all non-negative rational numbers.

4. Q−0 is the set of all non-positive rational numbers. 5. R+0 is the set of all non-negative real numbers.

6. R−0 is the set of all non-positive real numbers.

Exercise 39. Which of the sets introduced above are finite?

Remark 10. The three pieces of notation introduced on this page (+ , − , and 0 ) aren’t terribly important or widely used. I give them a quick mention only because they’re listed on p. 16 of your A-Level syllabus.

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The Empty Set ∅

4.12.

A set can contain any number of elements. Indeed, it can even contain zero elements: Definition 15. The empty set {} is the set that contains zero elements (and is often also denoted ∅).

Informally, the empty set {} = ∅ is the “container” with nothing inside. Hence the name.

The empty set {} or ∅.

Example 65. In 2016, the set of all Singapore Ministers who are younger than 30 is {} or ∅. This means there is no Singapore Minister who is younger than 30.

Example 66. The set of all even prime numbers greater than 2 is {} or ∅. This means there is no even prime number that is greater than 2. Example 67. The set of numbers that are greater than 4 and smaller than 4 is {} or ∅. This means there is no number that is simultaneously greater than 4 and smaller than 4. Example 68. The set {∅} is not the same as the set ∅.

{∅} is a set containing a single element, namely the empty set.

Informally, {∅} is a box containing an empty box—it is not empty. In contrast, ∅ is the empty set.

Informally, it is simply an empty box. We can also rewrite the two sets as {∅} = {{}}

and

∅ = {}.

Now it is perhaps clearer that {{}} ≠ {}.

Example 69. The set {∅, 3, {∅}} is the set containing exactly three elements, namely the empty set, the number 3, and a set containing the empty set. Take care to note that this set does not contain four elements.

{∅} = {{}}

∅ = {}

3

{∅, 3, {∅}}

Exercise 40. Tricky: Let S = {{{}} , ∅, {∅} , {}}. What is n (S)? (Answer on p. 1727.) 47, Contents

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4.13.

Intervals

Using left parenthesis (, right parenthesis ), left bracket [, and right bracket ], we can write intervals of real numbers: Example 70. Here are nine intervals:

A = (1, 3)

1

3

B = [1, 3] C = (1, 3] D = [1, 3) E = (1, ∞)

(a) The (b) The (c) The (d) The (e) The (f) The (g) The (h) The (i) The

F = [1, ∞) G = (−∞, −1) H = (−∞, −1] R = (−∞, ∞)

interval A = (1, 3) is the set of real numbers that are > 1 and < 3. interval B = [1, 3] “ ≥ 1 and ≤ 3. interval C = (1, 3] “ > 1 and ≤ 3. interval D = [1, 3) “ ≥ 1 and < 3. interval E = (1, ∞) “ > 1. interval F = [1, ∞) “ ≥ 1. interval G = (−∞, 1) “ < 1. interval H = (−∞, 1] “ ≤ 1. interval I = (−∞, ∞) denotes the set of all real numbers. That is, (−∞, ∞) = R.

Given below are the left and right endpoints of the above nine intervals: A B C D E F G H I Left endpoint 1 1 1 1 1 1 −∞ −∞ −∞ Right endpoint 3 3 3 3 ∞ ∞ 1 1 ∞

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Definition 16. Let a and b be real numbers with b ≥ a. An interval is any one of these nine sets:

(a) The (b) The (c) The (d) The (e) The (f) The (g) The (h) The (i) The

interval (a, b) is the set of real numbers that are > a and < b. interval [a, b] “ ≥ a and ≤ b. interval (a, b] “ > a and ≤ b. interval [a, b) “ ≥ a and < b. interval (a, ∞) “ > a. interval [a, ∞) “ ≥ a. interval (−∞, a) “ < a. interval (−∞, a] “ ≤ a. interval (−∞, ∞) is the set of all real numbers. That is, (−∞, ∞) = R.

Given below are the left and right endpoints of the above nine intervals:

a b c d e f g h i Left endpoint a a a a a a −∞ −∞ −∞ Right endpoint b b b b ∞ ∞ a a ∞

By the way, we call

• (a, b) an open interval. • [a, b] a closed interval.

• (a, b] a half-closed interval, a half-open interval, a left-half-open interval, or a right-half-closed interval.

• [a, b) a half-closed interval, a half-open interval, a left-half-closed interval, or a right-half-open interval. Exercise 41. Let A = [1, 1], B = (1, 1), C = (1, 1], D = [1, 1), and E = (1, 1.01). Find the number of elements in each of these five sets. Express each set in another way. It turns out that each of the sets A, B, C, and D is called a degenerate interval, while E is called a non-degenerate interval. Can you write down some possible definitions of the two terms degenerate interval and non-degenerate interval? (Answer on p. 1728.)

Exercise 42. Express R, R+ , R+0 , R− , and R−0 in interval notation. (Answer on p. 1728.)

Remark 11. Some writers (usually French-speakers) use reverse bracket notation: They write ]a, b[, ]a, b], or [a, b[ instead of (a, b), (a, b], or [a, b). We will not do so in this textbook.86

86

One good argument in favour of reverse bracket notation is that it avoids confusing the open interval (a, b) with the ordered pair (a, b) (we’ll learn more about ordered pairs in Ch. 6). However, by and large, the reverse bracket notation remains uncommon, except in continental Europe and especially France (where it was introduced by the Bourbaki group).

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4.14.

Subset Of ⊆

Definition 17. Let A and B be sets. If every element of A is an element of B, then we call A a subset of B and write A ⊆ B.

Otherwise, we say that A is not a subset of B and write A ⊈/ B.

Example 71. Let M = {1, 2}, N = {1, 2, 3}, and O = {1, 2, 4, 5}, and P = {3, 2, 1}. Then M is a subset of N , O, and P N is a subset of P , but not of M or O O is not a subset of M , N , or P P is a subset of N , but not of M or O

M ⊆ N , M ⊆ O, and M ⊆ P N ⊆ P , but N ⊈ M and N ⊈ O O ⊈ M , O ⊈ N , and O ⊈ P P ⊆ N , but P ⊈ M and P ⊈ O

Observe that N ⊆ P and P ⊆ N . Indeed, the sets N = P . We have the following fact: Fact 8. Two sets are equal ⇐⇒ They are subsets of each other.

Proof. Two sets A and B are equal

⇐⇒ Every element in A is in B and every element in B is in A (Definition 8) ⇐⇒ A is a subset of B and B is a subset of A (Definition 17). Exercise 43. Consider Z, Q, and R.

(a) Is Z a subset of Q or R? (b) Is Q a subset of Z or R?

(c) Is R a subset of Z or Q?

Exercise 44. Explain whether this statement is true:

“The set of current Singapore Prime Minister(s) is a subset of the set of current Singapore Minister(s).” Exercise 45. Let A and B be sets. Explain whether each of the following statements is generally true. (If false, give a counterexample.) (Answer on p. 1728.) (a) A ⊆ B Ô⇒ A = B. (b) B ⊆ A Ô⇒ A = B. (c) A = B Ô⇒ A ⊆ B.

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(d) A = B Ô⇒ B ⊆ A. (e) A = B ⇐⇒ A ⊆ B. (f) A = B ⇐⇒ B ⊆ A.

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4.15.

Proper Subset Of ⊂

Definition 18. Let A and B be sets. If A ⊆ B but A ≠ B, then we call A a proper subset of B and write A ⊂ B.

Otherwise, we say that A is and write A ⊂/ B.

Example 72. Let M = {1, 2}, N = {1, 2, 3}, and O = {1, 2, 4, 5}, and P = {3, 2, 1}. Then M is a proper subset of N , O, and P N is not a proper subset of M , O, or P O is not a proper subset of M , N , or P P is not a proper subset of M , N , or O

M ⊂ N , M ⊂ O, and M ⊂ P N ⊂/ M , N ⊂/ O, N ⊂/ P O ⊂/ M , O ⊂/ N , O ⊂/ P P ⊂/ M , P ⊂/ N , P ⊂/ O

Observe that N = P so that by Definition 18, N ⊂/ P and P ⊂/ N .

Remark 12. The A-Level syllabus (p. 16) uses the symbol ⊆ to mean “subset of” and ⊂ to mean “proper subset of”. So this textbook will do the same.

However, confusingly enough, some writers use the symbol ⊂ to mean “subset of” and ⊊ to mean “proper subset of”. We will not follow such practice in this textbook. This is just so you know, in case you get confused when reading other mathematical texts! Exercise 46. Let S be the set of all squares and R be the set of all rectangles. Is S ⊂ R? (Answer on p. 1729.) Exercise 47. Does A ⊆ B imply that A ⊂ B?

Exercise 48. Does A ⊂ B imply that A ⊆ B?

Exercise 49. Explain whether this statement is true: “If A is a subset of B, then A is either a proper subset of or is equal to B.” (Answer on p. 1729.)

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Remark 13. If A is a (proper) subset of B, then we are very much tempted to say that A is “(strictly) smaller than” B—or equivalently, that B is “(strictly) larger than” A. Informally and intuitively, this is perhaps permissible. However, we shall avoid speaking of one set as being “(strictly) smaller or larger than” (or “the same size as”) another. The reason is the “(proper) subset of” relation is merely one way by which we can compare the “sizes” of sets. There are at least two other ways, called cardinality and measure.

Example: Suppose A = [0, 1] and B = [0, 2]. Then A is a proper subset of B, so that in this sense, we may consider A to be “strictly smaller than” B. However, it turns out that • The cardinality of A is the same as that of B. So, in the sense of cardinality, A is “the same size as” B.

• The measure of A is strictly smaller than that of B. So, in the sense of measure, A is “strictly smaller than” B. Happily, for A-Level Maths, you needn’t know anything about the concepts of cardinality and measure. This remark is merely to explain to you why we avoid speaking of one set being “(strictly) smaller or larger than” (or “the same size as”) another. Example 73. Again, let M = {1, 2} and N = {1, 2, 3}, so that M is a (proper) subset of N. While tempting, we shall avoid saying that M is (strictly) smaller than N .

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4.16.

Union ∪

Definition 19. Let A and B be sets. The union of A and B, denoted A ∪ B, is the set of elements that are in A or B. The set A

Tip: “U” for Union.

The set B

A ∪ B is the yellow region.

Example 74. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then

T ∪ U = {1, 2, 3, 4}, T ∪ V = {1, 2, 3}, U ∪ V = {1, 2, 3, 4}, T ∪ U ∪ V = {1, 2, 3, 4}.

Exercise 50. Rewrite each set more simply: (a) [1, 2] ∪ [2, 3]

(b) (−∞, −3) ∪ [−16, 7)

(c) {0} ∪ Z+

Exercise 51. What is the union of each pair of sets?(Answer on p. 1729.) (a) The set of squares S and the set of rectangles R. (b) The set of rationals and the set of irrationals.

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4.17.

Intersection ∩

The set A

The set B

A ∩ B is the yellow region.

Definition 20. Let A and B be sets. The intersection of A and B, denoted A ∩ B, is the set of elements that are in A and B. Two sets intersect if their intersection contains at least one element. Equivalently, two sets A and B intersect if their intersection is non-empty, i.e. A ∩ B ≠ ∅.

Definition 21. Two sets are mutually exclusive or disjoint if they do not intersect (i.e. their intersection is empty). Example 75. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then T ∩ U = ∅,

T ∩ V = {1, 2},

U ∩ V = {3},

T ∩ U ∩ V = ∅.

The intersection of T and U is empty. Hence, T and U are mutually exclusive or disjoint. The intersection of the three sets T , U , and V is also empty. However, the three sets T , U , and V are not mutually exclusive or disjoint: Definition 22. Three or more sets are mutually exclusive or disjoint if every two of them are mutually exclusive or disjoint.

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Example 76. Let T = {1, 2}, U = {3, 4}, and W = {5, 6}. Then T ∩ U ∩ W = ∅. 1

Note though that = does not suffice for declaring that T , U , and W are mutually exclusive. 1

Instead and as per Definition 22, T , U , and W are mutually exclusive or disjoint because every two of them are mutually exclusive or disjoint: T ∩ U = ∅,

T ∩ W = ∅,

Exercise 52. Rewrite each set more simply: (a) (4, 7] ∩ (6, 9)

U ∩ W = ∅.

(b) [1, 2] ∩ [5, 6]

(c) (−∞, −3) ∩ [−16, 7)

Exercise 53. What is the intersection of each pair of sets?(Answer on p. 1729.) (a) The set of squares S and the set of rectangles R. (b) The set of rationals and the set of irrationals.

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Set Minus ∖

4.18.

The set minus (sometimes also called set difference) sign ∖ is very convenient. Sadly, it does not appear in the A-Level syllabus. Nonetheless, it’s worth a quick mention and I’ll sometimes use it in this textbook. Definition 23. Suppose A and B are sets. Then A set minus B, denoted A ∖ B, is the set of elements that are in A and not in B The set A

The set B

A ∖ B is the yellow region.

Example 77. Let T = {1, 2}, U = {3, 4}, and V = {1, 2, 3}. Then T ∖ U = T,

T ∖ V = ∅,

U ∖ V = {4}.

Exercise 54. In the above example, what are V ∖ T and V ∖ U ? (Answer on p. 1729.)

Examples of why the set minus notation is sometimes very convenient and helps us avoid ugly monstrosities: Example 78. Consider the set of all real numbers except 1.

Without the set minus sign, we’d write this as (−∞, 1) ∪ (1, ∞).

With the set minus sign, we can more simply write R ∖ {1}.

Example 79. Consider the set of all real numbers that aren’t integers.

Without the set minus sign, we’d write this as ⋅ ⋅ ⋅ ∪ (−3, −2) ∪ (−2, −1) ∪ (−1, 0) ∪ (0, 1) ∪ (1, 2) ∪ (2, 3) . . . .

With the set minus sign, we can more simply write R ∖ Z. 56, Contents

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4.19.

The Universal Set E

The universal set, denoted E (that’s a squiggly E), is the set of “all” elements.87 Note though that what we mean by “all” elements depends on the context: Example 80. In the context of a roll of a six-sided die, the universal set might be the set of all possible outcomes:

E = {1, 2, 3, 4, 5, 6} .

Example 81. In the context of a spin of a European-style roulette wheel, the universal set might be is the set of all possible outcomes:

E = {0, 1, 2, 3, . . . , 36} .

Note that in American-style roulette, there is a 38th possible outcome—double zero 00. And so in the American context, the universal set might instead be:

E = {00, 0, 1, 2, 3, . . . , 36} .

(It looks like Marina Bay Sands does American-style roulette.) Example 82. In the context of a game of chess, the universal set might be the set of all possible outcomes for White:

E = {Win, Lose, Draw} .

Remark 14. I give the universal set notation E a quick mention here only because it appears on your A-Level syllabus (p. 16). We will rarely (if ever) make use of this piece of notation in this textbook. Exercise 55. For each context, write down the universal set and its number of elements. (a) 4D. (b) Sex (as on a Singapore NRIC). (c) Singapore’s official language(s). (d) Singapore’s national language(s).

87

This is not a precise definition. Indeed, there can be no single precise definition of the universal set since what this is depends on the context. Note also that the “set of all objects” leads to a contradiction (see e.g. Russell’s paradox) and thus does not exist.

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4.20.

The Set Complement A′

Definition 24. Let A be a set. The set complement of A, denoted A′ , is the set of all elements that are not in A.

The set A

Equivalently: A′ = E ∖ A. That is, A′ is the universal set E minus those elements in A. In the figure (right), A′ is the yellow region.

To repeat, what the universal set E is (i.e. what we mean by “all” elements) depends on the context: Example 83. Let A = {2, 3}. If the relevant context is the roll of a die, then

E = {1, 2, 3, 4, 5, 6}

A′ = {1, 2, 3, 4, 5, 6} ∖ A = {1, 4, 5, 6}.

and

But if instead the relevant context is a European-style roulette wheel spin, then

E = {0, 1, 2, . . . , 36}

and

A′ = {0, 1, 2, . . . , 36} ∖ A = {0, 1, 4, 5, 6, . . . , 36}.

And if instead the relevant context is the positive integers smaller than 10, then

E = {1, 2, . . . , 9}

and

A′ = {1, 2, . . . , 9} ∖ A = {1, 4, 5, 6, 7, 8, 9}.

Example 84. Let B = {2, 4, 6, . . . }. If the relevant context is the positive integers, then

E = Z+

B ′ = Z+ ∖ {2, 4, 6, 8, . . . } = {1, 3, 5, 7, . . . }.

and

That is, B ′ is simply the set of positive odd numbers.

But if instead the relevant context is the set of all integers, then

E =Z

and

B ′ = Z ∖ {2, 4, 6, 8, . . . } = Z−0 ∪ {1, 3, 5, 7, . . . }.

That is, B ′ is the set of positive odd numbers and all non-positive integers.

Example 85. Let C = R+ = (0, ∞). If the relevant context is all real numbers, then

E =R

and

C ′ = R ∖ C = R−0 .

That is, C ′ is simply the set of non-positive reals.

But if instead the relevant context is the set of reals greater than or equal to −1, then

E = [−1, ∞)

and

C ′ = [−1∞) ∖ C = [−1, 0].

That is, C ′ is the set of reals between −1 and 0 (inclusive).

Remark 15. Just so you know, instead of A′ , some writers write Ac or A.

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Exercise 56. For each given context and set S, write down S ′ .

(a) Context: 4D; S = {3000, 3001, 3002, . . . , 3999}. (b) Context: Temperature (in degrees Celsius); S = (30, ∞).

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4.21.

De Morgan’s Laws

It turns out there’s a deep connection between logic and set theory. In particular, The intersection ∩ corresponds to the logical connective AND (the conjunction). The union ∪ corresponds to the logical connective OR (the disjunction).

Earlier, for logic, we had De Morgan’s Laws (Facts 1 and 2):

(a) Fact 1: The negation of the conjunction P AND Q is the disjunction NOT-P OR NOT-Q. (b) The negation of the disjunction P OR Q is the conjunction NOT-P AND NOT-Q. Correspondingly, for set theory, we also have these De Morgan’s Laws:

(a) The complement of the intersection P ∩ Q is the union P ′ ∪ Q′ .

(b) The complement of the intersection P ∩ Q is the union P ′ ∪ Q′ .

The set P

The set Q

The set P

The set Q

(P ∩ Q) = P ′ ∪ Q′ is in yellow.

(P ∪ Q) = P ′ ∩ Q′ is in yellow.

For future reference, let’s jot this down as a formal result: Fact 9. Suppose P and Q are sets. Then (a) (P ∩ Q) ′ = P ′ ∪ Q′

Proof. See p. 1529 (Appendices).

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and

(b) (P ∪ Q) ′ = P ′ ∩ Q′ .

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Example 86. Suppose that every animal is (i) either tall or short; and (ii) either fat or lean. So for example, the horse is tall and lean, the elephant is tall and fat, the pig is short and fat, and the mouse is short and lean. Let T be the set of tall animals and F be the set of fat animals.

Consider T ∩ F , the set of animals that are tall AND fat. Its complement is the set of animals that are short OR lean (yellow region below): (T ∩ F ) ′ = T ′ ∪ F ′ .

The set T

The set F

🐎 🐘 🐖 🐁 Consider T ∪ F , the set of animals that are tall OR fat. Its complement is the set of animals that are short AND lean (yellow region below): (T ∪ F ) ′ = T ′ ∩ F ′ .

The set T

The set F

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4.22.

Set-Builder Notation (or Set Comprehension)

Previously, we simply wrote out a set using the method of set enumeration. That is, we simply enumerated (i.e. listed out) all their elements: Example 87. The set of Singapore PMs (both past and present) is S = {Lee Kuan Yew, Goh Chok Tong, Lee Hsien Loong} .

Where a set has too many elements to list out, we can use the ellipsis “. . . ”. But this too counts as the method of set enumeration: Example 88. The set of integers greater than 100 is T = {101, 102, 103, 104, . . . } .

We now introduce a second method of writing out a set, called set-builder notation or set comprehension: Example 89. The set of Singapore PMs (both past and present) is S = {x ∶ x has ever been the Singapore PM} .

In set-builder notation, the mathematical punctuation mark colon “∶” means such that. Following the colon is the property or criterion that x must satisfy in order to be an element of the set. Hence, S is “the set of all objects x such that x has ever been the Singapore PM”. Note that the letter or symbol x is simply a dummy or placeholder variable. We could have replaced x with any other letter or indeed any other symbol and the set would have been unchanged. For example, we could’ve replaced x with the letter y: S = {y ∶ y has ever been the Singapore PM} .

Or even a smiley face ,:

S = {, ∶ , has ever been the Singapore PM} .

Remark 16. A dummy variable is also called a placeholder, dead, or bound variable.

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Example 90. Let T = {x ∶ x ∈ Z, x > 100}. (Recall that the comma “,” means AND.) Then T is the set of all objects x such that x is an integer AND x > 100.

Observe that this time, following the colon are two properties or criteria that x must satisfy in order to be an element of the set. We could also have written T as “the set of all integers x such that x > 100”: T = {x ∈ Z ∶ x > 100}.

Remark 17. In maths (as in natural language), there are often many ways of saying the same thing. Here we’ve written the set of Singapore PMs (past and present) in four different ways and the set of integers greater than 100 in three different ways. So, if there are many ways to say the same thing, then how should we say it? Well, in maths (as in natural language), you should always strive to express yourself as clearly and simply as possible. This will require using your judgment, experience, and wisdom. Example 91. Let S be the set of ASEAN members. Using set enumeration, we’d write S = {Brunei, Cambodia, Indonesia, Laos, Malaysia, Myanmar, the Philippines, Singapore, Thailand, Vietnam}.

Using set-builder notation or set comprehension, we’d write S = {x ∶ x is a member of ASEAN} .

This is “the set of all objects x such that x is a member of ASEAN”. Example 92. Let Familee be the set of individuals who have ever been members of Singapore’s Royal or First Family. Consider A = {x ∶ x has ever been the Singapore PM, x ∉ Familee} .

In words, A is “the set of all objects x such that x has ever been the Singapore PM AND x has never been a member of the Familee”. And so, A = {Goh Chok Tong} .

Example 93. Let B = {x ∶ x is a member of ASEAN, x has fewer than 5, 000 islands}.

In words, B is “the set of all objects x such that x is a member of ASEAN AND x has fewer than 5, 000 islands”. So, using set enumeration, we’d write B = {Brunei, Cambodia, Laos, Malaysia, Myanmar, Singapore, Thailand, Vietnam} .

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Exercise 57. Rewrite the sets S and T using set enumeration.

S = {x ∶ x is a child of Lee Kuan Yew} , T = {x ∶ x is a child of Lee Kuan Yew, x has never been the Singapore PM} .

Exercise 58. What is the following set?

{x ∶ x has ever been the Singapore PM} ∩ Familee.

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We’ve used set-builder notation to describe sets of prime ministers or countries. But more commonly, we’ll use it to describe sets of numbers: Example 94. Let A = {x ∶ x2 − 1 = 0}.

In words, A is “the set of all objects x such that x2 − 1 = 0”. A = {−1, 1}.

Hence,

Example 95. Let B = {x ∶ x2 − 1 = 0, x > 0}.

In words, B is “the set of all objects x such that x2 − 1 = 0 and x > 0”.

Hence,

We could also have written,

B = {1}.

B = {x ∈ R+ ∶ x2 − 1 = 0}.

This says that B is “the set of all positive real numbers x such that x2 − 1 = 0”.

Example 96. Consider C = {x ∈ Z ∶ 3.5 < x < 5.5}.

In words, C is “the set of all integers x such that 3.5 < x < 5.5. C = {4, 5}.

Hence,

Example 97. The set of positive reals may be written R+ = (0, ∞) = {x ∈ R ∶ x > 0} .

In words, this is “the set of all reals x such that x is greater than 0”. Similarly, the set of non-negative reals may be written as R+0 = [0, ∞) = {x ∈ R ∶ x ≥ 0} .

In words, this is “the set of all reals x such that x is greater than or equal to 0”. Similarly, we may write Q+ = {x ∈ Q ∶ x > 0}, Q+0 = {x ∈ Q ∶ x ≥ 0},

Z+ = {x ∈ Z ∶ x > 0}, Z+0 = {x ∈ Z ∶ x ≥ 0}.

Exercise 59. Write out, in words, what Q+ , Q+0 , Z+ , and Z+0 are. (Answer on p. 1729.)

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Example 98. Let S be the set of positive even numbers. Then we may write S = {2, 4, 6, 8, . . . } = {x ∶ x = 2k, k ∈ Z+ } .

In words, S is “the set of all objects x such that x equals 2k AND k is a positive integer”. Notice that here we’ve introduce a second dummy variable k to help us describe the set. Again, both symbols x and k could’ve been replaced by any other symbols and we’d still have the same set. For example, S = {2, 4, 6, 8, . . . } = {p ∶ p = 2q, q ∈ Z+ } = {⋆ ∶ ⋆ = 2⧫, ⧫ ∈ Z+ } = {, ∶ , = 2∎, ∎ ∈ Z+ } .

(Of course, letters like x, k, p, and q are customary and hence preferable to weird symbols like shapes and faces.) Here’s another way to write down S: S = {2, 4, 6, 8, . . . } = {x ∶ x/2 ∈ Z+ } .

In words, S is “the set of all elements x such that x divided by 2 is a positive integer”. And here’s another: S = {2, 4, 6, 8, . . . } = {2k ∶ k ∈ Z+ } .

In words, S is “the set of all elements 2k such that k is a positive integer”. Exercise 60. Let a ≤ b. Rewrite each set in set-builder notation. (Answer on p. 1730.) (a) (e) (i) (l) (n) (o) (p) (q) (r)

R− (b) Q− (c) − − Q0 (f) Z0 (g) (a, b] (j) [a, b) (k) (−∞, 1] ∪ (2, 3) ∪ (3, ∞) (m) The set of negative even numbers The set of positive odd numbers The set of negative odd numbers. {π, 4π, 7π, 10π, . . . } {−3π, π, 4π, 7π, 10π, . . . }.

Z− (d) R−0 (a, b) (h) [a, b] (−∞, −3) ∪ (5, ∞) (−∞, 3) ∩ (0, 7)

Remark 18. Following the A-Level syllabus (p. 16), in set-builder notation, we use the colon “∶” to mean such that. Note though that some writers use the pipe “∣” instead.

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4.23.

Chapter Summary

• Informally, a set is a container or box whose objects we call its elements. Whenever we talk about a set, we refer to both the container and the objects inside. • ∈ means in and ∉ means not in.

• The ellipsis “. . . ” means continue in the obvious fashion. • The order of the elements doesn’t matter and repeated elements don’t count: {1, 2, 3} = {3, 3, 2, 1, 3, 2, 1, 1, 1, 1, 2, 3} .

• R, Z, and Q are the sets of reals, integers, and rationals.88 • R+ , R− , R+0 , and R−0 are the sets of positive reals, negative reals, non-negative reals, and non-positive reals. Similarly, we have Q+ , Q− , Q+0 , and Q−0 ;

Z+ , Z− , Z+0 , and Z−0 .

and

• The empty set {} = ∅ is the set with zero elements. • Interval notation. Let a < b. Then 1. 2. 3. 4.

(a, b) is the set of reals between a (excluded) and b (excluded). [a, b] is the set of reals between a (included) and b (included). (a, b] is the set of reals between a (excluded) and b (included). [a, b) is the set of reals between a (included) and b (included).

• A is a subset of B—denoted A ⊆ B—if every element in A is also in B. • A is a proper subset of B—denoted A ⊂ B—if A ⊆ B and A ≠ B.

• A ∪ B, the union of A and B, is the set of elements that are in A OR B. • A ∩ B, the intersection of A and B, is the set of elements that are in A AND B.

• A ∖ B (or A − B), A set minus B, is the set of elements that are in A AND not in B. • The universal set E is the set of “all” elements (where “all” depends on the context).

• The set complement A′ is the set of elements that are not in A. • De Morgan’s Laws: (P ∩ Q) ′ = P ′ ∪ Q′ and (P ∪ Q) ′ = P ′ ∩ Q′ .

• Set-builder notation (or set comprehension). The set of objects satisfying the property P is denoted {x ∶ P (x)}. Exercise 61. Write each set more simply. (a) R ∖ R+ . (c) [1, 6] ∖ ((3, 5) ∩ (1, 4)). (e) {2, 5, 8, 11, . . . } ∩ {2, 4, 6, 8, . . . }. 88

(b) R ∖ (Q ∪ Z). (d) {1, 5, 9, 13, . . . } ∩ {2, 4, 6, 8, . . . }. (f) (0, 5] ∩ ([1, 8] ∩ [5, 9)) ′ .

Not on your syllabus: N is the set of natural numbers.

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5. 5.1.

O-Level Review

Some Mathematical Vocabulary

Example 99. Consider the equation 1 + 4 = 2 + 3.

LHS expression RHS expression ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ 1 + 4 = 2 + 3 ↓ ↓ ↓ ↓ Term Term Term Term

The expression on the equation’s left-hand-side (LHS) is 1 + 4 and contains the two terms 1 and 4.

The expression on the right-hand-side (RHS) is 2 + 3 and contains the two terms 2 and 3.

Informally, the difference between an equation and an expression is this:

An equation contains an equals sign; an expression does not.

Variable

Example 100. Consider the equation y = 5x + 6.

Constant term

y = 5x + 6 ↙ ↘ Coefficient Variable

This equation contains two variables: x and y.

The coefficient on x is 5. The coefficient on y is 1. The constant term or more simply constant is 6. (The constant term is simply any real number89 that does not involve any variables.) Example 101. Consider the inequality 8x2 + 4x + 3 > y − 2.

Constant term ↗ ↖ 8x2 + 4x + 3 > y − 2 ↙ ↘ ↙ ↓ Coefficient Variable

This inequality contains two variables: x and y.

The coefficients on x2 and x are 8 and 4. The coefficient on y is 1.

The constant terms on the LHS is 3. The constant terms on the RHS −2.

89

When we study complex numbers in Part IV, constants will also include complex numbers.

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5.2.

The Absolute Value or Modulus Function

Definition 25. The absolute value (or modulus) function, denoted ∣⋅∣, is defined for all x ∈ R by ⎧ ⎪ ⎪ ⎪x, ∣x∣ = ⎨ ⎪ ⎪ ⎪ ⎩−x,

Example 102. ∣5∣ = 5, ∣−5∣ = 5, ∣0∣ = 0.

for x ≥ 0,

for x < 0.

Each of the following three equations is not generally true. a ∣a∣ ∣ ∣= . 7 b b

a a ∣ ∣= . 7 b ∣b∣

a a ∣ ∣= . 7 b b

Example 103. ∣

6 ∣6∣ 6 ∣6∣ 6 ∣≠ because ∣ ∣ = ∣3∣ = 3 but = = −3. −2 −2 −2 −2 −2

Example 105. ∣

−6 −6 −6 −6 ∣≠ because ∣ ∣ = ∣3∣ = 3 but = −3. 2 2 2 2

Example 104. ∣

−6 −6 −6 −6 −6 ∣≠ because ∣ ∣ = ∣3∣ = 3 but = = −3. 2 ∣2∣ 2 ∣2∣ 2

What’s true is this:

a ∣a∣ Fact 10. If a, b ∈ R with b ≠ 0, then ∣ ∣ = . b ∣b∣

a 0 ∣0∣ ∣a∣ Proof. If a = 0, then ∣ ∣ = ∣0∣ = 0 = = = . b ∣b∣ ∣b∣ ∣b∣ Now suppose a ≠ 0. Case 1.

Case 2.

a and b have the same signs. Then either ∣a∣ / ∣b∣ = a/b or ∣a∣ / ∣b∣ = (−a) / (−b) = a/b. Moreover, a/b > 0 so that ∣a/b∣ = a/b. a and b have opposite signs.

Then either ∣a∣ / ∣b∣ = (−a) /b or ∣a∣ / ∣b∣ = a/ (−b). Moreover, a/b < 0 so that ∣a/b∣ = −a/b.

If the above proof doesn’t convince you, hopefully the following examples do: 6 ∣6∣ 6 ∣6∣ 6 Example 106. ∣ ∣ = because ∣ ∣ = ∣3∣ = 3 and = = 3. 2 ∣2∣ 2 ∣2∣ 2 69, Contents

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Example 107. ∣

−6 ∣−6∣ −6 ∣−6∣ 6 ∣= because ∣ ∣ = ∣−3∣ = 3 and = = 3. 2 ∣2∣ 2 ∣2∣ 2

Example 109. ∣

−6 ∣−6∣ −6 ∣−6∣ 6 ∣= because ∣ ∣ = ∣3∣ = 3 and = = 3. −2 ∣−2∣ −2 ∣−2∣ 2

Example 108. ∣

6 ∣6∣ 6 ∣6∣ 6 ∣= because ∣ ∣ = ∣−3∣ = 3 and = = 3. −2 ∣−2∣ −2 ∣−2∣ 2

We also have this similar result:

Fact 11. If a, b ∈ R, then ∣ab∣ = ∣a∣ ∣b∣.

Proof. If a, b ≥ 0, then ∣ab∣ = ab = ∣a∣ ∣b∣. 3 If a, b < 0, then ∣ab∣ = ab = ∣a∣ ∣b∣. 3 If a ≥ 0 and b < 0, then ∣ab∣ = −ab = ∣a∣ ∣b∣. 3 If a < 0 and b ≥ 0, then ∣ab∣ = −ab = ∣a∣ ∣b∣. 3 Example 110. ∣6 × 2∣ = ∣6∣ ∣2∣ because ∣6 × 2∣ = ∣12∣ = 12 and ∣6∣ ∣2∣ = 6 × 2 = 12.

Example 111. ∣−6 × 2∣ = ∣−6∣ ∣2∣ because ∣−6 × 2∣ = ∣−12∣ = 12 and ∣−6∣ ∣2∣ = 6 × 2 = 12.

Example 112. ∣6 × (−2)∣ = ∣6∣ ∣−2∣ because ∣6 × (−2)∣ = ∣−12∣ = 12 and ∣6∣ ∣−2∣ = 6 × 2 = 12.

Example 113. ∣(−6) (−2)∣ = ∣−6∣ ∣−2∣ because ∣(−6) (−2)∣ = ∣12∣ = 12 and ∣−6∣ ∣−2∣ = 6 × 2 = 12.

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5.3.

The Factorial n!

Definition 26. Let n ∈ Z+0 . Then n-factorial, denoted n!, is the number defined by

Or equivalently,90 And so,

⎧ ⎪ ⎪ ⎪1, n! = ⎨ ⎪ ⎪ ⎪ ⎩1 × 2 × ⋅ ⋅ ⋅ × n, ⎧ ⎪ ⎪ ⎪1, n! = ⎨ ⎪ ⎪ ⎪ ⎩(n − 1)! × n, 0! 1! 2! 3! 4! 5! 6! ⋮

= = = = = = =

1, 0! × 1 1! × 2 2! × 3 3! × 4 4! × 5 5! × 6 ⋮

= = = = = =

for n = 0, for n > 0.

for n = 0,

for n > 0.

1, 1 × 2, 1 × 2 × 3, 1 × 2 × 3 × 4, 1 × 2 × 3 × 4 × 5, 1 × 2 × 3 × ⋅ ⋅ ⋅ × 6, ⋮

You may be wondering, “Why is 0! defined as 1?” It turns out this is the definition that causes us the least overall inconvenience.91

90 91

This latter equivalent definition is an example of a recursive definition. For example, with the Maclaurin series (Ch. 98) and combinatorics (Part VI).

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5.4.

Exponents

Definition 27. Suppose b ≠ 0 and x ∈ Z. Then b to the power of x, denoted bx , is the number defined by ⎧ ⎪ ⎪ b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b, ⎪ ⎪ ⎪ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ⎪ ⎪ x times ⎪ ⎪ ⎪ ⎪ bx = ⎨1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ , ⎪ ⎪ ⎩ b∣x∣

for x > 0, for x = 0,

for x < 0.

Given the number bx , we call b its base and x its exponent. Example 114. 21 = 2, 22 = 2 ⋅ 2 = 4, 23 = 2 ⋅ 2 ⋅ 2 = 8, 24 = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16,

220 = 2 ⋅ 2 ⋅ ⋅ ⋅ ⋅ ⋅ 2 = 1 048 576. ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ 20 times

Example 115. 2−1 =

2−4 =

1 1 1 1 1 1 1 1 = 0.5, 2−2 = ⋅ = = 0.25, 2−3 = ⋅ ⋅ = = 0.125, 2 2 2 4 2 2 2 8

1 1 1 1 1 ⋅ ⋅ ⋅ = = 0.0625, 2 2 2 2 16 2−20 =

1 1 1 1 ⋅ ⋅ ⋅⋅⋅ ⋅ = = 0.000 000 953 674 316 406 25. 2 2 2 1 048 576 ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ 20 times

1 1 1 1 2 1 1 1 1 3 1 1 1 1 1 4 1 1 1 1 1 Example 116. ( ) = , ( ) = ⋅ = , ( ) = ⋅ ⋅ = , ( ) = ⋅ ⋅ ⋅ = , 2 2 2 2 2 4 2 2 2 2 8 2 2 2 2 2 16 20 1 1 1 1 1 ( ) = ⋅ ⋅ ⋅⋅⋅ ⋅ = . 2 2 2 2 1 048 576 ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ 20 times

1 −1 1 −2 1 −3 1 −4 Example 117. ( ) = 2, ( ) = 2 ⋅ 2 = 4, ( ) = 2 ⋅ 2 ⋅ 2 = 8, ( ) = 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16, 2 2 2 2 −20 1 ( ) = 2 ⋅ 2 ⋅ ⋅ ⋅ ⋅ ⋅ 2 = 1 048 576. ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ 2 20 times

1 0 Example 118. 2 = 1, ( ) = 1. 2 0

Note well that Definition 27 covers only the case where the base b is non-zero and the exponent x is an integer. We now cover also the case where the base b is 0 and the exponent x is any real number: 72, Contents

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Definition 28. Suppose x ∈ R. Then 0 to the power of x, denoted 0x , is the number defined by ⎧ ⎪ ⎪ 0, ⎪ ⎪ ⎪ ⎪ 0x = ⎨1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩Undefined,

Example 119. 03 = 0, 0π = 0, 0100 = 0.

for x > 0, for x = 0,

for x < 0.

0−3 , 0−π , and 0−100 are undefined.

00 = 1. There’s actually nothing “obvious” or “natural” about defining 00 = 1. Similar to 0! = 1, this is merely the definition that will cause us the least inconvenience.92

Next, we’ll define roots. But to do so, we’ll need to make use of this result:

Theorem 1. Suppose b > 0 and x is a non-zero integer. Then there exists a unique positive real number a such that ax = b.

Proof. Omitted.93

Example 120. Let b = 18.73 and x = 39. Then Theorem 1 says that there exists a > 0 such that a39 = 18.73.

Definition 29. Let b > 0 and x be a non-zero integer. By Theorem 1, there exists a unique positive real number a such that ax = b. √ x We call a the xth root of b and write a = b.

We’ll also call a b to the power of 1/x and write a = b1/x .

Remark 19. To be clear, thing.

√ x

b and b1/x are simply two different ways to write the exact

One convenience this definition affords is this: For all b ∈ R, we simply have b0 = 1. Some writers argue that we should simply leave 00 undefined. But in my judgment, 00 = 1 is probably the definition that will cause us the least inconvenience in H2 Maths. 93 See e.g. Rudin (Principles of Mathematical Analysis, 1976, p. 10, Theorem 1.21). Note that our Theorem 1 also allows for the possibility that the exponent x is a negative integer. 92

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Example 121. By Definition 29, √ √ 2 4 = 4 = 4 = 2, 83 = 1

16 4 = 1

Also, And,

√ 3 8 = 2,

√ 4 16 = 2,

√ 1 √ 1 2 = 0.25 = 0.25 = 0.5 = , 4 2 √ 1 1 1 3 1 √ 1 3 3 ( ) = 0.125 3 = = 0.125 = 0.5 = , 8 8 2 √ 1 1 1 4 1 1 √ 3 4 = 0.0625 = 0.5 = . ( ) = 0.0625 4 = 16 8 2 1

1 1 2 ( ) = 0.25 2 = 4

1 2

1 048 576 20 = 1

2

√ 1 048 576 = 2.

20

√ 1 20 1 1 1 20 ( ) = 0.000 000 953 674 316 406 25 20 = 1 048 576 1 048 576 √ 1 20 = 0.000 000 953 674 316 406 25 = 0.5 = . 2

Note the requirement in Definition 29 that b > 0. If b < 0, then the definition does not apply: Example 122. There is no positive number (or indeed any real number) a such that

So, for now, we’ll leave −11/2 =

a2 = −1.

√ −1 undefined.

(Spoiler: Later on in Part IV (Complex Numbers), we will define i = −11/2 =

Note also the requirement in Definition 29 that x ≠ 0. If x = 0, then undefined. 1

−1.)

√ 1 0 b or b 0 is simply

Separately, given b x , in some special cases, we have these customary notation and terminology: √ 1 • If x = 1, we will not write a = b. Instead, we will simply write a = b. √ √ 2 • If x = 2, we will not write a = b. Instead, we will simply write a = b. Moreover, while it’s OK to call a the second root of b, we will more commonly call it the square root of b. √ 3 • While it’s OK to call a = b the third root of b, we will more commonly call it the cube root of b. We next extend our definition of bx to the case where b is positive and x is any rational number:

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Definition 30. Let b > 0 and x ∈ Q. Suppose x = m/n for some integers m and n ≠ 0. Then b to the power of x is denoted bx and is the number defined as bx = (b n ) . 1

m

√ m 1 m m n bx = b n = (b n ) = ( b) .

So,

Example 123. By Definition 30, 21.5 = 2 2 3

213.83 = 2 100

1 383

22.3 = 2 3 7

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√ 3 3 = ( 2) = (1.414 . . . ) = 2.828 . . . , √ 1383 100 1383 = ( 2) = (1.007 . . . ) = 14 562.798 . . . , √ 7 3 7 = (1.259 . . . ) = 5.039 . . . . = ( 2)

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Proposition 1. (Laws of Exponents) Suppose a, b > 0 and x, y ∈ R. Then (a) b b = b x y

x+y

−x

(b) b

.

1 = x. b

bx (c) y = bx−y . b

(d) (bx ) = bxy . y

(e) (ab) = ax bx . x

Remark 20. Note well the requirement in Proposition 1 that the base b be positive. Otherwise we can “prove” that −1 = 1: 2× 12 ⋆

1 2

= [(−1) ] = 1 2 = 1.

−1 = −1 = (−1) 1

2

1

At =, we applied Proposition 1(d). But this is an error because the base −1 is not positive, so that Proposition 1 cannot be applied.

Our partial proof below covers only the case where x, y ∈ Z+ (exponents are positive integers). (This partial proof can easily be extended to cover also the case where x and/or y are non-positive integers.) For a more complete proof covering also the case where x and y are any real numbers, see p. 1700 (Appendices). ⋆

Proof (partial). Suppose x, y ∈ Z+ . Below, = will denote the use of Definition 27. ⋆

(a) bx by = b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b × b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b = b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b = bx+y . ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ y times

x times

x+y times

(b) Since x ≥ 0, we have ∣−x∣ = x and hence b−x = (c) If x ≥ y so that x − y ≥ 0, then bx−y

1

b∣−x∣

1 . bx

=

y times

x times

y times

y times

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b ⋆ b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b bx ⋆ = b ⋅ b ⋅ ⋅⋅⋅ ⋅ b = b ⋅ b ⋅ ⋅⋅⋅ ⋅ b × = = . ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b by x−y times x−y times ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶

If instead x < y so that x − y < 0 and ∣x − y∣ = y − x, then ⋆

bx−y =

1

b∣x−y∣

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ 1 1 b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b bx = = × = = . b ⋅ b ⋅ ⋅⋅⋅ ⋅ b b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b by ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ∣x−y∣=y−x times

y−x times

x times

x times

x times

y times

y times

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ⋆ x y ⋆ y ⋆ (d) (b ) = (b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b) = (b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b) × ⋅ ⋅ ⋅ × (b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b) = b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b = bxy . ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹xy times x ⋆

x times

x times

x times ⋆

(e) (ab) = (ab) (ab) . . . (ab) = a ⋅ a ⋅ ⋅ ⋅ ⋅ ⋅ a × b ⋅ b ⋅ ⋅ ⋅ ⋅ ⋅ b = ax bx . ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹x¹ ¹ ¹ ¹ ¹ ¹times x times

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Exercise 62. Let x ∈ R. Simplify each expression. (a)

54x ⋅ 251−x . 52x+1 + 3 ⋅ 25x + 17 ⋅ 52x

√ 8x+2 − 34 ⋅ 23x 2 √ 2x+1 . 8

Exercise 63. Let b > 0 and x, y ∈ Z+ . Explain whether each equation is generally true. (a) b(x ) = bxy . y

(b) (bx ) = bxy . y

Remark 21. We have defined bx to cover several cases. However, we still have not defined bx in these two cases: 1. The base b is positive and the exponent x is irrational. 2. The base b is negative and the exponent x is a non-integer. Unfortunately, defining these is somewhat beyond the scope of H2 Maths. Yet at the same time, in H2 Maths, we will often deal with the above two cases of bx , which is a little discomforting. One comfort is that with Case 1, Proposition 1 (the Laws of Exponents) still holds. With Case 2, those Laws of Exponents don’t always hold and we should be a little more careful. Nonetheless, for H2 Maths, there is probably little danger in simply assuming that these Laws always hold.

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5.5.

The Square Root Refers to the Positive Square Root

By Definition 29,

The square root refers to the positive square root. Example 124. It is true that the equation b2 = 25 has two solutions, namely, b=5

and

b = −5.

However, each of the following four statements is false: 25 2 = ±5. 7 1

By Definition 29, 251/2 or

25 2 = −5. 7 1

√ 25 = ±5. 7

√ 25 is a positive number. So, 251/2 =

√ 25 = −5. 7

√ 25 = 5.

3

To talk about the negative square root −5, (simply) stick a minus sign in front: √ −251/2 = − 25 = −5

3

In general, given any b > 0, the square root of b is the positive number denoted √ b > 0.

The negative square root of b is the negative number denoted with a minus sign in front, like this: √ − b < 0.

This seems like a trivial and “obvious” point. You may find it strange that I have devoted this subchapter to this point. But failure to appreciate this point is a common source of error: √ Exercise 64. True or false: “ x2 = x for all x ∈ R.” (Answer on p. 1731.) √ √ x2 1 x2 2 Exercise 65. True or false: “For all x ∈ R ∖ {0}, = √ = 1.” (Answer on p. 1731.) x x2 √ x 1 x2 2 Exercise 66. True or false: “For all x ∈ R ∖ {0}, √ = √ = 1.” (Answer on p. 1731.) x2 x2

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√ Remark 22. Because of the “ambiguity” just discussed, some writers refer to 25 as the principal square root of the number √ 25. More generally, given any non-negative number x, they call the positive number x the principal square root of x.

We shall however have no use for the term principal in this textbook. The reason is that there is in √ positive √ fact no ambiguity at all—we have in this textbook clearly defined the number 25 to be the square root of 25. There is thus no need to refer to 25 as the principal square root of 25.

Remark 23. Definition 29 is to some extent an arbitrary convention, similar to whether we drive on the left or right side. Singapore drives on the left side of the road (because Singapore was part of the British Empire). But Singapore could just as easily drive on the right side of the road (which is what nearly all of the world except the former British Empire does). Except it doesn’t and it would be quite unwise to try to go against the accepted convention. Here similarly, Definition 29 says that the square root (of a positive number b) is itself a positive number. But mathematicians could just as easily have chosen the opposite convention. Except they didn’t and it would be quite unwise to try to go against the accepted convention. It’s not really important which convention a community chooses. What’s important is that everyone in that community knows what the convention is and agrees to use it, hence reducing confusion and accidents (whether in maths or on the road).

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In Exercise 64, we gave this false statement: √

x2 = x for all x ∈ R.

7

The correct statement should instead be this: Fact 12.

√ x2 = ∣x∣ for all x ∈ R.

Proof. If x ≥ 0, then

√ x2 = x = ∣x∣.

Here’s another result we’ll find useful: Given any x ≠ 0, the expression ∣x∣ x

gives us its sign. That is, ⎧ ⎪ x x ⎪ ⎪1, Fact 13. = =⎨ ∣x∣ ∣x∣ ⎪ ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

or

x ∣x∣

.

Proof. If x > 0, then x/ ∣x∣ = x/x = 1.

If x < 0, then x/ ∣x∣ = x/ (−x) = −1.

Example 125. In Exercise 65, we gave this false statement:

√ √ x2 x2 = √ = 1. For all x ∈ R ∖ {0}, x x2

7

Using the above two facts, the correct statement should instead be this: √ ⎧ ⎪ x2 ∣x∣ ⎪ ⎪1, For all x ∈ R ∖ {0}, = =⎨ x x ⎪ ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

3

Example 126. Similarly, in Exercise 66, we gave this false statement: √ x x2 For all x ∈ R ∖ {0}, √ = √ = 1. x2 x2

7

Using the above two facts, the correct statement should instead be this:

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⎧ ⎪ x ⎪ x ⎪1, For all x ∈ R ∖ {0}, √ = =⎨ x2 ∣x∣ ⎪ ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

3

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5.6.

Rationalising the Denominator with a Surd

The term surd usually refers to a square root √ √ 1 TOT 1 2 2 Example 127. √ = √ × √ = . 2 2 2 2

√ ⋅.

At = , we used what I’ll call the Times One Trick (TOT). It is of course always legitimate to multiply any quantity by 1, though one might wonder why we’d ever want to do that. The above is one example of when the TOT comes in handy. TOT

Definition 31. The conjugate of a + b is a − b. We call a + b and a − b a conjugate pair.

Given a denominator with a surd, we can often rationalise it by using the TOT, the conjugate, and the identity (a + b) (a − b) = a2 − b2 : √ √ √ √ 1 TOT 1 1− 2 1− 2 1− 2 1− 2 √ √ = √ × √ = Example 128. = = 2 − 1. = √ 1−2 −1 1+ 2 1 + 2 1 − 2 12 − ( 2)2

We take this opportunity to talk about the ± and ∓ notation:

√ √ √ 2− 3 2− 3 1 TOT 1 √ × √ = 2 Example 129. We have √ = = 2 − 3. 2+ 3 2+ 3 2− 3 2 − 3 √ √ √ 1 TOT 1 2+ 3 2+ 3 √ = √ × √ = 2 = 2 + 3. And, 2− 3 2− 3 2+ 3 2 − 3

%

Using plus-minus ± and minus-plus ∓ notation, we can combine the two statements ⋆ and % into this single statement: √ √ √ 1 1 2∓ 3 2∓ 3 √ = √ √ = 2 = 2∓ 3. 2± 3 2± 3 2∓ 3 2 − 3

The minus-plus notation ∓ indicates the “opposite” of ±. So here, • If ± is +, then ∓ is − and we have ⋆. And

• If ± is −, then ∓ is + and we have %.

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Remark 24. We only ever use ∓ after we’ve already used ± and have a need to indicate that the signs go the opposite way. So for example, we’d never write x2 = 17

because we can simply write x2 = 17

Ô⇒

Ô⇒

√ x = ∓ 17,

√ x = ± 17.

Following this last line, if we subsequently have a need to discuss −x, then that’s when we might use ∓: √ −x = ∓ 17.

Exercise 67. Let x, y ∈ R, with y ≠ 0. Prove the following. √ 1 x x2 =− ± + 1. √ 2 2 x x y y ± +1 y

y2

Exercise 68. Let a, b, c ∈ R with a ≠ 0 and b2 − 4ac > 0. Prove the following.94 √ −b ± b2 − 4ac 2c √ = . (Answer on p. 1733.) 2a −b ∓ b2 − 4ac

94

The LHS is the quadratic formula in familiar form. The RHS is also the quadratic formula, but in an alternative and less familiar form.

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5.7.

Logarithms

Informally, logarithms are simply the inverse of exponents. A bit more formally, Definition 32. Let x ∈ R and b, n > 0 with b ≠ 1. If bx = n, then we call x the base-b logarithm of n and write Example 130. 23 = 8 ⇐⇒ 3 = log2 8.

x = logb n.

Example 131. 34 = 81 ⇐⇒ 4 = log3 81.

Example 132. 45 = 1 024 ⇐⇒ 5 = log4 1 024.

You should find these Laws of Logarithms familiar: Proposition 2. (Laws of Logarithms) Let x ∈ R and a, b, c > 0 with b ≠ 1. Then (a)

(b) (c) (d) (e) (f)

logb 1 = 0 logb b = 1

logb bx = x blogb a = a

c logb a = logb ac logb

1 = − logb a a

(Logarithm of Reciprocal)

(g) logb (ac) = logb a + logb c a (h) logb = logb a − logb c c logc a (i) logb a = , for c ≠ 1 logc b (j)

logab c =

1 loga c, b ⋆

(Sum of Logarithms) (Difference of Logarithms) (Change of Base)

for a ≠ 1

Proof. In this proof, Ô⇒ denotes the use of Definition 32. (a)

(b) (c) 83, Contents

b0 = 1 ⇐⇒ logb 1 = 0. ⋆

b1 = b ⇐⇒ logb b = 1. ⋆

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Let y = logb a ⇐⇒ by = a ⇐⇒ blogb a = a.

(d)

(e) First use Proposition 1(d):

c (d)9758 MATHEMATICS ⋆ GCE ADVANCED LEVEL H2 SYLLABUS (2019)

bc logb a = (blogb a ) = ac ⇐⇒ c logb a = logb ac . 4. Functions1 logb a f

(f)

f(x) (g) First use Proposition 1(a):

b

logb a+logb c

= logb a−1 = − logb a. (e)

the function f

the value of the function f at x

f: A →B (d) =f:bxlogbya blogb c = –1

ac

f is a function under which each element of set A has an image in ⋆ ⇐⇒ logb (ac) = log the function f maps the element x btoc. the element y b a + log

the inverse of the function f a f 1 (g) 1 (f) 1 logb = log ⋅ ) = logb athe + log = function logb a −of log b (a b b g .which is defined by o f, gf g composite f and c c c c (g o f)(x) or gf(x) = g(f(x))

(h)

(i) Let y = logb a or by = a. Let z = logc a or cz = a. lim f(x)

the limit of f(x) as x tends to a

∆x ; δx

an increment of x

Hence, b = c or z = logc b = y logc b or logc a = logb a logc b or logb a = y

(k)

z

y

x→ a

dy

log d x ab c =

(i)

dn y

logc a . logc b

loga c (c) the loga c 1 of y with respect to x = loga c. = derivative b loga a b b

the nth derivative of y with respect to x

dx expression equals 2. Exercise 69. Show that each (Answer on p. 1733.) (n) f'(x), f''(x), …, f (x) nth derivatives of f(x) with respect to x the first, second, √ 1 4 (b) log3 45 − log9 25 (c) log16 768 − log2 3 (a) log2 32 + log3 27 indefinite integral of y with respect to x ∫ y dx n

Remark 25. Some (including TI84) writethelog x tointegral meanofthe of dx definite y withbase-10 respect to xlogarithm for values of x between a ∫a yyour x. Still others write log x to mean the natural logarithm of x. (Note that at this point, x& , &x& , … the first, second, derivatives x withwill respect we still haven’t discussed what the natural logarithm function is.of We do tosotime only in Ch. 25.) b

5. Exponential and Logarithmic Functions

We have thus a rather confused situation with different writers using different notation. e

base of natural logarithms

e , exp x

exponential function of x

log a x

logarithm to the base a of x

ln x

natural logarithm of x

lg x

logarithm of x to base 10

In this textbook, we’ll stickx strictly to this notation on your H2 Maths syllabus (p. 18):

We will only ever write logb x and even then fairly rarely. We will never write log x. 6. Circular Functions and Relations sin, cos, tan, cosec, sec, cot sin–1, cos–1, tan–1 cosec–1, sec–1, cot–1

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} the circular functions } the inverse circular functions

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5.8.

Polynomials

Definition 33. Let c0 , c1 , . . . , cn be constants, with cn ≠ 0. We call the following expression an nth-degree polynomial (in one variable x): We also call

cn xn + cn−1 xn−1 + cn−2 xn−2 + ⋅ ⋅ ⋅ + c1 x + c0 .

• Each ci xi the ith-degree term (or more simply the ith term) • Each ci the ith coefficient on xi (or the ith-degree coefficient, or the ith coefficient) • The 0th coefficient c0 the constant term (or, more simply, the constant) • The following equation a (nth-degree) polynomial equation (in one variable x): cn xn + cn−1 xn−1 + cn−2 xn−2 + ⋅ ⋅ ⋅ + c1 x + c0 = 0.

As usual, in the above definition, x is merely a dummy variable that can be replaced with any other symbol, like y, z, ,, or ☀. Example 133. The expression 7x − 3 is a 1st-degree or linear polynomial.

The equation 7x − 3 = 0 is a 1st-degree polynomial (or linear) equation.

1st-degree term 0th-degree term ↖ ↗ ¬« 7 x −3 ¯ ↙ ↘ 1st-degree coefficient 0th-degree coefficient Term Coefficients 0th-degree −3 −3 1st-degree 7x 7

The 1st-degree term is 7x. The 1st-degree coefficient is 7. The 0th-degree term, the constant term, the constant, and the 0th-degree coefficient is −3.

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Example 134. The expression 3x2 + 4x − 5 is a 2nd-degree or quadratic polynomial. The equation 3x2 + 4x − 5 = 0 is a 2nd-degree polynomial (or quadratic) equation.

1st-degree term ↑ 2nd-degree term ↖ ↗ 0th-degree term ¬ ¬« 2 3 x + 4 x −5 ↙ ¯ ↓ 2nd-degree coefficient ↘ 0th-degree coefficient 1st-degree coefficient Term Coefficient 0th-degree −5 −5 1st-degree 4x 4 2nd-degree 3x2 3

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Example 135. The expression −5x3 + 2x + 9 is a 3rd-degree or cubic polynomial. The equation −5x3 + 2x + 9 = 0 is a 3rd-degree polynomial (or cubic) equation.

When a particular coefficient is 0, we usually don’t bother writing out that term (as is the case here with the 2nd-degree term).

2nd-degree term 3rd-degree term ↑ ↗ 1st-degree term ¬ « ª 3rd-degree coefficient ← −5x3 + 0x2 + 2x + 9 ® ↙ ↓ 0th-degree term & coefficient 2nd-degree coefficient 1st-degree coefficient 0th-degree 1st-degree 2nd-degree 3rd-degree

Term Coefficient 9 9 2x 2 2 0x 0 2 −5x −5

You get the idea. We also have 4th-, 5th-, 6th-, . . . degree (or quartic, quintic, sextic, . . . ) polynomials and equations. Example 136. Technically, the expression 7 could be regarded as a 0th-degree polynomial, because 7 = 7x0 . Indeed, in certain contexts, it may be convenient to refer to the expression 7 as such. But more commonly, we’ll simply call the expression 7 a constant.

In this textbook, we’ll almost always consider only polynomials in one variable. So unless otherwise stated, when we say polynomial, we’ll always mean a polynomial in one variable.95

95

But in case you were wondering, an example of a polynomial in two variables is Ax + Bxy + Cy. In general, the degree of each term in a polynomial is the sum of the exponents on the variables. And the polynomial’s degree is simply the highest such degree. So here, the term Ax has degree 1, Bxy has degree 2, and Cy has degree 1. So this polynomial is of degree 2. Another example of a polynomial in two variables is Ax2 + Bxy + Cy 2 + Dx + Ey + F . Despite looking more complicated than the previous example, this polynomial also has degree 2 because the greatest sum of exponents on any term is again 2. And by the way, as Ch. 138.19 (Appendices) discusses, the conic section is, in general, described by this 2nd-degree polynomial equation in two variables:

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

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Part I. Functions and Graphs

Revision in progress (May 2020). And hence messy at the moment. Appy polly loggies for any inconvenience caused.

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[A] large part of mathematics which became useful developed with absolutely no desire to be useful, and in a situation where nobody could possibly know in what area it would become useful; and there were no general indications that it ever would be so. By and large it is uniformly true in mathematics that there is a time lapse between a mathematical discovery and the moment when it is useful; and that this lapse of time can be anything from thirty to a hundred years, in some cases even more ... This is true for all of science. Successes were largely due to forgetting completely about what one ultimately wanted, or whether one wanted anything ultimately; in refusing to investigate things which profit, and in relying solely on guidance by criteria of intellectual elegance; it was by following this rule that one actually got ahead in the long run, much better than any strictly utilitarian course would have permitted. — John von Neumann (1954).

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6. 6.1.

Graphs Ordered Pairs

Recall that with sets, the order of the elements doesn’t matter: Example 137. {Cow, Chicken} = {Chicken, Cow}. Example 138. {−5, 4} = {4, −5}.

We now introduce a new mathematical object called an ordered pair (a, b). Like the sets {Cow, Chicken} and {−5, 4}, you can think of an ordered pair as a container with two objects. But unlike sets, with ordered pairs, the order matters (hence the name).96

Definition 34. Given the ordered pair (a, b), we call a its first or x-coordinate and b its second or y-coordinate.

Two ordered pairs are equal if and only if both their x- and y-coordinates are equal.

Fact 14. Suppose a, b, x, and y are objects; and (a, b) and (x, y) are ordered pairs. Then (a, b) = (x, y)

Proof. See p. 1530 in the Appendices.

96

⇐⇒

a = x AND b = y.

For the formal definition of an ordered pair, see p. 1530 in the Appendices.

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Example 139. Consider these two ordered pairs: (Cow, Chicken)

and

(Chicken, Cow).

The ordered pair (Cow, Chicken) has x-coordinate Cow and y-coordinate Chicken. The ordered pair (Chicken, Cow) has x-coordinate Chicken and y-coordinate Cow.

Since these two ordered pairs have different x- and y-coordinates, they are not equal: (Cow, Chicken) ≠ (Chicken, Cow) ,

This is in contrast to what we saw above with sets, where:

{Cow, Chicken} = {Chicken, Cow} .

To distinguish an ordered pair from a set with two elements, we use parentheses (instead of braces). Be very clear that the ordered pair (a, b) is a completely different mathematical object from the set {a, b}: and,

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(Cow, Chicken) ≠ {Cow, Chicken};

(Chicken, Cow) ≠ {Chicken, Cow}.

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Example 140. Consider these two ordered pairs: (-5,4)

and

(4,-5).

The ordered pair (−5, 4) has x-coordinate −5 and y-coordinate 4.

In contrast, the ordered pair (4, −5) has x-coordinate 4 and y-coordinate −5.

Since these two ordered pairs have different x- and y-coordinates, they are not equal: (−5, 4) ≠ (4, −5)

This is in contrast to what we saw above with sets, where {−5, 4} = {4, −5}.

Again, be very clear that the ordered pair (a, b) and the set {a, b} are two completely different mathematical objects: (−5, 4) ≠ {−5, 4}

and

(4, −5) ≠ {4, −5}.

It is possible that in an ordered pair (a, b), we have a = b:

Example 141. Here are four ordered pairs that are distinct (or not equal): (Cow, Chicken),

(Chicken, Cow),

(Cow, Cow), and (Chicken, Chicken).

The ordered pair (Cow, Cow) has x-coordinate Cow and y-coordinate Cow.

The ordered pair (Chicken, Chicken) has x-coordinate Chicken and y-coordinate Chicken.

Example 142. Here are four distinct ordered pairs: (1,-5),

(-5,1),

(1,1), and (-5,-5).

The ordered pair (1, 1) has x-coordinate 1 and y-coordinate 1.

The ordered pair (−5, −5) has x-coordinate −5 and y-coordinate −5.

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Remark 26. Confusingly, (−5, 4) can denote two entirely different things:

• In Ch. 4.13, we learnt that (−5, 4) is an interval—in particular, it denotes the set of real numbers between −5 and 4.

• Here we learn that (−5, 4) denotes the ordered pair with the x-coordinate −5 and the y-coordinate 4.

This is an unfortunate and confusing situation. But don’t worry.

In the Oxford English Dictionary, the word run has 645 different meanings.97 But with a little experience, one rarely has trouble telling from the context which of these is meant when someone says uses that word.

Likewise, you’ll rarely have trouble telling from the context whether by (−5, 4), the writer means a set of real numbers or an ordered pair.

97

According to the NYT, the OED editor Peter Gilliver spent nine months working on the word run. Previously, the word set was said to have the most different meanings, at 430 (see e.g. Guinness Book of World Records).

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6.2.

The Cartesian Plane

Rather than ordered pairs of cows and chickens, we’ll usually be concerned with ordered pairs of real numbers: Definition 35. We call an ordered pair of real numbers a point. Example 143. A = (−5, 4), B = (1, 1), and C = (2, −3) are points D = (Cow, Chicken) and E = (3, Chicken) are not.

Definition 36. The cartesian plane is the set of all points: {(x, y) ∶ x, y ∈ R} . Fun Fact

The cartesian plane98 is named after René Descartes (1596–1650), who’s also the same dude who came up with “Cogito ergo sum” (“I think, therefore I am”). Definition 37. The origin is the point O = (0, 0).

98

There’s some disagreement over whether to capitalise cartesian here—see e.g. . Indeed, in your H2 Maths syllabus, it is used to be capitalised on p. 19 but not on pp. 7–8! (My guess is that while pp. 1–15 were written by the local Singapore authorities, pp. 16–20 were simply copy-pasted from some standard Cambridge notation template.) [2020 update: They’ve now corrected this in the latest 2021 syllabus. Now cartesian on p. 19 is also not capitalised.] My personal preference is to capitalise cartesian, but it seems that the A-Level exams do not do so. I shall therefore follow the sacred A-Level exams by not capitalising cartesian.

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Example 144. Depicted below is the cartesian plane, centred on the origin O = (0, 0) (x-coordinate 0 and y-coordinate 0). Note that the cartesian plane stretches out to ±∞ in both the x- and y-directions. Also depicted are three points:

• A = (−5, 4)—x-coordinate −5 and y-coordinate 4 • B = (1, 1)—x-coordinate 1 and y-coordinate 1

• C = (2, −3)—x-coordinate 2 and y-coordinate −3

y

A = (−5, 4)

4 2

-6

-4

O = (0, 0)

-2

-2 -4 -6

B = (1, 1)

2

x

4

C = (2, −3)

When depicting the cartesian plane, it is customary (and helpful) to draw the x-axis (or horizontal axis) and y-axis (or vertical axis). The point at which these two axes intersect is the origin O = (0, 0).

Remark 27. For now, we’ll be concerned only with the cartesian plane, which is a twodimensional space.99 And so, for now, whenever we say point, it should be clear that we’re talking about an ordered pair of real numbers. Note though that more generally, it is also to talk points in a one-dimensional space and a three-dimensional space: • A point (in one-dimensional space) is simply any real number. • As we’ll learn in Part III (Vectors), a point in (three-dimensional space) is any ordered triple of real numbers. By the way and just so you know, a point (in any-dimensional space) is itself a zerodimensional object.

99

Formally, we may define a two-dimensional space to be any subset of the cartesian plane R2 . Similarly, a one-dimensional space is any subset of the real number line R and a three-dimensional space is any subset of R3 . See Definition 267 (Appendices).

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6.3.

A Graph is Any Set of Points

You’re probably used to thinking of a graph (or a curve) as a “drawing”. But formally, Definition 38. A graph (or curve) is any set of points. And so equivalently, a graph is any subset of the cartesian plane. Example 145. The set G = {(−5, 4) , (1, 1) , (2, −3)} contains three points. And so by definition, G is also a graph.

We’ve defined graph as a noun (it is a set of points). But at the slight risk of confusion, we’ll also use graph as the verb meaning to draw a graph. So, we can either say, “The graph G is drawn below,” or, “G is graphed below”. y

(−5, 4)

-5

The graph G = {(−5, 4) , (1, 1) , (2, −3)} contains three points. -4

-3

-2

3 (1, 1)

1 -1

x

-1

1

-3

2

(2, −3)

-5

Example 146. The set H = {(−5, 4) , (2, −3)} contains two points. And so by definition, H is also a graph. y

(−5, 4)

3

The graph H = {(−5, 4) , (2, −3)} contains two points. -5

-4

-3

-2

1 -1

-1 -3

x 1

2

(2, −3)

-5

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Example 147. The set I = {(1, 1)} contains one point. And so by definition, I is also a graph. y

3

The graph I = {(1, 1)} contains one point. -5

-4

-3

-2

1 -1

-1

(1, 1) x 1

2

-3 -5

If a set contains at least one element that isn’t a point, then it isn’t a graph: Example 148. Consider J = {(−5, 4) , Love}.

The set J contains two elements—the point (−5, 4) and the abstract concept called Love. Since J contains at least one element that isn’t a point (i.e. an ordered pair of real numbers), J is not a graph. Example 149. Consider K = {(−5, 4) , (1, 1) , 1}.

The set K contains three elements—the points (−5, 4) and (1, 1), and the number 1. Since K contains at least one element that isn’t a point (i.e. an ordered pair of real numbers), K is not a graph.100 Example 150. The set R contains at least one element that isn’t a point (for example, 5). Indeed, R contains no points at all and infinitely many elements that are not points. Thus, R is not a graph. (The same is true of Q and Z.) You may be used to thinking of a graph as a “drawing”. But you should now think of a graph as being simply a set of points. A “drawing” of a graph is not the graph itself, but merely a visual aid.101 Remark 28. Your A-Level exams seem to use the terms graph and curve interchangeably (i.e. as entirely equivalent synonyms), so that’s what this textbook will do too. As explained in Remark 27, a real number can also be regarded as a point in one-dimensional space. However, as stated earlier, for now, whenever we say point, we mean a point in two-dimensional space. That is, a point is strictly an ordered pair of real numbers. And so, following this usage, the number 1 here is not a point. 101 Albeit a tremendously helpful one. Indeed, analytic or cartesian geometry was one of the major milestones in the history of mathematics. The idea of combining algebra and geometry is today “obvious” even to the secondary school student, but wasn’t always so to mathematicians.

100

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6.4.

The Graph of An Equation

In the previous subchapter, we looked at graphs that were simply sets of random, isolated points. But going forward, we’ll be looking mostly at graphs of equations (and shortly, also of functions): Definition 39. The graph of an equation is the set of points (x, y) for which the equation is true. Example 151. Let G be the graph of the equation y = x + 2. Then G = {(x, y) ∶ y = x + 2} .

In words, G is the set of points (x, y) for which the equation y = x + 2 is true. And so, G contains • (5, 7), because 7 = 5 + 2.

• (1, 3), because 3 = 1 + 2.

We can say, “Below we’ve graphed the equation y = x + 2” or more simply, “Below is graphed y = x + 2,” or, “Below is graphed G”. 8

y

(5, 7)

6 4

(1, 3)

2

-3

-2

-1

-2

1

2

The graph of the equation y = x + 2 is the set G = {(x, y) ∶ y = x + 2}, which contains (5, 7) and (1, 3). 3

4

5

6

x

Informally, we say that the equation y = x + 2 describes a line—specifically, the line with gradient 1 and which passes through the point (0, 2).

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Example 152. Let H be the graph of the equation y = x2 . Then H = {(x, y) ∶ y = x2 } .

In words, H is the set of points (x, y) for which the equation y = x2 is true. And so, H contains (1, 1), because 1 = 12 ;

(2, 4), because 4 = 22 .

and

y

8 6

-5

The graph of the equation y = x2 is the set G = {(x, y) ∶ y = x2 }, which contains (1, 1) and (2, 4). -4

-3

-2

(2, 4)

4 2

-1

1

(1, 1)

2

x

Informally, we say that the equation y = x2 describes a parabola.

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6.5.

Graphing with the TI84

Our first examples of using the TI84 (see “Use of Graphing Calculators”, p. xxxvii): Example 153. Graph the equation y = x2 . 1. Press ON to turn on your calculator. 2. Press Y= to bring up the Y= editor.

3. Press X,T,θ,n to enter “X”; then x2 to enter the squared “2 ” symbol.

4. Now press GRAPH and the calculator will graph y = x2 . Step 1.

Step 2.

Step 3.

Step 4.

(Screenshots are of the end of each Step.) Example 154. XXX Exercise 70. Graph these equations. (a) y = ex .

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(b) y = 3x + 2.

(Answer on p. 1734.) (c) y = 2x2 + 1.

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6.6.

The Graph of An Equation with Constraints

Example 155. Let G1 be the graph of the equation y = x + 2 with the constraint x ≥ 3. Equivalently, let G1 be the graph of y = x + 2, x ≥ 3. Then G1 = {(x, y) ∶ y = x + 2, x ≥ 3} .

In words, G1 is the set of points (x, y) for which “y = x + 2 AND x ≥ 3” is true. And so, G1 contains • (5, 7), because 7 = 5 + 2 AND 5 ≥ 3. • (3, 5), because 5 = 3 + 2 AND 3 ≥ 3.

In contrast, G1 does not contain (1, 3), because 1 ≥/ 3.

11

y

G1 = {(x, y) ∶ y = x + 2, x ≥ 3}

9 7 5 3

-1

(3, 5)

(1, 3)

1 -1

1

(5, 7)

2

3

4

5

6

7

8

x

Above we labelled our graph as G1 = {(x, y) ∶ y = x + 2, x ≥ 3}. But going forward, we’ll be a little lazy/sloppy and simply label it as y = x + 2, x ≥ 3 (as done below), with the understanding that this is the graph that satisfies the equation and constraint contained in that label. Nonetheless, you should always remember:

11

A graph is a set of points.

y

9 7 5 3 1 -1

-1

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1

(5, 7)

(3, 5)

(1, 3)

2

3

y = x + 2, x ≥ 3

4

5

6

7

8

x

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Example 156. Let G2 be the graph of y = x + 2, x > 3. Then G2 = {(x, y) ∶ y = x + 2, x > 3} .

The set G2 is exactly the same as G1 but with one difference—the constraint (inequality) is now strict, so that this time, G2 does not contain (3, 5).

11

y

y = x + 2, x > 3

9 7 5 3

-1

-1

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(3, 5)

(1, 3)

1 1

(5, 7)

2

3

4

5

6

7

8

x

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Example 157. Let H1 be the graph of y = x2 , x ≤ 2. Then H1 = {(x, y) ∶ y = x2 , x ≤ 2} .

In words, H1 is the set of points (x, y) for which “y = x2 AND x ≤ 2” is true. And so, H1 contains • (1, 1), because 12 = 1 AND 1 ≤ 2. • (2, 4), because 22 = 4 AND 2 ≤ 2.

In contrast, H1 does not contain (3, 9), because 3 ≤/ 2.

y

y = x2 , x ≤ 2

(3, 9)

8 6 4 2

-4

-3

-2

-1

1

(1, 1)

(2, 4)

2

x

3

Example 158. Let H2 be the graph of y = x2 , x < 2. Then H2 = {(x, y) ∶ y = x2 , x < 2} .

The set H2 is exactly the same as H1 but with one difference—the constraint (inequality) is now strict, so that this time, H2 does not contain (2, 4).

y

(3, 9)

8

y = x ,x < 2 2

6 4 2

-4

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-3

-2

-1

1

(1, 1)

(2, 4)

2

3

x

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1 Example 159. Let I1 be the graph of x2 + y 2 = 1, x ≥ − . Then 2 1 I1 = {(x, y) ∶ x2 + y 2 = 1, x ≥ − } . 2

1 In words, I1 is the set of points (x, y) for which “x2 + y 2 = 1 AND x ≥ − ” is true. And 2 so, I1 contains

• (1, 0), because 12 + 02 = 1 AND 1 ≥ −0.5. √ 2 √ 2 √ √ √ 1 2 2 2 2 2 , ), because ( ) +( ) = 1 AND ≥− . • ( 2 2 2 2 2 2 √ √ 2 3 1 3 1 2 1 1 • (− , ), because (− ) + ( ) = 1 AND − ≥ − . 2 2 2 2 2 2 √ √ 2 3 1 2 3 1 1 1 ), because (− ) + (− ) = 1 AND − ≥ − . • (− , − 2 2 2 2 2 2

√ 1 3 (− , ) 2 2

y

x2 + y 2 = 1, x ≥ −

1 2

√ √ 2 2 ( , ) 2 2

(1, 0)

x

√ 1 3 ) (− , − 2 2

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1 Example 160. Let I2 be the graph of x2 + y 2 = 1, x > − . Then 2 1 I2 = {(x, y) ∶ x2 + y 2 = 1, x > − } . 2

The set I2 is exactly the same as I1 but with one difference—the constraint √ √ (inequality) 1 3 1 3 is now strict, so that this time, I2 does not contain (− , ) or (− , − ). 2 2 2 2

√ 1 3 (− , ) 2 2

y

x2 + y 2 = 1, x > −

1 2

√ √ 2 2 ( , ) 2 2

(1, 0)

x

√ 1 3 ) (− , − 2 2

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More generally, a graph can also be of multiple equations with multiple constraints: Example 161. Let J be the graph of

⎧ ⎪ ⎪ ⎪−x, y=⎨ ⎪ 2 ⎪ ⎪ ⎩x ,

for x ≤ 0,

for x > 0.

In words, J is the set of points for which either “y = −x AND x ≤ 0” or “y = x2 AND x > 0” is true.

J is the graph of ⎧ ⎪ ⎪ ⎪−x, for x ≤ 0 y=⎨ ⎪ 2 ⎪ ⎪ ⎩x , for x > 0.

y

J

x We can actually write down J using set-builder notation and the union operator: J = {(x, y) ∶ y = −x, x ≤ 0} ∪ {(x, y) ∶ y = x2 , x > 0} .

But this is cumbersome and difficult to read. And so, we’ll usually simply specify J as was done above.

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Example 162. Let K be the graph of

⎧ ⎪ ⎪ ⎪x, y=⎨ ⎪ ⎪ ⎪ ⎩0,

for x ≠ 2, for x = 2.

In words, K is the set of points for which either “y = x AND x ≠ 2” or “y = 0 AND x = 2” is true.

y

K

K is the graph of ⎧ ⎪ ⎪ ⎪x, for x ≠ 2 ⎨ y=⎪ ⎪ ⎪ ⎩0, for x = 2.

x 2

Exercise 71. Graph each of the following. (a) y = ex , −1 ≤ x < 2. (b) y = 3x + 2, −1 ≤ x < 2.

(c) y = 2x2 + 1, −1 ≤ x < 2.

Exercise 72. Graph each of the following.

⎧ ⎪ ⎪ ⎪x + 1, (a) y = ⎨ ⎪ ⎪ ⎪ ⎩x − 1, ⎧ ⎪ ⎪ ⎪x + 1, (b) y = ⎨ ⎪ ⎪ ⎪ ⎩x − 1,

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for x ≤ 0, for x > 0.

for x < 0, for x ≥ 0.

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6.7.

Intercepts and Roots

In secondary school, you learnt about x-intercepts, y-intercepts, and solutions (or roots). We now give their formal definitions: Definition 40. Let G be a graph. • If the point (0, b) is in G, then we call (0, b) a vertical or y-intercept of G.

• If the point (a, 0) is in G, then we call (a, 0) a horizontal or x-intercept of G. If, moreover, G is the graph of an equation (or function), then we also call a a solution or root of that equation (or function). Example 163. The graph of the equation y = x + 2 has horizontal or x-intercept (−2, 0) and vertical or y-intercept (0, 2).

Or more simply, “The equation y = x + 2 has horizontal or x-intercept (−2, 0) and vertical or y-intercept (0, 2).” We also say that −2 is a solution (or root) of the equation y = x + 2.

y

y =x+2

6 4 2

-3

(−2, 0)

-2

-1

(0, 2)

1

2

3

4

5

6

x

-2 Remark 29. Just so you know, some writers (including your TI84) also call a solution (or root) a zero. So in the above example, they’d say that the zero of the equation y = x + 2 is −2. However, we will avoid using the term zero in this textbook because it does not appear on your A-Level exams or syllabus.

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Example 164. Consider the equation y = x2 . The point (0, 0) is both its (only) xintercept and its (only) y-intercept. Also, the equation has solution (or root) 0.

y

8

y = x2

6 4 2 (0, 0) -3

-2

-2

-1

1

1

2

x

Example 165. The equation x2 + y 2 = 1 has two x-intercepts (−1, 0) and (1, 0), two y-intercepts (0, −1) and (0, 1), and two solutions (or roots) −1 and 1.

y

(0, 1)

(−1, 0)

{(x, y) ∶ x2 + y 2 = 1}

(1, 0)

x

(0, −1)

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Example 166. The equation y = x2 − 1 has two x-intercepts (−1, 0) and (1, 0), one y-intercept −1, and two solutions (or roots) −1 and 1.

y

3

y = x2 − 1

2 1 (1, 0)

(−1, 0) -2

-1 -1

(0, −1)

1

x

Exercise 73. For each equation, write down any x-intercept(s), y-intercept(s), and solutions (or roots). (Answer on p. 1736.) (a) y = 2. (b) y = x2 − 4.

(c) y = x2 + 2x + 1. (d) y = x2 + 2x + 2.

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7.

Lines

You already know what a line is, at least informally and intuitively. Here’s the formal definition:102 Definition 41. A line is the graph of any equation ax + by + c = 0,

where a, b, c ∈ R and at least one of a or b is non-zero.

You may find this definition a little puzzling—didn’t we always simply write lines as y = dx + e?

It turns out that if b ≠ 0, then ax + by + c = 0 and y = dx + e are equivalent, as we now prove: ax + by + c = 0

⇐⇒

by = −ax − c

⇐⇒

a c y= − x − . b b ° ¯ e

d

Writing a line as y = dx + e has two advantages—it immediately tells us that its gradient103 is d

y-intercept is (0, e).

and

But writing a line as y = dx + e also has one big disadvantage—it can’t describe the case where b = 0, i.e. vertical lines: Example 167. The line x + y + 1 = 0 can also be written as y = −x − 1.

In contrast, the vertical line x − 1 = 0 cannot be written in the form y = dx + e.

The line x + y + 1 = 0 can also be written as y = −x − 1.

-4

-3

-2

-1

3

y

1 -1

1

The vertical line x − 1 = 0 cannot be written in the form y = dx + e.

2

3

x

-3 -5 Here’s what we’ll do in this textbook: If we know for sure that a line isn’t vertical, then we’ll write it in the form y = dx + e, because of the two advantages. Otherwise, we’ll write it as ax + by + c = 0.

Definition 41 covers only lines in 2D space. In Part IV (Vectors), we’ll learn a more general definition (Definition 138) that covers also lines in higher-dimensional spaces and supersedes Definition 41. 103 We’ll formally define the gradient of a line in Ch. 7.3. 102

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7.1.

Horizontal, Vertical, and Oblique Lines

Definition 42. A graph is (a) Horizontal if any two points in that graph have the same y-coordinate; (b) Vertical if any two points have the same x-coordinate; and (c) Oblique (or slanted) if it is neither horizontal nor vertical. Example 168. The line y = 1 is horizontal, the line x = 2 is vertical, and the line y = x+1 is oblique.

y

x=2

y =x+1

x y = −1

Fact 15. Suppose ax + by + c = 0 describes a line. Then

(a) The line is horizontal ⇐⇒ a = 0. (b) The line is vertical ⇐⇒ b = 0.

Proof. See p. 1533 in the Appendices.

Example 169. The line y = 1 is horizontal because the coefficient on x is zero.

The line x = 2 is vertical because the coefficient on y is zero.

The line y = x + 1 is oblique because the coefficients on x and y are both non-zero.

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7.2.

Another Equation of the Line

Fact 16. Given the distinct points (x1 , y1 ) and (x2 , y2 ), the unique line that contains both points is (x2 − x1 ) (y − y1 ) = (y2 − y1 ) (x − x1 ) .

Proof. See p. 1534 in the Appendices.

Example 170. The line containing the points (1, 2) and (−1, 3) is (−1 − 1) (y − 2) = (3 − 2) (x − 1)

or

y

−2y = x − 5

(4, 5)

5 1 y =− x+ 2 2 (−1, 3)

(1, 2)

5 1 y =− x+ . 2 2

or

5 y = x−5 2

x (2, 0)

The line containing the points (2, 0) and (4, 5) is (4 − 2) (y − 0) = (5 − 0) (x − 2)

or

2y = 5x − 10

or

If x1 ≠ x2 and y1 ≠ y2 , then we can rewrite the equation in Fact 16 as

5 y = x − 5. 2

y − y1 ⋆ x − x 1 = . y2 − y1 x2 − x1

This latter equation is perhaps more familiar and easier to remember. But of course, it is 113, Contents

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illegal if x1 = x2 or y1 = y2 , which is why we may sometimes prefer the equation given in Fact 16. We reuse the last example:

Example 171. The line containing the points (1, 2) and (−1, 3) is y−2 ⋆ x−1 = 3 − 2 −1 − 1

or

1 y − 2 = − (x − 1) 2

The line containing the points (2, 0) and (4, 5) is y−0 ⋆ x−2 = 5−0 4−2

or

y=

or

1 5 y =− x+ . 2 2

5 5 (x − 2) = x − 5. 2 2

Exercise 74. Find the line that contains each pair of points.

(a) (4, 5) and (7, 9)

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(b) (1, 2) and (−1, −3)

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7.3.

Informally, the gradient (or slope) of a line is • “Rise over Run”. • ∆y/∆x or “Change in y over change in x”. (∆ is the upper-case Greek letter delta.) • The answer to this question: “If we move 1 unit to the right while remaining along the line, by how many units will we move up?” Example 172. The gradient of the line y = 2x + 1 is

y

∆y 2 = = 2. ∆x 1

y = 2x + 1

∆y = 2 ∆x = 1

x

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Example 173. The gradient of the line y = −x − 1 is ∆y −1 = = −1. ∆x 1

y = −x − 1 ∆y = −1

y

∆x = 1

x

Remark 30. Slope is a perfectly good synonym for gradient. However, your A-Level syllabus and exams do not use the word slope. And so we’ll stick to using only the word gradient. In general, to find a line’s gradient, pick two distinct points on the line (x1 , y1 ) and (x2 , y2 ), then compute Gradient =

Rise ∆y y2 − y1 = = . Run ∆x x2 − x1

Figure to be inserted here.

One caveat is that if the line is vertical, then x1 = x2 so that our last expression has denominator 0 and is thus undefined. And so, if a line is vertical, we’ll simply say that its gradient is undefined. 116, Contents

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Let’s jot the above down as a formal definition: Definition 43. Suppose a non-vertical line contains the distinct points (x1 , y1 ) and (x2 , y2 ). Then its gradient is the number defined as y2 − y1 . x2 − x1

If a line is vertical, then its gradient is simply left undefined. Fact 17. Suppose ax + by + c = 0 is a non-vertical line. Then its gradient is −a/b.

Proof. Since the line ax + by + c = 0 is non-vertical, b ≠ 0 (Fact 15). c c Case 1. If a = 0, then the line contains the points (0, − ) and (1, − ), so that by Definition b b 43, its gradient is − cb − (− cb ) a =0=− . 1−0 b

3

c a+c Case 2. If a ≠ 0, then the line contains the points (0, − ) and (1, − ), so that by b b Definition 43, its gradient is

Example 174. XXX

c − a+c a b − (− b ) =− . 1−0 b

Example 175. XXX Exercise 75. XXX

A75.

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7.4.

Yet Another Equation of the Line

Fact 18. The line that contains the point (x1 , y1 ) and has gradient m is y − y1 = m (x − x1 ) .

Proof. Let the line be ax + by + c = 0. 1

c 3 = mx.104 b c c 4 3 Also, the line contains (x1 , y1 ) — plugging this into = yields y1 + = mx1 or = mx1 − y1 . b b

The line’s gradient is m = −a/b. Plugging = into = yields y + 2

2

1

Plug = into = to get y + mx1 − y1 = mx or y − y1 = m (x − x1 ). 4

3

Example 176. The line that contains the point (−1, 2) and has gradient 3 is105 y − 2 = 3 [x − (−1)]

y

−2

or

y = 3x + 5.

y = 3x + 5

1 y = −2x + 19

3

(7, 5)

1

x

(−1, 2)

The line that contains the point (7, 5) and has gradient −2 is y − 5 = −2 (x − 7)

104 105

or

y = −2x + 19.

Since the line’s gradient is defined, b ≠ 0. It looks like Wolfram Alpha understands slope but not gradient.

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Exercise 76. Find the line that has the given point and gradient.

(a) Point (4, 5) and gradient 3 p. 1737.)

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(b) Point (1, 2) and gradient −2

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7.5.

Perpendicular Lines

Definition 44. Two lines are perpendicular if (a) Their gradients are negative reciprocals of each other; or (b) One line is vertical while the other is horizontal.

If two lines l and m are perpendicular, then we will also write l ⊥ m.

In Part IV (Vectors), we’ll give Definition 145, which is a more general definition of when two lines are perpendicular and which will supersede the above Definition. Example 177. XXX Example 178. XXX Exercise 77. XXX

A77.

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7.6.

Lines vs Line Segments vs Rays

Example 179. Let A and B be points. The line AB contains both A and B. It has infinite length, extending “forever” in both directions.

A

B

A

B

In contrast, the line segment AB has finite length—it contains A, B, and every point between, but no other point.

With lines and line segments, the order in which we write A and B doesn’t matter. The line AB is the same as the line BA. And the line segment AB is the same as the line segment BA. But with rays, the order in which we write A and B does matter. The ray AB is the “half-infinite line” that starts at A, passes through B, and also contains every point “beyond” B.

A

B

A

B

In contrast, the ray BA is the “half-infinite line” that starts at B, passes through A, and also contains every point “beyond” A.

The rays AB and BA are distinct. Both have infinite length. It makes no sense to speak of the length of a line or a ray (because these have infinite length). We can however speak of a line segment’s (finite) length.106 Remark 31. Confusingly, some other writers use ray to mean a (finite) line segment. But this textbook will strictly reserve the word ray to mean a “half-infinite line”. Remark 32. According to ISO 80000-2: 2009,107 AB may be used to denote the line segment from point A to point B. This notation is convenient and allows us to distinguish between the line AB and the line segment AB. However, your A-Level syllabus and exams (see e.g. N2007/I/6) do not seem to use this notation.108 And so, this textbook shall not do so either. For us, the only way to tell whether “AB” refers to a line or a line segment is to see if we say “the line AB” or “the line segment AB”. Thus, we must always be absolutely clear if we’re talking about a line or a line segment.

For the formal definitions of a line, a line segment, and a ray, see Definition 269 in the Appendices. See Item No. 2-8.4. The 40-something-page PDF costs a mind-blowing 158 Swiss Francs or about S\$214 at the ISO store. As always, you may or may not be able to find free versions of this document elsewhere on the interwebz. 108 Indeed, they don’t seem to be very careful about distinguishing between lines and line segments.

106 107

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8. 8.1.

Distance

The Distance between Two Points on the Real Number Line

On the real number line, the distance between two points (or real numbers) is simply the magnitude of their difference. Formally, Definition 45. Let A and B be two real numbers. The distance between A and B, denoted ∣AB∣, is the number defined by ∣AB∣ = ∣B − A∣ .

Example 180. The distance between A = 3 and B = 5 is

∣AB∣ = ∣B − A∣ = ∣5 − 3∣ = ∣2∣ = 2.

Example 181. The distance between C = 7 and D = −2 is

∣CD∣ = ∣D − C∣ = ∣−2 − 7∣ = ∣−9∣ = 9.

Of course, the distance between A and B is the same as the distance between B and A: ∣AB∣ = ∣B − A∣ = ∣A − B∣ = ∣BA∣ .

Exercise 78. Let A = 5, B = −7, and C = 1. What are ∣AB∣, ∣AC∣, and ∣BC∣?(Answer on p. 1738.)

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8.2.

The Distance between Two Points on the Cartesian Plane

Let A = (a1 , a2 ) and B = (b1 , b2 ) be two points on the cartesian plane. How should we define the distance between A and B? To motivate this definition, let’s review the Pythagoras’ Theorem:

Theorem 2. (Pythagoras’ Theorem) Suppose a right triangle has legs of lengths a and b, and hypotenuse of length c. Then a2 + b2 = c2 .

Proof. Construct the triangle ABC, where the vertices A, B, and C are opposite the sides of lengths a, b, and c, respectively. Let D be the point on AB that is the base of the perpendicular from the point C:

c

B

D

d

A

c−d

b

a

C

a c−d 1 = or a2 = c (c − d). c a b d 2 The triangles ABC and ACD are similar. Hence, = or b2 = cd. c b

The triangles ABC and CBD are similar. Hence,

Now, = + = yields a2 + b2 = c (c − d) + cd = c (c − d + d) = c2 . 1

2

So, if we draw A = (a1 , a2 ) and B = (b1 , b2 ) as two vertices of a right triangle, then by the √ 2 2 Pythagoras’ Theorem, the length of the hypotenuse AB should simply be (a1 − b1 ) + (a2 − b2 ) :

√ 2 2 (a1 − b1 ) + (a2 − b2 )

B = (b1 , b2 )

a1 − b1

A = (a1 , a2 )

a2 − b2 C = (a1 , b2 )

The foregoing discussion suggests this formal definition of distance:

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Definition 46. Let A = (a1 , a2 ) and B = (b1 , b2 ) be points in the cartesian plane. The distance between A and B, denoted ∣AB∣, is the number defined by √ 2 2 ∣AB∣ = (a1 − b1 ) + (a2 − b2 ) .

Also, we define the length of the line segment AB to be ∣AB∣.

Example 182. The distance between A = (3, 6) and B = (5, −2) is √ √ √ 2 2 ∣AB∣ = (3 − 5) + [6 − (−2)] = 22 + 82 = 68.

Example 183. The distance between C = (1, 7) and D = (0, −2) is √ √ √ 2 2 ∣CD∣ = (1 − 0) + [7 − (−2)] = 12 + 92 = 82.

Of course, the distance between A and B is the same as the distance between B and A: √ √ 2 2 2 2 ∣AB∣ = (a1 − b1 ) + (a2 − b2 ) = (b1 − a1 ) + (b2 − a2 ) = ∣BA∣ .

(We say that the distance or length operator ∣⋅∣ is commutative, just as addition and multiplication are: a + b = b + a and ab = ba.) Remark 33. Later in Part IV (Vectors), we’ll again use the Pythagoras’ Theorem to motivate a similar definition for the distance between two points in three-dimensional space.

Exercise 79. Let A = (5, −1), B = (−7, 0), and C = (1, 2). What are ∣AB∣, ∣AC∣, and ∣BC∣? (Answer on p. 1738.)

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8.3.

Closeness (or Nearness)

Definition 47. Let A, B, and C be points.

If ∣AB∣ < ∣AC∣, then we say that B is closer (or nearer) to A than C. If ∣AB∣ = ∣AC∣, then we say that B is as close (or near) to A as C. If ∣AB∣ > ∣AC∣, then we say that B is further from A than C.

Let S be a set of points. We say that D ∈ S is the point in S that’s closest to A if ∣AD∣ ≤ ∣AE∣ for every point E ∈ S.

Example 184. Let A = (3, 6), B = (5, −2), and C = (1, 7). We have √ √ √ 2 2 ∣AB∣ = (3 − 5) + [6 − (−2)] = 22 + 82 = 68, √ √ √ 2 2 ∣AC∣ = (3 − 1) + (6 − 7) = 22 + 12 = 5, √ √ √ 2 2 ∣BC∣ = (5 − 1) + (−2 − 7) = 42 + 92 = 117.

Since ∣AC∣ < ∣AB∣, C is closer to A than B. Equivalently, B is further from A than C.

Since ∣AB∣ = ∣BA∣ < ∣BC∣, A is closer to B than C. Equivalently, C is further from B than A.

Example 185. The point on the line y = 0 closest to the point (1, 2) is the point (1, 0).

Exercise 80. Let A = (5, −1), B = (−7, 0), and C = (1, 2). Fill in the blanks:(Answer on p. 1738.)

(a) A is _____ B than C. (b) C is _____ A than B.

Exercise 81. Find the point on the line x = 3 that is closest to the point (5, 6).(Answer on p. 1738.)

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9.

Circles

The unit circle centred on the origin (the word unit means that the circle has radius 1) is graphed below. This is the set of points (x, y) whose distance from the origin (0, 0) is 1, i.e. √ 2 2 (x − 0) + (y − 0) = 1.

Or equivalently and more elegantly,

x2 + y 2 = 1.

The above discussion motivates this definition:

Definition 48. The unit circle centred on the origin is the graph of x2 + y 2 = 1.

That is, the unit circle centred on the origin is this set: G = {(x, y) ∶ x2 + y 2 = 1} .

Figure to be inserted here.

√ √ 2 2 ,− ) is on the unit circle because it satisfies the Example 186. The point (− 2 2 equation x2 + y 2 = 1: √

√ 2 2 2 2 2 2 (− ) + (− ) = + = 1. 2 2 4 4

√ Example 187. The point (0.1, − 0.99) is on the unit circle because it satisfies the equation x2 + y 2 = 1: √ 2 0.12 + (− 0.99) = 0.01 + 0.99 = 1.

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Next, the circle of radius r centred on (a, b) is graphed below.

This is the set of points (x, y) whose distance from the point (a, b) is r, i.e. √ 2 2 (x − a) + (y − b) = r. Or equivalently,

(x − a) + (y − b) = r2 . 2

2

The above discussion motivates this definition:

Definition 49. Let (a, b) be a point in the cartesian plane and r ≥ 0. The circle of radius r centred on (a, b) is the graph of (x − a) + (y − b) = r2 . 2

2

The circle’s radius is the number r, its centre is the point (a, b), and its diameter that is the number 2r.

Equivalently, the circle of radius r centred on (a, b) is this set: G = {(x, y) ∶ (x − a) + (y − b) = r2 } . 2

2

Figure to be inserted here.

Example 188. XXX Example 189. XXX Example 190. XXX

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Exercise 82. (a) What is the graph of each of these three equations? (i) (x − 5) + (y − 1) = 16 2

2

(ii) (x + 3) + (y + 2) = 2 2

2

(iii) (x + 1) + y 2 = 81 2

(b) Write down the equation that corresponds to each of these three circles:

(i) The circle of radius 5 centred on (−1, 2). (ii) The circle of radius 1 centred on (3, −2). √ (iii) The circle of radius 3 centred on (0, 0).

For each of the six circles given above in (a) and (b), (c) Does it contain the point (8, 2)?

(d) What are its centre, radius, and diameter?

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9.1.

Graphing Circles on the TI84

Example 191. Graph the equation x2 + y 2 = 1.

Unfortunately, you cannot directly input this equation, because the piece of junk that is the TI84 requires that you enter equations with y on the left √ hand side. And so√here, we’ll have to tell the TI84 to graph two separate equations: y = 1 − x2 and y = − 1 − x2 . 1. Press ON to turn on your calculator.

2. Press Y= to bring up the Y= editor.

Most buttons on the TI84 have three different functions. Simply pressing a button executes the function that’s printed on the button itself. Pressing the 2ND button and then a button executes the function that’s printed in blue above the button. And pressing the ALPHA and then a button executes the function that’s printed in green above the button. √ 2 3. Press √ the 2ND button and then the x button to execute the √ function and enter “ ”. Next press 1 − X,T,θ,n x2 . Altogether you’ve entered 1 − x2 .

Warning: Confusingly, the TI84 has two different minus buttons: − and (-) . The − is used as the minus operation in the middle of an expression, as was just done. In contrast, the (-) is used to denote negative numbers at the beginning of an expression, as we’ll do in Step 5 (below). Unfortunately, you really have to key in the correct button, or else you’ll get a frustrating error message. 4. Now press ENTER and the blinking cursor will move down, to the right of “Y2 =”. Step 1.

Step 2.

Step 3.

Step 4.

(Example continues on the next page ...)

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(Example continues on the next page ...) We’ll now enter the second equation. 5. Press the (-) button. Now repeat what we did in step 3 above: Press the blue 2ND √ √ 2 button and then (which corresponds to the x button) to enter “ √ ”. Next press 2 2 1 − X,T,θ,n x to enter “1 − X ”. Altogether you will have entered − 1 − x2 . √ √ 2 6. Now press GRAPH and the calculator will graph both y = 1 − x and y = − 1 − x2 . Notice the graphs are very small. To zoom in,

7. Press the ZOOM button to bring up a menu of ZOOM options. 8. Press 2 to select the Zoom In option. Nothing seems to happen. But now press ENTER and the TI84 will zoom in a little for you.

We expected to see a perfect circle—instead, we get an ellipse. Hm, what’s going on? The reason is that by default, the x- and y- axes are scaled differently. To set them to the same scale, 9. Press the ZOOM button again to bring up the ZOOM menu of options. Press 5 to select the ZSquare option. The TI84 will adjust the x- and y- axes so that they have the same scale and thus give us a perfect circle. Step 5.

Step 6.

Step 8.

Step 9.

Step 7.

√ P.S. An alternative to Step 5 is to enter “−Y1 ” instead of “− 1 − X 2 ”. To do so, replace Step 5 with these instructions: First press (-) to enter the minus sign, as was done in Step 5. Next press VARS to bring up the VARS menu. Then press ⟩ to go to the Y-VARS menu. Now press ENTER to select “1: Function...”. Press ENTER again to select “1: Y1 ”. Altogether, we will have entered “Y2 = −Y1 ”. Now go to Step 6. Exercise 83. Graph (x − 3) + (y + 2) on your TI84. 2

2

A83. TBD.

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9.2.

The Number π (optional)

Theorem 3. The ratio of a circle’s area to the square of its radius is constant. Proof. Omitted. See e.g. Euclid’s Elements (c. 300 BC, Book XII, Proposition 2, [Joyce, 1997]).109 Definition 50. The circumference of a circle is its length. Remark 34. In writing down the above definition, we’re actually cheating a little: So far we’ve only defined what the length of a line segment is; we have not defined the “length” of any curve (e.g. a circle). It turns out that the question of the “length” of a curve is a little harder than one might think. In particular, it is beyond the scope of H2 Maths (and is dealt with in H2 Further Maths, but only cursorily and as a mindless formula that students are to mug and apply). Nonetheless, just to screw students over, the question of length abruptly appeared as a curveball question in the 2018 A-Level H2 Maths exam (see Exercise 672). Theorem 4. The area of a circle with radius r and circumference C is Cr/2. Proof. Omitted. This result was first proven by Archimedes (c. 250 BC, Measurement of a Circle, Proposition 1, [Heath, 1897]). Corollary 1. The ratio of a circle’s circumference to its radius is constant. Proof. Suppose two circles have circumferences C1 and C2 and radii r1 and r2 . Then by Theorem 4, their areas are C1 r1 /2 and C2 r2 /2.

And by Theorem 3,

C1 r1 /2 C2 r2 /2 = r12 r22

or

C1 C2 = . r1 r2

We’ve just shown that the ratio of a circle’s circumference to its radius is constant. Given Corollary 1, we can define the number π: Definition 51. The real number π is the ratio of a circle’s circumference to its diameter. Remark 35. The above is the definition of π that should be familiar to you from primary school. Note though that more advanced mathematical texts will usually adopt some other definition of π—for example, after defining the sine function sin, they might define π to be the smallest positive number such that sin π = 0.

109

But this result was probably proven before Euclid.

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Fact 19. A circle with radius r has circumference 2πr and area πr2 . Proof. Let C be the circle’s circumference.

By Definition 51, π = C/ (2r). Rearranging, C = 2πr.

By Theorem 4, the circle’s area is Cr/2 = (2πr) r/2 = πr2 . Fun Fact

There is a a movement to replace π = 3.14 . . . with τ = 6.28 . . . , the latter being the ratio of a circle’s circumference to its radius. See “The Tau Manifesto” or this 2020-03-13 Wall Street Journal story. (In my opinion, what’s more urgent is for Americans to replace their silly convention of writing short dates as MM-DD with the DD-MM used by the rest of the world.)

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10.

Tangent Lines, Gradients, and Stationary Points 10.1.

Tangent Lines

Here’s the (very) informal definition of tangent lines you probably learnt in secondary school: Definition 52 (informal). A tangent line (to a graph at a point) is a line that “just” touches that graph at that point. Later on in Part V (Calculus), we’ll define tangent lines more formally and precisely (see Definition XXX). But for now, the above definition will be good enough. Example 192. XXX Example 193. XXX Example 194. XXX

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10.2.

The Gradient of a Graph at a Point

Definition 53 (informal). The gradient (of a graph at a point) is the gradient of the tangent line (to that graph at that point). Example 195. XXX Example 196. XXX Example 197. XXX

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10.3.

Stationary Points

Definition 54 (informal). A point of a graph is called a stationary point if the gradient of the graph at that point is zero. Equivalently, a point of a graph is called a stationary point if the tangent line (to that graph at that point) is horizontal. Example 198. XXX Example 199. XXX Example 200. XXX

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11.

Maximum, Minimum, and Turning Points

In secondary school, you learnt about maximum points and minimum points. We’ll now relearn these, but in a little more depth.

11.1.

Maximum Points

In particular, for maximum points, we’ll learn to make these two distinctions: • Global vs local maximum points; • Strict vs non-strict maximum points. Informally, a global maximum (point) of a graph is a point that’s at least as high as any other point (that’s also in that graph).110 And a strict global maximum (point) of a graph is a point that’s strictly higher than every other point (that’s also in that graph). Example 201. Consider y = −x2 . The point D = (0, 0) is a global maximum (of the graph of y = −x2 ), because it is at least as high as any other point (in that graph).

y

y = −x2

D = (0, 0)

x

Indeed, it is also a strict global maximum, because it is strictly higher than every other point.

110

Here, we’ll merely give the informal definitions of the eight types of extrema introduced. For their formal definitions, see Definition 273 (Appendices).

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Example 202. Consider y = sin x and these three points: A = (−

3π , 1), 2

π B = ( , 1), 2

and

C=(

5π , 1). 2

Each of A, B, and C is a global maximum (of the graph of y = sin x), because each is at least as high as any other point (in that graph).

A = (−

y

3π , 1) 2

π B = ( , 1) 2

C=(

5π , 1) 2

y = sin x

x

Indeed, y = sin x has infinitely many global maxima: For every integer k, the following point is a global maximum: 1 ((2k + ) π, 1) . 2

Note though that y = sin x has no strict global maximum, because no point is strictly higher than every other point.

Obviously, if a point is strictly higher than any other point, then it must also be at least as high as any other point. And so, Fact 20. Every strict global maximum is also a global maximum. However, the converse is not true. That is, a global maximum need not be strict. (In the above example, each of A, B, and C is a global maximum, but none is a strict global maximum. Indeed, the graph of y = sin x has no strict global maximum.) We next introduce the concept of a local maximum:

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Informally, a local maximum (point) of a graph is a point that’s at least as high as any “nearby” point (that’s also in that graph). Example 203. Consider the graph of y = 6x5 − 15x4 − 10x3 + 30x2 .

Neither E = (−1, 19) nor F = (1, 11) is a global maximum, because neither is at least as high as any other point.

E = (−1, 19)

y

F = (1, 11)

x y = 6x5 − 15x4 − 10x3 + 30x2 However, each of E and F is a local maximum, because each is at least as high as any “nearby” point. Informally, a strict local maximum (point) of a graph is a point that’s strictly higher than any “nearby” point (that’s also in that graph). Example 204. In the last example, the points E and F are also strict local maxima, because each is strictly higher than any “nearby” point. Again and obviously, if a point is strictly higher than any “nearby” point, then it must also be at least as high as any “nearby” point. And so in general, Fact 21. Every strict local maximum is also a local maximum. Again, the converse is not true. That is, a local maximum need not be strict:

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Example 205. Consider y = 3. The point G = (1, 3) is a local maximum, because it’s at least as high as any “nearby” point. However, it is not a strict local maximum, because it is not strictly higher than any “nearby” point.

Actually, every point in y = 3 is a local maximum (though not a strict local maximum)! (This statement may initially seem puzzling, but you should take a moment to convince yourself that it is true.)

y=3

y G = (1, 3)

x

Indeed, every point in y = 3 is a global maximum (though not a strict global maximum)! (Ditto.)

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11.2.

Minimum Points

Similarly, for minimum points, we can make the same two distinctions: • Global vs local minimum points; • Strict vs non-strict minimum points. Informally, a global minimum (point) of a graph is a point that’s at least as low as any other point (that’s also in that graph). And a strict global minimum (point) of a graph is a point that’s (strictly) lower than any other point (that’s also in that graph). Example 206. Consider y = x2 . The point K = (0, 0) is a global minimum (of the graph of y = x2 ), because it is at least as low as any other point (in that graph).

y=x

2

y

K = (0, 0)

x

Indeed, it is also a strict global minimum, because it is (strictly) lower than every other point.

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Example 207. Consider y = sin x and these three points: H = (−

5π , −1), 2

π I = (− , −1), 2

and

J =(

3π , −1). 2

Each of H, I, and J is a global minimum, because each is at least as low as every other point (in the graph of y = sin x).

y

y = sin x

x

H = (−

5π , −1) 2

π I = (− , −1) 2

J =(

3π , −1) 2

Indeed, y = sin x has infinitely many global minima: For every integer k, the following point is a global maximum: 1 ((2k − ) π, 1) . 2

Note though that y = sin x has no strict global minimum, because no point is strictly lower than every other point.

Obviously, if a point is strictly lower than any other point, then it must also be at least as low as any other point. And so, Fact 22. Every strict global minimum is also a global minimum. However, the converse is not true. That is, a global maximum need not be strict. (In the above example, each of H, I, and J is a global minimum, but none is a strict global minimum. Indeed, the graph of y = sin x has no strict global minimum.) 141, Contents

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Informally, a local minimum (point) of a graph is a point that’s at least as low as any “nearby” point (that’s also in that graph). Example 208. In the graph of y = 6x5 − 15x4 − 10x3 + 30x2 , neither L = (0, 0) nor M = (2, −8) is a global minimum, because neither is at least as low as every other point (in that graph). However, each is a local minimum, because each is at least as low as any “nearby” point.

y

y = 6x5 − 15x4 − 10x3 + 30x2

L = (0, 0)

x M = (2, −8)

Informally, a strict local minimum (point) of a graph is a point that’s strictly lower than any “nearby” point (that’s also in that graph). Example 209. In the last example, the points L and M are also strict local minima, because each is strictly lower than any “nearby” point. Again and obviously, if a point is strictly lower than any “nearby” point, then it must also be at least as low as any “nearby” point. And so, Fact 23. Every strict local minimum is also a local minimum. However, the converse is not true. That is, a local minimum need not be strict:

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Example 210. Consider y = 3. The point G = (1, 3) is a local minimum, because it’s at least as low as any “nearby” point. However, it is not a strict local minimum, because it is not strictly lower than any “nearby” point.

Actually, every point in y = 3 is a local minimum (though not a strict local minimum)! (This statement may initially seem puzzling, but you should take a moment to convince yourself that it is true.)

y=3

y G = (1, 3)

x

Indeed, every point in y = 3 is a global minimum (though not a strict global minimum)! (Ditto.)

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11.3.

Extrema

It will be convenient to have a word for all maximum and minimum points. Definition 55. An extremum (plural: extrema) is any maximum or minimum point. A strict extremum is any strict maximum or minimum point. So, so far in this chapter, we’ve learnt about eight types of extrema, now summarised: (a) A global maximum is (b) The strict global maximum is (c) A local maximum is (d) A strict local maximum is (e) A global minimum is (f) The strict global minimum is (g) A local minimum is (h) A strict local minimum is

at least as high as higher than at least as high as higher than at least as low as lower than at least as low as lower than

any any any any any any any any

other point. other point. “nearby” point. “nearby” point. other point. other point. “nearby” point. “nearby” point.

Exercise 84. Explain whether each statement is true.

(a) A global maximum must also be a strict local maximum. (b) A global maximum must also be a local maximum. (c) A strict global maximum must also be a global maximum, strict local maximum, and local maximum. (d) A global maximum cannot be a local minimum.

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11.4.

Turning Points

Definition 56 (informal). A turning point (of a graph) is a point that’s both a strict local extremum and stationary point (of that graph). Remember turning points? We’ll formally define these in Definition 110. For now, we’ll rely on your intuitive understanding of what turning points are. We’ll also mention that every turning point is a strict extremum (i.e. either a strict maximum or minimum), but not every strict extremum is a turning point. Example 211. The graph of y = x (left) has no extrema.

y

x y=x

I = (−1, −1)

y

y = x, x ≥ −1

x

In contrast, the graph of y = x with the constraint x ≥ −1 (right) has one extremum, namely I = (−1, −1), which is a global minimum, strict global minimum, local minimum, and strict local minimum. Observe though that I is not a turning point.

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Example 212. Consider the graph of

⎧ ⎪ ⎪ −81, ⎪ ⎪ ⎪ ⎪ y = ⎨0.75x5 − 3.75x4 − 5x3 + 30x2 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩125,

y

C = (−2, 76)

B = (−3, −81)

for − 3 < x < 5, for x ≥ 5.

G = (5, 125)

H = (6, 125)

E = (2, 44) D = (0, 0)

A = (−4, −81)

for x ≤ −3,

x F = (4, −32)

The following table says, for example, that A = (−8, −81) is a local maximum, a global minimum, and a local minimum, but is not any of the other five types of extrema and is not a turning point. You should verify that the entire table is correct. A B C D E F

G H 3 3

Global maximum Strict global maximum Local maximum 3 3 3 3 3 Strict local maximum 3 3 Global minimum 3 3 Strict global minimum Local minimum 3 3 3 3 3 Strict local minimum 3 3 Turning point 3 3 3 3 Besides the eight points A–H, does this graph have any other extrema?111 111

Yes. In fact, there are infinitely many other extrema: For every p < −3, the point (p, −81) is—like A—a global minimum, local maximum, and local minimum. And for every q > 5, the point (q, 125) is—like H—a global maximum, local maximum, and local

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Example 213. Consider the graph G = {A, B, C}, where A = (−5, 4),

B = (1, 1),

and

C = (2, −3).

Informally, the three points A, B, and C are not “close” to each other. (A bit more formally, we’d say that they are isolated points.)112 And so, interestingly, each of A, B, and C is a strict local maximum of G. This is because each of A, B, and C has no “nearby” point. It is thus trivially or vacuously true that each of A, B, and C is strictly higher than any “nearby” point. By the same token, it is likewise trivially or vacuously true that each of A, B, and C is a strict local minimum of G.

y

A = (−5, 4)

4 2

-6

-4

-2

B = (1, 1)

2

x

4

-2 -4

C = (2, −3)

Note though that there’s only one strict -6 global maximum or global maximum, namely A. Likewise, there is only one strict global minimum or global minimum, namely C. Observe also that none of A, B, or C is a turning point. Exercise 85. Identify each graph’s extrema and turning points. (a) y = x2 + 1. (c) y = cos x.

112

(b) y = x2 + 1, −1 ≤ x ≤ 1. (d) y = cos x, −1 ≤ x ≤ 1.

minimum. Informally, a point of a set is isolated if no other point in the same set is “nearby”. For the formal definition, see Definition 272 (Appendices).

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12. 12.1.

Reflection and Symmetry

The Reflection of a Point in a Point

Example 214. Let P = (−2, 0) and Q = (0, 1) be points.

Suppose we reflect P in Q to produce the point R. Then what would R be? Well, intuition suggests that the reflection R should satisfy these two properties: (a) ∣P Q∣ = ∣QR∣

(b) R is on the line P Q.

(Of course, we should also have R ≠ P .)

+1

y

+1

+2

+2

R = (2, 2)

Q = (0, 1)

P = (−2, 0)

x

How can properties (a) and (b) be satisfied? Common-Sense Reasoning suggests that

• Since Qx − Px = +2, we must also have Rx − Qx = +2. • Since Qy − Py = +1, we must also have Ry − Qy = +1. Hence,

+2

+2

P = (−2, 0) in Q = (0, 1) is R = (2, 2). +1

+1

Intuitively, it seems “obvious” that given points P and Q, there should be exactly one point R that satisfies both properties (a) and (b). Happily, it turns out that this piece of 148, Contents

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intuition is correct.113 And so, we are allowed to write down this formal definition of a reflection: Definition 57. Let P and Q be distinct points. Then the reflection of P in Q is the unique point R ≠ P that satisfies these two properties: (a) ∣P Q∣ = ∣QR∣.

(b) R is on the line P Q.

According to our Common-Sense Reasoning, if P = (a, b) and Q = (c, d), then the reflection of P in Q should be the point R = (c+ (c − a), d+ (d − b)) = (2c − a, 2d − b) .

For future reference, let’s jot this down as a formal result:

Fact 24. Let P = (a, b) and Q = (c, d) be distinct points. Then the reflection of P in Q is the point

Proof. See Exercise 87.

R = (2c − a, 2d − b) .

Rather than mug Fact 24, it’s probably easier to find reflection points by directly using our above Common-Sense Reasoning:

+4

+4

Example 215. The reflection of (1, 2) in (5, 8) is (9, 14).

+6

+6

Exercise 86. What is the reflection of (8, 5) in (−2, 4)?

Exercise 87. Prove Fact 24—to do so, simply verify that the point R = (2c − a, 2d − b) satisfies properties (a) and (b). (Answer on p. 1742.)

113

See Proposition 26 (Appendices).

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12.2.

The Reflection of a Point in a Line

Definition 58. The reflection of a point P in a graph G is defined as the reflection of P in Q, where Q is the point in G that’s closest to P . The above definition is general in that G can be any sort of graph. But to keep things simple, we’ll look only at cases where G is a line: Example 216. Consider the point P = (−3, 2) and the line y = 2x + 3. You are given that the point on the line that’s closest to P is Q = (−1, 1). So, by Definition 58, the reflection of P in the line is the reflection of P = (−3, 2) in Q = (−1, 1), or R = (1, 0).

P = (−3, 2)

y = 2x + 3

2

Q = (−1, 1)

-4

-3

-2

y

1 R = (1, 0)

-1

1

2

3

x

-1

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Two simple but useful results: Fact 25. The reflection of the point (p, q) in the line y = x is the point (q, p).

Proof. See p. 1542 in the Appendices.

Example 217. The reflection of (−2, 1) in y = x is (1, −2). The reflection of (0, 1) in y = x is (1, 0).

y

(−2, 1)

-4

-3

1

y=x

(0, 1) (1, 0)

-2

-1

1

2

x

3

-1 (1, −2)

-2

Fact 26. The reflection of the point (p, q) in the line y = −x is the point (−q, −p).

Proof. See p. 1542 in the Appendices.

Example 218. The reflection of (−2, 1) in y = −x is (−1, 2). The reflection of (0, 1) in y = x is (−1, 0).

(−1, 2)

(−2, 1)

-5

-4

-3

2 1

(−1, 0)

-2

-1

y

(0, 1)

1 -1

2 y = −x

Exercise 88. Find the reflections of (3, 2) in y = x and y = −x. 151, Contents

3

4

x

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12.3.

Lines of Symmetry

Definition 59. Let G and L be graphs. The reflection of G in L is the set of points obtained by reflecting every point in G in L. Again, the above definition is general in that L can be any graph. But again, to keep things simple, we’ll look only at cases where L is a line: Example 219. By informal observation,114 the reflection of y = x2 in the line y = 2 is y = 4 − x2 .

y

y=x

2

7 5 3

y=2

-3

y = 4 − x2

1 -2

-1

-1

1

2

x

Definition 60. Suppose the reflection of the graph G in the line l is G itself. Then we say that G is symmetric in l, or that that l is a line (or axis) of symmetry for G. Example 220. By informal observation,115 the graph of y = x2 is symmetric in the line x = 0 (also the y-axis). Equivalently, x = 0 is a line or axis of symmetry for y = x2 .

Figure to be inserted here.

Here’s a formal proof: Pick any point (a, a2 ) in the graph of y = x2 . It is possible to show that the closest point on the line y = 2 is (a, 2). The reflection of the point (a, a2 ) in (a, 2) is the point (a, 4 − a2 ). Hence, the reflection of y = x2 in the line y = 2 is y = 4 − x2 . 115 Formal proof: Pick any point (a, a2 ) in the graph of y = x2 . It is possible to show that the closest point on the line x = 0 is (0, a2 ). The reflection of the point (a, a2 ) in (0, a2 ) is the point (−a, a2 ), which is also a point in the graph of y = x2 . We’ve just shown that the reflection of any point in the graph of y = x2 in the line x = 0 is itself also in the same graph. Hence, the graph of y = x2 is its own reflection in the line x = 0.

114

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1 Example 221. By informal observation,116 the graph of y = is symmetric in the lines x y = x and y = −x.

y

y = −x

y=

116

1 x

y=x

x

Formal proof: Pick any point (a, 1/a) in the graph of y = 1/x.

a − 1/a 1 a − 1/a , +2 )= 2 a 2 1 ( , a), which is in the graph of y = 1/x. We’ve just shown that the reflection of any point in the graph a of y = 1/x in the line y = x is itself also in the same graph. Hence, the graph of y = 1/x is its own reflection in the line y = x. a + 1/a 1 a + 1/a 1 Similarly, the reflection of (a, 1/a) in the line y = −x is (a − 2 , −2 ) = (− , −a), which 2 a 2 a is in the graph of y = 1/x. We’ve just shown that the reflection of any point in the graph of y = 1/x in the line y = −x is itself also in the same graph. Hence, the graph of y = 1/x is its own reflection in the line y = −x. By Fact 231 (Appendices), the reflection of (a, 1/a) in the line y = x is (a − 2

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Example 222. By informal observation,117 the graph of x2 + y 2 = 1 is symmetric in every line through the origin. For example, it is symmetric in the lines y = 2x and y = −x.

y = −x

y

y = 2x

x2 + y 2 = 1

x

Exercise 89. Find any line(s) of symmetry for y = x2 + 2x + 2.

117

√ Formal proof: Pick any point (a, ± 1 − a2 ) in the graph of x2 + y 2 = 1. √ By Fact 231 (Appendices), the reflection of (a, ± 1 − a2 ) in the line y = kx (for k ∈ R) is √ √ √ √ √ a − k (± 1 − a2 ) √ a − k (± 1 − a2 ) ⎞ ⎛ ak 2 − a ± 2k 1 − a2 ± 1 − a2 + 2ak ∓ k 2 1 − a2 2 a−2 , ± 1 − a + 2k =( , ), 1 + k2 1 + k2 1 + k2 1 + k2 ⎝ ⎠ which we can verify is on the graph of x2 + y 2 = 1: √ √ √ 2 2 ak 2 − a ± 2k 1 − a2 ± 1 − a2 + 2ak ∓ k 2 1 − a2 ( ) +( ) 1 + k2 1 + k2 = =

2 √ (( √ (( √((( 2 2 k 2 + k 4 (1−a 2) ( 2 k 2 ±4ak    (((1( (((1( [ ∓4ak − a2 ] + [1 −a2 +  4a ±4ak − a2 − 2k 2 (1 − a2 k4 + a2 + 4k −a2 )  −2a  (((3( 1 − a2(  (1

(1 + k 2 )

2

2k 2 + 1 + k 4 (1 + k 2 )

2

= 1.

We’ve just shown that the reflection of any point in the graph of x2 + y 2 = 1 in the line y = kx (for k ∈ R) is itself also in the same graph. Hence, the graph of x2 + y 2 = 1 is its own reflection in the line y = kx (for k ∈ R).

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13.

Solutions and Solution Sets

To solve an equation is to find all its solutions (or roots). Equivalently, to solve an equation is to find that equation’s solution set. Example 223. Solve the equation x − 1 = 0 (x ∈ R).

Here are three perfectly good (and equivalent) answers to the above problem: • “x = 1.” • “x ∈ {1}.”

• “The equation’s solution set is {1}.”

Example 224. Solve the equation x + 5 = 8 (x ∈ R).

Here are three perfectly good (and equivalent) answers to the above problem: • “x = 3.”

• “x ∈ {3}.” • “The equation’s solution set is {3}.”

The above two examples were easy enough to solve. In the next chapter, we’ll review the solution to the quadratic equation ax2 + bx + c = 0. For now, here’s a quick example:

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Example 225. Consider the equation x2 − 1 = 8 (x ∈ R). Let’s see how graphs can help us solve this equation.

Graph y = x2 − 1 and y = 8. We find that these two graphs intersect at x = −3 and x = 3.

(−3, 8)

y=8

y

(3, 8)

y = x2 − 1

x

Thus, the equation x2 − 1 = 8 (x ∈ R) has two solutions: −3 and 3. Equivalently, its solution set is {−3, 3} or {±3}.

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We won’t be learning to solve the cubic equation. Nonetheless, here’s a quick example: Example 226. Consider the equation x3 − 3x = 3x − x2 (x ∈ R).

Graph y = x3 − 3x and y = 3x − x2 . We find that these two graphs intersect at x = −3, x = 0, and x = 2.

y

(2, 2)

y = x3 − 3x

x

(0, 0) y = 3x − x2 (−3, −18)

Thus, the equation x3 −3x = 3x−x2 (x ∈ R) has three solutions: −3, 0, and 2. Equivalently, its solution set is {−3, 0, 2}. By the way, don’t worry. We don’t know how to solve cubic equations and we won’t be learning to do so.

However, we are required to know how to use a graphing calculator to find the solutions to this and indeed just about any equation. We’ll learn how to do so later.

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Like the above examples, most equations we’ll encounter will have only finitely many solutions. But this isn’t always the case: Example 227. Consider the equation sin x = 0 (x ∈ R).

It has infinitely many solutions—for every k ∈ Z, kπ is a solution. The equation’s solution set is {kπ ∶ k ∈ Z}.

y = sin x

(−2π, 0)

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(−π, 0)

y

(0, 0)

(π, 0)

(2π, 0)

x

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We now look at inequalities. To solve an inequality is to find all its solutions (or roots). Equivalently, to solve an inequality is to find its solution set. Inequalities usually have infinitely many solutions: Example 228. Solve x − 1 > 0 (x ∈ R). 1

As before, here are three perfectly good (and equivalent) answers to the above problem: • “x > 1.”

• “x ∈ (1, ∞).”

• “The solution set of > is (1, ∞).” 1

Example 229. Solve x + 5 ≥ 8 (x ∈ R). 2

Three perfectly good (and equivalent) answers to the above problem: • “x ≥ 3.”

• “x ∈ [3, ∞).”

• “The solution set of ≥ is [3, ∞).” 2

The above two inequalities were easy enough to solve. In Ch. 37.2, we’ll learn the general solution to the quadratic inequality ax2 + bx + c > 0. For now, here’s a quick example:

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Example 230. Solve x2 − 1 > 8 (x ∈ R).

Again, let’s see how graphs can help us solve this inequality. Graph y = x2 − 1 and y = 8. We find that the graph of y = x2 −1 is above that of y = 8 in the light-red regions indicated below (excluding the dotted red lines).

Thus, three perfectly good (and equivalent) answers to the above problem are • “x < −3, x > 3.” • “x ∈ (−∞, −3) ∪ (3, ∞) = R ∖ [−3, 3].”

• “The solution set is (−∞, −3) ∪ (3, ∞) = R ∖ [−3, 3].”

Also, we may say in words that every real number except those between −3 and 3 (inclusive) is a solution.

(−3, 8)

y=8

y (3, 8)

y = x2 − 1

x

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Example 231. Solve x3 − 3x ≥ 3x − x2 .

Graph y = x3 − 3x and y = 3x − x2 . We find that the graph of y = x3 − 3x is above that of y = 3x − x2 in the light-red regions (including the solid red lines) indicated below. Thus, three perfectly good (and equivalent) answers to the above problem are • “−3 ≤ x ≤ 0, x ≥ 2.”

• “x ∈ [−3, 0] ∪ [2, ∞).”

• “The solution set is [−3, 0] ∪ [2, ∞).”

Also, we may say in words that every real number between −3 and 0 (inclusive) and every real number greater than or equal to 2 is a solution.

y (2, 2) y = x3 − 3x

x

y = 3x − x2

(−3, −18)

Here are the formal definitions of a solution and a solution set: Definition 61. Given an equation (or inequality) in one variable, any number that satisfies the equation (or inequality) is called a solution (or root) of that equation (or inequality). And the set of all such solutions is called the solution set of that equation (or inequality). Remark 36. Note that historically, the A-Level exams have never mentioned the concept of a solution set. Nonetheless, it is quite convenient and this textbook will use it. For now, we’ll be dealing only with real numbers. And so for now, we’ll be looking only at real solutions and real solution sets. But in Part III (Complex Numbers), we’ll learn also that solutions can include non-real or 161, Contents

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complex numbers. Here’s a quick preview: Example 232. The equation x2 +1 = 0 (x ∈ R) has no solution. (We say that its solution set is ∅, the empty set.) But this is only because we’ve specified that x ∈ R.

In contrast, the equation x2 + 1 = 0 (x ∈ C) has two solutions: −i and i. (We say that its solution set is {−i, i}.)

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14.

O-Level Review: The Quadratic Equation y = ax2 + bx + c

In this chapter, we’ll review the quadratic equation: 0 = ax2 + bx + c

(If a = 0, then the equation is not quadratic but linear: 0 = bx + c.)

(a ≠ 0).

Let’s find the solutions (or roots) of this quadratic equation:

Divide both sides by a ≠ 0: b2 Add and subtract 2 : 4a

b c 0 = x2 + x + . a a

b b2 b2 c 0=x + x+ 2 − 2 + . a 4a 4a a 2

By the way, the above step is an example of the Plus Zero Trick:118 It is always perfectly legitimate to add zero to any expression. So, given any quantity q (in this case q = b2 /4a2 ), it is also always legitimate to add +q − q = 0 to any expression. Complete the square: Rearrange:

Take the square root:

0 = (x +

c b 2 b2 ) − 2+ . 2a 4a a

b 2 b2 c b2 − 4ac (x + ) = 2 − = . 2a 4a a 4a2 b x+ =± 2a

√ b2 − 4ac ± b2 − 4ac = 4a2 2a

Rearrange to get the quadratic formula (i.e. the two solutions or roots of the quadratic equation):

√ −b ± b2 − 4ac . x= 2a

The quadratic formula is not on the List of Formulae (MF26) and so sadly, you’ll have to memorise it. Unfortunately, I’ve never come across a good mnemonic that works (for me). Look for one that works for you. (Lemme know if you think you’ve found a good one!)

118

Tao (2016) calls this the “middle-man trick”.

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One mnemonic is to sing the quadratic formula to the tune of Pop Goes the Weasel.119 Arranged by a musical genius so that each syllable matches each note: Moderato

x e-quals to ne-ga-tive b,

Plus or mi-nus square root,

b squared mi-nus four a c,

All!

O-ver two a.

Definition 62. The discriminant of the quadratic equation 0 = ax2 + bx + c is b2 − 4ac.

Remark 37. Some writers denote the discriminant by the symbol D or ∆.120 (Your H2 Maths syllabus and exams don’t seem to do so, so neither shall we.) On the next page, Fact 27 summarises the key features of the quadratic equation and also reviews some of the concepts we’ve gone through in previous chapters. Six examples follow.

I checked out about ten versions on YouTube and unfortunately I found all of them to be very annoying and cannot recommend them. Maybe I’ll make one—don’t worry, I won’t be the one singing. 120 The symbol ∆ is the upper-case Greek letter delta. 119

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The equation y = ax2 + bx + c is also the quadratic equation, but in two variables. Fact 27. Consider the graph of the quadratic equation y = ax2 + bx + c.

(a) The only y-intercept is (0, c).

(b) There are two, one, or zero x-intercepts, depending on the sign of b2 − 4ac: (i) If b2 − 4ac > 0, then there are two x-intercepts: √ −b ± b2 − 4ac x= . 2a √ √ 2 − 4ac −b + b b2 − 4ac −b − And ax2 + bx + c = a (x − ) (x − ). 2a 2a

(ii) If b2 − 4ac = 0, then there is one x-intercept x = −b/2a (where the graph just touches the x-axis). And

b 2 ax + bx + c = a (x + ) . 2a 2

(iii) If b2 − 4ac < 0, then there are no x-intercepts. There is also no way to factorise the quadratic polynomial ax2 + bx + c (unless we use complex numbers).

(c) It is symmetric in the vertical line x = −b/2a. (d) The only turning point is (−b/2a, −b2 /4a + c), which is a strict global (i) minimum if a > 0; or (ii) maximum if a < 0.

Proof. See p. 1543 in the Appendices.

Corollary 2 (informal). Consider the graph of the quadratic equation y = ax2 + bx + c.

(a) If a > 0, then the graph is ∪-shaped. (b) If a < 0, then the graph is ∩-shaped.

“Proof.” “Clearly”, the graph is either ∪-shaped or ∩-shaped.121

(a) By Fact 27(d)(i), if a > 0, then the graph has a strict global minimum. So, the graph must be ∪-shaped.

(b) By Fact 27(d)(ii), if a < 0, then the graph has a strict global maximum. So, the graph must be ∩-shaped. Remark 38. In mathematics, a corollary is a statement that follows readily from another result. (Here for example, Corollary 2 follows readily from Fact 27.)

121

It is this bit here that makes the result informal. We have to more precisely define the terms “∪-shaped” and “∩-shaped”. One simple possibility is to replace these two terms with the terms convex and concave. However, we’ll only be introducing these latter two terms in Part V (Calculus). So, for now, we’ll just stick with the informal “∪-shaped” and “∩-shaped”.

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We can distinguish between six cases of the quadratic equation, depending on whether of a ≷ 0 and whether b2 − 4ac ⪋ 0. Here are six examples to illustrate the six cases:

Example 233. Consider the quadratic equation y = x2 + 3x + 1.

x=−

y

3 2

y = x2 + 3x + 1

D = (0, 1)

√ −3 + 5 C=( , 0) 2

√ −3 − 5 , 0) A=( 2

x

3 5 B = (− , − ) 2 4

1. The only y-intercept is D = (0, 1).

2. The equation 0 = x2 + 3x + 1 has two real solutions or roots: √ √ √ −b ± b2 − 4ac −3 ± 32 − 4 × 1 × 1 −3 ± 5 x= = = . 2a 2×1 2 √ √ −3 − 5 −3 + 5 So, the two x-intercepts are A = ( , 0) and C = ( , 0). 2 2 And we may write

√ √ −3 − 5 −3 + 5 x + 3x + 1 = (x − ) (x − ). 2 2 2

3. The graph is symmetric in the line x = −

3 3 b =− =− . 2a 2×1 2

b b2 3 32 3 5 4. The only turning point is B = (− , c − ) = (− , 1 − ) = (− , − ). 2a 4a 2 4×1 2 4 Since a = 1 > 0, B is the strict global minimum and the graph is ∪-shaped.

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Example 234. Consider the quadratic equation y = x2 + 2x + 1.

x = −1

y

y = x2 + 2x + 1

B = (0, 1)

A = (−1, 0)

x

1. There is one y-intercept: B = (0, 1).

2. The equation y = x2 + 2x + 1 has one real root: √ √ √ −b ± b2 − 4ac −2 ± 22 − 4 × 1 × 1 −2 ± 0 = = = −1. x= 2a 2×1 2

And so, there is one x-intercept, where the graph just touches the x-axis: A = (−1, 0).

We can factorise the quadratic polynomial:

x2 + 2x + 1 = [x − (−1)] = (x + 1) . 2

2

b 2 =− = −1. 2a 2×1 4. In general, if a quadratic equation has only one real root, then the turning point is also the x-intercept: 3. There is one (vertical) line of symmetry: x = −

b b2 22 A = (− , c − ) = (−1, 1 − ) = (−1, 0) . 2a 4a 4×1

5. Since the coefficient 1 on x2 is positive, the graph is ∪-shaped, with the turning point being the strict global minimum.

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Example 235. Consider the quadratic equation y = x2 + x + 1.

1 x=− 2

1 3 A = (− , ) 2 4

y

y = x2 + x + 1

B = (0, 1)

x

1. There is one y-intercept: B = (0, 1).

2. The equation y = x2 + x + 1 has no real roots: √ √ √ 2 −b ± 2 − 4ac −2 ± 12 − 4 × 1 × 1 −2 ± −3 = = . x= 2a 2×1 2

And so, there are no x-intercepts. We cannot factorise the quadratic polynomial (unless we use complex numbers). b 1 1 3. There is one (vertical) line of symmetry: x = − = − =− . 2a 2×1 2 4. The one turning point is: b b2 1 12 1 3 A = (− , c − ) = (− , 1 − ) = (− , ) . 2a 4a 2 4×1 2 4

5. Since the coefficient 1 on x2 is positive, the graph is ∪-shaped, with the turning point being the strict global minimum.

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Example 236. Consider the quadratic equation y = −x2 + 3x − 1.

y

x= 3 5 C=( , ) 2 4

A = (0, 1)

√ 3− 5 , 0) B=( 2

3 2 √ 3+ 5 D=( , 0) 2

x

y = −x2 + 3x − 1 1. There is one y-intercept: A = (0, 1).

2. The equation y = −x2 + 3x − 1 has two real roots: √ √ √ √ 2 −b ± b − 4ac −3 ± 32 − 4 × (−1) × (−1) −3 ± 5 3 ∓ 5 = = = . x= 2a 2 × (−1) −2 2 And so, there are two x-intercepts: √ √ 3+ 5 3− 5 B=( , 0) and D = ( , 0). 2 2 We can factorise the quadratic polynomial:

√ √ 3 − 5 5 3 + −x2 + 3x − 1 = − (x − ) (x − ). 2 2

3. There is one (vertical) line of symmetry: x = −

4. The one turning point is:

b 3 3 =− = . 2a 2 × (−1) 2

b b2 3 32 3 5 C = (− , c − ) = ( , −1 − ) = ( , ). 2a 4a 2 4 × (−1) 2 4

5. Since the coefficient −1 on x2 is negative, the graph is ∩-shaped, with the turning point being the strict global maximum. 169, Contents

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Example 237. Consider the quadratic equation y = −x2 + 2x − 1.

y

x=1

A = (0, −1)

B = (1, 0)

x

y = −x2 + 2x − 1 1. There is one y-intercept: A = (0, −1).

2. The equation y = −x2 + 2x − 1 has one real root: √ √ √ −b ± b2 − 4ac −2 ± 22 − 4 × (−1) × (−1) −2 ± 0 = = = 1. x= 2a 2 × (−1) −2

And so, there is one x-intercept, which is also where it just touches the x-axis: B = (1, 0).

We can factorise the quadratic polynomial:

−x2 + 2x − 1 = − (x − 1) . 2

2 b =− = 1. 2a 2 × (−1) 4. In general, if a quadratic equation has only one real root, then the turning point is also the x-intercept:

3. There is one (vertical) line of symmetry: x = − B = (−

b b2 22 , c − ) = (1, 1 − ) = (1, 0) . 2a 4a 4 × (−1)

5. Since the coefficient −1 on x2 is negative, the graph is ∩-shaped, with the turning point being the strict global maximum. 170, Contents

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Example 238. Consider the quadratic equation y = −x2 + x − 1.

y

x=

A = (0, −1)

1 2

x 1 3 B = ( ,− ) 2 4

1. There is one y-intercept: A = (0, −1).

2. The equation y = −x2 + x − 1 has no real roots: √ √ √ −b ± 22 − 4ac −1 ± 12 − 4 × (−1) × (−1) −1 ± −3 = = . x= 2a 2 × (−1) −2

And so, there are no x-intercepts. We cannot factorise the quadratic polynomial (unless we use complex numbers). b 1 1 3. There is one (vertical) line of symmetry: x = − = − = . 2a 2 × (−1) 2 4. The one turning point is: b2 1 12 1 3 b B = (− , c − ) = ( , −1 − ) = ( ,− ). 2a 4a 2 4 × (−1) 2 4

5. Since the coefficient −1 on x2 is negative, the graph is ∩-shaped, with the turning point being the strict global maximum.

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The last six examples in one figure:

x2 + x + 1

x2 + 3x + 1

x2 + 2x + 1

y

−x2 + 3x − 1

x

−x2 + 2x − 1

−x2 + x − 1 Exercise 90. Sketch each graph, identifying any intercepts, lines of symmetry, turning points, and extrema. (a) y = 2x2 + x + 1. (b) y = −2x2 + x + 1. (c) y = x2 + 4x + 4.

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15.

Functions

Undoubtedly the most important concept in all of mathematics is that of a function—in almost every branch of modern mathematics functions turn out to be the central objects of investigation. — Michael Spivak (1994). In secondary school, you probably learnt to describe functions like this: “Let f (x) = 2x be a function.”

Strictly speaking,122 the above description of functions is incorrect. In particular, it suffers from these two big problems: 1. It fails to make any mention of the domain and the codomain. Whenever we specify a function, we must also specify the domain and codomain.123 2. It incorrectly suggests that a function must always be some sort of a “formula”. But it needn’t be. A function simply maps or assigns every element in the domain to (exactly) one element in the codomain. There need be nothing logical or formulaic about how this mapping or assignment is done. We will illustrate this important point with many examples below. Here is one correct way to describe functions:124 Definition 63 (informal). A function consists of these three objects: 1. A set called the domain; 2. A set called the codomain; and 3. A mapping rule that maps (or assigns) every element in the domain to (exactly) one element in the codomain. Two very simple examples of functions:

Pedagogical note: Many bits of H2 Maths are equivalent to first- and even second-year university courses in many countries. (Well, at least on paper. See my Preface/Rant.) My view is that while the above incorrect description of a function may have been suitable at earlier stages of a student’s maths education, it is no longer so at this stage. Definition 63 (below) is not at all difficult (especially when compared with a lot of the junk that’s already in H2 Maths). At the cost of very little additional pain and time, the student gains a far better understanding of what functions are and how they work, and thus saves herself more grief in the long run. 123 Things become so much simpler if we make it clear from the outset and indeed insist that the very definition of a function includes the specification of a domain and codomain. Instead we have the present rigmarole where we ask students to explain which values “to exclude” from a function’s domain, as if we were doing some ad hoc repair to make the function “work”. 124 Definition 63 will serve us very well. Note though that it is a little informal. For the formal definition, see Definition 274 (Appendices). 122

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Example 239. Let f be the function with 1. Domain: The set {1, 5, 3}. 2. Codomain: The set {100, 200}. 3. Mapping rule: f (1) = 100, f (5) = 100, f (3) = 200.

The function f is well-defined, because every element in the domain maps to or “hits”125 (exactly) one element in the codomain: 1 “hits” 100,

5 “hits” 100,

3 “hits” 200.

function There is no apparent logic or formula behind how theThe mapping rulef works. Why should 1 “hit” 100, 5 “hit” 100, and 3 “hit” 200? No matter. To qualify as a well-defined function, all that matters is that every element in the domain is mapped to or “hits” (exactly) one element in the codomain. Which is true of f . So, f is well-defined.

125

1

100 5

3 Domain f

200

Codomain f

The term “hits” is informal (hence the scare quotes). The formal term is maps to. But “hits” is probably clearer, more evocative, and easier to understand.

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Example 240. Let g be the function with 1. Domain: The set {1, 5, 3}. 2. Codomain: The set {100, 200}. 3. Mapping rule: g (1) = 100, g (5) = 100, g (3) = 100.

Observe that the element 200 in the codomain is not “hit”. Nonetheless, g is again a welldefined function, because it satisfies the requirement that every element in the domain maps to or “hits” (exactly) one element in the codomain: 1 “hits” 100, 5 “hits” 100, 3 “hits” 100. For a function to be welldefined, there is no requirement that every element in the codomain be “hit”.

The function g

1

100 5

3 Domain g

200

Codomain g

And so here, even though 200 in the codomain isn’t “hit”, g is a perfectly well-defined function.

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The concept of a function is of great generality. To qualify as a function, all we require is that every element in the domain maps to or “hit” (exactly) one element in the codomain. In each of our above examples, both the domain and codomain were sets of real numbers. But in general, they could be any sets whatsoever. We’ll use these six symbols in this chapter and again in Ch. 22: Cow Chicken Dog Produces eggs Guards the home Produces milk

🐄

🐔

Example 241. Let h be the function with

🐄

1. Domain: The set { , 🐔}. 2. Codomain: The set { , , }. 3. Mapping rule: “Match the animal to its role.” Here the mapping rule is informally stated in words. Nonetheless, it is clear enough. If we wanted to, we could write it out formally, like this: h(

🐄) =

h (🐔) =

🐄

The function h

, .

🐔

The function h is welldefined, because it satisfies the requirement that every Domain h element in the domain “hits” (exactly) one element in the codomain.

Codomain h

In the above example, the mapping rule “makes sense”—we simply map each animal to its role. In the next example, the mapping rule “makes no sense”. Nonetheless and all the same, our function is perfectly well-defined:

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Example 242. Let i be the function with

🐄

1. Domain: The set { , 🐔}. 2. Codomain: The set { , , }. ) = , i (🐔) = . 3. Mapping rule: i ( The function i This time, the mapping rule “makes no sense”—it maps to and 🐔to . Nonetheless, this is a welldefined function, because it maps every element in the domain to (exactly) one element in the codomain.

🐄

🐄

🐄 🐔

Domain i

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Codomain i

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Another pair of examples: Example 243. Let j be the function with 1. Domain: The set of UN member states. 2. Codomain: The set of ISO three-letter country codes. 3. Mapping rule: “Match the state to its code.” The domain, codomain, and mapping rule are all informally stated in words. Nonetheless, they are clear enough. The domain contains 193 elements, namely Afghanistan, Albania, Algeria, ..., Zambia, and Zimbabwe (list). The codomain contains 249 elements, namely ABW, AFG, AGO, AIA, ALA, ..., ZMB, and ZWE (list).126 And the mapping rule maps every element in the domain to (exactly) one element in the codomain. For example, j (The United States of America) = USA, j (China) = CHN, j (Singapore) = SGP.

Altogether then, j is a well-defined function, because it satisfies the requirement that every element in the domain “hits” exactly one element in the codomain. Again, a very similar example, but this time the mapping rule “makes no sense”: Example 244. Let k be the function with 1. Domain: The set of UN member states. 2. Codomain: The set of ISO three-letter country codes. 3. Mapping rule: For every x, k (x) = SGP.

The function k simply maps every UN member state to the three-letter code SGP. This is strange and seems to make no sense. Nonetheless, k is a well-defined function, because it satisfies the requirement that every element in the domain “hits” exactly one element in the codomain. So! Specifying a function is jolly simple. Simply write down the 1. Domain (could be any set); 2. Codomain (could be any set); 3. Mapping rule (maps every element in the domain to exactly one element in the 126

There are more ISO three-letter country codes than UN members because not every “country” is a UN member. For example, Greenland is assigned the ISO code GRL, but is a territory of Denmark and is not a UN member state. The Holy See (or Vatican City) is a fully sovereign state and is assigned the ISO code VAT, but is not a UN member state. Taiwan is, for all intents and purposes, a fully sovereign state and is assigned the ISO code TWN, but is not a UN member state.

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codomain).

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Now, a function consists of the above three pieces. Hence—and here’s a somewhat subtle point—two functions are identical (i.e. equal) if and only if they have the same domain, codomain, and mapping rule.

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Example 245. Recall from above the function h:

🐄

The function h

🐔 Codomain h

Domain h Now consider the very similar-looking function h1 :

🐄

The function h1

🐔 Codomain h1

Domain h1

The functions h and h1 look very similar. Indeed, they have the same domain and mapping rule. However, their codomains are different. And so h and h1 are distinct:127 h ≠ h1 .

One might think, “Aiyah, the codomain not very important wat. Both h and h1 map to and 🐔to . So just call them the same function lah!” But this thinking is wrong.

🐄

Again, to reiterate, stress, and emphasise, a function consists of three pieces: the domain, the codomain, and the mapping rule. So, Two functions are identical ⇐⇒ They have the same domain, codomain, and mapping rule.

127

Distinct is a synonym for not equal.

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Example 246. Let l be the function with 1. Domain: The set {1, 2}. 2. Codomain: The set {1, 2, 3, 4}. 3. Mapping rule: For every x, l (x) = 2x.

The function l

1

1

2 3

2

4 Codomain l

Domain l

Now consider the very similar-looking function l1 , which has 1. Domain: The set {1, 2}. 2. Codomain: The set {1, 2, 4}. 3. Mapping rule: For every x, l1 (x) = 2x.

The function l1

1

1

2

2

4 Codomain l1

Domain l1

Again, the functions l and l1 look very similar—they have the same domain and mapping rule. However, their codomains are different. And so, l and l1 are distinct: l ≠ l1 .

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Exercise 91. Fill in the blanks.

A function consists of ____ objects: namely, the ____, the ____, and the ____. Exercise 92. Fill in the blanks with “can be any set”; “must be R”; or “must be a subset of R”. (Answer on p. 1744.) In general, the domain ______; and the codomain _____. Exercise 93. Fill in the blanks with “at least one element”; “every element”; or “exactly one element”. (Answer on p. 1744.) A function maps _____ in its domain to _____ in its codomain.

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15.1.

What Functions Aren’t

To better understand what functions are, we’ll now look at what functions aren’t. That is, we’ll now look at examples of non-functions: Example 247. It is alleged that f is the function with 1. Domain: The set {1, 5, 3}. 2. Codomain: The set {100, 200}. 3. Mapping rule: f (1) = 100, f (3) = 200.

The “function” f

1

100 5

3 Domain f

200

Codomain f

Unfortunately, f is not a function, because f fails to map every element in the domain to (exactly) one element in the codomain—in particular, f fails to map 5 to any element in the codomain.

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Example 248. It is alleged that g is the function with 1. Domain: The set {1, 5, 3}. 2. Codomain: The set {100, 200}. 3. Mapping rule: g (1) = 100, g (1) = 200, g (5) = 100, g (3) = 200.

The “function” g

1

100 5 200

3 Domain g

Codomain g

Unfortunately, g is not a function, because g fails to map every element in the domain to (exactly) one element in the codomain—in particular, g maps 1 to two elements, namely 100 and 200. Example 249. It is alleged that h is the function with 1. Domain: The set of ISO three-letter country codes. 2. Codomain: The set of UN member states. 3. Mapping rule: “Match the code to the corresponding state.” Unfortunately, h is not a function, because h fails to map every element in the domain to (exactly) one element in the codomain. For example, h fails to map VAT in the domain to any element in the codomain. This is because VAT corresponds to the Holy See (or the Vatican City), which is not a UN member state and thus not an element of the codomain. Example 250. It is alleged that i is the function with 1. Domain: The set of real numbers R. 2. Codomain: The set of real numbers R. √ 3. Mapping rule: For all x, i (x) = x.

Unfortunately, i is not a function, because i fails to map every element in the domain to (exactly) one element in the codomain.

For example, i fails to map −1 in the domain to any element in the codomain. This is √ because −1 is not a real number.

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Exercise 94. Below are four alleged functions. Which are well-defined? (a) The function a has domain {Cow, Chicken, Dog}, codomain {Produces eggs, Guards the home, Produces milk}, and mapping rule “match the animal to its role”. (b) The function b has domain {Cow, Chicken, Dog}, codomain {Produces eggs, Produces milk}, and mapping rule “match the animal to its role”. (c) The function c has domain the set of UN member states, codomain the set of cities, and mapping rule “match the state to its most splendid city”. (d) The function d has domain the set of UN member states, codomain the set of cities, and mapping rule “match the state to a city with over 10M people”. (Answer on p. 1744.)

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15.2.

Notation for Functions

We first introduce some shorthand notation for “the domain of the function f ” and “the codomain of the function f ”: Definition 64. Given a function f , the symbol Domain f denotes its domain, while the symbol Codomain f denotes its codomain. Example 251. Let f be the function with 1. Domain: The set {3, 4}. 2. Codomain: The set {5, 6, 7, 8}. 3. Mapping rule: “Double it.”

The function f

5

3

6 7

4

8

Domain f We may write

Codomain f

Domain f = {3, 4}, Codomain f = {5, 6, 7, 8}.

Remark 39. The notation Domain f and Codomain f are not standard. For example, in place of Domain f , others may instead write Domain (f ), Dom f , Dom (f ), D, Df , D (f ), D, Df , or D (f )—among countless other variations.

Your A-Level maths syllabus and exams do not seem to have any shorthand notation at all for “the domain of the function f ” and “the codomain of the function f ”. Nevertheless, this textbook will use Domain f and Codomain f because they are convenient. We now discuss how the mapping rule is written. To write down a mapping rule, there is usually more than one way. In the above example, the mapping rule was written informally as “Double it”. But we could also have written it more formally as “For every x ∈ Domain f , we have f (x) = 2x.”

Alternatively, we could’ve written the mapping rule explicitly, stating what each and every element in the domain is to be mapped to: 187, Contents

“f (3) = 6,

f (4) = 8.”

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The mathematical punctuation mark ↦ means maps to. And so, here’s yet another way to write the mapping rule: “f ∶ 3 ↦ 6,

f ∶ 4 ↦ 8.”

So, altogether, in the above example alone, we’ve given four different but entirely equivalent ways to write out the mapping rule. You can choose to write the mapping rule however you like. What’s important is that you make clear how the mapping rule maps each element in the domain to (exactly) one element in the codomain. If you haven’t made it sufficiently clear, then you have failed to communicate to others what your function is and your function is not well-defined. Here are eight ways to say aloud “f (4) = 8” or “f ∶ 4 ↦ 8”: “f of 4 is 8.”

“f of 4 equals 8.”

“f of 4 is equal to 8.”

“f maps 4 to 8.”

“The value of f at 1 is 8.”

“f evaluated at 4 is 8.”

“The value of f when applied to 4 is 8.”

“The image of 4 under f is 8.”

If x ∉ Domain f , then f (x) is simply undefined. So in the last example, 0 ∉ Domain f ; and thus, f (0) is simply undefined.

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Example 252. Let g be the function with 1. Domain: The set of positive integers Z. 2. Codomain: The set of real numbers R. 3. Mapping rule: “Double it.” Then we have, for example, Or equivalently,

g (1) = 2,

g (2) = 4,

and

g (3) = 6.

g ∶ 1 ↦ 2, g ∶ 2 ↦ 4, and g ∶ 3 ↦ 6.

Since −1, 1.5, π, 0 ∉ Domain g, g (−1), g (1.5), g (π), and g (0) are undefined.

Likewise, since

🐄, 🐔 ∉ Domain g, g (🐄) and g (🐔) are undefined.

The mapping rule, “Double it,” was somewhat informal. We could’ve written it more formally as For every x ∈ Z, we have g (x) = 2x.

With the aid of an ellipsis, we could also have written it down explicitly:128

Or equivalently,

g (1) = 2, g (2) = 4, g (3) = 6, g (4) = 8, . . . g ∶ 1 ↦ 2, g ∶ 2 ↦ 4, g ∶ 3 ↦ 6, g ∶ 4 ↦ 8, . . .

So far, we’ve written down functions with the aid of tables and/or figures. But going forward, we’ll want to write down functions more concisely and without the aid of tables or figures. Here are nine entirely equivalent and formal ways to fully write down the function g from the last example: 1. Let g be the function that maps every element x in the domain Z to the element 2x in the codomain R. 2. Let g be the function that has domain Z, codomain R, and maps every x ∈ Z to 2x ∈ R. 3. Let g ∶ Z → R be the function defined by g (x) = 2x. 4. Let g ∶ Z → R be the function defined by g ∶ x ↦ 2x. 5. Let g ∶ Z → R be defined by g (x) = 2x. 6. Let g ∶ Z → R be defined by g ∶ x ↦ 2x. 7. Define g ∶ Z → R by g (x) = 2x. 8. Define g ∶ Z → R by g ∶ x ↦ 2x. 128

But in general, this may not be possible.

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9. Define g ∶ Z → R by x ↦ 2x.

In Statements 3–9, the domain comes after the colon “∶”, the codomain after the right arrow “→”, and the mapping rule at the end of the statement. The general version of Statement 9 is

Codomain f Domain f ↖ ↗ Define f ∶ A → B by x ↦ f (x) . ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ Mapping rule

In words, the function f is defined to have domain A, codomain B, and mapping rule x ↦ f (x). Example 253. Define h ∶ R+ → R by h (x) = 2x.

The domain of h is R+ (the set of positive real numbers). The codomain is R (the set of real numbers). Expressed informally in words, the mapping rule is “Double it”.

So for example, Or equivalently, Observe that −1, 0,

h (1) = 2, h (2.3) = 4.6, and h (−3) = −6. h ∶ 1 ↦ 2, h ∶ 2.3 ↦ 4.6, and h ∶ −3 ↦ −6.

🐄, 🐔 ∉ R

+

= Domain h. Thus, the following are all undefined:

h (−1), h (0), h (

🐄), and h (🐔).

Example 254. Define i ∶ Z → R by i (x) = x2 .

The domain of i is Z (the set of integers). The codomain is R (the set of real numbers). Expressed informally in words, the mapping rule is “Square it”. So for example, Or equivalently, Observe that 1.5, π,

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i (1) = 1, i (7) = 49, and i (−6) = 36. i ∶ 1 ↦ 1, i ∶ 7 ↦ 49, and i ∶ −6 ↦ 36.

🐄, 🐔 ∉ Z = Domain i. Thus, the following are all undefined: h (1.5), i (π), i (🐄), and i (🐔).

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Example 255. Define j ∶ R+ → R by j (x) = x2 .

The domain of j is R+ (the set of positive real numbers). The codomain is R (the set of real numbers). Expressed informally in words, the mapping rule is “Square it”. So for example, Or equivalently, Observe that −1, −3.2,

j (1) = 1, j (2.2) = 4.84, and j (−6) = 36. j ∶ 1 ↦ 1, j ∶ 2.2 ↦ 4.84, and j ∶ −6 ↦ 36.

🐄, 🐔 ∉ Z = Domain j. Thus, the following are all undefined: j (−1), j (−3.2), j (🐄), and j (🐔).

Note that as usual, the symbol x here is merely a dummy variable that can be replaced by any other symbol, such as y, z, ,, or ☀. So, we could also have written this example’s first sentence as either “Define j ∶ R+ → R by j (y) = y 2 ”

or

“Define j ∶ R+ → R by j (,) = ,2 ”.

(Consider the last four examples. Is g = h? Is i = j?)129

In the above examples, we actually cheated a little. Or rather, we took it for granted that the specified mapping rule applied to all elements in the domain. If we wanted to be extra careful/pedantic, then here’s what we should “really” have written: Example 256. Define g ∶ Z → R by g (x) = 2x for all x ∈ Z.

Example 257. Define h ∶ R+ → R by h (x) = 2x for all x ∈ R+ . Example 258. Define i ∶ Z → R by i (x) = x2 for all x ∈ Z.

Example 259. Define j ∶ R+ → R by j (x) = x2 for all x ∈ R+ .

This is because we can have a piecewise function, where there are different mapping rules for different “sections”, “pieces”, or intervals of the domain:

129

The functions g and h have different domains and so g ≠ h. Likewise, the functions i and j have different domains and so i ≠ j.

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Example 260. Define k ∶ Z → R by

⎧ ⎪ ⎪ ⎪2x, k (x) = ⎨ ⎪ ⎪ ⎪ ⎩x + 1,

for x ≤ 5, for x > 5.

In words, the function k doubles integers that are less than or equal to 5, but adds one to those that are greater than 5. So for example, But,

k (−10) = −20, k (0) = 0, and k (4) = 8.

k (7) = 8, k (15) = 16, and k (100) = 101.

Example 261. Define l ∶ R+ → R by

⎧ ⎪ ⎪ ⎪x2 , l (x) = ⎨ ⎪ 3 ⎪ ⎪ ⎩x ,

for x ≤ 5, for x > 5.

In words, the function l squares positive real numbers that are less than or equal to 5, but cubes those that are greater than 5. So for example, But,

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l (1) = 1, l (1.1) = 1.21, and l (4) = 64.

l (5.3) = 148.877, l (7) = 343, and l (9) = 729.

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Exercise 95. Evaluate each function at 1. Define

a∶R→R b ∶ [−1, 1] → R c ∶ Z+ → R d ∶ Z− → R e∶R→R

Exercise 96. Four functions are given below.

by by by by by

a (x) b (x) c (x) d (x) e (x)

= x + 1; = 17x; = 3x ; = 3x ; = 17.

(a) Verify that each is well-defined. (b) Which (if any) of them are equivalent? Function a b c d

Domain {1, 2} {1, 2, 3} {x ∈ Z+ ∶ x < 4} [0, 4) ∩ Z

Codomain {1, 2, 3, 4} {1, 2, 3, 4, 5, 6} {1, 2, 3, 4, 5, 6} (−3, 6] ∩ Z+0

Mapping “Double “Double “Double “Double

rule it” it” it” it”

Exercise 97. Define f ∶ R+ → R by “round off to the nearest integer (and round up half-integers)”. (Answer on p. 1744.)

(a) What are f (3), f (π), f (3.5), f (3.88), and f (0)?

Would the function still be well-defined if we changed , respectively,

(b) Its domain to R and codomain Z? (c) Its domain to Z and codomain R?

Exercise 98. Below, the functions f , g, and h are described in words. Write down a formal and concise statement that defines each. (Answer on p. 1744.) (a) The function f has domain the set of positive integers, codomain the set of integers, and mapping rule, “Triple it”. (b) The function g has domain the set of negative real numbers, codomain the set of non-zero real numbers, and mapping rule, “Square it”. (c) The function h has domain the set of even integers, codomain the set of positive even integers, and mapping rule, “Square it”.

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15.3.

Warning: f and f (x) Refer to Different Things

If you cannot say what you mean, Your Majesty, you will never mean what you say and a gentleman should always mean what he says. — Peter O’Toole as Reginald Johnson, in The Last Emperor (1987). A common mistake is to believe that f (x) denotes a function. But this is wrong.

f and f (x) refer to two different things.

f denotes a function.

f (x) denotes the value of f at x.

In the next two examples, S is the set of human beings.

Example 262. Let h ∶ S → R be the height function. That is, h gives each human being’s height (rounded to the nearest centimetre). So for example, h (Joseph Schooling) = 184.

h (Joseph Schooling) is not a function.

Instead, h (Joseph Schooling) is the number 184.

It would be silly to say that h (Joseph Schooling) is a function, because it isn’t— h (Joseph Schooling) is the number 184.

And in general, for any human being x, h (x) is her height (rounded to the nearest cm). Again, h (x) is not a function—it’s x’s height (rounded to the nearest cm), which is a number.

Example 263. Let w ∶ S → R be the weight function. That is, w gives each human being’s weight (rounded to the nearest kilogram). So for example, w (Joseph Schooling) = 74.

w (Joseph Schooling) is not a function.

Instead, w (Joseph Schooling) is the number 74.

It would be silly to say that w (Joseph Schooling) is a function, because it isn’t— w (Joseph Schooling) is the number 74.

And in general, for any human being x, w (x) is her weight (rounded to the nearest kg). Again, w (x) is not a function—it’s x’s weight (rounded to the nearest kg), which is a number.

This may seem like an excessively pedantic distinction. But maths is precise and pedantic. In maths, we are gentlemen (and ladies) who say what we mean and mean what we say. There is no room for ambiguity or alternative interpretations. 194, Contents

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15.4.

Nice Functions

Example 264. Consider the function f ∶ [1, 5] → R defined by f (x) = x2 .

The domain of f contains only real numbers. Hence, f is a function of a real variable. The codomain of f contains only real numbers. Hence, f is a real-valued function. Both the domain and codomain of f contain only real numbers. Hence, f is a nice function.

Definition 65. We call a function whose 1. Domain contains only real numbers a function of a real variable; 2. Codomain contains only real numbers a real-valued function (or more simply, real function); 3. Domain and codomain both contain only real numbers a nice function. Remark 40. The terms function of a real variable and real-valued function (and real function) are standard and widely used. However, the term nice function is completely non-standard and used only in this textbook. We use it nonetheless, just so we don’t have to keep saying “real-valued function of a real variable”. Example 265. Define the function g ∶ {

🐄, 🐔} → Z by g (🐄) = 0.

Domain g contains objects that aren’t real numbers. So, g is not a function of a real variable. Codomain g contains only real numbers. So, g is a real-valued function.

Either Domain g or Codomain g contains objects that aren’t real numbers. So, g is not a nice function. Example 266. Define the function h ∶ (−2, 5) → {

🐄,🐔} by h (x) = 🐄.

Domain h contains only real numbers. So, h is not a function of a real variable. Codomain h contains objects that aren’t real numbers. So, h is a real-valued function.

Either Domain h or Codomain h contains objects that aren’t real numbers. So, h is not a nice function. Happily, most functions we’ll encounter in H2 Maths are nice functions. That is, we’ll often encounter functions like f , but not functions like g or h.

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Exercise 99. Explain whether each function is a (i) function of a real variable; (ii) real-valued function; and (iii) nice function. (Answer on p. 1744.)

🐄,🐔, 2} → [1, 3] defined by i (x) = 3. (b) j ∶ [1, 3] → {🐄,🐔, 2} defined by j (x) = 2. (a) i ∶ {

(c) k ∶ {2} → [1, 3] defined by k (x) = 1. } defined by l (x) = 1. (d) l ∶ {2} → [1, 3] ∪ {

🐄

Exercise 100. What’s wrong with the following statement?

“Let f (x) be a function.”

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15.5.

Graphs of Functions

Given a function f , its graph is simply the graph of the equation y = f (x) with the constraint x ∈ Domain f . A little more formally, Definition 66. Given a function f , its graph is this set of points: {(x, y) ∶ x ∈ Domain f, y = f (x)} .

Example 267. Define f ∶ R → R by f (x) = 2x.

The graph of f is the set {(x, y) ∶ x ∈ R, y = 2x}.

y

x

By the way, strictly speaking, we should say “The point (3, 6) is in the graph of f .”

That is, we should explicitly state both the x- and y-coordinates of any point we are talking about. However, since the x-coordinate is sufficient for identifying a point on the graph of a function (why?),130 we will sometimes be lazy and say things like

Or even,

130

“The point x = 3 is on the graph of f .” “The point 3 is on the graph of f .”

A function maps each x-coordinate to exactly one y-coordinate. So, given only an x-coordinate, we immediately also know what the corresponding y-coordinate is.

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Example 268. Define f1 ∶ R+ → R by f1 (x) = 2x.

The graph of f1 is the set {(x, y) ∶ x ∈ R+ , y = 2x}.

y

x Note that the graph of f1 does not contain the point (0, 0), because 0 ∉ R+ . Example 269. Define f2 ∶ Z → R by f2 (x) = 2x.

The graph of f2 is the set {(x, y) ∶ x ∈ Z, y = 2x}. 8

y

4

-6

-4

-2

-4

2

4

x

-8 -12 The graph of f2 is simply this set of isolated points: {. . . , (−3, −3) , (−2, −2) , (−1, −1) , (0, 0) , (1, 1) , (2, 2) , (3, 3) , . . . }

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Example 270. Define g ∶ R → R by g (x) = x2 .

The graph of g is the set y {(x, y) ∶ x ∈ R, y = x2 }.

x Example 271. Define g1 ∶ R+ → R by g1 (x) = x2 .

The graph of g1 is the set {(x, y) ∶ x ∈ R+ , y = x2 }.

y

x Note that again, the graph of g1 does not contain the point (0, 0), because 0 ∉ R+ .

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Example 272. Define g2 ∶ Z → R by g2 (x) = x2 .

The graph of g2 is the set {(x, y) ∶ x ∈ Z, y = x2 }.

y

x

The graph of g2 is simply this set of isolated points: {. . . , (−3, 9) , (−2, 4) , (−1, 1) , (0, 0) , (1, 1) , (2, 4) , (3, 9) , . . . }

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15.6.

Piecewise Functions

Informally, a piecewise function is a function that has different mapping rules for different intervals (or “pieces”): Example 273. Define h ∶ R → R by

h

⎧ ⎪ ⎪ ⎪x2 , h (x) = ⎨ ⎪ ⎪x, ⎪ ⎩

for x ≤ 2,

y

for x > 2.

x

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Example 274. Define i ∶ R → R by

⎧ ⎪ ⎪ ⎪x, i (x) = ⎨ ⎪ ⎪ ⎪ ⎩0,

for x ∉ Z,

y

for x ∈ Z.

i

x −3

−2

1

−1

2

3

An important piecewise function is the absolute value function (definition reproduced from Ch. 5.2): Definition 25. The absolute value (or modulus) function, denoted ∣⋅∣, is defined for all x ∈ R by ⎧ ⎪ ⎪ ⎪x, ∣x∣ = ⎨ ⎪ ⎪ ⎪ ⎩−x,

for x ≥ 0, for x < 0.

Figure to be inserted here.

Another important piecewise function is the sign function: 202, Contents

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Definition 67. The sign function sgn ∶ R → R is defined by ⎧ ⎪ ⎪ −1, ⎪ ⎪ ⎪ ⎪ sgn x = ⎨0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩1,

for x < 0, for x = 0, for x > 0.

Figure to be inserted here.

Example 275. sgn 7 = 1, sgn − 5 = −1, sgn 0 = 0, sgn 100 = 1, sgn − 531 = −1.

The absolute value function is closely related to the sign function: Example 276.

7 7 −5 −5 = = 1 = sgn 7 and = = −5 = sgn − 7. ∣7∣ 7 ∣−5∣ 5

In Ch. 5.5, we gave this result: ⎧ ⎪ x x ⎪ ⎪1, Fact 13. = =⎨ ∣x∣ ∣x∣ ⎪ ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

.

Given the sign function, the above result may be restated thus: ⎧ ⎪ ⎪ x x ⎪1, = = sgn x = ⎨ Fact 28. ⎪ ∣x∣ ∣x∣ ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

And so, Examples 125 and 126 from Ch. 5.5 may be restated thus: Example 277. In Exercise 65 (Ch. 5.5), we gave this false statement: √ √ x2 x2 For all x ∈ R ∖ {0}, = √ = 1. x x2

7

The correct statement should instead be this:

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√ ⎧ ⎪ ⎪ x2 ∣x∣ ⎪1, For all x ∈ R ∖ {0}, = = sgn x = ⎨ ⎪ x x ⎪ ⎪ ⎩−1,

for x > 0, for x < 0.

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Example 278. Similarly, in Exercise 66 (Ch. 5.5), we gave this false statement: √ x2 x For all x ∈ R ∖ {0}, √ = √ = 1. x2 x2

7

The correct statement should instead be this:

⎧ ⎪ ⎪ x x ⎪1, For all x ∈ R ∖ {0}, √ = = sgn x = ⎨ ⎪ x2 ∣x∣ ⎪ ⎪ ⎩−1,

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for x > 0, for x < 0.

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Exercise 101. Let A = {Lion, Eagle} and B = {Fat, Tall}. Can we construct a welldefined function using A as the domain and B as the codomain? (Answer on p. 1745.) Exercise 102. What do we call a function whose ...

(a) Domain contains only real numbers? (b) Codomain contains only real numbers? (c) Domain and codomain contain only real numbers?

Exercise 103. Below are 17 alleged functions named a through q, with a given domain, codomain, and mapping rule (the last being given informally in words). (i) Explain whether each alleged function is actually well-defined. And if it is, (ii) Write down a statement that formally and concisely defines it.(Answer on p. 1745.) Function a b c d e f g h i j k l m n o p q

Domain {5, 6, 7} {5, 6, 7} {5, 6, 7} {5.4, 6, 7} {5.5, 6, 7} {3} {3, 3.1} {0, 3} {3, 4} {2, 4} {1} {1} {1, 2} R R R [0, 1]

Codomain Z Z+ Z− Z Z {3, 4} {3, 4} {3, 4} {3, 4} {3, 4} {1} {1, 2} {1} R R [0, 1] R

Mapping rule “Double it” “Double it” “Double it” “Double it” “Double it” “Any larger number” “Any larger number” “Any larger number” “Any larger number” “Any smaller number” “Stay the same” “Stay the same” “Stay the same” “Take the square root” “Take the reciprocal” “Add one” “Add one”

Exercise 104. Continuing with the above exercise, how can we change the domains of n and o so that n and o become well-defined?131 (Answer on p. 1746.) 131

The sharp student may have noticed that one trivial answer here is to simply change the domain to the empty set ∅. Then it is trivially or vacuously true that every element in the domain is mapped to exactly one element in the codomain. If you’ve noticed and are bothered by this, please change the question to, “What are the ‘largest’ subsets of R to which the domains of n and o can be changed, so

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15.7.

Range

Informally, the range (of a function) is the set of elements in the codomain that are “hit”. Formally, Definition 68. The range of a function f , denoted Range f , is the set defined by Range f = {f (x) ∶ x ∈ Domain f } .

Example 279. Define f ∶ [1, 2] → R by f (x) = x + 3. Then • Codomain f = R (the set of reals).

• Range f = [4, 5] (the set of reals between 4 and 5, inclusive).

Figure to be inserted here.

So, Range f ⊂ Codomain f and, in particular, Range f ≠ Codomain f .132

Example 280. Define g ∶ R → R by g (x) = x + 3. Then Codomain g = R (the set of reals).

Every element in the codomain R is “hit”. And so, the range is the same as the codomain: Range g = R.

Figure to be inserted here.

So, Range g = Codomain g (and Range g ⊆ Codomain g).

The range will always be a subset of the codomain. Equivalently, it will always be either a proper subset of the codomain (as in Example 279) or equal to the codomain (Example 280). Example 280 shows that the range can sometimes be identical to the codomain. But as Example 279 shows, this is not generally true—in general, not every element in the codomain will be “hit”. Because this is such a common point of confusion, let me repeat: 132

that they become well defined?” Informally and intuitively, we are tempted to say that the range of f is “strictly smaller than” its codomain. However, as explained in Remark 13, we shall avoid such language.

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♡ The range is not the same thing as the codomain. ♡

Remark 41. Unfortunately and very confusingly, some writers (especially in older books) use the term range to mean what we call the codomain in this textbook. The standard modern usage of the two terms is as given in this textbook and is also what I believe your A-Level examiners have in mind.133 Example 281. Define h ∶ R → R by h (x) = ex . Then Codomain h = R

and

Range h = R+ .

Figure to be inserted here.

So, Range h ⊂ Codomain h and, in particular, Range h ≠ Codomain h. Example 282. Define i ∶ R+ → R by i (x) = ln x. Then Codomain i = R

and

Range i = R.

Figure to be inserted here.

So, Range i = Codomain i (and Range i ⊆ Codomain i).

133

Of course, I can’t be sure because your A-Level syllabus and exams are never clear about what they mean by the term range and never use the term codomain.

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Example 283. Define j ∶ {1, 2} → R+0 by j (x) = x + 3. Then Codomain j = R+0

and

Range j = {4, 5}.

Figure to be inserted here.

So, Range j ⊂ Codomain j and, in particular, Range j ≠ Codomain j. Example 284. Define k ∶ {1, 2} → {4, 5} by k (x) = x + 3. Then Codomain j = {4, 5}

and

Range j = {4, 5}.

Figure to be inserted here.

So, Range k = Codomain k (and Range k ⊆ Codomain k).

Is the function k the same as the function j in the previous example?134

Definition 69. A function whose range and codomain are equal is called onto.

134

No. Although k and j have the same domain and mapping rule (indeed, they even have the same range), their codomains differ. Hence, k ≠ j.

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Example 285. In the above six examples, we defined these six functions:

f ∶ [1, 2] → R h∶ R →R j ∶ {1, 2} → R+0

Not onto by f (x) = x + 3, Range f = [4, 5]; by h (x) = ex , Range h = R+ ; by j (x) = x + 3, Range j = [4, 5];

Onto g∶ R →R by g (x) = x + 3, + i∶ R →R by i (x) = ln x, k ∶ {1, 2} → {4, 5} by k (x) = x + 3,

For each of the functions f , h, and j, the range is a proper subset of—and, in particular, is not equal to—the codomain. Hence, these three functions are not onto. For each of the functions g, i, and k, the range equals the codomain. Hence, these three functions are onto. Remark 42. An exact synonym for onto is surjective. (And a surjection is a function that’s onto or surjective.)

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Exercise 105. Find the range of each function. Which (if any) of these functions is onto? (Answer on p. 1746.) √ (a) Define a ∶ R+0 → R by a (x) = x. (b) Define b ∶ Z → R by b (x) = x2 . (c) Define c ∶ Z → Z by c (x) = x2 . (d) Define d ∶ Z → Z by d (x) = x + 1. (e) Define e ∶ Z → R by e (x) = x + 1. (f) The function f

1

100 5 200

3

Codomain f

Domain f (g)

The function g

1

100 5

3 Domain g

200

Codomain g

Exercise 106. Let f be a function. Of the following six statements, which must be true? (a) Range f ⊆ Domain f . (b) Range f ⊆ Codomain f . (c) Range f ⊂ Domain f .

(d) Range f ⊂ Codomain f . (e) Range f = Domain f . (f) Range f = Codomain f .

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16.

An Introduction to Continuity

Informally, if you can draw a function’s graph without lifting your pencil, then that function is continuous.135 Most functions we’ll encounter in A-Level maths will be continuous: Example 286. The function f ∶ R → R defined by f (x) = x5 − x2 − 1 is continuous, because you can draw its entire graph without lifting your pencil.

y

x f is continuous because you can draw its entire graph without lifting your pencil.

Example 287. The sine function sin is continuous.

y

sin is continuous because you can draw its entire graph without lifting your pencil.

x

135

The converse though is false—that is, a function can be continuous even if its graph cannot be drawn without lifting your pencil. For the formal definition of continuity, see Ch. 142.6 (Appendices).

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Example 288. The exponential function exp is continuous.

y

exp is continuous because you can draw its entire graph without lifting your pencil.

x Example 289. The absolute value function ∣⋅∣ is continuous.

y

∣⋅∣ is continuous because you can draw its entire graph without lifting your pencil.

x

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As stated, most functions we’ll encounter are continuous. However, we’ll occasionally encounter functions that aren’t continuous everywhere. For example, functions with “holes” in them are not continuous: Example 290. Define i ∶ R → R by

⎧ ⎪ ⎪ ⎪x i (x) = ⎨ ⎪ ⎪ ⎪ ⎩0

for x ≠ 1, for x = 1.

The function i is not continuous. In Part V (Calculus), we’ll learn why precisely this is so. For now, an intuitive understanding will suffice.

y i

x

Note though that i is continuous on (−∞, 1) because you can draw this portion of the graph without lifting your pencil.

Similarly, i is continuous on (1, ∞) because you can draw this portion of the graph without lifting your pencil. And so, we can actually say that i is continuous on (−∞, 1) ∪ (1, ∞).

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This isn’t something you need to know, but just to illustrate, here’s a somewhat exotic function whose domain is R but which is nowhere-continuous. Example 291. The Dirichlet function136 d ∶ R → R is defined by ⎧ ⎪ ⎪ ⎪1, d (x) = ⎨ ⎪ ⎪ ⎪ ⎩0,

for x ∈ Q, for x ∉ Q.

Then d is not continuous everywhere, because to draw its entire graph, you must lift your pencil. In fact, it’s worse than that—d is nowhere-continuous! Informally, this means you can’t draw more than one point of the graph without lifting your pencil.137

The Dirichlet function d ∶ R → R is defined by: ⎧ ⎪ ⎪ ⎪1 for x ∈ Q, d(x) = ⎨ ⎪ ⎪ ⎪ ⎩0 for x ∉ Q.

y

The graph of d contains the point (x, 1) for every x ∈ Q.

The graph of d contains the point (x, 0) for every x ∉ Q.

x

Or the characteristic function of the rationals. Named after the German mathematician Peter Gustav Lejeune Dirichlet (1805–59). 137 We’ll revisit this example in Part V (Calculus). 136

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17.

When a Function Is Increasing or Decreasing

Informally, we know what it means for a function to be increasing, decreasing, strictly increasing, and/or strictly decreasing: Example 292. Consider the function f ∶ R → R defined by f (x) = x2 .

Figure to be inserted here.

The function f is decreasing—and indeed, strictly decreasing—on R−0 = (−∞, 0].

Also, it is increasing—and indeed, strictly increasing—on R+0 = [0, ∞).

Formal definitions:

Definition 70. Let f ∶ D → R be a nice function and S ⊆ D. We say that f is

(a) Increasing on S if for any a, b ∈ S with a < b, we have f (a) ≤ f (b). (b) Strictly increasing “ f (a) < f (b). (c) Decreasing “ f (a) ≥ f (b). (d) Strictly decreasing “ f (a) > f (b). If f is increasing on D, then f is an increasing function.

If f is strictly increasing on D, then f is a strictly increasing function. If f is decreasing on D, then f is a decreasing function.

If f is strictly decreasing on D, then f is a strictly decreasing function.

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Example 293. Consider the function g ∶ R → R

defined by g (x) = x3 .

Figure to be inserted here.

The function g is increasing on R, because if we pick any a, b ∈ R with a < b, we have a3 ≤ b3

or

a3 < b3

or

g (a) ≤ g (b).

Since g is increasing on its domain R, it is an increasing function. Indeed,

g (a) < g (b),

so that g is also strictly increasing on its domain R and is also a strictly increasing function.

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Example 294. Consider the function h ∶ [0, 2π] → R Ch. ??, we will review trigonometric functions.)

defined by h (x) = cos x. (In

Figure to be inserted here.

The function h is decreasing on [0, π], because if we pick any a, b ∈ [0, π] with a < b, we have Indeed,

cos a ≤ cos b

cos a < cos b

or

or

so that h is also strictly decreasing on [0, π].

h (a) ≤ h (b).

h (a) < h (b),

Similarly, h is increasing on [π, 2π], because if we pick any a, b ∈ [π, 2π] with a < b, we have Indeed,

cos a ≥ cos b

cos a > cos b

or

or

so that h is also strictly increasing on [π, 2π].

h (a) ≥ h (b).

h (a) > h (b),

Fact 29. If a function is strictly increasing on a set, then it is also increasing on that set. Similarly, if a function is strictly decreasing on a set, then it is also decreasing on that set.

Proof. If a function f is strictly increasing on a set S, then by Definition 70(a), f (a) < f (b) for all a, b ∈ S with a < b. It must also be true that f (a) ≤ f (b) for all a, b ∈ S with a < b, so that by Definition 70(b), f is increasing on S. The proof for the case where f is instead strictly decreasing on S is similar and thus omitted.

Fact 29 says that “strictly increasing implies increasing” and “strictly decreasing implies decreasing”. However, the converses of these two statements are false, as the next example shows:

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Example 295. Consider the function i ∶ R → R

defined by i (x) = 1.

Figure to be inserted here.

The function i is increasing on R, because if we pick any a, b ∈ R with a < b, we have 1≤1

or

1 / 1

or

i (a) ≤ i (b).

However, i is not strictly increasing on R, because if we pick any a, b ∈ R with a < b, we have i (a) / i (b).

Altogether, i is both an increasing function and a decreasing function. However, i is neither a strictly increasing function nor a strictly decreasing function. Exercise 107. Consider the function j ∶ R → R defined by j (x) = x4 . 1748.)

Figure to be inserted here.

Explain why j is (a) strictly decreasing on (−∞, 0]; (b) strictly increasing on [0, ∞).

Hence, conclude (fill in the blanks): (c) By _____, j is also decreasing on _____ and increasing on _____. 218, Contents

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18.

Arithmetic Combinations of Functions

In this chapter, we’ll take given functions and create new ones through arithmetic combinations.

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Example 296. Define f, g ∶ R → R by f (x) = 7x + 5 and g (x) = x3 . Let k = 2. Then we can also define these six functions: 1. (f + g) ∶ R → R

2. (f − g) ∶ R → R 3.

4. 5.

(f ⋅ g) ∶ R → R

(kf ) ∶ R → R

by (f + g) (x) = f (x) + g (x) = 7x + 5 + x3 ;

by (f − g) (x) = f (x) − g (x) = 7x + 5 − x3 ; by (f ⋅ g) (x) = f (x) g (x) = (7x + 5) x3 ; by

f ( ) ∶ R ∖ {0} → R by g

6. (f + k) ∶ R → R

(kf ) (x) = kf (x) = 2 (7x + 5) = 14x + 10; f f (x) 7x + 5 ( ) (x) = = ; g g (x) x3

by (f + k) (x) = f (x) + k = 7x + 5 + 2 = 7x + 7.

For lack of any standard names, I’ll call the above six functions the sum, difference, product, constant multiple, quotient, and translated functions, respectively. Evaluating each of these six functions at 1, we have 1. (f + g) (1) = 7 ⋅ 1 + 5 + 13 = 13;

2. (f − g) (1) = 7 ⋅ 1 + 5 − 13 = 11;

3.

(f ⋅ g) (1) = (7 ⋅ 1 + 5) × 13 = 12;

4. (kf ) (1) f 5. ( ) (1) g

= 14 ⋅ 1 + 10 = 24;

=

7⋅1+5 = 12; 13

6. (f + k) (1) = 7 ⋅ 1 + 7 = 14.

By the way, as usual, parentheses can be helpful for making things clear. Here for example, they make clear that f + g is a single function—it’d be confusing and unclear if we wrote f + g (1) instead of (f + g) (1). Note the domain of each of the above six functions: For kf and f + k, the domain is simply Domain f .

For each of f − g, and f ⋅ g, the domain is Domain f ∩ Domain g (i.e. the set of numbers that are in both the domains of f and g).

For f /g, it’s a little trickier figuring out what the domain is. Observe that g (0) = 0. And so, in order for f /g to be well-defined, we must restrict the domain by removing the element 0. Otherwise, (f /g) (0) would be undefined and f /g would fail to be a well-defined function. Thus, the domain of f /g is Domain f ∩ Domain g ∖ {x ∶ g (x) = 0} = R ∖ {0} .

In words, the domain of f /g is the set of numbers x that are in both the domains of f and g, except those for which g (x) equals zero. 220, Contents

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Example 297. Define h, i ∶ [−1, ∞) → R by h (x) = x + 1 and i (x) =

Then we can also define these six functions:

√ x + 1. Let l = 5.

1. (h + i) ∶ [−1, ∞) → R by (h + i) (x) = h (x) + i (x) = x + 1 +

x + 1;

√ 2. (h − i) ∶ [−1, ∞) → R by (h − i) (x) = h (x) − i (x) = x + 1 − x + 1; √ 3. (h ⋅ i) ∶ [−1, ∞) → R by (h ⋅ i) (x) = h (x) i (x) = (x + 1) x + 1; 4.

5.

(lh) ∶ [−1, ∞) → R by

h ( ) ∶ (−1, ∞) → R by i

(lh) (x) = lh (x) = 5 (x + 1) = 5x + 5;

√ h h (x) x+1 ( ) (x) = =√ = x + 1; i i (x) x+1

6. (h + l) ∶ [−1, ∞) → R by (h + l) (x) = h (x) + l = x + 1 + 5 = x + 6.

Evaluating each of these six functions at 1, we have 1. (h + i) (1) = 1 + 1 +

2. (h − i) (1) = 1 + 1 − 3.

√ √ 1 + 1 = 2 + 2;

√ √ 1 + 1 = 2 − 2;

√ √ (h ⋅ i) (1) = (1 + 1) 1 + 1 = 2 2;

4. (lh) (1)

h 5. ( ) (1) i

= 5 (1 + 1) = 10; =

1+1=

2;

6. (h + l) (1) = 1 + 1 + 5 = 7.

Again, note the domain of each of the above six functions: For each of lh and h + l, the domain is simply Domain h.

For each of h + i, h − i, and h ⋅ i, the domain is Domain h ∩ Domain i (the set of numbers that are in both the domains of h and i).

But for h/i, it’s again a little trickier figuring out the domain. Observe that i (−1) = 0. And so, in order for h/i to be well-defined, we must restrict the domain by removing the element 0. Otherwise, (h/i) (0) would be undefined and h/i would fail to be a welldefined function. Thus, the domain of h/i is Domain h ∩ Domain i ∖ {x ∶ i (x) = 0} = [−1, ∞) ∖ {−1} = (−1, ∞) .

In words, the domain of h/i is the set of numbers x that are in both the domains of h and i, except those for which i (x) equals zero.138

138

√ Here the sharp student may wonder, “But isn’t x + 1 perfectly well-defined for all x ∈ [−1, ∞) and, in particular, even for x = −1? So couldn’t the domain of h/i instead be [−1, ∞)?” Great point! But the thing is, we really want to think of (h/i) (x) as being equal to h (x) divided by i (x) (without doing any simplification beforehand and without thinking of h/i as being simply the √ “formula” x + 1). So, if i (x) is undefined, then we want (h/i) (x) to also be undefined.

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Formal definitions of the six functions: Definition 71. Let k ∈ R and f ∶ D → R and g ∶ E → R be nice functions. Then we may define 1. The sum function (f + g) ∶ D ∩ E → R by (f + g) (x) = f (x) + g (x);

2. The difference function (f − g) ∶ D ∩ E → R by (f − g) (x) = f (x) − g (x); 3. The product function (f ⋅ g) ∶ D ∩ E → R by (f ⋅ g) (x) = f (x) ⋅ g (x);

4. The constant multiple function (kf ) ∶ D → R by (kf ) (x) = kf (x); f f (x) f ; 5. The quotient function ( ) ∶ D ∩ E ∖ {x ∶ g (x) = 0} → R by ( ) (x) = g g g (x) 6. The translated function (f + k) ∶ D → R by (f + k) (x) = f (x) + k.

Remark 43. Again, there are no standard names for the six functions just defined. They are often just left nameless and may only occasionally be collectively referred to as “arithmetic combinations of functions” (hence the name of this chapter). The names given above are simply those that I thought make a little sense, though some (especially translated function) are a bit awkward. Remark 44. As we’ll learn in the next chapter, f g refers to a function that’s entirely different from f ⋅ g. So take great care to write f ⋅ g if that’s what you mean.139 Exercise 108. Let k = 2 and l = 5. Define

Evaluate the following. (a) (f + g) (2)

(b) (g − f ) (1) (c) (g ⋅ f ) (2)

f ∶R→R g∶R→R h ∶ [−1, ∞) → R i ∶ [−1, ∞) → R

(d) (kg) (1) g (e) ( ) (1) f (f) (f + k) (1)

by by by by

f (x) g (x) h (x) i (x)

= 7x + 5, = x3 , = x + 1, √ = x + 1.

(g) (h + i) (2)

(h) (i − h) (1) (i) (i ⋅ h) (2)

(j) (li) (1) i (k) ( ) (1) h (l) (h + l) (1)

(k) Write down the domain of each of the functions f + h, f − h, f ⋅ h, and f /h. Then define each function.

139

Unfortunately and very confusingly, a minority of writers do use f g to mean f ⋅ g. In the A-Level syllabus and exams and in this textbook, we will follow majority practice by insisting that f g ≠ f ⋅ g.

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19.

Domain Restriction

In this chapter, we’ll take given functions and create new ones through domain restrictions. Example 298. Define f ∶ R → R by f (x) = 2x.

Then the restriction of f to [1, 2] is the function f ∣[1,2] ∶ [1, 2] → R defined by f ∣[1,2] (x) = f (x).

Figure to be inserted here.

Note well that the functions f and f ∣[1,2] are distinct: f ≠ f ∣[1,2] . Example 299. Define g ∶ R+0 → R by g (x) =

√ x.

Then the restriction of g to [2, 3] is the function g∣[2,3] ∶ [2, 3] → R defined by g∣[2,3] (x) = g (x).

Figure to be inserted here.

Note well that the functions g and g∣[2,3] are distinct: g ≠ g∣[2,3] .

Definition 72. Let f ∶ A → B be a function and C ⊆ A. The restriction of f to C is the function f ∣C ∶ C → B defined by f ∣C (x) = f (x).

Remark 45. The notation f ∣C is not in your A-Level maths syllabus or exams. Nonetheless, it is sufficiently convenient and commonly used that we’ll use it in this textbook.

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Exercise 109. For each given function, what are its restrictions to [0, 1] and R+ ? Draw the graphs of the given function and also the two restrictions. (Answer on p. 224.)

(a) xxx (b) xxx A109.

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20.

Composite Functions

Let f and g be nice functions. We just learnt to construct the product function f ⋅ g (read aloud as “f times g”).

We now learn to construct a new function f g (read aloud simply as “f g”), called the composite function. The functions f g and f ⋅ g look similar but are completely different.

Example 300. Define f, g ∶ R → R by f (x) = 2x and g (x) = x + 1. Then the composite function f g ∶ R → R is defined by (f g) (x) = f (g (x)) = f (x + 1) = 2 (x + 1) = 2x + 2.

(To get the composite function f g, plug g into f .)

In contrast, the product function f ⋅ g ∶ R → R is defined by

(f ⋅ g) (x) = f (x) g (x) = (2x) (x + 1) = 2x2 + 2x.

(To get the product function f ⋅ g, multiply f and g.)

The functions f g and f ⋅ g are different. For example, the values of each at 0 are (f g) (0) = 2 × 0 + 2 = 2

and

(f ⋅ g) (0) = 2 × 02 + 2 × 0 = 0.

Note that f ⋅ g = g ⋅ f , but f g ≠ gf . We say that the product (of two functions) is commutative, but the composition (of two functions) is not (generally) so. Here for example, the composite function gf ∶ R → R is defined by (gf ) (x) = g (f (x)) = g (2x) = 2x + 1.

(To get the composite function gf , plug f into g.)

The value of gf at 0 is (gf ) (0) = 2 × 0 + 1 = 1, which is different from (f g) (0) = 2.

Note also that for the composite function f g, we plug g into f . So for example, to compute f g (7), we first compute g (7) = 7 + 1 = 8, then compute f (g (7)) = f (8) = 2 ⋅ 8 = 16. (A common mistake is to do the opposite because one goes instinctively from left to right.)

Conversely, for the composite function gf , we plug f into g. So for example, to compute gf (7), we first compute f (7) = 2 ⋅ 7 = 14, then compute g (f (7)) = g (14) = 14 + 1 = 15. (Ditto.)

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Example 301. Define h, i ∶ R → R by h (x) = x2 − 1 and i (x) = x/2. Then the composite function hi ∶ R → R is defined by

(hi) (x) = h (i (x)) = h (x/2) = x2 /4 − 1.

In contrast, the product function h ⋅ i ∶ R → R is defined by

(h ⋅ i) (x) = h (x) i (x) = (x2 − 1) x/2 = x3 /2 − x/2.

The functions hi and h ⋅ i are different. For example, the values of each at 0 are (hi) (0) = 02 /4 − 1 = −1

and

(h ⋅ i) (0) = 03 /2 − 0/2 = 0.

Again, note that h ⋅ i = i ⋅ h, but hi ≠ ih. The composite function ih ∶ R → R is defined by (ih) (x) = i (h (x)) = i (x2 − 1) = x2 /2 − 1/2.

The value of ih at 0 is (ih) (0) =

02 1 1 − = − , which is different from (hi) (0) = −1. 2 2 2

Formal definition of a composite function:

Definition 73. Let f and g be functions with Range g ⊆ Domain f . Then the composite function f g is the function with Domain: Domain g; Codomain: Codomain f ; and Mapping rule: (f g) (x) = f (g (x)).

So, the domain of a composite function will be the same as the innermost (or rightmost) function. And the codomain of a composite function will be the same as the outermost (or leftmost) function. Remark 46. The composite function f g may also be written f ○ g (read aloud as “f circle g”). This alternative piece of notation is useful for making clear that we do not mean f ⋅ g.

The condition Range g ⊆ Domain f is important.140 It ensures that for every x ∈ Domain g, we have g (x) ∈ Domain f and hence that f (g (x)) is well-defined.

If this condition fails, then we simply say that the composite function f g does not exist 140

Just so you know, many writers take a different approach: They define the composite function f ○ g if and only if Codomain g = Domain f . In contrast, in Definition 73, we use the weaker requirement Range g ⊆ Domain f . (Many other writers also do the same as us—see e.g. MathWorld).

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or is undefined:

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Example 302. Define f ∶ R+ → R by f (x) = ln x and g ∶ R → R by g (x) = x + 1. Observe Range g = R is not a subset of Domain f = R+ . And so for example, (f g) (−5) = f (g (−5)) = f (−5 + 1) = f (−4)

would be undefined. Hence, the composite function f g simply does not exist. The rest of this example is explicitly excluded from your syllabus.141 Nonetheless, spending a minute or two reading it will earn you a better understanding of composite functions. (Plus, this is the sort of thing that your A-Level examiners could easily include as a curveball question.) Consider the restriction of g to (−1, ∞), i.e. the function g∣(−1,∞) . Observe that Range g∣(−1,∞) = R+ is a subset of Domain f = R+ .

So, we have the composite function f g∣(−1,∞) ∶ (−1, ∞) → R defined by

(f g∣(−1,∞) ) (x) = f (g∣(−1,∞) (x)) = f (x + 1) = ln (x + 1) .

Example 303. Define i ∶ R+0 → R by i (x) =

√ x and j ∶ R → R by j (x) = x − 3.

Observe Range j = R is not a subset of Domain i = R+0 . And so for example, (ij) (−5) = i (j (−5)) = i (−5 − 3) = i (−8)

would be undefined. Hence, the composite function ij simply does not exist. Again, the rest of this example is explicitly excluded from your syllabus. Consider the restriction of j to [3, ∞), i.e. the function j∣[3,∞) . Observe that Range j∣[3,∞) = R+0 is a subset of Domain i = R+0 .

So, we have the composite function ij∣[3,∞) ∶ [3, ∞) → R defined by √ (ij∣[3,∞) ) (x) = i (j∣[3,∞) (x)) = j (x − 3) = x − 3.

141

Your syllabus (p. 5) states, “Exclude ... restriction of domain to obtain a composite function.”

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We can use a single function to build a composite function. Example 304. Define f ∶ R → R by f (x) = 2x.

Observe that Range f = R is a subset of Domain f = R. Hence, the composite function f f ∶ R → R exists and is defined by

So for example,

(f f ) (x) = f (f (x)) = f (2x) = 4x. (f f ) (1) = 4 and (f f ) (3) = 12.

The composite function f f is usually simply written as f 2 . So, the above line can also be written as f 2 (1) = 4

and

f 2 (3) = 12.

Next, observe that Range f 2 = R is also a subset of Domain f = R. Hence, the composite function f f 2 = f 3 ∶ R → R exists and is defined by So for example,

f 3 (x) = f f 2 (x) = f (f 2 (x)) = f (4x) = 8x.

We also have f 4 , f 5 , etc.:

f 3 (1) = 8

and

f 3 (3) = 24.

f 4 ∶ R → R is defined by f 4 (x) = 16x,

f 5 ∶ R → R is defined by f 5 (x) = 32x, ⋮

Following common practice (and also your syllabus and exams), we’ll use f 2 to mean the composite function f f , f 3 to mean f f 2 , f 4 to mean f f 3 , etc. For future reference, let’s jot this down formally: Definition 74. Let f be a function. The symbol f 2 denotes the composite function f ○f . For any integer n ≥ 3, the symbol f n denotes the composite function f ○ f n−1 .

Remark 47. Later on in Part V (Calculus), we’ll use f ′ , f ′′ , f ′′′ , f (4) , . . . , f (n) , . . . to denote the “first derivative of”, “second derivative of”, “third derivative of”, “fourth derivative of”, . . . , “nth derivative of”, . . . . Do not confuse the composite function f n with the nth derivative f (n) .

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Example 305. Define g ∶ R → R by g (x) = 1 − x/2.

Observe that Range g = R is a subset of Domain g = R. Hence, the composite function g 2 ∶ R → R exists and is defined by So for example,

1 − x/2 1 x x = + . g 2 (x) = g (g (x)) = g (1 − ) = 1 − 2 2 2 4 3 5 g 2 (1) = and g 2 (3) = . 4 4

Next, observe that Range g 2 = R is also a subset of Domain g = R. Hence, the composite function g 3 ∶ R → R also exists and is defined by 1/2 + x/4 3 x 1 x = − . g 3 (x) = gg 2 (x) = g (g 2 (x)) = g ( + ) = 1 − 2 4 2 4 8 5 5 So for example, g 3 (1) = and g 3 (3) = . 8 8 Exercise 110 continues with this example.

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Exercise 110. Continue with the above example.

(a) Let n = 4.

(i) Explain why the composite function g n exists. (ii) Write down the function g n . (iii) Evaluate g n (1) and g n (3).

(b) Repeat (a), but now for n = 5. (c) Repeat (a), but now for n = 6.

The remainder of this question is a little harder, but is also the sort of curveball that the A-Level examiners might throw at you: So far, we have g (x) =

1 1 + (− ) x, 1 2

g 4 (x) =

? + (?) x, ?

g 2 (x) =

g 5 (x) =

1 1 + ( ) x, 2 4 ? + (?) x, ?

g 3 (x) =

g 6 (x) =

3 1 + (− ) x, 4 8 ? + (?) x, ?

where the question marks in the second line can be filled in by your answers to parts (a)–(c). As the above suggests, it turns out that for each positive integer n, we have g n (x) =

an + (cn ) x. bn

You are given that a general expression for an is

2n − (−1) an = . 3 (d) Write down a general expression for bn and cn . Hence, write down a general expression for g n (x). n

(e) Explain why as n approaches ∞, each g n (x) approaches 2/3.

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Exercise 111. For each of (a)–(d), determine if the composite functions f g and gf exist, then compute f g (1), f g (2), gf (1), and gf (2) (where these exist). (Answer on p. 1751.)

(a) Define f, g ∶ R → R by f (x) = ex and g (x) = x2 + 1. 1 1 (b) Define f, g ∶ R ∖ {0} → R by f (x) = and g (x) = . x 2x 1 (c) Define f ∶ R ∖ {0} → R by f (x) = and g ∶ R → R by g (x) = x2 + 1. x 1 (d) Define f ∶ R ∖ {0} → R by f (x) = and g ∶ R → R by g (x) = x2 − 1. x Exercise 112. For each of (a)–(d), determine if the composite function f 2 exists. If it does, write it down and compute f 2 (1) and f 2 (2). (a) Define f ∶ R → R by f (x) = ex . (b) Define f ∶ R → R by f (x) = 3x + 2.

(c) Define f ∶ R → R by f (x) = 2x2 + 1. (d) Define f ∶ R+ → R by f (x) = ln x.

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21.

One-to-One Functions

Recall that a function is onto (or surjective) if every element in its codomain is “hit” at least once: Example 306. XXX Example 307. XXX We now introduce a closely related concept: A function is one-to-one (or injective) if each element in its codomain is “hit” at most once. More precisely, Definition 75. A function f is one-to-one if for all y ∈ Codomain f , there is at most one x ∈ Domain f such that f (x) = y.

Equivalently and graphically,142

Fact 30. (Horizontal Line Test) A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once. Proof. See p. 1549 (Appendices). Example 308. Consider f ∶ R → R defined by f (x) = x2 + 2x.

Observe that f (−4) = 8 and f (2) = 8—the element 8 in Codomain f is “hit” twice.143 So, by Definition 75, f is not one-to-one.

Figure to be inserted here.

Equivalently, the horizontal line y = 8 intersects the graph of f twice.144 So, by Fact 30, f is not one-to-one. 142

In the book Mathematics Education in Singapore (2019), an unnamed JC maths teacher is quoted as claiming that, to ‘proof’ [sic] that a function is 1-1, one uses the horizontal line test which is incorrect.

The above quote is false. The horizontal line test is perfectly correct. The only possible objection is that the horizontal line test is informal as it relies on geometry. But even this objection can be dismissed because the horizontal line test can be stated and proven analytically (as is done here and in the Appendices). 143 Indeed, every element in Codomain f except −1 is “hit” twice. 144 Indeed, for every c > −1, the horizontal line y = c intersects the graph of f twice. 233, Contents

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Example 309. Consider g ∶ R → R defined by g (x) = 2x + 1.

Observe that every element in Codomain g is “hit” at most once.145 So, by Definition 75, g is one-to-one.

Figure to be inserted here.

Observe also that no horizontal line intersects the graph of f more than once.146 So, by Fact 30, g is one-to-one. Example 310. Consider the absolute value function ∣⋅∣.

Observe that ∣−2∣ = 2 and ∣2∣ = 2—the element 2 in Codomain ∣⋅∣ is “hit” twice.147 So, by Definition 75, ∣⋅∣ is not one-to-one.

Figure to be inserted here.

Equivalently, the horizontal line y = 2 intersects the graph of ∣⋅∣ twice.148 So, by Fact 30, ∣⋅∣ is not one-to-one. Remark 48. The name one-to-one is apt because every element in the codomain is “hit” by exactly one element in the domain. In contrast, if a function isn’t one-to-one, then it is many-to-one—at least one element in the codomain is “hit” by more than one (or “many”) elements in the domain. Is it possible that a function is one-to-many?149 Example 311. XXX Example 312. XXX Indeed, every element in Codomain g is “hit” exactly once. Indeed, for every c ∈ R, the horizontal line y = c intersects the graph of f exactly once. 147 Indeed, every element in Codomain ∣⋅∣ except 0 is “hit” twice. 148 Indeed, for every c > 0, the horizontal line y = c intersects the graph of ∣⋅∣ twice. 149 Nope. A one-to-many function would be one that maps an element in the domain to more than one (or “many”) elements in the codomain. But this violates our definition of a function—a function maps each element in the domain to exactly one element in the codomain. 145

146

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Remark 49. Just so you know, an exact synonym for one-to-one is injective. And a one-to-one (or injective) function may be called an injection. In this textbook, we will not use the terms injective or injection. Exercise 113. XXX

A113.

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21.1.

Two More Characterisations of One-to-One

The following result gives two more characterisations150 of one-to-one functions: Fact 31. The function f is one-to-one if and only if for all x1 , x2 ∈ Domain f ,

(a) If x1 ≠ x2 , then f (x1 ) ≠ f (x2 ). (b) If f (x1 ) = f (x2 ), then x1 = x2 .

Proof. Let P , Q, and R be, respectively, these statements: • The function f is one-to-one.

• For all x1 , x2 ∈ Domain f , if x1 ≠ x2 , then f (x1 ) ≠ f (x2 ). • For all x1 , x2 ∈ Domain f , if f (x1 ) = f (x2 ), then x1 = x2 .

(a) The claim is “P ⇐⇒ Q” is equivalent to “NOT − P ⇐⇒ NOT − Q”, which we’ll now prove: ⇐⇒

⇐⇒

NOT − P “There exists some y ∈ Codomain f such that there are more than one x ∈ Domain f (call them x1 and x2 ) with f (x1 ) = y and f (x2 ) = y” NOT − Q

(b) In (a), we showed that P ⇐⇒ Q. But by the contrapositive,151 Q is equivalent to R. Hence, we also have P ⇐⇒ R. Example 313. Consider f ∶ R → R defined by f (x) = x2 + 2x.

Observe that −4 ≠ 2, but f (−4) = f (2). So, by Fact 31(a), f is not one-to-one.

Figure to be inserted here.

Equivalently, f (−4) = f (2), but −4 ≠ 2. So, by Fact 31(a), f is not one-to-one. A characterisation (of a property) is a condition that’s equivalent (to that property). Here, the property is “one-to-one” and the (characterising) condition is “distinct elements are mapped to distinct elements”. Earlier the horizontal line test gave a graphical characterisation of one-to-one: Again, the property was “one-to-one”, but the (characterising) condition there was “no horizontal line intersects the graph more than once”. 151 If you don’t know what the contrapositive is, read Ch. 3.10. 150

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Example 314. Consider the absolute value function ∣⋅∣.

Observe that −2 ≠ 2, but f (−2) = f (2). So, by Fact 31(a), ∣⋅∣ is not one-to-one.

Figure to be inserted here.

Equivalently, f (−2) = f (2), but −2 ≠ 2. So, by Fact 31(b), ∣⋅∣ is not one-to-one. Example 315. Consider g ∶ R → R defined by g (x) = 2x + 1.

To prove that g is one-to-one, we can use either Fact 31(a) or (b):

1. Suppose x1 , x2 ∈ Domain g = R with x1 ≠ x2 . Then 2x1 ≠ 2x2 or 2x1 + 1 ≠ 2x2 + 1 or g (x1 ) ≠ g (x2 ). So, by Fact 31(a), g is one-to-one.

2. Suppose g (x1 ) = g (x2 ). Then 2x1 + 1 = 2x2 + 1 or 2x1 = 2x2 or x1 = x2 . So, by Fact 31(b), g is one-to-one.

Figure to be inserted here.

Example 316. Consider h ∶ R → R defined by h (x) = x3 .

To prove that h is one-to-one, we can use either Fact 31(a) or (b):

1. Suppose x1 , x2 ∈ Domain h = R with x1 ≠ x2 . Then x31 ≠ x32 or h (x1 ) ≠ h (x2 ). So, by Fact 31(a), h is one-to-one.

2. Suppose h (x1 ) = h (x2 ). Then x31 = x32 or x1 = x2 . So, by Fact 31(b), h is one-to-one.

Figure to be inserted here.

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Example 317. XXX Example 318. XXX Exercise 114. XXX

A114.

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21.2.

A Strictly Increasing/Decreasing Function Is One-to-One

Fact 32. If a function is strictly increasing (or strictly decreasing), then it is also oneto-one. Proof. Let f be a function. Suppose f is strictly increasing (or strictly decreasing).

Pick any distinct a, b ∈ Domain f with a < b.

Then f (a) < f (b) (or f (a) > f (b))—and, in particular, f (a) ≠ f (b).

Hence, by Fact 31(a), f is one-to-one. Example 319. XXX Example 320. XXX

Unfortunately, the converse of Fact 32 is false: Example 321. XXX Happily, Fact 32 has this partial converse. This converse is only partial because it adds two assumptions: The function is (a) continuous; and (b) defined on an interval (i.e. its domain is an interval). Proposition 3. Suppose a nice function is continuous and its domain is an interval. If this function is one-to-one, then it is also either strictly increasing or strictly decreasing. Proof. See p. 1549 (Appendices). Combining Fact 32 and Proposition 3, we have Corollary 3. Suppose a nice function is continuous and its domain is an interval. Then that function is One-to-one Example 322. XXX

⇐⇒

It is strictly increasing or strictly decreasing.

Example 323. XXX Exercise 115. XXX

A115.

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22.

Inverse Functions

Example 324. Consider the function f ∶ { animal to its role”:

🐄

🐄, 🐔} → {

,

The function f

} defined by “match the

🐔 Codomain f

Domain f

Its inverse function (or more simply inverse) is the function f −1 ∶ { , } → { defined by “match the role to the corresponding animal”:

The function f −1

🐄

🐄,🐔},

🐔 Domain f −1

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Codomain f −1

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Example 325. Consider the function g ∶ { the animal to its role”:

🐄

🐄, ൸, 🐔} → {

, ,

The function g

} defined by “match

🐕 🐔

Codomain g

Domain g Its inverse is the function g −1 ∶ { , , to the corresponding animal”:

}→{

🐄, ൸, 🐔}, defined by “match the role

The function g −1

🐄 🐕 🐔

Domain g −1

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Codomain g −1

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Example 326. Consider the function h ∶ R → R defined by h (x) = x + 2:

Figure to be inserted here.

Its inverse is the function h−1 ∶ R → R defined by h−1 (y) = y − 2:

Figure to be inserted here.

Remark 50. In the above example, we use the symbols x and y as our dummy variables for the functions h and h−1 , respectively. As usual, these being dummy variables, the symbols x and y can be replaced by any other variable. For example, all of the following would also have been perfectly correct: 1. Define h ∶ R → R by h (,) = , + 2 and h−1 ∶ R → R by h−1 (◻) = ◻ − 2.

2. Define h ∶ R → R by h (⋆) = ⋆ + 2 and h−1 ∶ R → R by h−1 (\$) = \$ − 2. 3. Define h ∶ R → R by h (x) = x + 2 and h−1 ∶ R → R by h−1 (x) = x − 2. 4. Define h ∶ R → R by h (y) = y + 2 and h−1 ∶ R → R by h−1 (y) = y − 2.

In #3, we use the same symbol x twice as the dummy variable for both h and h−1 . There is nothing wrong with this. (Indeed, we’ve always been doing this—e.g., “Define f, g ∶ R → R by f (x) = 3x and g (x) = 5x.”)

Ditto for #4, where we use the same symbol y twice as the dummy variable for both h and h−1 .

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Example 327. Consider the function i ∶ [0, 1] → [0, 2] defined by i (x) = 2x:

Figure to be inserted here.

Its inverse is the function i−1 ∶ [0, 2] → [0, 1] defined by i−1 (y) = y/2:

Figure to be inserted here.

We’ll formally define the inverse function only on p. 246. But for now, try this exercise: Exercise 116. XXX

A116.

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22.1.

Only a One-to-One Function Can Have an Inverse

If a function isn’t one-to-one, then it does not have an inverse. Let’s see why:

🐄

Example 328. Consider the function f ∶ { , ൸, 🐔} → {Yum, Yuck} defined by ) = Yum, f (൸) = Yuck, and f (🐔) = Yum: f(

🐄

Figure to be inserted here.

The element Yum in the codomain is “hit” twice. So, f is not one-to-one. Let’s see what goes wrong if we try to find f −1 , the inverse of f :

Figure to be inserted here.

The “function” f −1 is not well-defined because an element in the domain, Yum, is mapped to two distinct elements in the codomain. We may simply say, “f has no inverse,” or “the inverse of f does not exist,” or “the inverse of f is undefined”.

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Example 329. Define g ∶ {0, 1, 2} → {3, 4} by g (0) = 4, g (1) = 4, and g (2) = 3.

Figure to be inserted here.

The element 4 in the codomain is “hit” twice. So, g is not one-to-one. Let’s see what goes wrong when we try to find g −1 , the inverse of g:

Figure to be inserted here.

The “function” g −1 is not well-defined because an element in the domain, 4, is mapped to two distinct elements in the codomain. We may simply say, “g has no inverse,” or “the inverse of g does not exist,” or “the inverse of g is undefined”. We’re now ready to give our formal definition of the inverse:

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22.2.

Formal Definition of the Inverse

Definition 76. Given a one-to-one function f , its inverse function (or inverse), denoted f −1 , is the function with152

Example 330. XXX

Domain: Range f . Codomain: Domain f . Mapping rule: y = f (x) Ô⇒ f −1 (y) = x.

Example 331. XXX Remark 51. To have an inverse, a function must be one-to-one (every element in its codomain is “hit” at most once), but need not be onto (every element in its codomain is “hit” at least once).153 Nonetheless, just to keep things simple, we’ll assume that you’ll only ever be asked to find the inverse of a function that is also onto.154 This ensures that the inverse’s domain is simply the function’s codomain. (Why?)155 Again, if a function is not one-to-one, then we need not bother finding its non-existent inverse: Example 332. The function f ∶ R → R defined by f (x) = 0 is not one-to-one.

So, f does not have an inverse. (Or, “the inverse of f does not exist or is undefined.”)

Example 333. The function g ∶ R → R defined by g (x) = x2 is not one-to-one.

So, g does not have an inverse. (Or, “the inverse of g does not exist or is undefined.”) Exercise 117. XXX

A117.

Actually, Definition 76 is not quite the standard definition of an inverse function. It turns out under the standard definition, to have an inverse, a function must also be onto. We discuss this in Ch. 138.14 (Appendices). But this isn’t something you need worry about. For A-Level Maths, Definition 76 will perfectly good and will be used throughout this textbook (except in Ch. 138.14, Appendices). 153 But see n. 152. 154 In your A-Level exams, no distinction is ever made between the codomain and the range — and the implicit assumption seems to be that they are the same (“usually” or “whenever we need them to be”), so that any function is onto (“usually” or “whenever we need them to be”). 155 Let f be a one-to-one function. By Definition 76, Domain f = Range f . If f is also onto, then Range f = Codomain f , so that Domain f = Codomain f . 152

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22.3.

A Method for Showing a Function Is One-to-One and Finding Its Inverse

Let f ∶ A → B be a function with range C. Suppose f is one-to-one. Then for each x ∈ C, we can find a unique y ∈ A such that f (y) = x—so, we can define a new function f −1 ∶ C → A, where for every x ∈ C, we have y = f (x)

Ô⇒

(By Definition 76, f −1 is called the inverse of f .)

f −1 (y) = x.

Let’s now consider the converse of the above argument. That is, let’s suppose we don’t know whether f is one-to-one, but we’re able to find some function g ∶ C → A, where for every x ∈ C, we have y = f (x)

Ô⇒

g (y) = x.

The function g looks very much like it must be the inverse of f . And so, we ask two questions: 1. Is f even one-to-one? 2. Is g the inverse of f ? Happily, the answer to both of the above questions is affirmative: Fact 33. Suppose f ∶ A → B is a function with range C. Then f is one-to-one if and only if there exists a function g ∶ C → A where for every y ∈ C, we have y = f (x)

Ô⇒

g (y) = x.

(Moreover, if g exists, then by Definition 76, it is the inverse of f .) Proof. See p. 1551 (Appendices). As the following examples show, Fact 33 provides a method for showing a function (1) is one-to-one; and (2) finding its inverse. In fact, this method will allow us to skip all the work that usually goes into showing that a function (1) is one-to-one. Example 334. Define f ∶ R → R by f (x) = x + 1. Let y ∈ Range f = R.

Since y ∈ Range f , there must exist some x ∈ Domain f = R such that y = f (x) or y = x+1. Do the algebra to get x = y − 1.

So, let’s construct the function f −1 ∶ R → R defined by f −1 (y) = y − 1. Observe that for each y ∈ Range f = R, we do indeed have y = f (x) = x + 1

Ô⇒

f −1 (y) = y − 1 = x.

Hence, by Fact 33, f is one-to-one and its inverse is f −1 . 247, Contents

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Example 335. Define g ∶ R → R by g (x) = 2x. Let y ∈ Range g = R.

Since y ∈ Range g, there must exist some x ∈ Domain g = R such that y = g (x) or y = 2x.

Do the algebra to get x = y/2.

So, let’s construct the function g −1 ∶ R → R defined by g −1 (y) = y/2. Observe that for each y ∈ Range g = R, we do indeed have y = g (x) = 2x

Ô⇒

g −1 (y) = y/2.

Hence, by Fact 33, g is one-to-one and its inverse is g −1 .

In the above two examples, we were very careful in spelling out each step of the argument. Going forward, we shall be a little lazier and distill the argument into this Three-Step Method: Three-Step Method for Showing a Function Is One-to-One and Finding Its Inverse Consider the function f ∶ A → B defined by x ↦ f (x) and with range C. 1. Write y = f (x). 2. Do the algebra to get x = f −1 (y).

3. Conclude: f is one-to-one and its inverse is the function f −1 ∶ C → A defined by y ↦ f −1 (y).

Let’s redo the above two examples:

Example 336. Define f ∶ R → R by f (x) = x + 1. 1. Write y = f (x) = x + 1. 2. Do the algebra: x = y − 1.

3. Conclude: f is one-to-one and its inverse is the function f −1 ∶ R → R defined by f −1 (y) = y − 1.

Example 337. Define g ∶ R → R by g (x) = 2x. 1. Write y = g (x) = 2x.

2. Do the algebra: x = y/2. 3. Conclude: g is one-to-one and its inverse is the function g −1 ∶ R → R defined by g −1 (y) = y/2.

More examples:

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Example 338. Define h ∶ R ∖ {0} → R ∖ {0} by h (x) = 1/x. 1. Write y = h (x) = 1/x.

2. Do the algebra: x = 1/y. 3. Conclude: h is one-to-one and its inverse is the function h−1 ∶ R ∖ {0} → R ∖ {0} defined by h−1 (y) = 1/y.

By the way, observe that h = h−1 . So, h is an example of a self-inverse function (more on this in Ch. 22.7). Example 339. Define i ∶ R+0 → R+0 by i (x) = x2 . 1. Write y = i (x) = x2 .

√ 2. Do the algebra: x = ± y.

+ Since Domain √ i = R0 , we have x ≥ 0. So, we discard the negative value and are left with x = y. 3. √ Conclude: i is one-to-one and its inverse is the function i−1 ∶ R → R defined by g −1 (y) = y.

Example 340. Define j ∶ R−0 → R+0 by j (x) = x2 . 1. Write y = j (x) = x2 .

√ 2. Do the algebra: x = ± y.

− Since Domain √ j = R0 , we have x ≤ 0. So, we discard the positive value and are left with x = − y. 3. Conclude: √j is one-to-one and its inverse is the function j −1 ∶ R → R defined by j −1 (y) = − y.

Example 341. Define k ∶ R+ → R+ by k (x) = 1/x2 . 1. Write y = k (x) = 1/x2 .

√ 2. Do the algebra: x = ±1/ y.

+ Since Domain √ k = R , we have x > 0. So, we discard the negative value and are left with x = 1/ y. √ 3. Conclude: k is one-to-one and its inverse is k −1 ∶ R+ → R+ defined by k −1 (y) = 1/ y.

Example 342. Define l ∶ R− → R+ by l (x) = 1/x2 . 1. Write y = l (x) = 1/x2 .

√ 2. Do the algebra: x = ±1/ y. Since Domain l = R− , we have x < 0. So, we discard the positive value and are left with √ x = −1/ y. √ 3. Conclude: l is one-to-one and its inverse is l−1 ∶ R+ → R− defined by l−1 (y) = −1/ y.

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Exercise 118. Show that each function is one-to-one and find its inverse. (a) (b) (c) (d) (e) (f)

a∶ b∶ c∶ d∶ e∶ f∶

R R R+0 R−0 R+0 R−0

→ R defined by → R defined by → [1, ∞) defined by → [1, ∞) defined by → (0, 1] defined by → (0, 1] defined by

a (x) b (x) c (x) d (x) e (x) f (x)

= 5x. = x3 . = x2 + 1. = x2 + 1. = 1/ (x2 + 1). = 1/ (x2 + 1).

Exercise 119. Is each function one-to-one? If it is, find its inverse. (a) (b) (c) (d)

a∶R→R defined b ∶ R → [−1, ∞) defined c ∶ R+0 → [−1, ∞) defined d ∶ R−0 → [−1, ∞) defined

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by by by by

a (x) = x2 − 1. b (x) = x2 − 1. c (x) = x2 − 1. d (x) = x2 − 1.

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22.4.

Inverse Cancellation Laws

Fact 34. Suppose the function f has inverse f −1 . Then y = f (x)

⇐⇒

Proof. ( Ô⇒ ) By Definition 76 (of inverses).

f −1 (y) = x.

( ⇐Ô ) See p. 1551 (Appendices).

Proposition 4. (Inverse Cancellation Laws) Suppose the function f has inverse f −1 . Then

(a) f −1 (f (x)) = x for all x ∈ Domain f ; and (b) f (f −1 (y)) = y for all y ∈ Domain f −1 .

Proof. (a) Let x ∈ Domain f and y = f (x). By Fact 34 ( Ô⇒ ), f −1 (y) = x or f −1 (f (x)) = x. (b) Let y ∈ Domain f −1 and x = f −1 (y). By Fact 34 ( ⇐Ô ), f (x) = y or f (f −1 (y)) = y. Example 343. Define f ∶ R → R by f (x) = x + 1.

The inverse of f is the function f −1 ∶ R → R defined by f −1 (y) = y − 1.

And so, by the Cancellation Laws for Inverses (Proposition 4), we have f −1 (f (x)) = x, for all x ∈ Domain f = R,

f (f −1 (x)) = x, for all x ∈ Domain f −1 = R.

For example, f −1 (f (3)) = f −1 (4) = 3 and f (f −1 (1)) = f (0) = 1. Exercise 120. XXX

A120.

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The Graphs of f and f −1 Are Reflections in the Line y = x

22.5.

We reproduce Fact 34 from Ch. 22.4:

Fact 34. Suppose the function f has inverse f −1 . Then y = f (x)

⇐⇒

Fact 34 yields this corollary:

f −1 (y) = x.

Corollary 4. Suppose the function f has inverse f −1 . Then (x, y) is in the graph of f

⇐⇒

(y, x) is in the graph of f −1 .

Hence, by Fact 25, the graphs of f and f −1 are reflections in the line y = x:

Fact 35. Suppose the nice function f has inverse f −1 . Then the graphs of f and f −1 are reflections of each other in the line y = x. Example 344. Define f ∶ R → R by f (x) = x + 1.

The inverse of f is the function f −1 ∶ R → R defined by f −1 (y) = y − 1.

The graphs of f and f −1 are reflections of each other in the line y = x:

y

=

x

y

f

x f −1

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Example 345. Define g ∶ R → R by g (x) = 2x.

The inverse of g is the function g −1 ∶ R → R defined by g −1 (y) = y/2.

The graphs of g and g −1 are reflections of each other in the line y = x:

y

=

x

y

g

g −1

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Example 346. Define h ∶ R ∖ {0} → R ∖ {0} by h (x) = 1/x.

The inverse of h is the function h−1 ∶ R ∖ {0} → R ∖ {0} defined by h−1 (y) = 1/y. The graphs of h and h−1 are reflections of each other in the line y = x:

y

=

x

y

−1

x

h, h

Indeed, h = h−1 . (Again, h is an example of a self-inverse function—more on this in Ch. 22.7).

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Example 347. Define i ∶ R+0 → R+0 by i (x) = x2 .

The inverse of i is the function i−1 ∶ R+0 → R+0 defined by i−1 (y) =

√ y.

The graphs of i and i−1 are reflections of each other in the line y = x:

y

=

x

y

i−1 i

x Example 348. Define j ∶ R−0 → R+0 by j (x) = x2 .

√ The inverse of j is the function j −1 ∶ R+0 → R−0 defined by j −1 (y) = − y. The graphs of j and j −1 are reflections of each other in the line y = x:

Figure to be inserted here.

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Example 349. Define k ∶ R+ → R+ by k (x) = 1/x2 .

√ The inverse of k is the function k −1 ∶ R+ → R+ defined by k −1 (y) = 1/ y.

The graphs of k and k −1 are reflections of each other in the line y = x:

Figure to be inserted here.

Example 350. Define l ∶ R− → R+ by l (x) = 1/x2 .

√ The inverse of l is the function l−1 ∶ R+ → R− defined by l−1 (y) = −1/ y.

The graphs of l and l−1 are reflections of each other in the line y = x:

Figure to be inserted here.

Sometimes, we may know that a function is one-to-one and hence has an inverse; however, we may be unable to find the mapping rule of the inverse. In such cases, Fact 35 is particularly useful:

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Example 351. Define m ∶ R → R defined by m (x) = x3 + x.

Since m is strictly increasing (can you show this?),156 it is one-to-one and has an inverse m−1 ∶ R → R

Unfortunately, not having learnt to solve cubic equations, we don’t know how to find m−1 (x).

Nonetheless, even though we have no idea what m−1 (x) is, if we already have the graph of m, we can use Fact 35 to sketch the graph of m−1 :

y

m y=x

m−1

x

In the above example, it’s actually possible to show that

¿ ¿ √ √ Á Á 2 3 3 x x 1 x x2 1 Á Á À À m−1 (x) = + + + − + . 2 4 27 2 4 27

We just don’t know how to show the above, because we haven’t learnt to solve cubic equations. In the next example, it’s impossible to do so. Nonetheless, we can again use Fact 35 to sketch the graph of the inverse.

156

Method 1. If x1 < x2 , then x31 < x32 and x1 < x2 , so that m (x1 ) = x31 + x1 < x32 + x2 = m (x2 ). Method 2 (calculus). Since m′ (x) = 3x2 + 1 > 0 for all x ∈ R, m is strictly increasing on R. (We’ll learn more about this method in Ch. 39 and Part V).

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Example 352. Define n ∶ R → R defined by n (x) = x5 + x.

Since n is strictly increasing (can you show this?),157 it is one-to-one and has an inverse n−1 ∶ R → R

Unfortunately, it is impossible to write down an algebraic expression for n−1 (x).158

Nonetheless, even though we have no idea what n−1 (x) is, if we already have the graph of n, we can use Fact 35 to sketch the graph of n−1 :

y n

y=x

n−1

x

Exercise 121. Find each function’s inverse, then graph both the function and its inverse. (Answers on p. 1755.) (a) f ∶ (0, 1] → (1, 2] (b) g ∶ (0, 1] → (0, 2] (c) h ∶ (0, 1] → [1, ∞) (d) i ∶ (0, 1] → (0, 1]

defined defined defined defined

by by by by

f (x) g (x) h (x) i (x)

= x + 1. = 2x. = 1/x. = x2 .

Method 1. If x1 < x2 , then x51 < x52 and x1 < x2 , so that n (x1 ) = x51 + x1 < x52 + x2 = n (x2 ). Method 2 (calculus). Since n′ (x) = 5x4 + 1 > 0 for all x ∈ R, n is strictly increasing on R. (We’ll learn more about this method in Ch. 39 and Part V). 158 We can find algebraic expressions for the roots of polynomial equations of degree 2 (quadratic formula), 3 (not taught in A-Level Maths), and 4 (ditto). However, it is impossible to do the same for polynomial equations of degree 5 or higher—this is Abel’s Impossibility Theorem. 157

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22.6.

When Do the Graphs of f and f −1 Intersect?

Let f be a function with inverse f −1 . By Fact 35, f and f −1 are reflections of each other in the line y = x.

Observe then that if f intersects the line y = x at some point, then f −1 must also intersect y = x at the very same point. y Hence, any point at which f intersects y = x is also a point at which f and f −1 intersect: Example 353. Define f ∶ R → R by f (x) = 2x − 1.

∶ R → R defined by f

−1

The graph of f intersects the line y = x at the point (1, 1).

Example 354. Define g ∶ R+0 → R+0 by g (x) = x2 . The inverse of g is the function g −1 ∶ R+0 → R+0 √ defined by g −1 (x) = x.

x

f

y 45 ○ = x lin or e

So, f and f −1 must also intersect at (1, 1).

(1, 1)

−11 1 (x)f= x + . 2 2

y

The graph of f intersects the line y = x at the points (0, 0) and (1, 1).

So, g and g −1 must also intersect at (0, 0) and (1, 1).

g −1

y 45 ○ = x lin or e

The inverse of f is the function f

−1

g (1, 1)

x

For future reference, let’s jot down the above observation: Fact 36. Let the function f have inverse f −1 . If f intersects the line y = x at some point P , then f also intersects f −1 at P .

Or equivalently, any point at which f intersects y = x is also a point at which f intersects f −1 . Also equivalently and in more concise notation, 259, Contents

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f (a) = a

Ô⇒

f (a) = f −1 (a).

Now, consider the converse of Fact 36. That is, consider this next statement:

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“If f intersects f −1 at some point P , then f also intersects the line y = x at P .” 7 −1 Or equivalently, any point at which f intersects f is also a point at which f intersects y = x. 7 Also equivalently and in more concise notation, “f (a) = f −1 (a)

Ô⇒

f (a) = a.”

7

The above statement sounds perfectly plausible. But unfortunately, it is false. And even more unfortunately, those writing your A-Level exams incorrectly assumed it to be true in N2008-II-4 (iv), (Exercise 562).159 y Here are two counterexamples to the above false statement.

h = h−1

Example 355. Define h ∶ R ∖ {0} → R ∖ {0} by h (x) = 1/x. The inverse of h is the function h−1 ∶ R ∖ {0} → R ∖ {0} 1 defined by h−1 (x) = . x

Observe that h = h−1 . So, h and h−1 share infinitely many intersection points.

Consider then the function i ∶ R → R defined by i (x) = −x3 . It is (a) defined on R (and moreover continuous everywhere); and (b) not its own inverse. Its inverse is the function i−1 ∶ R → R √ defined by i−1 (x) = − 3 x.

The graphs of i and i−1 intersect at the three points (0, 0), (−1, 1), and (1, −1). But only the first point is on the line y = x, while the other two are not.

159

45 ○ = x lin or e

(−1, 1)

45 ○ = x lin or e

i−1

y

i

y

Example 356. One might object that the function in the above counterexample is unusual because it is (a) not defined on R; and (b) its own inverse.

(−1, −1)

x

y

However, only (−1, −1) and (1, 1) are on the line y = x. Every other intersection point is not. For example, h and h−1 intersect at the point (3, 1/3), but this point is not on y = x.

(1, 1)

(3, 1/3)

(0, 0)

x

(1, −1)

Similarly, one set of published TYS answers incorrectly states, “As y = f (x) is a reflection of y = f −1 (x) about the line y = x, the point of intersection of the two curves must meet on y = x.”

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Our last two examples show that the following statement is false: “If f intersects f −1 at some point P , then f also intersects the line y = x at P .”

Or equivalently,

f (a) = f −1 (a)

Ô⇒ /

f (a) = a.

7 3

Nonetheless, here are two results that come kinda close:

Fact 37. Let D be an interval and f ∶ D → R be a continuous function with inverse f −1 . If f and f −1 intersect at least once, then at least one of their intersection points is on the line y = x.

Proof. See p. 1558 (Appendices).

Fact 37 is illustrated by the functions f , g, and i from the last four examples. (How?)160

Our second result says that by adding the peculiar assumption that f and f −1 intersect at an even number of points, we can obtain the stronger result that all of the intersection points are on the line y = x. Fact 38. Let D be an interval and f ∶ D → R be a continuous function with inverse f −1 . If f and f −1 intersect at an even number of points, then all of their intersection points are on the line y = x.

Proof. See p. 1558 (Appendices).

Fact 38 is illustrated by the function g (Example 354).

160

In each case, the given function was continuous on an interval and intersected its inverse at least once. And so sure enough, in each case, the function intersected its inverse on the line y = x at least once.

Fact 37 is also illustrated by the functions h∣ 262, Contents

R+

and h∣ − . R

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22.7.

Self-Inverses

Your A-Level examiners like to write questions about self-inverses:161 Definition 77. The function f is self-inverse if f (f (x)) = x,

Example 357. XXX

for all x ∈ Domain f .

Example 358. XXX Your A-Level examiners also like to use this result: Fact 39. Let f be a function and a ∈ Domain f . Suppose f has inverse f −1 . Then f (f (a)) = a

Proof. ( Ô⇒ ) Suppose f (f (a)) = a.

⇐⇒

f −1 (a) = f (a).

Since a ∈ Range f = Domain f −1 , we can apply f −1 to get f −1 (f (f (a))) = f −1 (a). 1

But by Proposition 4(a) (Cancellation Law), we also have f −1 (f (f (a))) = f (a). 2

Now plug = into = to get f −1 (a) = f (a). 1

2

( ⇐Ô ) Suppose f −1 (a) = f (a).

Since f −1 (a) ∈ Codomain f −1 = Domain f , we can apply f to get f (f −1 (a)) = f (f (a)). 3

But by Proposition 4(b) (Cancellation Law), we also have f (f −1 (a)) = a. 4

Now plug = into = to get f (f (a)) = a. 3

4

Example 359. XXX Example 360. XXX

Remark 52. A synonym for self-inverse is involution, which is actually the more commonly used term. But we’ll stick with the term self-inverse, which is simpler and more readily understood. Example 361. XXX Exercise 122. XXX

A122. 161

See e.g. N2018/I/5, N2017/II/3, N2016/I/10, N2012/I/7 (respectively, Exercises 530, 535, 538, 550). Indeed, N2012/I/7 explicitly uses the term self-inverse.

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22.8.

If f Is Strictly Increasing/Decreasing, Then So Is f −1

In Ch. 21.2, we learnt that if a function f is strictly increasing (or decreasing), then it is also one-to-one—and so has some inverse f −1 . It turns out that f −1 must also be strictly increasing (or decreasing): Proposition 5. (a) A strictly increasing function’s inverse is strictly increasing. (b) A strictly decreasing function’s inverse is strictly decreasing. Proof. (a) Let f be a strictly increasing function with inverse f −1 . Pick any distinct a, b ∈ Domain f −1 with a < b.162

Let c = f −1 (a) and d = f −1 (b), so that f (c) = a and f (d) = b. Since f is strictly increasing and a < b, it must be that c < d.

We’ve just shown that a < b implies f −1 (a) < f −1 (b)—hence, f −1 is strictly increasing. (b) Similar, omitted.

Example 362. XXX Example 363. XXX Exercise 123. XXX

A123.

162

If no such a, b exist, then it is vacuously true that f −1 is strictly increasing.

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22.9.

A Function Is the Inverse of Its Inverse −1

Let f be a function with inverse f −1 . Suppose f −1 itself has inverse (f −1 ) .163 −1

So, (f −1 )

is the inverse of the inverse of f .

−1

One question we might ask is this: How are f and (f −1 ) Here’s one possible answer:

related? −1

Fact 40. Let f be a function with inverse f −1 . Suppose (f −1 ) Then −1

f = (f −1 )

Proof. See p. 1551 (Appendices).

⇐⇒

is the inverse of f −1 .

f is onto.

As discussed in Remark 51, we’ll assume that you’ll only ever be asked to find the inverse −1 of a function that is onto. And so, by Fact 40, we may assume that f = (f −1 ) always holds (even though it actually fails to hold if f is not onto). Example 364. Define f ∶ R → R by f (x) = 3x − 1. (Is f onto?)164 First, we find f −1 :

Write y = f (x) = 3x − 1. Do the algebra: x = (y + 1) /3.

So, f is one-to-one and its inverse is the function f −1 ∶ R → R defined by f −1 (y) = (y + 1) /3. −1

Next, we find (f −1 ) :

Write x = f −1 (y) = (y + 1) /3. Do the algebra: y = 3x − 1.

−1

So, f −1 is one-to-one and its inverse is the function (f −1 ) −1 −1

(f ) (x) = 3x − 1.

−1

Now, observe that f and (f −1 ) −1 −1

Hence, f = (f ) .

∶ R → R defined by

have the same domain, codomain, and mapping rule.

It turns out that (f −1 ) must exist because f −1 must be one-to-one—Fact 233 (Appendices). 164 Yes, because every element in Codomain f = R is “hit” at least once. (Or equivalently, “Yes, because Range f = R = Codomain f .

163

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Example 365. Define g ∶ R → R by g (x) = 5x + 2. (Is g onto?)165 First, we find g −1 :

Write y = g (x) = 5x + 2. Do the algebra: x = (y − 2) /5.

So, g is one-to-one and its inverse is the function g −1 ∶ R → R defined by g −1 (y) = (y − 2) /5. −1

Next, we find (g −1 ) :

Write x = g −1 (y) = (y − 2) /5. Do the algebra: y = 5x + 2. −1

So, g −1 is one-to-one and its inverse is the function (g −1 ) 5x + 2. −1

Now, observe that g and (g −1 ) −1 −1

Hence, g = (g ) .

−1

∶ R → R defined by (g −1 ) (x) =

have the same domain, codomain, and mapping rule.

Exercise 124. For each function given below, find (i) its inverse; and (ii) the inverse of its inverse. Also, determine whether the function is (iii) onto; (iv) and equal to the inverse of its inverse? (Answer on p. 1756.) (a) h ∶ (0, 1] → [1, ∞) defined by h (x) = 1/x. (b) i ∶ (0, 1] → (0, 1] defined by i (x) = x2 .

165

Yes, because every element in Codomain g = R is “hit” at least once. (Or equivalently, “Yes, because Range g = R = Codomain g.

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22.10.

Domain Restriction to Create a One-to-One Function

Again, if a function is not one-to-one, then it does not have an inverse. But it turns out that even if a function is not one-to-one, we can always 166 restrict its domain to create a brand new one-to-one function that does have an inverse: Example 366. Define f ∶ R → R+0 by f (x) = x2 .

We can show167 that f is not one-to-one and so has no inverse.

But consider the restriction of f to the set of non-positive reals, i.e. the function f ∣R+ . 0

We can show that f ∣R+ is strictly increasing168 and hence one-to-one. 0

Figure to be inserted here.

We now find the inverse of f ∣R+ using the usual Three-Step Method: 0

1. Write y = f ∣R+ (x) = x2 . 0 √ 2. Do the algebra: x = ± y. Since x ∈ Domain f ∣R+ = R+0 , we may discard the negative 0 √ value and be left with x = y. −1

−1

3. So, f ∣R+ is one-to-one and its inverse is the function f ∣R+ ∶ R+0 → R+0 defined by f ∣R+ (y) = 0 0 0 √ y.

Exercise 125. Continue to define f ∶ R → R+0 by f (x) = x2 .

(a) In words, what is f ∣R− ? 0

(b) Show that f ∣R− is one-to-one. 0

(c) Find the inverse of f ∣R− . 0

(d) Are there any other sets S such that the function f ∣S is also one-to-one? 166

Given any function f , we can, in the worst case, consider its restriction to the empty set, i.e. the function f ∣ . Observe that the range of f ∣ is the empty set. So, it is vacuously true that no element ∅

in the codomain of f ∣ is “hit” more than once. Hence, f ∣ is is one-to-one. ∅

The element 4 in its codomain is “hit” twice: f (2) = 22 = 4 and f (−2) = (−2) = 4. 168 Pick any distinct x1 , x2 ∈ Domain f ∣R+ = R+0 . Suppose x1 < x2 . Then f ∣R+ (x1 ) = x21 < x22 = f ∣R+ (x2 ).

167

0

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Example 367. Define g ∶ R → R+0 by g (x) = ∣x∣.

We can show169 that g is not one-to-one and so has no inverse. But consider the restriction of g to the set of non-positive reals, i.e. the function g∣R+ . 0

We can show that g∣R+ is strictly increasing170 and hence one-to-one. 0

Figure to be inserted here.

We now find the inverse of g∣R+ using the usual Three-Step Method: 1. Write y = g∣R+ (x) = ∣x∣.

0

0

2. Do the algebra: x = ±y. Since x ∈ Domain g∣R+ = R+0 , we may discard the negative value 0 and be left with x = y. −1

−1

3. So, g∣R+ is one-to-one and its inverse is the function g∣R+ ∶ R+0 → R+0 defined by g∣R+ (y) = 0 0 0 y.

169 170

The element 2 in its codomain is “hit” twice: g (2) = ∣2∣ = 2 and g (−2) = ∣−2∣ = 2. Pick any distinct x1 , x2 ∈ Domain g∣R+ = R+0 . Suppose x1 < x2 . Then g∣R+ (x1 ) = ∣x1 ∣ = x1 < x2 = ∣x2 ∣ = g∣R+ (x2 ).

0

0

0

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Exercise 126. Continue to define g ∶ R → R+0 by g (x) = x2 .

(a) In words, what is g∣R− ? 0

(b) Show that g∣R− is one-to-one. 0

(c) Find the inverse of g∣R− . 0

(d) Are there any other sets S such that the function g∣S is also one-to-one? Exercise 127. Define h ∶ R ∖ {1} → R+0 by h (x) =

(a) Explain whether h is one-to-one. (b) In words, what is h∣(1,∞) ?

1

(x − 1)

2.

(c) Show that h∣(1,∞) is one-to-one.

(d) Find the inverse of h∣(1,∞) . (e) In words, what is h∣(−∞,1) ?

(f) Show that h∣(−∞,1) is one-to-one.

(g) Find the inverse of h∣(−∞,1) .

(h) Are there any other sets S such that the function h∣S is also one-to-one?

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23.

Asymptotes and Limit Notation

Informally, an asymptote is a line that a graph “gets ever closer to”. 1 Example 368. Consider the function f ∶ (1, 3] → R defined by f (x) = + 2. Observe x−1 that As x approaches 1 (from the right), f (x) approaches ∞.

,

As x → 1+ , f (x) → ∞.

,

The limit of f (x) as x approaches 1 (from the right) is ∞.

,

lim f (x) = ∞.

,

Or in equivalent but more compact mathematical notation,

Equivalently, we may say that

Or in equivalent but more compact mathematical notation, x→1+

The above four statements labelled , are entirely equivalent and are perfectly formal and rigorous. The thing is, they serve as shorthand for a precise but long-winded statement that is given only in Ch. 142.3 (Appendices) and which you need not know for A-Level Maths. For A-Level Maths, you need only know what the statement , means informally and intuitively, as depicted in this figure.

Figure to be inserted here.

Statement , implies that

The vertical line x = 1 is a vertical asymptote of the function f .

Note that the parenthetical “(from the right)”s and superscript +s above are optional and may be omitted.

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1 Example 369. Consider the function g ∶ [−1, 1) → R defined by g (x) = +2. Observe x−1 that As x approaches 1 (from the left), g (x) approaches −∞.

-

The limit of g (x) as x approaches 1 (from the left) is −∞.

-

As x → 1− , g (x) → −∞. lim g (x) = −∞.

x→1−

-

Again, for A-Level Maths, you need only know what the statement - means informally and intuitively, as depicted in this figure:

Figure to be inserted here.

Statement - implies that

The vertical line x = 1 is a vertical asymptote of the function g.

Again, the parenthetical “(from the left)”s and superscript −s above are optional and may be omitted.

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1 Example 370. Consider the function h ∶ (3, ∞) → R defined by h (x) = +2. Observe x−1 that As x approaches ∞, h (x) approaches 2 (from above).

The limit of h (x) as x approaches ∞ is 2+ .

As x → ∞, h (x) → 2+ . lim h (x) = 2+ .

x→∞

☼ ☼

Again, for A-Level Maths, you need only know what the statement ☼ means informally and intuitively, as depicted in this figure:

Figure to be inserted here.

Statement ☼ implies that

The horizontal line y = 2 is a horizontal asymptote of the function h.

Again, the parenthetical “(from above)”s and superscript +s above are optional and may be omitted.

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1 Example 371. Consider the function i ∶ (−∞, 1) → R defined by i (x) = +2. Observe x−1 that As x approaches −∞, i (x) approaches 2 (from below).

\$

The limit of i (x) as x approaches −∞ is 2− .

\$

As x → −∞, i (x) → 2− . lim i (x) = 2− .

x→−∞

\$ \$

Again, for A-Level Maths, you need only know what the statement \$ means informally and intuitively, as depicted in this figure:

Figure to be inserted here.

Statement \$ implies that

The horizontal line y = 2 is a horizontal asymptote of the function i.

Again, the parenthetical “(from below)”s and superscript −s above are optional and may be omitted.

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Let’s now combine the last four examples: Example 372. Consider the function j ∶ R ∖ {1} → R defined by i (x) =

1 + 2. x−1

Figure to be inserted here.

, lim+ j (x) = ∞

We have

x→1

lim− j (x) = −∞.

and

x→1

, Either statement = or = (by itself) implies that

The vertical line x = 1 is a vertical asymptote of the function g.

We also have

☼ lim j (x) = 2+

x→∞

and

\$ lim j (x) = 2− .

x→−∞

☼ \$ Either statement = or = (by itself) implies that

The horizontal line y = 2 is a horizontal asymptote of the function i.

Again, the superscript +s and −s above are optional and may be omitted.

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Example 373. Consider the function f ∶→ R defined by f (x) = ex + 1. Observe that As x approaches −∞, f (x) approaches 1 (from above).



The limit of f (x) as x approaches −∞ is 1+ .



As x → −∞, f (x) → 1+ . lim f (x) = 1+ .

x→−∞

 

Again, the above four statements  are equivalent and perfectly formal and rigorous. But for A-Level Maths, you need only know what  means informally and intuitively, as depicted in this figure:

Figure to be inserted here.

Statement  implies that

The horizontal line y = 1 is a horizontal asymptote of the function f .

Again, the parenthetical “(from above)”s and superscript +s above are optional and may be omitted.

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Example 374. Consider the tangent function tan. Observe that As x approaches π/2 (from the left), tan x approaches ∞. −

As x → (π/2) , tan x → ∞.

The limit of tan x as x approaches (π/2) is ∞. lim

x→(π/2)

tan x = ∞.

♀ ♀ ♀ ♀

Again, the above four statements ♀ are equivalent and perfectly formal and rigorous. But for A-Level Maths, you need only know what ♀ means informally and intuitively, as depicted in this figure:

Figure to be inserted here.

Similarly, observe that As x approaches π/2 (from the right), tan x approaches −∞. +

As x → (π/2) , tan x → −∞.

+

The limit of tan x as x approaches (π/2) is −∞. lim

+

x→(π/2)

tan x = −∞.

♂ ♂ ♂ ♂

Either statement ♀ or ♂ (by itself) implies that

The vertical line x = π/2 is a vertical asymptote of the function tan.

Again, the parenthetical “(from above)”s, “(from below)”s, and superscript +s and −s above are optional and may be omitted.

Of course, tan actually has infinitely many vertical asymptotes—for every positive integer 1 k, the vertical line x = (k + ) π is a vertical asymptote for tan. (We’ll learn more about 2 this when we formally review trigonometric functions in Ch. ??.)

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Exercise 128. Explain why x = −π/2 is also a vertical asymptote for tan.(Answer on p. 1759.)

Exercise 129. Consider the function f ∶ R ∖ {0} → R defined by f (x) = 1/x. Find all asymptotes of f . (Answer on p. 1759.)

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23.1.

Oblique Asymptotes

Asymptotes need not only be vertical or horizontal—they can also be oblique (or slanted). Example 375. Consider the function f ∶ R ∖ {0} → R defined by f (x) = We note in passing that a vertical asymptote for f is x = 1. (Why?)171 Now, observe that

1 + x. x−1

As x approaches −∞, f (x) approaches x (from below).

,

The limit of f (x) as x approaches −∞ is x− .

,

As x → −∞, f (x) → x− . lim f (x) = x− .

x→−∞

, ,

Figure to be inserted here.

Similarly, As x approaches ∞, f (x) approaches x (from above).

,

The limit of f (x) as x approaches ∞ is x+ .

,

As x → ∞, f (x) → x+ . lim f (x) = x+ .

x→∞

Either statement , or △ (by itself) implies that

, ,

The oblique line y = x is an oblique asymptote for the function f .

Remark 53. For a formal definition of oblique asymptotes, see Definition 314 (Appendices).

171

As x approaches 1 (from the left or right), f (x) approaches −∞ or ∞. Hence, x = 1 is a vertical asymptote for f .

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Exercise 130. XXX

A130.

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24.

Transformations 24.1.

y = f (x) + a

The graph of y = f (x) + a is simply the graph of f translated (or shifted) upwards by a units. (If a < 0, then we have a “negative upward” shift—or more simply, a downward shift.) Example 376. Graphed below in black is the function f ∶ R → R defined by f (x) = x3 −1.

Translate f upwards by 2 units to get the graph of y = f (x) + 2 = x3 + 1.

y

(0, 1)

f

y = f (x) + 2

(0, −1) (0, −4)

+2

x −3

y = f (x) − 3

Translate f downwards by 3 units to get the graph of y = f (x) − 3 = x3 − 4.

When a graph is translated upwards or downwards, any y-intercepts, lines of symmetry, turning points, and asymptotes simply shift along with it.172 Here, f has y-intercept (0, −1). And so, y = f (x) + 2 and y = f (x) − 3 simply have y-intercepts (0, 1) and (0, −4).

Remark 54. In general, when a graph shifts upwards or downwards, we can’t say anything about how the x-intercepts change.

172

This assertion is formally stated as Fact 235(a) (Appendices).

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Example 377. The function g ∶ R ∖ {0} → R defined by g (x) = 1/x is graphed below (black). The graph of y = g (x) + 1 = 1/x + 1 is simply that of g translated upwards by 1 unit.

Since g has horizontal asymptote y = 0 (the x-axis), y = g (x)+1 has horizontal asymptote y = 1.

y = −x − 1

+1

y =x+1

y

y =x+1

y=x y = g (x) + 1

y=1

g

x

x = 0 is a vertical asymptote for both g and y = g (x) + 1

Note that with an upward or downward shift, any vertical asymptotes remain unchanged, because a vertical line translated upwards or downwards is simply the same vertical line. And so here, both g and y = g (x) + 1 have the vertical asymptote x = 0 (the y-axis).

The two lines of symmetry for g are y = x and y = −x. Thus, the two lines of symmetry for y = g (x) + 1 are simply the same, but translated upwards by 1 unit—y = x + 1 and y = −x + 1.

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24.2.

y = f (x + a)

The graph of y = f (x + a) is simply the graph of f translated leftwards by a units. (If a < 0, then we have a “negative leftward” shift, or more simply a rightward shift.)

Why leftwards (and not rightwards as one might expect)? The reason is that in order for f (x1 ) and f (x2 + a) to “hit” the same value, it must be that x2 = x1 − a. That is, x2 must be a units to the left of x1 . Example 378. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

Also graphed are these two equations:

y = f (x + 2) = (x + 2) − 1 = x3 + 6x2 + 12x + 7, 3 y = f (x − 1) = (x − 1) − 1 = x3 − 3x2 + 3x − 2. 3

The first equation is simply f translated leftwards by 2 units. The second is simply f translated rightwards by 1 unit.

y +1 −2

(−1, 0) y = f (x + 2)

f

(1, 0)

(2, 0)

x

y = f (x − 1)

When a graph is translated leftwards or rightwards, any x-intercepts, lines of symmetry, turning points, and asymptotes simply shift along with it.173 In this example, f has x-intercept (1, 0). And so, y = f (x + 2) and y = f (x − 1) simply have x-intercepts (−1, 0) and (2, 0).

173

This assertion is formally stated as Fact 235(b) (Appendices).

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Remark 55. In general, when a graph shifts leftwards or rightwards, we can’t say anything about how the y-intercepts change.

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Example 379. The function g ∶ R ∖ {0} → R defined by g (x) = 1/x is graphed below (black). The graph of y = g (x + 1) is simply that of g translated leftwards by 1 unit.

Since g has vertical asymptote x = 0 (the y-axis), y = g (x + 1) has vertical asymptote x = −1.

x = −1

y = 0 is a horizontal asymptote for both g and y = g (x + 1)

y = g (x + 1)

y

g

y =x+1 y=x

x

y = −x − 1

y = −x

Note that with a leftward or rightward shift, any horizontal asymptotes remain unchanged, because a horizontal line translated leftwards or rightwards is simply the same horizontal line. And so here, both g and y = g (x + 1) have the horizontal asymptote y = 0 (the x-axis).

The two lines of symmetry for g are y = x and y = −x. Thus, the two lines of symmetry for y = g (x + 1) are simply the same, but translated leftwards by 1 unit—y = x + 1 and y = − (x + 1) = −x − 1.

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24.3.

y = af (x)

Let a > 0. The graph of y = af (x) is simply that of f stretched vertically (outwards from the x-axis) by a factor of a. (If a < 1, then the graph is compressed rather than stretched.) Example 380. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

The graph of y = 2f (x) = 2x3 − 2 is simply that of f stretched vertically (outwards from the x-axis) by a factor of 2.

y

←Ð

1 (0, − ) 2

Compress 2× 1 y = f (x) 2 f

Stretch 2×

(0, −1)

(1, 0)

x

(0, −2)

y = 2f (x)

The graph of y = 0.5f (x) = 0.5x3 − 0.5 is simply that of f compressed vertically (inwards towards the x-axis) by a factor of 2. When a graph is stretched vertically, any y-intercepts, lines of symmetry, turning points, and asymptotes stretch along with it.174 In this example, f has y-intercept (0, −1). And so, y = 2f (x) and y = 0.5f (x) simply have y-intercepts (0, −2) and (0, −0.5).

Under a vertical stretch, any x-intercepts remain unchanged. Here, all three graphs have the same x-intercept (1, 0).

174

This assertion is formally stated as Fact 235(c) (Appendices).

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To get y = −af (x) (where a > 0), first reflect f in the x-axis to get y = −f (x), then stretch vertically by a factor of a. Example 381. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

Suppose we want to graph y = −2f (x).

To do so, first reflect f in the x-axis to get y = −f (x) = −x3 + 1.

y

y = −2f (x) y = −f (x)

(0, 2) (0, −1)

f

(0, −1)

(1, 0)

x

Then stretch vertically by a factor of 2, to get y = −2f (x) = −2x3 + 2.

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24.4.

y = f (ax)

Let a > 0. The graph of y = f (ax) is simply that of f compressed horizontally (inwards towards the y-axis) by a factor of a.

Why compressed inwards (and not stretched outwards as one might expect)? The reason is that in order for f (x1 ) and f (ax2 ) to “hit” the same value, we must have x2 = x1 /a. Example 382. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

The graph of y = f (2x) = (2x) − 1 = 8x3 − 1 is simply that of f compressed horizontally (inwards towards the y-axis) by a factor of 2. 3

y y = f (2x)

(0, −1)

1 ( , 0) (1, 0) 2

f x y=f( ) 2 (2, 0)

x

Compress 2×

Stretch 2×

The graph of y = f (0.5x) = (0.5x) − 1 = 0.125x3 − 1 is simply that of f stretched horizontally (outwards from the y-axis) by a factor of 2. 3

When a graph is stretched horizontally, any x-intercepts, lines of symmetry, turning points, and asymptotes stretch along with it.175

In this example, f has x-intercept (1, 0). And so, y = f (2x) and y = f (0.5x) simply have x-intercepts (0.5, 0) and (2, 0). Under a horizontal stretch, any y-intercepts remain unchanged. Here, all three graphs have the same y-intercept (0, −1).

175

This assertion is formally stated as Fact 235(d) (Appendices).

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To get y = f (−ax) (where a > 0), first reflect f in the y-axis to get y = f (−x), then compress horizontally by a factor of a. Example 383. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

Say we want to graph y = f (−2x).

To do so, first reflect f in the y-axis to get y = f (−x) = (−x) − 1 = −x3 − 1. 3

y

Compress 2×

(−1, 0)

1 (− , 0) 2 (0, −1)

(1, 0)

f y = f (−2x)

x

y = f (−x)

Then compress horizontally by a factor of 2, to get y = f (−2x) = (−2x) − 1 = −8x3 − 1. 3

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24.5.

Combinations of the Above

Example 384. Some unknown function f is graphed below. You are told only that points A, B, and C are (approximately) (−1.4, 0), (0.8, −1.1), and (1.4, 0).

Armed only with this knowledge, we will try to graph four equations: y = 2f (x + 1), y = 2f (x + 1), y = f (2x) + 1, and y = f (2x + 1).

y

f x A ≈ (−1.4, 0)

B ≈ (0.8, −1.1)

C ≈ (1.4, 0)

First stretch f vertically by a factor of 2 to get y = 2f (x), then translate upwards by 1 unit to get y = 2f (x) + 1.

y = 2f (x) + 1

y

y = 2f (x) f

x

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(... Example continued from the previous page.)

First translate f leftwards by 1 unit to get y = f (x + 1), then stretch vertically by a factor of 2 to get y = 2f (x + 1).

y

y = 2f (x + 1)

y = f (x + 1)

f

x

First compress f horizontally by a factor of 2 to get y = f (2x), then translate upwards by 1 unit to get y = f (2x) + 1.

y

y = f (2x) + 1

y = f (2x)

x

f

(Example continues on the next page ...)

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(... Example continued from the previous page.)

First translate f leftwards by 1 unit to get y = f (x + 1), then compress horizontally by a factor of 2 to get y = f (2x + 1).

y

f y = f (x + 1)

x

y = f (2x + 1) Summary: Fact 41. Let f be a nice function; a, b > 0; and c, d ∈ R. To get from the graph of f to the graph of y = af (bx + c) + d, follow these steps: 1. Translate leftwards by c units, to get (the graph of) y = f (x + c). 2. Compress inwards towards y-axis by a factor of b, to get y = f (bx + c).

3. Stretch outwards from x-axis by a factor of a, to get y = af (bx + c). 4. Translate upwards by d units, to get y = af (bx + c) + d.

Proof. See p. 1559 (Appendices).

Figure to be inserted here.

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Exercise 131. Continue with f from the last example. Graph the following eight equations. (Hint: You can make use of what was already shown in the above example.) (a) (c) (e) (g)

y y y y

= −2f (x) − 1. = −2f (x + 1). = −f (2x) + 1. = −f (2x + 1).

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(b) (d) (f) (h)

y y y y

= 2f (−x) + 1. = 2f (−x + 1). = f (−2x) + 1. = f (−2x + 1).

on on on on

p. p. p. p.

1760.) 1761.) 1762.) 1763.)

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24.6.

y = ∣f (x)∣

Given the graph of f , we can easily graph y = ∣f (x)∣:

1. Where f ≥ 0 (i.e. above the x-axis), the graphs of f and y = ∣f (x)∣ coincide.

2. But where f < 0 (i.e. below the x-axis), they’re reflections of each other in the x-axis.

Example 385. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black).

The red dotted graph is of y = ∣f (x)∣.

y = ∣f (x)∣

Where f < 0, the two graphs are reflections of each other in the x-axis.

y

Where f ≥ 0, the two graphs coincide.

x

f

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Example 386. The function g ∶ R ∖ {0} → R defined by g (x) = 1/x is graphed below (black). The red dotted graph is of y = ∣g (x)∣.

y = ∣g(x)∣

y Where g ≥ 0, the two graphs coincide.

x g Where g < 0, they’re reflections of each other in the x-axis.

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24.7.

y = f (∣x∣)

Given the graph of f , we can easily graph y = f (∣x∣):

1. The portion east of the y-axis is exactly the same. 2. To get the left portion, reflect the above right portion in the y-axis. Example 387. Graphed in black is the function f ∶ R → R defined by f (x) = x3 − 1. To get the graph of y = f (∣x∣), follow the above two steps:

y

y = f (∣x∣)

1. The right portions coincide. 2. To get the left portion, simply reflect the right portion in the y-axis.

x

f

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Example 388. The black graph below is the function g ∶ R ∖ {0} → R defined by g (x) = 1/x. y = g (∣x∣) is the red dotted graph.

y = g (∣x∣) g

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y

1. The right portions coincide. 2. To get the left portion, simply reflect the right portion in the y-axis.

x

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24.8.

1 f

Given the graph of f , we can easily graph 1/f : 1. Small (near 0) becomes big (near ±∞). 2. Big (near ±∞) becomes small (near 0).

3. Intersect at f (x) = ±1. 4. y-intercept (0, a) in f becomes y-intercept (0, 1/a) in f −1 .

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Example 389. The function f ∶ R → R defined by f (x) = x3 −1 is graphed below (black). To graph 1/f ,

1. Small becomes big:

• Where f → 0− , 1/f → −∞. • Where f → 0+ , 1/f → ∞. • So, x = 1 is a vertical asymptote of 1/f .

2. Big becomes small:

• Where f → −∞, 1/f → 0− . • Where f → ∞, we have 1/f → 0+ . • So, y = 0 is a horizontal asymptote of 1/f .

√ 3 3. Intersect at f (x) = ±1: So, f and 1/f intersect at ( 2, 1) and (0, −1).

4. y-intercept (0, −1) in f becomes y-intercept (0, 1/ − 1) = (0, −1) in f −1 (so here, the y-intercept happens to be unchanged).

y 1 f

Horizontal asymptote y=0

f

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√ 3 ( 2, 1) (0, −1)

x

Vertical asymptote x=1

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Example 390. The function g ∶ R → R defined by g (x) = x2 +2 is graphed below (black). To graph 1/g, 1. Small becomes big: Not applicable here. 2. Big becomes small:

• Where g → ∞, 1/g → 0+ . • So, y = 0 is a horizontal asymptote of 1/g.

3. Intersect at g (x) = ±1: But here there is no x at which g (x) = ±1. So, g and 1/g do not intersect at all. 4. y-intercept (0, 2) in f becomes y-intercept (0, 1/2) in f −1 .

y

g

(0, 2) 1 g

(0, 0.5) Horizontal asymptote y=0

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Exercise 132. Graphed below is some unknown function f . You are told only that points A, B, and C are (approximately) (−1.4, 0), (0.8, −1.1), and (1.4, 0). Graph (a) y = ∣2f (2x)∣; and (b) y = f (∣x − 1∣) + 2 (Answer on p. 1764.)

y

f x A ≈ (−1.4, 0)

B ≈ (0.8, −1.1)

C ≈ (1.4, 0)

Exercise 133. Describe a sequence of transformations that would transform the graph of y=

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1 x

onto

y =3−

1 . 5x − 2

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24.9.

When a Function Is Symmetric in Vertical or Horizontal Lines

Figure to be inserted here.

Fact 42. Suppose f is a nice function with domain D. Then ⇐⇒

f is symmetric in the vertical line x = a If x ∈ D, then 2a − x ∈ D and f (x) = f (2a − x).

Proof. See p. 1561 (Appendices).

Taking the above result and setting a = 0, we get this result:

Corollary 5. Suppose f is a nice function with domain D. Then ⇐⇒

f is symmetric in the y-axis (i.e. the vertical line x = 0) If x ∈ D, then −x ∈ D and f (x) = f (−x).

Fact 43. Let f be a nice function. If f is not constant, then f is not symmetric in any horizontal line. Proof. See p. 1561 (Appendices).

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25.

ln, exp, and e

In secondary school, your teachers probably went in this order: 1. First introduce Euler’s number e = 2.718 281 828 459 . . .

2. Then define the natural logarithm ln as the logarithm with base e. In this textbook, we’ll go the other way round. We’ll

1. First define the natural logarithm function ln. 2. Then define Euler’s number e as being the number such that ln e = 1.

This may seem strange but will pay off in Part V (Calculus).

Definition 78 (informal). The natural logarithm function ln ∶ R+ → R maps each t ∈ R+ to

y

ln t = Area bounded by the graph of y = 1/x, the x-axis, and the vertical lines x = 1 and x = t.

y=

ln t is defined as this shaded area 1

t

1 x

x

The above definition is considered (slightly) informal because the mapping rule is described using geometry (and in particular, with reference to an “area” bounded by a curve and three lines). After we’ve learnt about the definite integral in Part V (Calculus), we’ll give a formal definition of the natural logarithm function (Definition 227) that supersedes the above informal definition. Remark 56. In Singapore and some other Britishy bits of the world, ln is usually read aloud as lawn. In the US, it’s usually read out loud as “el en”. The notation ln was probably first published in Steinhauser (1875, p. 277), where it stood for the Latin Logarithmus naturalis.

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Example 391. We have ln 4 = 1.386 . . . and ln 5 = 1.609 . . .

y

1

ln 4 = 1.386 . . .

y=

1 x

x

4

y

1

y=

ln 5 = 1.609 . . .

5

1 x

x

Note that right now, we don’t yet know how to calculate ln 4 or ln 5.

One possibility is to draw an extremely precise and large graph of y = 1/x on some graph paper, then slowly count up the squares. This sounds like a ridiculous idea, but as we’ll learn in Part V, this isn’t too different from how integration is actually done.

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Example 392. By definition, ln 1 is the area under y = 1/x, between x = 1 and x = 1. Hence, ln 1 = 0.

y

y=

ln 1 = 0 because there’s no area!

1 x

1

x

Note that if x < 1, then ln x is negative: Example 393. ln 0.9 = −0.105 . . .

y

0.9 1

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ln 0.9 = −0.105 . . .

y=

1 x

x

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Example 394. ln 0.5 = −0.693 . . .

y

ln 0.5 = −0.693 . . . 0.5

y= 1

1 x

x

Note that since Domain ln = R+ , if x ≤ 0, then ln x is simply undefined: Example 395. ln 0 and ln (−2.5) are undefined.

Below is the graph of ln.

• Since ln 1 = 0, the x-intercept of ln is (1, 0). • For x ∈ (0, 1), ln x < 0. • For x ≤ 0, ln x is simply undefined.

y

ln

1

As x → 0+ , ln x → −∞. Thus, ln has the vertical asymptote x = 0 (also the y-axis).

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25.1.

The Exponential Function exp

For x > 0, the graph of y = 1/x is strictly positive. But ln is defined as the area under the graph of y = 1/x. So, ln is strictly increasing. Hence, by Fact 32, ln is one-to-one. That is, ln has an inverse. Now, we don’t know what exactly this inverse function is, but we know it exists. So, let’s simply name it the exponential function: Definition 79. The exponential function, denoted exp, is the inverse of the natural logarithm function. By Definition 76 (of inverse functions),

1. Domain exp = Range ln = R; 2. Codomain exp = Domain ln = R+ ;

3. Mapping rule: If y = ln x, then exp y = x. Example 396. Earlier, we saw that ln 4 ≈ 1.386,

ln 5 ≈ 1.609,

ln 1 = 0,

Since exp is the inverse of ln, we have exp 1.386 ≈ 4,

exp 1.609 ≈ 5,

exp 0 = 1,

ln 0.9 ≈ −0.105, exp (−0.105) ≈ 0.9,

ln 0.5 ≈ −0.693. exp (−0.693) ≈ 0.5.

Since exp is the inverse of ln, by Fact 35, their graphs are reflections of each other in the line y = x:

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y

exp

ln

1

1

x

We next introduce Euler’s number e.

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25.2.

Euler’s Number

Definition 80. Euler’s number, denoted e, is the number that satisfies ln e = 1

or equivalently,

e = exp 1.

Remark 57. Euler is pronounced oiler (and not yuler).

y

And so by definition of ln, Euler’s number176 e is the number such that the area under y = 1/x, between x = 1 and x = e is equal to 1.

Right now we have no idea how to compute e. But I’ll cheat by telling you that e = exp 1 ≈ 2.718 281 828 459 . . .

y=

We define e so that this area is 1. 1

e

1 x

x

In secondary school, you were probably taught that ex is simply another way to write exp x. It turns out that this is actually a result that is to be proven using the definitions of the exponential function (just given above) and exponentiation (given only (Appendices)): Fact 44. For every x ∈ R, ex = exp x.

Proof. See p. 1700 (Appendices).

So, it is not that ex is simply another way to write exp x. Instead, it takes some work to prove that ex = exp x—i.e. that the number e raised to the power of x is equal to the value of the exponential function at x. Remark 58. While interesting, Euler’s number e is by itself not particularly important. What’s really important are the natural logarithm and exponential functions.177

For this reason, we will in this textbook often write exp x rather than ex . This is to emphasise that we’re usually thinking more about the value of the exponential function at x than about some number e raised to the power of x. It was Leonhard Euler (1707–1783) himself who first used the letter e to denote this number. Presumably he did not do this to honour himself. Calling e Euler’s number is simply an honour conferred by posterity. Confusingly, there’s also another number called Euler’s constant γ ≈ 0.577 215 664 . . . that, fortunately, we will not encounter in A-Level maths. 177 Indeed, one mathematician Walter Rudin (Real and Complex Analysis, 1966 [1987, 3e, p. 1]) says that the exponential function “is the most important function in mathematics”. 176

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We also have these two lovely results: Theorem 5. e =

1 1 1 1 + + + + .... 0! 1! 2! 3!

Proof. Happily, we’ll learn to prove this in Part V (Calculus)—see p. 986. 1 n Theorem 6. e = lim (1 + ) . n→∞ n

Proof. Happily, we’ll learn to prove this in Part V (Calculus)—see p. 1130. Although we can’t prove either of the above theorems right now, we can nonetheless numerically “verify” that they are plausible: 1 1 To “verify” Theorem 5, define f ∶ Z+0 → R by f (n) = + ⋅ ⋅ ⋅ + . 0! n! Then write f (0) =

1 1 = = 1. 0! 1

f (1) = f (0) +

f (2) = f (1) +

1 1 = 1 + = 2. 1! 1

1 1 = 2 + = 2.5. 2! 2

f (3) = f (2) +

1 1 = 2.5 + = 2.6. 3! 6

f (5) = f (4) +

1 1 = 2.716 + = 2.718 . . . 5! 120

f (4) = f (3) +

1 1 = 2.6 + = 2.716. 4! 24

We see that f rapidly converges towards e = 2.718 281 828 459 . . . By f (6), we have e correct to three decimal places. Lovely. 1 n Similarly, to “verify” Theorem 6, define g ∶ R → R by g (n) = (1 + ) . n Then write 1 1 g (1) = (1 + ) = 2. 1

1 2 g (2) = (1 + ) = 2.25. 2 1 3 g (3) = (1 + ) = 2.6. 3

1 4 g (4) = (1 + ) = 2.708 3. 4 1 5 g (5) = (1 + ) = 2.716. 5

g (10) =

1 10 (1 + ) 10

g (100) = (1 +

= 2.593 742 . . .

1 100 ) = 2.704 813 . . . 100

1 1 000 g (1 000) = (1 + ) = 2.716 923 . . . 1 000 6

1 10 6 g (10 ) = (1 + 6 ) 10

9

1 10 9 g (10 ) = (1 + 9 ) 10

= 2.718 280 . . . = 2.718 281 . . .

We see that g also converges towards e = 2.718 281 828 459 . . . , though much less rapidly than f —for example, g (100) gets e correct only to two decimal places, while even g (106 ) gets e correct to only five decimal places. 309, Contents

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26.

Trigonometry: Arcs (of a Circle)

From this chapter through Ch. 32, we’ll learn about trigonometry. You will already have seen a good deal of this stuff in secondary school. But as usual, we’ll go into slightly greater depth. Informally, an arc of a circle is any connected178 subset of that circle. A little more precisely,179 Definition 81 (informal). Given two points A and B on a circle, the arc AB (of the circle) is the set of points travelling anticlockwise on the circle between A and B (inclusive). Example 397. Graphed below is a circle containing the points A and B. Suppose we simply said that the arc AB is the subset of the circle connecting the points A and B. Then we’d be faced with an ambiguity: Are we referring to the red arc or the blue arc?

Figure to be inserted here.

To resolve this ambiguity, in the above definition, we adopt the convention that we “travel anticlockwise” from the first point to the second. So, • Arc AB goes anticlockwise from A to B. • Arc BA goes anticlockwise from B to A. Remark 59. Just so you know, more generally, the word arc can be used to refer to any “smooth” curve. But in this textbook, we use arc only in the context of circles. So, in this textbook, arc will always refer to a connected subset of a circle.

178 179

The term connected can actually be formally defined but here it will suffice to grasp its intuitive meaning. This definition is still a little imprecise, because we use the phrase “travelling anticlockwise”. For a more precise version of this definition, see Definition 282 (Appendices).

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Example 398. Graphed below is a circle containing the points C and D. Suppose we simply said that the arc CD is the subset of the circle connecting the points C and D. Then we’d be faced with an ambiguity: Are we referring to the red arc or the blue arc?

Figure to be inserted here.

To resolve this ambiguity, in the above definition, we adopt the convention that we “travel anticlockwise” from the first point to the second. So, • Arc CD goes anticlockwise from C to D. • Arc DC goes anticlockwise from D to C. Remark 60. The usual western or European convention is to go clockwise (when viewed from above), be it with clocks (which began as sundials in the northern hemisphere)180 or card games. (In contrast, Chinese games such as mahjong are usually played anticlockwise.) It may thus seem a bit puzzling that here we (or rather European mathematicians from centuries ago) adopt the convention of travelling anticlockwise. Later on, when we learn about the unit circle definitions of sine and cosine, we’ll see why this convention might have come about (see p. 335). In any case, this is to some extent an arbitrary convention (Remark 23 again applies here). Note that like a line, an arc is a one-dimensional object. (In contrast, a point is a zero-dimensional object, an area is a two-dimensional object, and a volume is a threedimensional object.) And similar to line segments, given an arc AB, we will denote by ∣AB∣ its length:181

180 181

Sundials in the southern hemisphere go anticlockwise. Here we may repeat the warning given earlier in Remark 34.

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Example 399. ∣AB∣ =???, ∣CD∣ =???

Figure to be inserted here.

Exercise 134. Name the labelled arcs using the labelled points.

Figure to be inserted here.

A134.

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27.

Trigonometry: Angles

You’ve known about angles since primary school. It turns out though that that writing down a precise definition of angle is tricky. In this subchapter, we’ll give two informal definitions of angle. In Part IV (Vectors), Definition 140 will then give this textbook’s official, formal definition of angle. Here’s our first informal definition of angle (reminder: a ray is a “half-infinite line”—see Ch. 7.6): Definition 82 (informal). Let a and b be rays. The angle between a and b is the anticlockwise rotation a must undergo to coincide with b (provided we first placed each ray’s endpoint at the same position).182 Example 400. Let A, B, C, and D be points on a circle with centre O. The angle α between the rays OA and OB is the anti-clockwise rotation OA must undergo to coincide with AB. We will also say that α is the angle subtended by the arc AB. Similarly, the angle β between the rays OC and OD is the anti-clockwise rotation OC must undergo to coincide with OD. We will also say that β is the angle subtended by the arc CD.

Figure to be inserted here.

(Example continues below ...) Definition 83 (informal). The full angle is the minimum positive anti-clockwise rotation a ray must undergo to coincide with itself. Remark 61. Just so you know, the full angle is also called the complete angle, round angle, or perigon. (This textbook will stick with using only the term full angle.) In primary school, you probably learnt the ancient Babylonian convention of assigning to the full angle the value of 360 degrees (also written 360○ ).

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(... Example continued from above.) Since α is the angle subtended by the arc AB, the angle subtended by the arc BA is 360○ − α.

Since β is the angle subtended by the arc CD, the angle subtended by the arc DC is 360○ − β.

Figure to be inserted here.

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27.1.

Angles, Informally Defined Again; Also, the Radian

We now give our second informal definition of angle, which supersedes Definition 83 (our first informal definition of angle): Definition 84 (informal). Suppose a circle of radius r with centre O contains the points A and B. Then the angle between the rays OA and OB or the angle subtended by the arc AB is this number:

where ∣AB∣ is the length of the arc AB.

∣AB∣ , r

Figure to be inserted here.

Example 401. Suppose r = 2, ∣AB∣ = 2, and ∣CD∣ = 3. Then by Definition 84, the angles α and β are α=

∣AB∣ 2 = =1 r 2

and

β=

∣CD∣ 3 = . r 2

Figure to be inserted here.

By the above definition, an angle is defined as the ratio of two lengths. And so, an angle is a “pure”, “unitless”, or “dimensionless” number. Hence, strictly speaking, angles should not have units. Nonetheless and perhaps slightly confusingly, angles are given units.

For example, the ancient Babylonians used degrees (○ ). This is what you were taught in primary school. In secondary school, you will also have learnt a second unit, called the radian (rad). This unit exactly corresponds to Definition 84 and is the SI unit for angles. It is also what we’ll be using in this textbook. So, in the last example, we could also have written 315, Contents

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α=

∣AB∣ 2 = = 1 rad r 2

β=

and

∣CD∣ 3 = rad. r 2

It is entirely optional whether we want to include the “rad” following the value of the angle. Indeed, going forward, we will almost always omit it. Let’s now consider what the full angle is in radians. As mentioned earlier, the ancient Babylonians assigned it the value of 360○ .

Well, by Definition 84, the full angle should be the ratio of the circle’s circumference to the circle’s radius. As we learnt in primary school (and is also stated in Fact 19), this ratio is Circumference 2πr = = 2π. Radius r

Hence, the full angle is 2π (radians). Equivalently, the full angle is the number 2π. We’ll jot this down as a formal definition in Definition 85 (and this will supersede our earlier informal Definition 83). We can now find the conversion rate between radians and degrees. Since

we must have

360○ ≈ 57.3○ . 2π

By Definition 84, 1 (rad) ≈ 57.3○ is the angle subtended by an arc whose length is equal to the circle’s radius r:

B

A

r

O

0

π 6

π 4

π 3

π 2

π

Degree(s) 0○ 30○ 45○ 60○ 90○ 180○ 360○ 316, Contents

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Going forward in this textbook, we’ll avoid using degrees and try to use only radians. (Indeed, repeating what we already said above, since angles are really “unitless”, we will also almost always not even bother to write “radians” or “rad”.) One reason for this is that radians work out much more nicely in calculus: If x is in radians, then

But if x is in degrees, then

d sin x = cos x. dx d 2π sin x = cos x. dx 360○

In the everyday world, laypersons are still more likely to use the degree than the radian, probably because the degree has been in use for about four millennia longer.183 So, you shouldn’t forget about the degree! Remark 62. Following convention, we’ll often denote angles by lower-case Greek letters α (alpha), β (beta), γ (gamma), and θ (theta). Your List of Formulae (MF26, p. 3) also uses upper-case Latin letters like A, B, P , and Q, and so we’ll use those too.

183

The Babylonians have been using degrees since c. 2000 BC. In contrast, the term radian was invented only in the late 19th century.

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27.2.

Names of Angles

We already gave 2π = 360○ the special name of full angle: Full angle = 2π = 360○ .

Special names are also given to six other angles (or types thereof): Definition 85. We call the angle A 1. The zero 2. An acute 3. The right 4. An obtuse 5. The straight 6. A reflex 7. The full

angle angle angle angle angle angle angle

A

if if if if if if if

A = 0. A ∈ (0, π/2) = (0○ , 90○ ). A = π/2 = 90○ . A ∈ (π/2, π) = (90○ , 180○ ). A = π = 180○ . A ∈ (π, 2π) = (180○ , 360○ ). A = 2π = 360○ .

A π r 2

πr The right angle is π/2.

The straight angle is π.

O

O

B

B

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Definition 86. We call a triangle 1. An acute triangle if its largest angle is acute. 2. A right triangle if its largest angle is right. 3. An obtuse triangle if its largest angle is obtuse.

Acute

Right

Obtuse

A few more terms we’ll find convenient: Definition 87. Let A and B be angles. π • If A + B = , then A and B are complementary and A and B are each other’s comple2 ments.

• If A + B = π, then A and B are supplementary and A and B are each other’s supplements. • If A + B = 2π, then A and B are explementary (or conjugate) and A and B are each other’s explements (or conjugates).

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28.

Trigonometry: Sine and Cosine

The two basic trigonometric (or circular) functions are sine and cosine. In secondary school, you probably learnt these right-triangle definitions of sine and cosine: Definition 88 (informal). Let α ∈ (0, π/2) (i.e. let α be an acute angle). Construct a right triangle with angles α, π/2, and π/2 − α. Let the lengths of the three sides opposite these three angles be o, h, and a (for “Opposite”, “Adjacent”, and “Hypotenuse”).

h

α

o

a

(a) For α ∈ (0, π/2), we define the sine and cosine functions (denoted sin and cos) by sin α = 1

(b) For α = 0 or α = π/2, we define sin 0 = 0, 3

o h

cos 0 = 1, 4

and

sin

2 a cos α = . h

π 5 = 1, 2

cos

π 6 = 0. 2

We’ll discuss =–=in the next subchapter. But first, let’s illustrate = and = with some simple examples: 3 6

1

2

Example 402. XXX Example 403. XXX

Remark 63. For now, we’ll stick with Definition 88. But in Ch. 28.6, it will be superseded by the unit-circle definitions of sine and cosine. Example 404. XXX

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28.1.

The Values of Sine and Cosine at 0 and 1

In Definition 88, we defined sin 0 = 0,

cos 0 = 1,

3

4

sin

π 5 = 1, 2

cos

π 6 = 0. 2

These definitions were seemed a bit arbitrary. Let’s see why they might make sense: Observe that as α → 0,184 o → 0 and a → h.

h

α

o

a

Since sin α = o/h for all α ∈ (0, π/2), it makes sense to define sin 0 =

0 3 = 0. h

cos 0 =

h 4 = 1. h

And since cos α = a/h for all α ∈ (0, π/2), it makes sense to define Similarly, observe that as α → π/2, o → h and a → 0.

h

α

o

a

Since sin α = o/h for all α ∈ (0, π/2), it makes sense to define sin

π h 5 = = 1. 2 h

And since cos α = a/h for all α ∈ (0, π/2), it makes sense to define cos

184

π 0 6 = = 0. 2 h

As in our discussion on limits, the right arrow “→” here is read aloud as “approaches”.

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Fun Fact According to Katz (2009),185 sine means bosom or breast: The English word “sine” comes from a series of mistranslations of the Sanskrit ¯ jy¯a-ardha (chord-half). Aryabhata frequently abbreviated this term to jy¯a or its synonym j¯ıv¯a. When some of the Hindu works were later translated into Arabic, the word was simply transcribed phonetically into an otherwise meaningless Arabic word jiba. But since Arabic is written without vowels, later writers interpreted the consonants jb as jaib, which means bosom or breast. In the twelfth century, when an Arabic trigonometry work was translated into Latin, the translator used the equivalent Latin word sinus, which also meant bosom, and by extension, fold (as in a toga over a breast), or a bay or gulf. This Latin word has now become our English “sine”.

185

Victor Katz, A History of Mathematics: An Introduction (2009, p. 253).

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28.2.

Three Values of Sine and Cosine

From Definition 88(b), we already know that √ 0 sin 0 = 0 = , 2

√ 4 cos 0 = 1 = , 2

√ π 4 sin = 1 = , 2 2

√ π 0 cos = 0 = . 2 2

Using geometry (see the following proof), we can also find the values of sin and cos at π/6, π/4, and π/3: Fact 45. α

0

√ 0 sin α 2 √ 4 cos α 2

π 6 √ 1 2 √ 3 2

π 4 √ 2 2 √ 2 2

π 3 √ 3 2 √ 1 2

π 2 √ 4 2 √ 0 2

√ √ Remark 64. Of course, 0/2 = 0 and 4/2 = 1. Nonetheless, the above table is written as it is because that’s the way I find it easiest to remember these values—sin is half square root 0, 1, 2, 3, 4, while cos is half square root 4, 3, 2, 1, 0. But perhaps you’ll find it easier to remember this table:

α

0

sin α

0

cos α

1

π 6 √ 1 2 √ 3 2

π 4 √ 2 2 √ 2 2

π 3 √ 3 2 √ 1 2

π 2 1 0

The following proof is informal because it relies on geometry and our right-triangle definitions of sine and cosine. Proof. Consider √ the triangle √ below. √ Suppose α = π/4. Then o = a, so that by Pythagoras’ 2 2 Theorem, h = o + a = 2o = 2a and hence, √ √ π o o 1 2 π a a 1 2 sin α = sin = = √ = √ = and cos α = cos = = √ = √ = . 4 h 2 4 h 2 2o 2 2a 2

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Next, consider △ABC (below), the equilateral triangle√with sides of √ length√1. Let D be 2 2 the midpoint of BC. By Pythagoras’ Theorem, ∣AD∣ = 1 − 0.5 = 0.75 = 3/2.

A

π 6 √ 3 2

1

1

π 3

B

√ 3/2 3 Hence, = 1 2 √ π ∣BD∣ 0.5 1 Also, sin = = = 0.5 = 6 ∣AB∣ 1 2 π ∣AD∣ sin = = 3 ∣AB∣

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0.5

D and and

C

√ π ∣BD∣ 0.5 1 cos = = = 0.5 = . 3 ∣AB∣ 1 2 √ √ π ∣AD∣ 3/2 3 cos = = = . 6 ∣AB∣ 1 2

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Exercise 135. Our goal in this Exercise is to prove that (Answer on p. 1770.) √ π 1 1+ 5 . x cos = 5 4 (a) You are told that any polygon with n sides can be partitioned into n − 2 triangles. For example, a polygon with 7 sides (called a heptagon) can be partitioned into 5 triangles:

△1 △3 △4

△2

△5

(i) A triangle has three interior angles. What is the sum of these three interior angles? (ii) Hence, what is the sum of the interior angles in a polygon with n sides? (b) Now consider the regular pentagon ABCDE each of whose five sides has length 1:

A 1

1 Q

P

B

E

R

1

1

C

1

D

(i) What is the sum of the pentagon’s five interior angles? 325, Contents (ii) Find ∠BAE.

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28.3.

Confusing Notation

In Ch. 20, we learnt that

• f 2 denotes the composite function f ○ f ;

• f 3 denotes the composite function f ○ f 2 ; • Etc.

We’d thus expect that sin2 denotes the composite function sin ○ sin. That is, we’d expect “sin2 x = sin (sin x).”

7

But this is not the case! Very confusingly, sin2 denotes the product function sin ⋅ sin. That is, sin2 x = (sin x) = (sin x) (sin x). 2

Example 405. sin2

π π ≠ sin (sin ) because 2 2

π 2 π 2 = (sin ) = (1) = 1, sin 2 2 2

but

3

π sin (sin ) = sin 1 ≈ 0.845. 2

And in general, for any positive integer n, sinn does not denote sin ○ sin ○ ⋅ ⋅ ⋅ ○ sin. That is, sinn x ≠ sin (sin (sin (. . . (sin x)))).

Instead, sinn denotes sin ⋅ sin ⋅ ⋅ ⋅ ⋅ ⋅ sin:

sinn x = (sin x) = (sin x) (sin x) . . . (sin x). n

3

This is yet another annoying and confusing bit of notation you’ll have to learn to live with. The above remarks also apply to cosine and the other four trigonometric functions we’ll be looking at in the next two chapters.

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28.4.

Some Formulae Involving Sine and Cosine

Happily, List MF26 (p.3) contains the following formulae, so no need to mug: Fact 46. Suppose A, B ∈ R. Then

Addition and Subtraction Formulae for Sine and Cosine (a) sin (A ± B) = sin A cos B ± cos A sin B (b) cos (A ± B) = cos A cos B ∓ sin A sin B

Double Angle Formulae for Sine and Cosine (c) sin 2A = 2 sin A cos A (d) cos 2A = cos2 A − sin2 A = 2 cos2 A − 1 = 1 − 2 sin2 A Exam Tip for Towkays

Whenever you see a question with trigonometric functions, put MF26 (p. 3) next to you! Proof. (a) and (b) The figure below186 is constructed so that ∣P R∣ = 1.187

Q

cos (A + B)

sin A sin B

A+B

S

R cos A sin B sin B

1

sin (A + B)

T cos B sin A cos B

B A P Observe that ∣P T ∣ = cos B. So, 186 187

∣P U ∣ = cos A cos B

cos A cos B

and

U

∣T U ∣ = sin A cos B.

Credit to Blue. Construction details. Let A, B, A + B ∈ (0, π/2). Construct the horizontal line segment P U . Rotate Ray P U anticlockwise by A to produce Ray P T . Then rotate Ray P T anticlockwise by B to produce Ray P R. Pick R to be the point on Ray P R such that ∣P R∣ = 1. Pick T to be the point on Ray P T such that RT ⊥ P T . Now construct the rectangle P QSU so that R and T are on the line segments QS and SU , respectively.

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Next, ∠P RQ and ∠RP U are alternate angles (or “Z-angles”). So, ∠P RQ = ∠RP U = A + B. And hence, ∣P Q∣ = sin (A + B)

and

∣QR∣ = cos (A + B).

Next, ∠RT S is complementary188 to ∠P T U , which is in turn complementary to ∠T P U . So, ∠RT S = A. Since ∣RT ∣ = sin B, we have ∣ST ∣ = cos A sin B

and

and ∣RS∣ = sin A sin B.

Now, ∣P Q∣ = ∣T U ∣ + ∣T S∣ or sin (A + B) = sin A cos B + cos A sin B.

Similarly, ∣QR∣ = ∣P U ∣ − ∣RS∣ or cos (A + B) = cos A cos B − sin A sin B.

This completes the proof of the the Addition Formulae for Sine and Cosine.189 For the proof of the Subtraction Formulae for Sine and Cosine, see Exercise 136.

For (c) and (d), see Exercise 137. Exercise 136. Use the figure below190 to prove the Subtraction Formulae for Sine and Cosine (in the special case where A is acute and B < A). (Answer on p. 1771.) sin (A − B) = sin A cos B − cos A sin B, cos (A − B) = cos A cos B + sin A sin B.

Q

S

R

1 T A−B P

B U

Exercise 137. Use the Addition and Subtraction Formulae for Sine and Cosine to prove the Double-Angle Formulae for Sine and Cosine, i.e. Fact 46(c) and (d).(Answers on p. 1772.) See Definition 87. This proof is (i) informal because it relies on geometry; and (ii) incomplete because it applies in the special case where A, B, and A+B are acute. For a formal and complete proof, see p. 1563 (Appendices). 190 The construction of this figure is very similar to before, except now we start with Ray P Q, rotate it clockwise by A − B to get Ray P R, then rotate Ray P R clockwise by B to get Ray P T .

188 189

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Remark 65. To repeat, the Double-Angle Formulae are on List MF26. So strictly speaking, you needn’t memorise them. Nonetheless, it’s a good idea to have them committed to memory so you can solve problems that much more quickly. Here’s an identity you should find familiar from secondary school: Fact 47. (First Pythagorean Identity) Suppose A ∈ R. Then sin2 A + cos2 A = 1.

Proof. Apply Definition 88 and Pythagoras’ Theorem: o 2 a 2 o2 + a2 h2 sin A + cos A = ( ) + ( ) = = 2 = 1. h h h2 h 2

2

h

A

o

a

The above proof is (i) informal because it relies on geometry; and (ii) incomplete because it covers only the case where A is acute. For a more formal and general proof, see p. 992 in Part V (Calculus). Remark 66. In this textbook, the terms formula and identity are exact synonyms.191 Fact 48. (Triple-Angle Formulae for Sine and Cosine) Suppose A ∈ R. Then

(a) sin 3A = 3 sin A − 4 sin3 A; (b) cos 3A = 4 cos3 A − 3 cos A.

Proof. See Exercise 138.

Exercise 138. Prove Fact 48(a) and (b).

191

In logic, an identity is a specific kind of formula (hence, identities form a subset of formulae). But this subtle distinction will not matter in this textbook and so let’s simply treat formula and identity as exact synonyms.

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Remark 67. The Triple-Angle Formulae are not on List MF26. So if they ever come up on exams, you’ll want to be able to either derive them from scratch or recall them. For cos 3A = 4 cos3 A − 3 cos A, there’s the Hokkien mnemonic “\$1.30 = \$4.30 − \$3”:

=

khoo sam \$4.30

khoo \$3

. .

Happily, List MF26 (p.3) contains the following formulae, so no need to mug: Fact 49. Suppose P, Q ∈ R. Then

Sum-to-Product (S2P) or Product-to-Sum (P2S) Formulae

P +Q 2 P +Q (b) sin P − sin Q = 2 cos 2 P +Q (c) cos P + cos Q = 2 cos 2 P +Q (d) cos P − cos Q = −2 sin 2 (a) sin P + sin Q = 2 sin

P −Q 2 P −Q sin 2 P −Q cos 2 P −Q sin 2 cos

Proof. See Exercise 139.

Fun Fact The above S2P or P2S Formulae are also known as the Prosthaphaeresis Formulae. Sounds cheem, but that’s just the combination of the Greek words for addition and subtraction—prosthesis and aphaeresis. So yea, something you can totally use to impress your friends and family. Exercise 139. Prove Fact 49(a)–(d).192

Exercise 140. The P2S Formulae will be particularly useful when we do integration, because they allow us to rewrite an otherwise-difficult-to-integrate product into an easyto-integrate sum. Rewrite each expression using the P2S Formulae: (Answer on p. 1772.) (a) sin 2x cos 5x (c) cos 2x cos 5x.

192

Hint: Write P =

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(b) cos 2x sin 5x (d) sin 2x sin 5x

P +Q P −Q P +Q P −Q + and Q = − . 2 2 2 2

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28.5.

The Area of a Triangle; the Laws of Sines and Cosines

Proposition 6. Suppose △ABC has angles A, B, and C, and sides of lengths a, b, and c. Then (a) The area of △ABC is (b) The Law of Sines:

(c) The Law of Cosines:

1 ab sin C. 2

sin A sin B sin C = = a b c

c2 = a2 + b2 − 2ab cos C

B

B

c

a a sin C

A

C b − a cos C

A

D

a cos C

C

b

Proof. Let D be the point on AC such that BD ⊥ AC. Then sin C =

∣BD∣ a

and

cos C =

So, ∣BD∣ = a sin C, ∣CD∣ = a cos C, and ∣AD∣ = b − a cos C.

∣CD∣ . a

(a) △ABC has base b and height a sin C. Hence, its area is 0.5ab sin C.

(b) We can similarly show that the area of △ABC is also 0.5bc sin A or 0.5ac sin B. So, Area of △ ABC =

bc sin A ac sin B ab sin C = = 2 2 2

×2÷abc

⇐⇒

sin A sin B sin C = = . a b c

(c) Consider the triangle ABD. It has hypotenuse of length c and legs of lengths a sin C and b − a cos C. Now use Pythagoras’ Theorem and the identity sin2 C + cos2 C = 1: 331, Contents

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c2 = (a sin C) + (b − a cos C) = a2 sin2 C + b2 − 2ab cos C + a2 cos2 C 2

2

= a2 (sin2 C + cos2 C) + b2 − 2ab cos C = a2 + b2 − 2ab cos C.

We can use the Law of Cosines to prove this “obvious” result:

Corollary 6. The length of one side of a triangle is less than the sum of the lengths of the other two sides. Proof. Consider a triangle with sides of lengths a, b, c. Let C be the angle opposite the side of length c. By the Law of Cosines,

c2 = a2 + b2 − 2ab cos C

= a2 + b2 − 2ab + 2ab − 2ab cos C

(Plus Zero Trick)

> (a − b)

(2ab (1 − cos C) > 0 because a, b > 0 and cos C < 1).

= (a − b) + 2ab (1 − cos C) 2 2

(a2 + b2 − 2ab = (a − b) ) 2

Now, c2 > (a − b) implies c > a − b or, equivalently, a < b + c. 2

We’ve just proven that a is less than the sum of b and c.

We can likewise prove that b < a + c and c < a + b.

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28.6.

The Unit-Circle Definitions

So far, we’ve been using Definition 88, the right-triangle definitions of sine and cosine. Unfortunately, these define sin and cos only for θ ∈ [0, π].

To define sin and cos more generally, that is, for all θ ∈ R, we must turn to the following unit-circle definitions, which supersede Definition 88. Definition 89 (informal). Consider the ray that is the positive x-axis (it begins at the origin O and extends eastwards). Call this ray x+ . Let α be any angle. Rotate x+ anticlockwise by α to produce the ray a.

Let A = (Ax , Ay ) be the point at which a intersects the unit circle centred on the origin.

y

The ray a

1

A = (Ax , Ay ) Ay

α O

Ax

The ray x+

x

The sine and cosine functions, sin ∶ R → R and cos ∶ R → R, are defined by sin α = Ay

and

cos α = Ax .

Equivalently, sin α and cos α are the x- and y-coordinates of the point A. And so, here’s an entirely equivalent definition that gives us a slightly different way of thinking about sin and cos (and hence hopefully also a better understanding):

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Definition 90 (informal). A particle A travels anticlockwise along the unit circle centred on the origin at a constant speed of 1 unit per second. At time α = 0 s, A is at the point (1, 0).

Figure to be inserted here.

We define sin α and cos α to be the y- and x-coordinates of A at time α. Example 406. If α = 0, then A = (0, 1) = (sin 0, cos 0).

Figure to be inserted here.

Example 407. If α =

π , then by the reasoning given in the proof of Fact 45, 6 √ 1 3 π π A=( , ) = (sin , cos ) . 2 2 6 6

Figure to be inserted here.

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Example 408. If α =

π , then by the reasoning given in the proof of Fact 45, 4 √ √ 2 2 π π A=( , ) = (sin , cos ) . 2 2 4 4

Figure to be inserted here.

Remark 68. By the way, we can now see how the anticlockwise convention might have arisen. It is arguably “natural” to adopt these two conventions (as we do): 1. The zero angle “should” correspond to the positive x-axis. 2. Small but positive angles “should” be in the northeast quadrant (of the cartesian plane) where x and y are positive. Given the above two conventions,193 we must necessarily “travel” anticlockwise. Example 409. If α =

π , then by the reasoning given in the proof of Fact 45, 3 √ 3 1 π π A=( , ) = (sin , cos ) . 2 2 3 3

Figure to be inserted here.

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Example 410. If α =

π , then 2 π π A = (1, 0) = (sin , cos ) . 2 2

Figure to be inserted here.

The above five examples show that the following table of values for sine and cosine (from Fact 45) also holds under the unit-circle definitions of sine and cosine:

α

0

√ 0 sin α 2 √ 4 cos α 2

π 6 √ 1 2 √ 3 2

π 4 √ 2 2 √ 2 2

π 3 √ 3 2 √ 1 2

π 2 √ 4 2 √ 0 2

In fact, Fact 50 (informal). For α ∈ [0, π/2], Definitions 89 and 88 are the same.

Proof. Continue with the figure in Definition 89. Let A¯ = (Ax , 0).

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y

The ray a

1

A = (Ax , Ay ) Ay

α O

Ax

A¯ = (Ax , 0)

x

¯ Consider the right triangle OAA. We have “Hypotenuse” of length ∣OA∣ = 1, “Opposite” ¯ = Ay , and “Adjacent” of length ∣OA∣ ¯ = Ax . And so, by Definition 88(a), for of length ∣AA∣ α ∈ (0, π/2), sin α =

¯ Ay o ∣AA∣ = = = Ay h ∣OA∣ 1

which indeed coincide with Definition 89.

and

cos β =

¯ A a ∣OA∣ x = = = Ax , h ∣OA∣ 1

For α = 0, by Definition 88(b), sin 0 = 0 and cos 0 = 1, which coincides with Definition 89—see Example 406.

For α = π/2, by Definition 88(b), sin (π/2) = 1 and cos (π/2) = 0, which coincides with Definition 89—see Example 410.

If α > π/2, then our right-triangle definitions (Definition 88) are silent. Let’s see what our unit-circle definitions say:

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Example 411. If α =

2π π = π − , then 3 3 √ 3 1 2π 2π A=( , − ) = (sin , cos ) . 2 2 3 3

Figure to be inserted here.

Example 412. If α =

π 3π = π − , then 4 4 √ √ 2 2 3π 3π A=( ,− ) = (sin , cos ) . 2 2 4 4

Figure to be inserted here.

Example 413. If α =

5π π = π − , then 6 6 √ 1 3 5π 5π A = ( ,− ) = (sin , cos ) . 2 2 6 6

Figure to be inserted here.

Nonetheless, let’s see what happens if we forcibly try to extend those right-triangle definitions. Consider the right triangle XXX.

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Example 414. If α = π, then

A = (0, −1) = (sin π, cos π) .

Figure to be inserted here.

Example 415. If α =

7π π = π + , then 6 6 √ 3 7π 7π 1 ) = (sin , cos ) . A = (− , − 2 2 6 6

Figure to be inserted here.

Example 416. If α =

5π π = π + , then 4 4 √ √ 5π 5π 2 2 A = (− ,− ) = (sin , cos ) . 2 2 4 4

Figure to be inserted here.

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Example 417. If α =

4π π = π + , then 3 3 √ 3 1 4π 4π A = (− , − ) = (sin , cos ) . 2 2 3 3

Figure to be inserted here.

Example 418. If α =

3π , then 2 A = (−1, 0) = (sin

3π 3π , cos ) . 2 2

Figure to be inserted here.

Example 419. If α =

π 5π = 2π − , then 3 3 √ 3 1 5π 5π A = (− , ) = (sin , cos ) . 2 2 3 3

Figure to be inserted here.

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Example 420. If α =

7π π = 2π − , then 4 4 √ √ 2 2 7π 7π A = (− , ) = (sin , cos ) . 2 2 4 4

Figure to be inserted here.

Example 421. If α =

π 11π = 2π − , then 6 6 √ 1 3 11π 11π A = (− , ) = (sin , cos ). 2 2 6 6

Figure to be inserted here.

Example 422. If α = 2π, then we will have gone one full circle, with A = (0, 1) = (sin 2π, cos 2π) = (sin 0, cos 0) .

Figure to be inserted here.

As we’ll see in the next subchapter, each of sin and cos is periodic with period 2π. Let’s summarise our findings from the last 17 examples (Examples 406–422) in a single table:

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Fact 51. \ 0 √ 0 sin α √2 4 cos α 2 α

π √6 1 √2 3 2

π √4 2 √2 2 2

π 2π 3π 5π 7π 5π 4π 3π 5π π 6 4 3 2 3 √2 √3 √4 √6 √ √ √ √ √ √ 4 3 2 1 0 1 2 3 4 − − − − − 2 2 2 2 √2 √ √ √ √ √2 √2 √2 √2 √2 0 1 2 3 4 3 2 1 0 1 − − − − − − − 2 2 2 2 2 2 2 2 2 2

π √3 3 √2 1 2

Remark 69. As before,194 the above table is as written because that’s how I find it easier to remember. But you may prefer this (equivalent) table: π 6 1 sin α 0 √2 3 cos α 1 2 A 0

π √4 2 √2 2 2

π √3 3 2 1 2

π 2 1 0

2π 3π 5π 7π 5π 4π π 6 6 4 3 √3 √4 √ √ 3 2 1 1 2 3 0 − − − 2 2 2 2 √ √ √2 √2 1 2 3 3 2 1 − − − −1 − − − 2 2 2 2 2 2

3π 2

5π 7π 3 4 √ √ 3 2 −1 − − 2 √2 1 2 0 2 2

Remark 70. To repeat, henceforth, we will abandon the right-triangle definitions (Definition 88) and use only the unit-circle definitions (Definition 89 or equivalently, Definition 90). Note though that even the unit-circle definitions are considered informal, because they rely on geometry. To formally and rigorously define sine and cosine, there are (at least) three common approaches taken: 1. Use power series—see Part V (Calculus), Ch. 98.3, Definitions 217 and 218. 2. Use Euler’s formula—see Part III (Complex Numbers), Ch. 80.2, Remark 116. 3. Use differential equations—see Part V (Calculus), Ch. 106.5. But don’t worry about these formal definitions of sine and cosine. For A-Level Maths, the unit-circle definitions (Definition 89 or 90) will be more than good enough.

194

See Remark 64.

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28.7.

Graphs of Sine and Cosine

Fact 51 gave 17 points for each of sin and cos. Let’s graph these points:

y cos

sin

π 2 π π π 6 4 3

2π 3

7π 6

5π 6 3π 4

π

4π 3 5π 4

3π 2

7π 4 5π 3

11π 6

x 2π

Hm, it looks a lot like we can simply “connect the dots” to get these graphs of sin and cos on [0, 2π]:

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y cos

sin

π 2 π π π 6 4 3

2π 3

7π 6

5π 6 3π 4

π

4π 3 5π 4

3π 2

7π 4 5π 3

11π 6

x 2π

But is it proper for us to simply “connect the dots”? All we’ve done is find 17 points for each of sin and cos. How do we know that their actual graphs can be obtained by smoothly “connecting the dots”? Given the 17 points, why couldn’t the actual graphs be something crazy like the following?

Figure to be inserted here.

Here’s an informal argument for why this last figure is wrong (while the previous figure is correct). We’ll use our alternative “travelling particle definition” (Definition 90) of sin and cos: 1. At time α = 0, our particle A is at (1, 0). So, sin α = 0 and cos α = 1. 2. During α ∈ (0, π/2), A travels northwest.

(a) Travel upwards is initially fast, but then slows down. So, throughout α ∈ (0, π/2), sin is increasing at a decreasing rate. (b) Travel leftwards is initially slow, but then speeds up. So, throughout α ∈ (0, π/2),cos is decreasing at an increasing rate.

3. At time α = π/2, our particle A is at (0, 1). So, sin α = 1 and cos α = 0. 4. During α ∈ (π/2, π), A travels southwest. 344, Contents

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(a) Travel downwards is initially slow, but then speeds up. So, throughout α ∈ (0, π/2), sin is decreasing at an increasing rate. (b) Travel leftwards is initially fast, but then slows down. So, throughout α ∈ (0, π/2), cos is decreasing at a decreasing rate.

5. At time α = π, our particle A is at (−1, 0). So, sin α = 0 and cos α = −1. 6. During α ∈ (π, 3π/2), A travels southeast.

(a) Travel downwards is initially fast, but then slows down. So, throughout α ∈ (0, π/2),sin is decreasing at a decreasing rate. (b) Travel rightwards is initially slow, but then speeds up. So, throughout α ∈ (0, π/2), cos is increasing at an increasing rate.

7. At time α = 3π/2, our particle A is at (0, −1). So, sin α = −1 and cos α = 0.

8. During α ∈ (3π/2, 2π), A travels northeast.

(a) Travel upwards is initially slow, but then speeds down. So, throughout α ∈ (0, π/2), sin is increasing at an increasing rate. (b) Travel rightwards is initially fast, but then slows down. So, throughout α ∈ (0, π/2), cos is increasing at a decreasing rate.

Figure to be inserted here.

Lovely. So, we’re fairly confident about what the graphs of sin and cos look like on [0, 2π].

Let’s now consider what the graphs of sin and cos look like on [2π, 4π].

As mentioned in the last example of the previous subchapter (Example 422), we know that at 2π, we’ve gone a full circle. Thereafter, everything should exactly repeat. And so, the graphs of sin and cos on [2π, 4π] must be exactly the same as those on [0, 2π]: ,

Figure to be inserted here.

Indeed, repeating the same reasoning for each k ∈ Z+ , the graphs of sin and cos on [2kπ, (2k + 2) π] must be exactly the same as those on [0, 2π]: 345, Contents

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Figure to be inserted here.

Next, let’s consider what the graphs of sin and cos look like on [−2π, 0].

Well, what does sin α or cos α for negative α even mean? Well, we’ve adopted the convention that rotating the positive x-axis x+ anticlockwise is the positive direction. So, negative angles should correspond to rotating x+ clockwise. Or equivalently and using our alternative definition (Definition 90), negative α corresponds to our particle A travelling clockwise.

And whether we’re travelling anticlockwise or clockwise, by reasoning similar to ,, after every 2π seconds, we will have gone one full circle. Hence, the graphs of sin and cos on [−2π, 0] must be exactly the same as those on [0, 2π]:

Figure to be inserted here.

Indeed, repeating the same reasoning for each k ∈ Z− , the graphs of sin and cos on [(2k − 2) π, 2kπ] must be exactly the same as those on [0, 2π]:

Figure to be inserted here.

With this last figure, we’ve completed our construction of the graphs of sin and cos. To repeat, each of sin and cos is said to be periodic with period 2π. Informally, this means that each graph “repeats” every 2π.195 A bit more formally, Fact 52. For all x ∈ R, (a) sin (x + 2π) = sin x; and (b) cos (x + 2π) = cos x.

Proof. (a) By the Addition Formula for Sine (Fact 46) and Fact 51, 195

For a more formal definition of periodicity, see Definition 283 (Appendices).

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sin (x + 2π) = sin x cos 2π + cos x sin 2π = sin x.

(b) By the Addition Formula for Cosine (Fact 46) and Fact 51,

cos (x + 2π) = cos x cos 2π − sin x sin 2π = cos x.

Equivalently, the translation of sin or cos by any integer multiple of 2π is itself:

Figure to be inserted here.

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28.8.

Graphs of Sine and Cosine: Some Observations

Figure to be inserted here.

“Clearly”, sin is symmetric in the vertical line x = π/2. And more generally, for any odd integer k, sin is symmetric in x = kπ/2.

Similarly, cos is symmetric in the vertical line x = π. And more generally, for any integer k, cos is symmetric in x = kπ. For future reference, let’s jot the above observations down as a formal result: Fact 53. The lines of symmetry for

(a) sin are x = kπ/2, for odd integers k; (b) cos are x = kπ, for integers k.

Proof. See p. 1564 (Appendices).

Cosine is symmetric in the y-axis: Fact 54. For all x ∈ R, cos x = cos (−x).

Proof. By Fact 53(b), cos is symmetric in x = 0. So, by Corollary 5, for all x ∈ R, cos x = cos (−x).

Figure to be inserted here.

The reflection of sin in the y-axis is the reflection of sin in the x-axis: Fact 55. For all x ∈ R, − sin x = sin (−x).

Proof. By Fact 49 (S2P or P2S Formulae), for all x ∈ R, sin x + sin (−x) = 2 sin

Rearranging, sin x = − sin (−x).

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The translation of sin or cos by any odd integer multiple of π is its own reflection in the x-axis: Fact 56. For all x ∈ R and all odd integers k,

(a) sin (x + kπ) = − sin x

(b) cos (x + kπ) = − cos x.

Figure to be inserted here.

Proof. First, observe that for all odd integers k, cos (kπ/2) = 0. 1

(a) By Fact 49 (S2P or P2S Formulae), for all x ∈ R, sin (x + kπ) + sin x = 2 sin

x + kπ + x x + kπ − x π π 1 cos = 2 sin (x + k ) cos k = 0. 2 2 2 2

Rearranging, sin (x + kπ) = − sin x.

(b) By Fact 49 (S2P or P2S Formulae), for all x ∈ R, cos (x + kπ) + cos x = 2 cos

x + kπ + x x + kπ − x π π 1 cos = 2 cos (x + k ) cos k = 0. 2 2 2 2

Rearranging, cos (x + kπ) = − cos x.

Figure to be inserted here.

Observe that sin is cos shifted right by π/2. Equivalently, cos is sin shifted left by π/2. Formally, Fact 57. For all x ∈ R,

π (a) sin x = cos (x − ) 2 π (b) cos x = sin (x + ). 2

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π π π Proof. (a) cos (x − ) = cos x cos + sin x sin = 0 + sin x = sin x. 2 2 2 π π π (b) sin (x + ) = sin x cos + cos x sin = 0 + cos x = cos x. 2 2 2

About the prefix co-. You will probably have noticed that the word cosine is simply the prefix co- added in front of the word sine: cosine = co + sine.

In the context of trigonometric functions, the prefix co- has this meaning: If A and B are complementary angles (i.e. A + B = π/2), then Or equivalently, for any A ∈ R,

sin A = cos B.

π sin A = cos ( − A) . 2

Also equivalently, using what we learnt in Ch. 24 (Transformations), Sine is cosine shifted left by π/2, then reflected in the y-axis. And symmetrically, Cosine is sine shifted left by π/2, then reflected in the y-axis. As we’ll see in the coming chapters, the above will also be true of the functions tangent and cotangent, and the functions secant and cosecant.

Figure to be inserted here.

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29.

Trigonometry: Tangent

Definition 91. The tangent function, denoted tan, is defined by tan =

Example 423. XXX

sin . cos

Example 424. XXX Exercise 141. What are the domain and codomain of tan? (Hint: Review Ch. 18.) (Answer on p. 1773.) Under the right-triangle definitions, we had sin =

Opposite Hypotenuse

and

cos =

sin Opposite/Hypotenuse Opposite = = . cos Adjacent/Hypotenuse Adjacent In the past, Singaporean students were taught the Hokkien mnemonic “big leg woman”: Hence, tan =

tu¯a

kha

TOA

CAH

SOH

Tangent =

Cosine =

Sine =

Opposite . Hypothenuse

But these days, we probably prefer a more angmoh mnemonic like Soccer tour—SOH-CAH-TOA.196 Example 425. XXX Example 426. XXX

196

Credit to Tom J at TheStudentRoom.co.uk (warning: that page may contain many other NSFW mnemonics).

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29.1.

Some Formulae Involving Tangent

Happily, List MF26 (p.3) contains the following formulae for the tangent function, so no need to mug: Fact 58. Suppose the real numbers A and B are not odd integer multiples of π/2. (a) (Addition and Subtraction Formula for Tangent) If A±B is not an odd integer multiple of π/2,197 then tan (A ± B) =

tan A ± tan B . 1 ∓ tan A tan B

(b) (Double Angle Formula for Tangent) If 2A is not an odd integer multiple of π/2, then tan 2A =

Proof. (a) See Exercise 142.

2 tan A . 1 − tan2 A

(b) See Exercise 143. Exercise 142. Prove Fact 58(a). Exercise 143. Prove Fact 58(b).

We just learnt about the Addition and Subtraction Formulae for Tangent: tan (A ± B) =

tan A ± tan B . 1 ∓ tan A tan B

A simple rearrangement yields the Sum-to-Product (S2P) Formula for Tangent: 197

In each of Fact 58(a) and (b), there is the worrying possibility that the denominator is zero, in which case we’d be committing the Cardinal Sin of Dividing by Zero (see Ch. 2.2). It turns out that happily, in each case, our additional condition that A ± B or 2A is not an odd integer multiple of π/2, as we now show (via the contrapositives):

(a) Suppose 1 ∓ tan A tan B = 0. This is equivalent to

sin A sin B = ±1 cos A cos B 0 = sin A sin B ∓ cos A cos B = ∓ cos (A ± B)

tan A tan B = ±1

⇐⇒

so that A ± B is not an odd integer multiple of π/2.

⇐⇒

(b) Suppose 1 − tan2 A = 0. This is equivalent to tan2 A = 1

⇐⇒

tan A = ±1 ⇐⇒ A ∈ {(k ± 0.25) π ∶ k ∈ Z} ⇐⇒ 2A ∈ {0.5kπ ∶ k is an odd integer},

so that 2A is not an odd integer multiple of π/2. 352, Contents

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Fact 59. (S2P Formula for Tangent) Suppose the real numbers A and B are not odd integer multiples of π/2. If A ± B is not an odd integer multiple of π/2, then tan A ± tan B = tan (A ± B) (1 ∓ tan A tan B) .

Fact 60. (Triple Angle Formula for Tangent) {kπ/2 ∶ k is an odd integer}. If tan 3A is defined,198 then 3 tan A − tan3 A tan 3A = . 1 − 3 tan2 A

Proof. See Exercise 144. Exercise 144. Prove Fact 60.

Let

A, B

R ∖

Earlier, we had the Laws of Sines and Cosines (Proposition 6(a) and (b)). It turns out there is also a Law of Tangents: Fact 61. Suppose △ABC has angles A, B, and C, and sides of lengths a, b, and c. Then

Proof. See Exercise 145.

198

a + b tan A+B 2 = . A−B a − b tan 2

(Law of Tangents)

Similar to n. 197, we can show that the condition that tan 3A prevents the denominator 1 − 3 tan2 A from being zero: Suppose 1 − 3 tan2 A = 0. This is equivalent to √ tan2 A = 1/3 ⇐⇒ tan A = ± 3/3 ⇐⇒ A ∈ {(k ± 1/6) π ∶ k ∈ Z} Ô⇒ 3A ⊆ {kπ/2 ∶ k is an odd integer}, so that tan 3A is not defined.

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Exercise 145. This Exercise will guide you through a proof of the Law of Tangents (Fact 61):. By the Law of Sines,

(a) By considering

sin A sin B = a b

or

b sin A = a sin B. 1

a + b sin A sin B 1 × (Times One Trick) and using =, show that a − b sin A sin B a + b sin A + sin B = . a − b sin A − sin B

a + b tan A+B 2 199 . (b) Now complete the proof that = a − b tan A−B 2

199

Hint: Whenever you see a trigonometry question, what should you do?

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29.2.

Graph of Tangent

Fact 51 gave a table of 17 values for each of sin and cos. We can use that table to produce these values of tan: Fact 62. \ π 6 1 tan α 0 √ 3 α 0

π π π 2π 3π 5π 7π π 4 √3 2 3 4 6 6 √ 1 1 1 3 Undefined − 3 −1 − √ 0 √ 3 3

5π 4π 3π 5π 4 √3 2 3 √ 1 3 Undefined − 3

Since tan = sin / cos and cos α = 0 ⇐⇒ α = kπ/2 for odd integers k, we know about the vertical asymptotes for tan: Fact 63. For each odd integer k, the vertical line x = kπ/2 is a vertical asymptote for tangent.

Let’s graph the points in Fact 62:

Figure to be inserted here.

Hm, it looks a lot like we can simply “connect the dots” to get this graph of tan on [0, 2π]:

Figure to be inserted here.

As before, it is possible to make an informal argument as to why the above graph is correct and something crazy like the below is wrong. But we shall omit this.

Figure to be inserted here.

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Like sine and cosine, tangent is periodic. Unlike sine and cosine, tangent has period π (instead of 2π). Informally, this means that the graph of tan “repeats” every π. A bit more formally, Fact 64. For all x ∈ R, tan (x + π) = tan x.

Proof. By the Addition Formula for Tangent (Fact 58) and Fact 62 tan (x + π) =

tan x tan x + tan π = = tan x. 1 − tan x tan π 1 − 0

Equivalently, the translation of tan by any integer multiple of π is itself. And so, the complete graph of tan looks like this:

Figure to be inserted here.

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29.3.

Graph of Tangent: Some Observations

The reflection of tan in the y-axis is the reflection of tanin the x-axis: Fact 65. If the real number x is not an odd integer multiple of π/2, then − tan x = tan (−x).

Proof. Note that −x + x = 0, which is not an odd integer multiple of π/2. So, by Fact 59 (S2P Formula for Tangent), tan (−x) + tan x = tan (−x + x) [1 − tan (−x) tan x] = 0.

Rearranging, − tan x = tan (−x).

Fact 66. The graph of tan has no lines of symmetry.

Proof. See p. 1565 (Appendices).

Figure to be inserted here.

You may be wondering whether any single “branch” of tan might have a line of symmetry. For example, might the graph of tan ∣ have a line of symmetry (perhaps some (−π/2,π/2)

downward-sloping line)?

Figure to be inserted here.

Unfortunately, the answer is no: Fact 67. The graph of tan ∣

(−π/2,π/2)

has no lines of symmetry.

Proof. See p. 1565 (Appendices). 357, Contents

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30.

Trigonometry: Summary and a Few New Things

Graphs of sin, cos, and tan:

tan

y

cos

−2π

sin

1

3π 2 −π

π 2

3π 2

π 2

x 2π

π −1

To remember these three graphs, here are the key things to remember: • sin and cos both have – – – –

the same shapes; period 2π; max value 1; and min value −1.

• sin and tan both go through the origin. • cos has y-intercept (1, 0).

• tan goes from −∞ to ∞ and has period π.

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30.1.

The Signs of Sine, Cosine, and Tangent

Fact 68. (The Signs of Sine, Cosine, and Tangent) ⎧ ⎪ ≥ 0, for x ∈ [0, π] , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for x ∈ (0, π) , ⎪> 0, (a) sin x ⎨ ⎪ ⎪ ≤ 0, for x ∈ [π, 2π] , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for x ∈ (π, 2π) . ⎩< 0,

⎧ ⎪ ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪> 0, (b) cos x ⎨ ⎪ ⎪ ≤ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩< 0,

⎧ ⎪ ≥ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪> 0, (c) tan x ⎨ ⎪ ⎪ ≤ 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩< 0,

for x ∈ [−π/2, π/2] ,

for x ∈ (−π/2, π/2) , for x ∈ [π/2, 3π/2] ,

for x ∈ (π/2, 3π/2) . for x ∈ [0, π/2) ,

for x ∈ (0, π/2) ,

for x ∈ (−π/2, 0] ,

for x ∈ (−π/2, 0) .

Figure to be inserted here.

Of course, since sin and cos are periodic with period 2π, while tanis periodic with period π, we can rewrite the above result more generally as

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Corollary 7. (The Signs of Sine, Cosine, and Tangent) Suppose k is any integer. Then ⎧ ⎪ ≥ 0, for x ∈ [2kπ, (2k + 1) π] , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for x ∈ (2kπ, (2k + 1) π) , ⎪> 0, (a) sin x ⎨ ⎪ ⎪ ≤ 0, for x ∈ [(2k + 1) π, (2k + 2) π] , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for x ∈ ((2k + 1) π, (2k + 2) π) . ⎩< 0, ⎧ 1 1 ⎪ ⎪ ≥ 0, for x ∈ [(2k − ) π, (2k + ) π] , ⎪ ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ 1 1 ⎪ ⎪ ⎪ ) π, (2k + ) π) , > 0, for x ∈ ((2k − ⎪ ⎪ 2 2 (b) cos x ⎨ 1 3 ⎪ ⎪ ≤ 0, for x ∈ [(2k + ) π, (2k + ) π] , ⎪ ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ 1 3 ⎪ ⎪ ⎪ < 0, for x ∈ ((2k + ) π, (2k + ) π) . ⎪ ⎪ 2 2 ⎩ ⎧ 1 ⎪ ⎪ ≥ 0, for x ∈ [2kπ, (2k + ) π) , ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ > 0, for x ∈ (2kπ, (2k + ) π) , ⎪ ⎪ 2 (c) tan x ⎨ 1 ⎪ ⎪ ≤ 0, for x ∈ ((2k − ) π, 2kπ] , ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ < 0, for x ∈ ((2k − ) π, 2kπ) . ⎪ ⎪ 2 ⎩

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30.2.

Half-Angle Formulae

Fact 69. (Half-Angle Formulae √ ⎧ ⎪ 1 − cos x ⎪ ⎪ ⎪ , ⎪ x ⎪ 2 √ (a) sin = ⎨ 2 ⎪ 1 − cos x ⎪ ⎪ ⎪ , − ⎪ ⎪ 2 ⎩ √ ⎧ ⎪ 1 + cos x ⎪ ⎪ ⎪ , ⎪ x ⎪ (b) cos = ⎨ √ 2 2 ⎪ 1 + cos x ⎪ ⎪ ⎪ − , ⎪ ⎪ 2 ⎩

x sin x = , for all 2 1 + cos x x 1 − cos x (c)(ii) tan = , for all 2 sin x √ ⎧ ⎪ 1 − cos x ⎪ ⎪ ⎪ ⎪ x ⎪ 1 + cos x , (c)(iii) tan = ⎨ √ 2 ⎪ 1 − cos x ⎪ ⎪ ⎪ − , ⎪ ⎪ 1 + cos x ⎩ (c)(i) tan

for Sine, Cosine, and Tangent) x ∈ [0, π] , 2 x for ∈ [π, 2π] . 2 for

x π π ∈ [− , ] , 2 2 2 x π 3π for ∈ [ , ] . 2 2 2 for

x ∈ R. 2 x ∈ R. 2

x π ∈ [0, ) , 2 2 x π for ∈ (− , 0] . 2 2

for

Proof. By the Double Angle Formulae, for all x ∈ R, So, Or,

x x x x cos x = cos ( + ) = 2 cos2 − 1 = 1 − 2 sin2 . 2 2 2 2

x 1 − cos x = 2 2 √ x 1 − cos x sin = ± 2 2 sin2

and

and

x 1 + cos x = . 2 2 √ x 1 + cos x cos = ± . 2 2

cos2

To figure out what the correct sign is for each specific x/2, use Corollary 7: (a) By Corollary 7, sin

Hence,

x x ≥ 0 if ∈ [0, π] and 2 2 √ ⎧ ⎪ 1 − cos x ⎪ ⎪ ⎪ , ⎪ x ⎪ 2 √ sin = ⎨ 2 ⎪ 1 − cos x ⎪ ⎪ ⎪ − , ⎪ ⎪ 2 ⎩

(b) Similarly and again by Corollary 7,

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sin

x x ≤ 0 if ∈ [π, 2π]. 2 2

x ∈ [0, π] , 2 x for ∈ [π, 2π] . 2

for

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cos

Hence,

x x ≥ 0 if ∈ [−π/2, π/2] and 2 2 √ ⎧ ⎪ 1 + cos x ⎪ ⎪ ⎪ , ⎪ x ⎪ 2 √ cos = ⎨ 2 ⎪ 1 + cos x ⎪ ⎪ ⎪ − , ⎪ ⎪ 2 ⎩

(c) See Exercise 146.

Exercise 146. Prove Fact 69(c).

cos

x x ≤ 0 if ∈ [π/2, 3π/2]. 2 2

x π π ∈ [− , ] , 2 2 2 x π 3π for ∈ [ , ] . 2 2 2 for

Of course, since sin and cos are periodic with period 2π, while tanis periodic with period π, we can rewrite (a), (b), and (c)(iii) in the above result more generally as Corollary 8. Suppose k is any integer. Then √ ⎧ ⎪ 1 − cos x x ⎪ ⎪ ⎪ , for ∈ [2kπ, (2k + 1) π] , ⎪ x ⎪ 2 (a) sin = ⎨ √ 2 ⎪ 2 ⎪ x 1 − cos x ⎪ ⎪ − , for ∈ [(2k + 1) π, (2k + 2) π] . ⎪ ⎪ 2 2 ⎩ √ ⎧ ⎪ x 1 1 1 + cos x ⎪ ⎪ ⎪ , for ∈ [(2k − ) π, (2k + ) π] , ⎪ x ⎪ 2 2 2 2 (b) cos = ⎨ √ 2 ⎪ 1 + cos x x 1 3 ⎪ ⎪ ⎪ − , for ∈ [(2k + ) π, (2k + ) π] . ⎪ ⎪ 2 2 2 2 ⎩ √ ⎧ ⎪ 1 − cos x x π ⎪ ⎪ ⎪ , for ∈ [kπ, (2k + 1) ), ⎪ x ⎪ 1 + cos x 2 2 (c)(iii) tan = ⎨ √ 2 ⎪ x π 1 − cos x ⎪ ⎪ ⎪ − , for ∈ ((2k − 1) , kπ] . ⎪ ⎪ 1 + cos x 2 2 ⎩

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31.

Trigonometry: Three More Functions

The cosecant and secant functions are defined as the reciprocals of the sine and cosine functions: Definition 92. The secant and cosecant functions, denoted sec and cosec, are defined by sec =

1 cos

cosec =

and

1 . sin

Definition 93. The cotangent function, denoted cot, is defined by

Example 427. XXX

cot =

cos . sin

Remark 71. Your A-Level syllabus and exams denote the cosecant function by cosec—and so, this is what we’ll do too. Be aware though that many writers instead denote it by csc. Example 428. XXX Remark 72. Together, sin, cos, tan, cosec, sec, and cot are the six trigonometric functions (or circular200 functions) that you need to know for A-Level Maths. Other trigonometric functions that were historically used but are no longer taught include versine (or versed sine), coversine, haversine (or half versed sine), chord, exsecant, and excosecant. (These other functions may all be expressed in terms of sin and/or cos.) In a right triangle,

Hence,

sin =

Opposite , Hypotenuse

cos =

tan =

1 Hypotenuse 1 Hypotenuse = , sec = = , sin Opposite cos Adjacent cos Adjacent/Hypotenuse Adjacent cot = = = . sin Opposite/Hypotenuse Opposite

cosec = Example 429. XXX Example 430. XXX 200

On p. 18 of your H2 Maths syllabus, these six functions are simply called the circular functions. However, the term trigonometric functions is probably more common.

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Exercise 147. XXX

A147. The cotangent function cot is “nearly” but not quite the same as the function 1/ tan:

Figure to be inserted here.

Fact 70. cot ≠ 1/ tan

Proof. Since tan is not defined at π/2, neither is the function 1/ tan. In contrast, cot is defined at π/2 (with cot π/2 = 0). Hence, cot ≠ 1/ tan.

This was the First Pythagorean Identity (Fact 47): sin2 A + cos2 A = 1 (for all A ∈ R).

We can use the above identity to prove two related identities (with equally inspiring names): Fact 71. Suppose A ∈ R.

(a) If cos A ≠ 0, then 1 + tan2 A = sec2 A. (b) If sin A ≠ 0, then 1 + cot2 A = cosec2 A.

(Second Pythagorean Identity) (Third Pythagorean Identity)

1 sin2 A cos2 A + sin2 A = = = sec2 A. 2 2 2 cos A cos A cos A 2 2 2 cos A sin A + cos A 1 (b) 1 + cot2 A = 1 + = = = cosec2 A. 2 2 2 sin A sin A sin A

Proof. (a) 1 + tan2 A = 1 +

Remark 73. In (a) of the above result, the condition that cos A ≠ 0 is simply to ensure that tan A and sec A are well-defined.

Similarly, in (b), the condition that sin A ≠ 0 is simply to ensure that cot A and cosecA are well-defined. Remark 74. You should mug the three Pythagorean Identities, because they’re not on List MF26. In Ch. 24.8, we learnt to use the graph of f to graph 1/f . Here’s more practice: 364, Contents

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Exercise 148. Graph cosec, sec, and cot. For your convenience, here are the graphs of sin, cos, and tan again: (Answer on the next page.)

tan

y

cos

sin

−2π

1

3π 2 −π

π 2

−1

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3π 2

cot

sec

−π

x 2π

π

cosec

−2π

3π 2

π 2

π 2

1 −1

π π 2

3π 2 2π

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32.

Trigonometry: Three Inverse Trigonometric Functions

Suppose we want to construct the inverses of the sine, cosine, and tangent functions.

Unfortunately, none of them is one-to-one. For example, the horizontal line y = 1 intersects each graph more than once—so, by the Horizontal Line Test (HLT), none is one-to-one.

tan

y

sin

cos

y=1

x

(Indeed, for any c ∈ [−1, 1], the horizontal line y = c intersects each graph more than once. And any horizontal line intersects the graph of tan more than once.)

So, sin, cos, and tan are not one-to-one. But as we learnt in Ch. 19, by restricting the domain of each function, we can create a brand new function that is one-to-one. As usual, there are many (in fact infinitely many) ways we can do the domain restriction.

For example, the restriction of sin to [2, 4] is one-to-one. So too are the restrictions of cos to [−3, −2] and tan to [4, 5]:

Figure to be inserted here.

But of course, as usual, we want to pick restrictions that are as “large” as possible. So, we might for example instead pick the restrictions of sin to [π/2, 3π/2], cos to [−2π, 0], and tan to (π/2, 3π/2), which are one-to-one: 366, Contents

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Figure to be inserted here.

Indeed, any translation (or shift) of the above restrictions by any integer multiple of 2π would also work.

However, by convention, we’ll pick the restrictions of sin to [−π/2, π/2], cos to [0, π], and tan to (−π/2, π/2): Exercise 149. Graph sin ∣

[− π2 , π2 ]

, cos ∣

[0,π]

, tan ∣

one-to-one. Then write down their inverses.

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(− π2 , π2 )

. Explain why these functions are (Answer on the next page.)

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201

Below are the graphs of sin ∣

[− π2 , π2 ]

, cos ∣

sects by the HLT, each is one-to-one.

[0,π]

, tan ∣

(− π2 , π2 )

y

tan ∣

sin

π 2

(− π2 , π2 )

π 2

x

cos ∣

cos

tan

. “Clearly”, no horizontal line inter-

π

[0,π]

sin ∣

[− π2 , π2 ]

We now define the three inverse trigonometric functions—arcsine, arccosine and arctangent: Definition 94. The functions arcsine, arccosine, and arctangent—denoted sin−1 , cos−1 , and tan−1 —are defined as the inverses of sin ∣ π π , cos ∣ , and tan ∣ π π and are called the three inverse trigonometric functions.

[− 2 , 2 ]

[0,π]

(− 2 , 2 )

More explicitly, the domain, codomain, and mapping rule of each of sin−1 , cos−1 , and tan−1 are these: sin−1

cos−1

tan−1

Domain Codomain Mapping rule π π [−1, 1] [− , ] If y = sin x, thensin−1 y = x. 2 2 [−1, 1] [0, π] If y = cos x, thencos−1 y = x. π π If y = tan x, thentan−1 y = x. R (− , ) 2 2

Graphs of sin−1 , cos−1 , and tan−1 : 201

Note that the domain restrictions given here are somewhat arbitrary. For example, with sin, we could equally well have chosen to restrict the domain to [π/2, 3π/2] instead. Nonetheless, by convention, these domain restrictions are standard and so they are what we’ll use.

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y

π

cos−1 π 2

−1

sin−1 tan−1

x

1

π 2

We learnt earlier (Fact 233) that for an inverse function, the range and codomain coincide. This is true here—for each of sin−1 , cos−1 , and tan−1 here, the range and codomain coincide. Observe that

• sin−1 has endpoints (−1, −π/2) and (1, π/2), and is strictly increasing; • cos−1 has endpoints (−1, π) and (1, 0), and is strictly decreasing; and • tan−1 has no endpoints and is strictly increasing. Example 431. XXX Example 432. XXX Example 433. XXX Remark 75. We could, of course, also construct the arccosecant, arcsecant, and arccotangent functions (which are denoted cosec−1 , sec−1 , and cot−1 in your H2 Maths syllabus, p. 18). But happily, we needn’t bother with these because they aren’t on your A-Level Maths syllabus. Exercise 150. XXX

A150.

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32.1.

Confusing Notation

There are two separate and big points of potential confusion:

1. In Ch. 28.3, we noted that sin2 = sin ⋅ sin. We then noted that this is confusing and contrary to our earlier use of f 2 to denote the composite function f ○ f . Here, to add to our confusion, sin−1 doesn’t mean 1/ sin, as would be logical given that sin2 = sin ⋅ sin. Instead, sin−1 x denotes the inverse of the function sin ∣ π π ! [− 2 , 2 ]

2. The notation (sin−1 , cos−1 , and tan−1 ) and names (inverse trigonometric functions) may incorrectly suggest that sin−1 , cos−1 , and tan−1

are the inverses of

sin, cos, and tan.

3

But strictly and pedantically speaking, the above statement is false! Instead, sin−1 , cos−1 , and tan−1

are the inverses of

sin ∣

[− π2 , π2 ]

, cos ∣

[0,π]

, tan ∣

(− π2 , π2 )

.

7

Because of the above two points of confusion, many writers prefer to denote the arcsine, arccosine, and arctangent functions by arcsin, arccos, and arctan. Some other writers use Sin−1 , Cos−1 , and Tan−1 . Still others (especially computer programs) use asin, acos, and atan. However and unfortunately, your A-Level exams and syllabus insist on using the notation sin−1 , cos−1 , and tan−1 . And so that’s what we’ll have to do too.

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32.2.

Inverse Cancellation Laws, Revisited

We reproduce Proposition 4 from Ch. 22.4: Proposition 4. (Inverse Cancellation Laws) Suppose the function f has inverse f −1 . Then

(a) f −1 (f (x)) = x for all x ∈ Domain f ; and (b) f (f −1 (y)) = y for all y ∈ Domain f −1 .

Given the above cancellation laws, the following result is perhaps not surprising: Fact 72. (a) sin (sin−1 x) = x for all x ∈ Domain sin−1 = [−1, 1].

(b) cos (cos−1 x) = x for all x ∈ Domain cos−1 = [−1, 1]. (c) tan (tan−1 x) = x for all x ∈ Domain tan−1 = R.

Proof. See p. 1565 (Appendices).

Example 434. For x = 1, we have π (a) sin (sin−1 1) = sin = 1. 2 (b) cos (cos−1 1) = cos 0 = 1. π (c) tan (tan−1 1) = tan = 1. 4 Example 435. For x = 0, we have

(a) sin (sin−1 0) = sin 0 = 0. π (b) cos (cos−1 0) = cos = 0. 2 −1 (c) tan (tan 0) = tan 0 = 0.

Of course, if x ∉ [−1, 1], then sin−1 x and cos−1 x are undefined, so that sin (sin−1 x) and cos (cos−1 x) are also undefined: √ Example 436. For x = 3 ≈ 1.73, we have √ (a) sin (sin−1 3) is undefined. √ (b) cos (cos−1 3) is undefined. √ π √ (c) tan (tan−1 3) = tan = 3. 3

The composite functions sin ○ sin−1 , cos ○ cos−1 , and tan ○ tan−1 exist (why?)202 and their 202

π π The composite function sin ○ sin−1 exists because Range sin−1 = [− , ] ⊆ R = Domain sin. 2 2

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graphs are these:

Figure to be inserted here.

Similarly, cos ○ cos−1 exists because Range cos−1 = [0, π] ⊆ R = Domain cos. π π And tan ○ tan−1 exists because Range tan−1 = (− , ) ⊆ R ∖ {(2k + 1) π/2 ∶ k ∈ Z} = Domain tan. 2 2

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Perhaps surprisingly, the following three claims are false:

(a) “sin−1 (sin x) = x (b) “cos−1 (cos x) = x (c) “tan−1 (tan x) = x

for all x ∈ Domain sin = R.” for all x ∈ Domain cos = R.” for all x ∈ Domain tan = R ∖ {(2k + 1) π/2 ∶ k ∈ Z}.”

7 7 7

A simple counterexample to show that the above three claims are false: Example 437. For x = 4π, we have

(a) sin−1 (sin 4π) = sin−1 0 = 0 ≠ 4π . π (b) cos−1 (cos 4π) = cos−1 1 = ≠ 4π . 2 (c) tan−1 (tan 4π) = tan−1 0 = 0 ≠ 4π.

Here are the correct versions of the above three false claims: Fact 73. (a) sin−1 (sin x) = x (b) cos−1 (cos x) = x

(c) tan−1 (tan x) = x

⇐⇒

⇐⇒

Proof. See p. 1566 (Appendices).

⇐⇒

π π x ∈ [− , ]. 2 2

x ∈ [0, π]. π π x ∈ (− , ). 2 2

Example 438. XXX Example 439. XXX The composite functions sin−1 ○ sin, cos−1 ○ cos, and tan−1 ○ tan exist (why?).203

And their graphs are these:204

Figure to be inserted here.

The composite function sin−1 ○ sin exists because Range sin = [−1, 1] ⊆ [−1, 1] = Domain sin−1 . Similarly, cos−1 ○ cos exists because Range cos = [−1, 1] ⊆ [−1, 1] = Domain cos−1 . And tan−1 ○ tan exists because Range tan = R ⊆ R = Domain tan−1 . 204 See Fact 236 (Appendices). 203

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Exercise 151. Here’s the graph of cos−1 :

y

cos−1

π

π 2

−1

1

x

Clearly, the following claim is false: “If x ∈ [−1, 1), then cos−1 x = 0.”

7

Nonetheless, below is a seven-step “proof” of the above false claim. Identify any error(s). (Answer on p. 1774.) 1. Let x ∈ [−1, 1).

2. Let A = cos−1 x. 3. So, x = cos A. 1

4. But by Fact 54 (Cosine Is Symmetric in y-Axis), cos A = cos (−A). 5. So, x = cos (−A). 6. Hence, cos−1 x = −A. 2

7. Taking = + =, we get 2 cos−1 x = 0 and thus cos−1 x = 0. 1

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32.3.

More Compositions

√ Fact 74. (a) sin (cos−1 x) = 1 − x2 for all x ∈ [−1, 1]. √ (b) cos (sin−1 x) = 1 − x2 for all x ∈ [−1, 1]. x (c) sin (tan−1 x) = √ for all x ∈ R. 1 + x2 1 (d) cos (tan−1 x) = √ for all x ∈ R. 1 + x2 x (e) tan (sin−1 x) = √ for all x ∈ (−1, 1). 1 − x2 √ 1 − x2 −1 (f) tan (cos x) = for all x ∈ [−1, 1] ∖ {0}. x

Proof. For a formal and complete proof, see p. 1568 (Appendices). Here we give only an informal and incomplete proof-by-picture:205

Consider a right triangle √ with legs of lengths x > 0 and 1. By Pythagoras’ Theorem, its hypotenuse has length 1 + x2 . Let α be the angle opposite the leg of length x.

Figure to be inserted here.

Since tan α =

Opposite x = = x, we have α = tan−1 x. Hence, Adjacent 1

Opposite x =√ and Hypotenuse 1 + x2 Adjacent 1 cos (tan−1 x) = cos α = =√ . Hypotenuse 1 + x2

sin (tan−1 x) = sin α =

Next, consider a right triangle with hypotenuse √ of length 1 and a leg of length x. By Pythagoras’ Theorem, its other leg has length 1 − x2 .

205

It’s informal because it relies on geometry and incomplete because it covers only the special case where x > 0.

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Figure to be inserted here. √ Let β and γ be the angles opposite the legs of lengths x and 1 − x2 , respectively. Opposite x Since sin β = = = x, we have β = sin−1 x. Hence, Hypotenuse 1 √ 1 − x2 √ Adjacent cos (sin−1 x) = cos β = and = = 1 − x2 Hypotenuse 1 Opposite x tan (sin−1 x) = tan β = =√ . Adjacent 1 − x2 Adjacent x = = x, we have γ = cos−1 x. Hence, Hypotenuse 1 √ Opposite 1 − x2 √ −1 sin (cos x) = sin γ = = = 1 − x2 and Hypotenuse 1 √ Opposite 1 − x2 −1 = . tan (cos x) = tan γ = Adjacent x Similarly, since cos γ =

The four composite functions sin ○ cos−1 , sin ○ tan−1 , cos ○ sin−1 , and cos ○ tan−1 are welldefined. Here are their graphs:

Figure to be inserted here.

Unfortunately, the composite functions tan ○ sin−1 and tan ○ cos−1 are not well-defined. (Why?)206

Nonetheless, we may graph the equations y = tan (sin−1 x) for x ∈ (−1, 1) and y = tan (cos−1 x) for x ∈ [−1, 0) ∪ (0, 1]:

206

π π Range sin−1 = [− , ] ⊈ Domain tan. Hence, tan ○ sin−1 is not well-defined. 2 2 Range cos−1 = [0, π] ⊈ Domain tan. Hence, tan ○ cos−1 is not well-defined.

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Figure to be inserted here.

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32.4.

Addition Formulae for Arcsine and Arccosine

Fact 75. Suppose x ∈ [−1, 1]. Then Arccosine)

(a) cos−1 x + sin−1 x = π/2. (b) sin−1 x + sin−1 (−x) = 0. (c) cos−1 x + cos−1 (−x) = π.

Proof. For the formal and complete proofs, see p. 1566 (Appendices). Here we’ll give only informal proofs-by-picture:

(a) Construct a right triangle with base x ∈ (0, 1) and hypotenuse 1.

1

β

α x Let α and β be the labelled angles. Then cos α =

x Adjacent = =x Hypotenuse 1

and

sin β =

Opposite x = = x. Hypotenuse 1

So, cos−1 x = α and sin−1 x = β. But of course, α + β = π/2. Hence, cos−1 x + sin−1 x = π/2.

(b) Let x ∈ (0, 1). Consider the unit circle centred on the origin. Let A and B be the points on its right half with y-coordinates x and −x.

A 1 x γ −γ

1

−x B

Let γ be the angle corresponding to the point A. Then the angle corresponding to the point B is −γ. 378, Contents

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By the unit-circle definition of sine, sin γ = x

and

Hence, sin−1 x = γ, sin−1 (−x) = −γ, and

sin (−γ) = −x.

sin−1 x + sin−1 (−x) = γ + (−γ) = 0.

(c) Let x ∈ (0, 1). Consider the unit circle centred on the origin. Let C and D be the points on its top half with x-coordinates x and −x.

D

C

π−δ

−x

δ x

Let δ be the angle corresponding to the point C. Then the angle corresponding to the point D is π − δ.

By the unit-circle definition of cosine, cos δ = x

and

Hence, cos−1 x = δ, cos−1 (−x) = π − δ, and Example 440. XXX

cos (π − δ) = −x.

cos−1 x + cos−1 (−x) = δ + (π − δ) = π.

Example 441. XXX Remark 76. On p. 350, we learnt that the prefix co- in the word cosine simply means that if A and B are complementary angles (i.e. A + B = π/2), then sin A = cos B

and

cos A = sin B.

Fact 75(a) is simply another way of saying exactly the above. If A + B = π/2, then there must exist some x such that A = sin−1 x and B = cos−1 x (or A = cos−1 x and B = sin−1 x).

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32.5.

Fact 76. Suppose x ∈ R. Then

(a) tan−1 x + tan−1 (−x) = 0. ⎧ ⎪ ⎪π/2, for x > 0, 1 ⎪ −1 −1 (b) tan x + tan =⎨ x ⎪ ⎪ for x < 0. ⎪ ⎩−π/2, ⎧ −1 x + y ⎪ ⎪ tan , for xy < 1, ⎪ ⎪ ⎪ 1 − xy ⎪ ⎪ ⎪ x+y ⎪ −1 −1 + π, for xy > 1 AND x > 0, (c) tan x + tan y = ⎨tan−1 ⎪ 1 − xy ⎪ ⎪ ⎪ ⎪ −1 x + y ⎪ ⎪ − π, for xy > 1 AND x < 0. tan ⎪ ⎪ 1 − xy ⎩

Proof. For the formal and complete proofs, see p. 1571 (Appendices). Here we’ll give most but not all of the proof of (c) and only informal proofs-by-picture of (a) and (b):

(a) This is very similar to the earlier proof-by-picture of Fact 75(b) (sin−1 x+sin−1 (−x) = 0): √ Let x ∈ (0, 1). Consider the circle with radius 1 + x2 centred on the origin. On this circle are the points A = (1, x) and B = (1, −x).

A = (1, x)

√ 1 + x2 γ −γ

x

1 −x

√ 1 + x2

B = (1, −x)

Let γ be the angle corresponding to the point A. Then the angle corresponding to the point B is −γ. By the unit-circle definitions,

tan γ = x

Thus, tan−1 x = γ, tan−1 (−x) = −γ, and 380, Contents

and

tan (−γ) = −x.

tan−1 x + tan−1 (−x) = γ + (−γ) = 0.

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(b) This is very similar to the earlier proof-by-picture of Fact 75(a) (cos−1 x+sin−1 x = π/2):

Construct a right triangle with legs of lengths x ∈ (0, 1) and 1.

β x α

1

Let α and β be the labelled angles. Then tan α =

Opposite x = =x Adjacent 1

tan β =

and

Opposite 1 = . Adjacent x

1 1 = β. But of course, α + β = π/2. Hence, tan−1 x + tan−1 = π/2. x x (c) By the Addition Formula for Tangent and Fact 72(c), So, tan−1 x = α and tan−1

tan (tan x + tan y) = −1

−1

tan (tan−1 x) + tan (tan−1 y)

1 − tan (tan x) tan (tan x)

x+y = tan−1 (tan (tan−1 x + tan−1 y)). 1 − xy We now consider the three cases in turn: So, tan−1

−1

−1

=

x+y . 1 − xy

Case 1. xy < 1.

Then by Lemma 1(b) (below), tan−1 x + tan−1 y ∈ (−π/2, π/2).

Hence, by Fact 73(c), tan−1 (tan (tan−1 x + tan−1 y)) = tan−1 x + tan−1 y. Thus, tan−1

x+y = tan−1 x + tan−1 y. 1 − xy

The remainder of the proof of (c) continues on p. 1572 (Appendices). Lemma 1. Suppose x, y ∈ R. Then

(a) (b) (c) (d)

tan−1 x + tan−1 y = ±π/2 ⇐⇒ xy = 1. tan−1 x + tan−1 y ∈ (−π/2, π/2) ⇐⇒ xy < 1. tan−1 x + tan−1 y ∈ (π/2, π) ⇐⇒ xy > 1 AND x > 0. tan−1 x + tan−1 y ∈ (−π, −π/2) ⇐⇒ xy > 1 AND x < 0.

Proof. See p. 1572 (Appendices).

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32.6.

You are supposed to have mastered the following in secondary school:

• “expression of a cos θ + b sin θ in the forms R sin (θ ± α) and R cos (θ ± α)”.207

So, let’s review how this is done. Example 442. XXX Example 443. XXX Example 444. XXX Example 445. XXX More generally,

Theorem 7. Let θ ∈ R. Suppose a, b ≠ 0. Then √ a (a) a cos θ + b sin θ = sgn b a2 + b2 sin (θ + tan−1 ) b √ −b (b) = sgn a a2 + b2 cos (θ + tan−1 ). a √ −a (c) = sgn b a2 + b2 sin (θ − tan−1 ) b √ b (d) = sgn a a2 + b2 cos (θ − tan−1 ). a

Remark 77. It’s probably unwise to try to mug Theorem 7, which is provided only as an official reference. It is better to understand how the above examples were done and also be able to do the following exercises.

Proof. (a) Write R sin (θ + α) = a cos θ + b sin θ. We will find R and α. By the Addition Formulae for Sine,

³¹¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ·¹ ¹ ¹ ¹ µ R sin (θ + α) = R cos α sin θ + R sin α cos θ. b

Now,

a

a R sin α = = tan α. b R cos α

a So, α = tan−1 . b And since R cos α = b, by Fact 74(d), we have √ √ √ b b b ∣b∣ b√ 2 2 2 2 = sgn b b2 + a R= = = = b 1 + (a/b) = b 1 + (a/b) = b + a cos α cos (tan−1 ab ) √ 1 2 ∣b∣ ∣b∣ 1+(a/b)

207

H2 Maths syllabus (p. 14).

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(b) Write R cos (θ + α) = a cos θ + b sin θ. We will find R and α. By the Addition Formulae for Cosine,

³¹¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ µ R cos (θ + α) = R cos α cos θ−R sin α sin θ. a

Now,

b

b −R sin α = = − tan α. a R cos α

b So, α = tan−1 (− ). a And since R cos α = a, by Fact 74(d), we have √ √ 2 √ a a b ∣a∣ b2 a √ 2 2 a R= = = a 1 + = a 1 + = a + b = sgn a a2 + b = cos α cos (tan−1 (− ab )) √ 1 2 a2 ∣a∣ a2 ∣a∣ 1+(−b/a)

(c) Take (a) and apply Fact 76(a).

(d) Take (b) and apply Fact 76(a). Exercise 152. XXX

A152.

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32.7.

Even More Compositions (optional)

The four composite functions sin−1 ○ cos, cos−1 ○ sin, tan−1 ○ sin, and tan−1 ○ cos are welldefined. Here are their graphs:

Figure to be inserted here.

Unfortunately, the composite functions sin−1 ○ tan and cos−1 ○ tan are not well-defined. (Why?)208 Nonetheless, we may graph the equations y = sin−1 (tan x) and y = cos−1 (tan x), for x ∈ [−kπ/4, kπ/4] (for k ∈ Z).

Figure to be inserted here.

It is possible to write down algebraic expressions for sin−1 (cos x) and cos−1 (sin x)—see Fact 237 (Appendices). However and somewhat interestingly, it turns out to be impossible to write down algebraic expressions for tan−1 (sin x), tan−1 (cos x), sin−1 (tan x), and cos−1 (tan x).

208

Range tan = R ⊈ [−1, 1] = Domain sin−1 . Hence, sin−1 ○ tan is not well-defined. Range tan = R ⊈ [−1, 1] = Domain cos−1 . Hence, cos−1 ○ tan is not well-defined.

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33.

Elementary Functions

This brief chapter serves as a quick review of some of the functions we’ve encountered so far. We’ll also learn a new term: elementary functions. Definition 95. A polynomial function is any nice function defined by x ↦ a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an xn , where a0 , a1 , . . . an ∈ R and n ∈ Z+0 .

Example 446. The function f ∶ R → R defined by f (x) = 5x3 − x + 3 is a polynomial function. (What is f (1)? f (0)?)209

Example 447. The function g ∶ [1, 2] → R defined by g (x) = 5x3 − x + 3 is a polynomial function. (What is g (1)? g (0)?)210 Definition 96. An identity function is any function defined by x ↦ x.211

Example 448. The function f ∶ R → R defined by f (x) = x is an identity function. (What is f (1)? f (0)?)212

By the way, any nice identity function is also a polynomial function. (Why?)213 Here for example, f is also a polynomial function.

Example 449. The function g ∶ [1, 2] → R defined by g (x) = x is an identity function. (What is g (1)? g (0)?)214

🐄

🐄

Example 450. The function h ∶ { , 🐔} → { , 🐔} defined by h (x) = x is an 215 )? h (🐔)? h (1)?) identity function. (What is h (

🐄

Definition 97. A constant function is any function defined by x ↦ c, where c is any object.

f (1) = 7, f (0) = 3. g (1) = 7, g (0) is undefined. 211 Strictly speaking, we should distinguish between an identity function and an inclusion function. Strictly speaking, Definition 96 given here is not of an identity function but of an inclusion function. See Definitions 276 and 277 (Appendices). This is a subtle distinction that shall not matter at all for A-Level Maths. Indeed, the term identity function is never mentioned in your A-Level Maths syllabus or exams. But it is sufficiently convenient that we’ll sometimes use it anyway. 212 f (1) = 1, f (0) = 0. 213 Its mapping rule may be written as x ↦ a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an xn , where a0 , a1 , . . . an ∈ R and n ∈ Z+0 (in particular, a0 = 0, a1 = 1, and n = 1). 214 g (1) = 1, g (0) is undefined. 215 h( )= , h (🐔) = 🐔, h (1) is undefined.

209 210

🐄 🐄

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Example 451. The function f ∶ R → R defined by f (x) = 5 is a constant function. (What is f (1)? f (0)?)216

By the way, any nice constant function is also a polynomial function. (Why?) Here for example, f is also a polynomial function.

Example 452. The function g ∶ [1, 2] → R defined by g (x) = 5 is a constant function. (What is g (1)? g (0)?)217

🐄 🐄

🐄

Example 453. The function h ∶ { , 🐔} → { , 🐔} defined by h (x) = )? h (🐔)? h (1)?)218 constant function. (What is h (

🐄 is a

🐄

Example 454. The function i ∶ R → { , 🐔} defined by i (x) = 🐔 is a constant )? i (🐔)? i (1)? i (0)?)219 function. (What is i (

🐄

A special case of a constant function is a zero function:

Definition 98. A zero function is any function defined by x ↦ 0.

Example 455. The function f ∶ R → R defined by f (x) = 0 is a zero function. (What is f (1)? f (0)?)220 By the way, any zero function is also a constant function. (Why?)221 So here for example, f is also a constant function.

Example 456. The function g ∶ [1, 2] → R defined by f (x) = 0 is a zero function. (What is g (1)? g (0)?)222 Definition 99. A power function is any nice function defined by x ↦ xk , where k ∈ R.

Example 457. The function f ∶ R → R defined by f (x) = x17.3 is a power function. (What is f (1)? f (0)?)223 Example 458. The function g ∶ [1, 2] → R defined by g (x) = x−π is a power function. (What is g (1)? g (0)?)224

Definition 100. A trigonometric (or circular) function is any nice function with the same mapping rule as sin, cos, tan, cosec, sec, or tan. f (1) = 5, f (0) = 5. Its mapping rule may be written as x ↦ a0 + a1 x + a2 x2 + ⋅ ⋅ ⋅ + an xn , where a0 , a1 , . . . an ∈ R and n ∈ Z+0 (in particular, a0 is the constant to which each x is mapped and n = 0). 217 g (1) = 5, g (0) is undefined. 218 h( )= , h (🐔) = , h (1) is undefined.

215 216

219

🐄 🐄 🐄 i (🐄 ) and i (🐔) are undefined, i (1) = 🐔, and i (0) = 🐔.

f (1) = 0, f (0) = 0. It maps each x in its domain to the same object (in this case, 0). 222 g (1) = 0, g (0) is undefined. 223 f (1) = 1, f (0) = 0. 224 g (1) = 1, g (0) is undefined. 220

221

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Example 459. Of course, sin is a trigonometric function.

But so too is the function f ∶ [1, 2] → R defined by f (x) = sin x.

Figure to be inserted here.

Example 460. Of course, cos is a trigonometric function.

But so too is the function g ∶ [2, 3] → R defined by g (x) = cos x.

Figure to be inserted here.

Remark 78. As stated in Remark 72, there are actually other trigonometric functions (e.g. versine, haversine, coversine). But we don’t need to worry about these in A-Level maths. Definition 101. An inverse trigonometric (or circular) functions is any nice function with the same mapping rule as sin−1 , cos−1 , or tan−1 . Example 461. Of course, sin−1 is an inverse trigonometric function. But so too is the function f ∶ [0, 1] → R defined by f (x) = sin−1 x.

Figure to be inserted here.

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Example 462. Of course, cos−1 is an inverse trigonometric function.

But so too is the function g ∶ [0, 0.5] → R defined by g (x) = cos−1 x.

Figure to be inserted here.

Remark 79. As stated in Remark 75, there are other inverse trigonometric functions (e.g. cosec−1 , sec−1 , cot−1 ). But we don’t need to worry about these in A-Level maths.

In Ch. 25, we defined

• The natural logarithm function ln ∶ R+ → R; and

• Its inverse the exponential function exp ∶ R → R+ . We can also define

Definition 102. A natural logarithm function is any nice function with the same mapping rule as ln. Definition 103. An exponential function is any nice function with the same mapping rule as exp. Example 463. The function f ∶ [1, 2] → R defined by f (x) = ln x is a natural logarithm function.

Figure to be inserted here.

The function g ∶ [1, 2] → R defined by g (x) = exp x is an exponential function.

All of the functions discussed in this brief chapter are elementary functions. Also, any arithmetic combination (Ch. 18) or composition (Ch. 20) of two elementary functions is also an elementary function. Formally,

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Definition 104. An elementary function is a polynomial function, a trigonometric function, an inverse trigonometric function, a natural logarithm function, an exponential function, a power function, any arithmetic combination of two elementary functions, or any composition of two elementary functions. Most functions you’ll encounter in H2 Maths are elementary. Through arithmetic combinations and compositions, we can build ever functions that “look” ever more complicated but are nonetheless elementary: Example 464. Define f ∶ R+ → R by

f (x) = 1 + 2x + 3 sin [cos (1 + x3 − ln x) ]. 2

Although f “looks” complicated, it is nonetheless elementary.

Example 465. The absolute value function ∣⋅∣ ∶ R → R is defined by ⎧ ⎪ ⎪ ⎪x, ∣x∣ = ⎨ ⎪ ⎪ ⎪ ⎩−x,

for x ≥ 0, for x < 0.

At first glance, the absolute value function doesn’t seem to satisfy our definition of an elementary function (Definition 104 ). But in fact, it does. To see why, recall Fact 12: ∣x∣ =

Then define f ∶ R+0 → R by f (x) =

x2 , for all x ∈ R.

√ x and g ∶ R → R by g (x) = x2 .

Both f and g are elementary. Moreover, ∣⋅∣ is the composition of f and g: √ ∣x∣ = f (g (x)) = f (x2 ) = x2 . And so by Definition 104, ∣⋅∣ is an elementary function.

To repeat, most functions we’ll encounter in H2 Maths are elementary. Probably the only non-elementary function we’ll spend any time on is the cumulative density function of the normal distribution. We’ll learn about this in Part VI (Probability and Statistics). It is possible to prove that the derivative (if it exists) of any elementary function is also elementary. However and importantly, the integrals of many elementary functions are not. We’ll learn a little about this in Part V (Calculus). 389, Contents

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Remark 80. Unfortunately, there is no single standard definition of the term elementary function.225 This textbook will nonetheless use this term because it is a convenient one that includes most functions that A-Level Maths students will encounter.

225

Definition 104 is merely this textbook’s and also ProofWiki’s. Here are some differences between our definition and those of other writers:

• Some writers consider the inverse of any elementary function to also be elementary (we do not). • Most (including me) consider the composition of any two elementary functions to also be elementary. However, there is at least one person who does not (davidlowryduda). 390, Contents

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34.

Factorising Polynomials

Much of this chapter will be about methods for factorising polynomials. But first, let’s see how we can long divide polynomials:

34.1.

Long Division of Polynomials

We’ll continue using the terms dividend, divisor, quotient, and remainder: Example 466. Consider the expression (2x + 1) ÷ x. We have ­ © © © 2x + 1 = 2 ⋅ x + 1 p(x)

q(x) d(x)

r(x)

or

­ q(x) 2x + 1 © = 2 + x ®

p(x)

r(x)

d(x)

d(x)

The four polynomials labelled above have the same four names as before: The dividend p (x) = 2x + 1

The divisor d (x) = x

The quotient q (x) = 2

The remainder r (x) = 1

In the above example, it was kinda obvious that the quotient had to be q (x) = 2. In the next example, it’s a little less obvious and long division will be helpful:

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Example 467. Consider (x2 + 3)÷(x − 1). The dividend is p (x) = x2 +3 and the divisor is d (x) = x − 1. This time, it’s not so obvious what the quotient q (x) should be. But it turns out that just like with (simple) division (of two numbers), long division can help us here. With (simple) long division, going from right to left, the columns were 1s, 10s, 100s, etc. Here with polynomial long division, going from right to left, they’re the constant x0 term, the linear x1 term, the squared x2 term, the cubed term x3 , etc. Terms: x2 x−1 x x2 2

x1 x +0x −x x x

x0 +1 +3

Explanation x ⋅ (x − 1) (x2 + 3) − (x2 − x) 1 ⋅ (x − 1) (x + 3) − (x − 1)

+3 −1 4

= x2 − x =x+3 =x−1 =4

The quotient is q (x) = x + 1, while the remainder is r (x) = 4. Altogether, we have ­ ³¹¹ ¹ ¹ ·¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ·¹ ¹ ¹ ¹ µ © x2 + 3 = (x − 1) ⋅ (x + 1) + 4 p(x)

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d(x)

q(x)

r(x)

or

­ q(x) © x2 + 3 ¬ 4 = x+1+ . x−1 x−1 ± ± p(x)

r(x)

d(x)

d(x)

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Example 468. Consider (3x2 + x − 4) ÷ (2x − 3). The dividend is p (x) = 3x2 + x − 4 and the divisor is d (x) = 2x − 3. Long division: Terms:

x2

2x − 3 3x2 3x2

x1 x0 11 3 x + 2 4

Explanation

+x −4 9 − x 2 11 x −4 2 11 33 x − 2 4

3 9 x ⋅ (2x − 3) = 3x2 − x 2 2 9 11 (3x2 + x − 4) − (3x2 − x) = x − 4 2 2 11 11 33 ⋅ (2x − 3) = x − 4 2 4

17 4

(

Thus the quotient and remainder are 11 3 q (x) = x + 2 4

q(x)

d(x)

d(x)

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17 . 4

r(x)

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ¬ 2 11 17/4 3x + x − 4 3 = x+ + . 2x − 3 2 4 2x − 3 ² ´¹¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ² p(x)

Or,

r (x) =

¹¹¹¹¹¹¹¹¹µ ª ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ·¹¹ ¹ ¹ ¹ ¹ µ ³¹¹3¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹¹11 17 3x2 + x − 4 = (2x − 3) ⋅ ( x + ) + . 2 4 4 p(x)

We have

and

11 11 33 17 x − 4) − ( x − ) = 2 2 4 4

r(x)

q(x)

d(x)

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Example 469. Consider (4x3 + 2x2 + 1) ÷ (2x2 − x − 1). The dividend is p (x) = 4x3 + 2x2 + 1 and the divisor is d (x) = 2x2 − x − 1. Long division: Terms:

x3

x2

2x2 − x − 1 4x3 +2x2 4x3 −2x2 4x2 4x2

x1 2x +0x −2x 2x −2x 4x

x0 +2 +1

+1 −2 +3

Explanation 2x ⋅ (2x2 − x − 1) (4x3 + 2x2 + 1) − (4x3 − 2x2 − 2x) 2 ⋅ (2x2 − x − 1) (4x2 + 2x + 1) − (4x2 − 2x − 2)

Thus the quotient and remainder are q (x) = 2x + 2

and

= 4x3 − 2x2 − 2x = 4x2 + 2x + 1 = 4x2 − 2x − 2 = 4x + 3

r (x) = 4x + 3.

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ³¹¹ ¹ ¹ ¹ ¹ ·¹¹ ¹ ¹ ¹ ¹ µ ­ 4x3 + 2x2 + 1 = (2x2 − x − 1) ⋅ (2x + 2) + 4x + 3. p(x)

We have

d(x)

q(x)

r(x)

³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ ­ 4x + 3 4x3 + 2x2 + 1 = 2x + 2 + 2 . 2 ² 2x − x − 1 2x − x − 1 q(x) ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶

Or,

p(x)

r(x)

d(x)

d(x)

The above examples suggest the following theorem and definitions: Theorem 8. (Euclidean Division Theorem for Polynomials.) Let p (x) and d (x) be P - and D-degree polynomials in x with D ≤ P . Then there exists a unique polynomial q (x) of degree P − D such that r (x) = p (x) − d (x)q (x) is a polynomial with degree less than D.

Proof. See 1575 (Appendices).

Definition 105. Given polynomials p (x) and d (x) and the expression p (x) ÷ d (x), we call p (x) the dividend, d (x) the divisor, the unique polynomial q (x) given in the above theorem the quotient, and r (x) = p (x) − d (x)q (x) the remainder. Exercise 153. For each expression, do the long division and identify the dividend, divisor, quotient, and remainder. (Answer on p. 1775.) (a)

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16x + 3 . 5x − 2

(b)

4x2 − 3x + 1 . x+5

(c)

x2 + x + 3 . −x2 − 2x + 1

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34.2.

Factorising Polynomials

In this subchapter, we learn what it means to factorise a polynomial or to be a factor of (or to divide) a polynomial. Example 470. Consider the 2nd-degree (or quadratic) polynomial x2 + 4x + 3. It can be factorised (i.e. written as the product of lower-degree polynomials): x2 + 4x + 3 = (x + 1) (x + 3) .

We say that each of x + 1 and x + 3 is a factor of the given polynomial (or equivalently, divides it). Note that each of x + 1 and x + 3 is itself a 1st-degree (or linear) polynomial.

Example 471. The quadratic polynomial 6x2 + x − 2 can be factorised: 6x2 + x − 2 = (2x − 1) (3x + 2) .

Each of 2x − 1 and 3x + 2 is a factor of (or divides) the given polynomial.

By the way, any non-zero multiple of a factor is itself also a factor. So here,

• Since 2x − 1 is a factor of the given polynomial, so too are 4x − 2, −6x + 3, x − 0.5, and any k (2x − 1) (for k ≠ 0): 6x2 +x−2 = (2x − 1) (3x + 2) =

1 1 (4x − 2) (3x + 2) = − (−6x + 3) (3x + 2) = 2 (x − 0.5) (3x + 2) 2 3

6x2 +x−2 = (2x − 1) (3x + 2) =

1 1 (2x − 1) (6x + 4) = − (2x − 1) (−9x − 6) = 2 (2x − 1) (1.5x + 1 2 3

• Similarly, since 3x + 2 is a factor of the given polynomial, so too are6x + 4, −9x − 6, 1.5x + 1, and any k (3x + 2) (for k ≠ 0): Formal definitions of the terms factor of (or divides) and factorise:

Definition 106. Let p (x) and d (x) be polynomials. We say that d (x) is a factor of p (x) or divides p (x) if there exists some polynomial q (x) such that p (x) = d (x) q (x).

To factorise a polynomial is to express it as the product of polynomials, none of which is of higher degree and at least one of which is of strictly lower degree.

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34.3.

Compare Coefficients

We’ll call our first method for factorising polynomials Compare Coefficients (plus Guess and Check): Example 472. To factorise 3x2 + 5x + 2, write

3x2 + 5x + 2 = (ax + b) (cx + d) = acx2 + (bc + ad)x + bd,

where a, b, c, and d are unknown constants to be found.

Comparing coefficients on the x2 , x, and constant terms, we have, respectively, ac = 3, 1

bc + ad = 5, 2

and bd = 2. 3

The above was the “Compare Coefficients” bit. Next is the “Guess and Check” bit: Of course, a, b, c, and d could be any numbers. But if the person who wrote this problem is nice,226 they’ll probably be integers. So, given =, let’s guess/try a = 3, c = 1. 1

And given =, let’s guess/try b = 1 and d = 2. 3

Unfortunately, with this set of guesses, = doesn’t hold: 2

bc + ad = 1 + 6 = 7 ≠ 5. 7

Well, let’s try switching the values of b and d, i.e. try b = 2 and d = 1. With this second 2 set of guesses, = now holds: Yay! We’re done:

bc + ad = 2 + 3 = 5. 3

3x2 + 5x + 2 = (3x + 2) (x + 1) .

In the above example, we explicitly wrote out the unknown constants a, b, c, and d. When we look later at higher-degree polynomials, this can help reduce mistakes. But with quadratic polynomials, we can usually skip doing this and save ourselves some time:

226

We can confirm that he is.

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Example 473. To factorise 6x2 − 11x − 35, first try

(3x + 7) (2x − 5) = 6x2 − x − 35.

Nope doesn’t work. Second try:

(3x − 5) (2x + 7) = 6x2 + 11x − 35.

Nope still doesn’t work. Third try:

Yay works! We’re done!

(3x − 7) (2x + 5) = 6x2 − 11x − 35.

The Compare Coefficients (plus Guess and Check) method works well enough with quadratic polynomials. With higher-degree polynomials, sometimes we can get lucky and quickly arrive at the correct guess: Example 474. To factorise x3 − 7x − 6, write

x3 − 7x − 6 = (ax + b) (cx + d) (ex + f ) .

Comparing coefficients, we have ace = 1 and bdf = −6.

So, let’s guess a = c = e = 1 and b = 2, d = −3, and f = 1:

(x + 2) (x − 3) (x + 1) = (x2 − x − 6) (x + 1) = x3 − x2 − 6x + x2 − x − 6 = x3 − 7x − 6.

3

Wah! So “lucky”! Success on the very first try! We’re done!

But more often, it will usually take quite a lot of time (and pain), even with some lucky and intelligent guesses:

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Example 475. To factorise 15x3 − 8x2 − 9x + 2, write

15x3 −8x2 −9x+2 = (ax + b) (cx + d) (ex + f ) = acex3 +(acf + ade + bce) x2 +(adf + bcf + bde) x+bdf Comparing coefficients on the x3 , x2 , x, and constant terms, we have, respectively, ace = 15, 1

acf + ade + bce = −8, 2

Given =, let’s try a = 5, c = 3, and e = 1. 1

adf + bcf + bde = −9,

and bdf = 2.

3

4

Given =, let’s try b = 2, d = 1, and f = 1. Hm wait no, given that some coefficients are negative, it can’t be that a–f are all positive. 4

So, let’s instead try b = −2, d = −1, and f = 1:

acf + ade + bce = 15 − 5 − 6 = 4 ≠ −8.

7

Hm too big. Let’s try switching b and d around—i.e. try b = −1, d = −2, and f = 1: acf + ade + bce = 15 − 10 − 3 = 2 ≠ −8.

7

A little closer but still wrong. But ah, we now notice that the numbers 15, 10, and 3 look a lot like they can be made to add up to −8. Indeed, let’s see what happens if we try b = −1, d = 2, and f = −1: acf + ade + bce = −15 + 10 − 3 = −8.

3

Yay! = holds! Cross our fingers and check/hope that = also holds: 2

3

adf + bcf + bde = −10 + 3 − 2 = −9.

3

It does! And so we’re done:

15x3 − 8x2 − 9x + 2 = (5x − 1) (3x + 2) (x − 1) .

To factorise higher-degree polynomials, we’ll want to also make use of other tools, such as the Factor Theorem (Ch. 34.6) and Intermediate Value Theorem (Ch. 34.7). Exercise 154. XXX

A154.

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34.4.

For quadratic polynomials, there’s actually no need to use the Compare Coefficients method at all. Instead, we can simply use the quadratic formula (Ch. 14), which involves no guess and check. Consider a quadratic polynomial ax2 + bx + c. Its discriminant is b2 − 4ac. By Fact 27(b), (i) If b2 − 4ac > 0, then

ax2 + bx + c = a (x −

(ii) If b2 − 4ac = 0, then

−b −

√ √ b2 − 4ac −b + b2 − 4ac ) (x − ). 2a 2a

b 2 ax + bx + c = a (x + ) . 2a 2

(iii) If b2 − 4ac < 0, then ax2 + bx + c cannot be factorised (unless we use complex numbers, as we’ll learn in Part III). Example 476. The quadratic polynomial x2 − 3x + 2 has discriminant b2 − 4ac = (−3) − 4 (1) (2) = 1. So, √ √ 3 − 1 3 + 1 x2 − 3x + 2 = (x − ) (x − ) = (x − 1) (x − 2) . 2 2 2

Example 477. The quadratic polynomial 3x2 + 5x + 2 has discriminant b2 − 4ac = 52 − 4 (3) (2) = 1. So, √ √ −5 + 1 2 −5 − 1 2 ) (x − ) = 3 (x + 1) (x + ) = (x + 1) (3x + 2) . 3x + 5x + 2 = 3 (x − 2⋅3 2⋅3 3

Example 478. The quadratic polynomial 4x2 −12x+9 has discriminant b2 −4ac = (−12) − 4 (4) (9) = 0. So, 2

−12 2 3 2 2 4x − 12x + 9 = 4 (x + ) = 4 (x − ) = (2x − 3) . 2×4 2 2

Example 479. The quadratic polynomial 3x2 − 2x + 1 has discriminant b2 − 4ac = (−2) − 4 (3) (1) < 0. So, 3x2 − 2x + 1 cannot be factorised. Exercise 155. XXX

2

A155.

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34.5.

The Remainder Theorem

You’re supposed to to have learnt this result in O-Level ‘A’ Maths: Theorem 9. (Remainder Theorem) If p (x) is a polynomial and a is a constant, then p (x) ÷ (x − a) leaves a remainder of p (a).

Proof. By Definition 105, p (x) ÷ (x − a) leaves the remainder p (x) − (x − a) q (x) or—if we plug in x = a—p (a) − (a − a) q (a) = p (a). Example 480. Consider the quadratic polynomial p (x) = x2 − 5x + 1. By the Remainder Theorem, p (x) divided by x − 3 leaves a remainder of p (3) = 32 − 5 (3) + 1 = −5. We could’ve figured this out using long division (below), but clearly the Remainder Theorem is a lot quicker. Terms Squared Linear Constant x −2 x−3 x2 −5x +1 x2 −3x −2x +1 −2x +6 −5.

Example 481. Consider the quintic polynomial p (x) = 17x5 − 5x4 + x2 + 1. By the Remainder Theorem, p (x) divided by x − 1 leaves a remainder of p (1) = 17 ⋅ 15 − 5 ⋅ 14 + 12 + 1 = 14. We could’ve figured this out using long division (I didn’t bother but you can try this as an exercise), but clearly the Remainder Theorem is a lot quicker.

Historically, the Remainder Theorem rarely featured on the A-Level exams ... Which means, of course, that it made a sudden appearance in 2017 just to screw students over.227 So yea, it’s another thing you’ll want to remember. An exercise to help with that: Exercise 156. For each, find the remainder.

(a) (2x3 + 7x2 − 3x + 5) ÷ (x − 3)

(b) (−2x4 + 3x2 − 7x − 1) ÷ (x + 2).

For A-Level Maths, the Remainder Theorem will have little use (except to screw over unsuspecting students). Instead, its corollary—the Factor Theorem—will be more useful for factorising polynomials.

227

See Exercise 533(a) (9758 N2017/I/5).

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34.6.

The Factor Theorem

Another result you’re supposed to have learnt in O-Level ‘A’ Maths: Theorem 10. (Factor Theorem) If p (x) is a polynomial and a is a constant, then x − a is a factor for p (x)

⇐⇒

p (a) = 0.

Proof. By the Remainder Theorem, p (x)÷(x − a) leaves a remainder of p (a). By Definition 106, x − a is a factor for p (x) if and only if the remainder is zero, i.e. p (a) = 0. Example 482. To factorise p (x) = x2 − 3x + 2, we can use the method I’ll call Factor Theorem (plus Guess and Check): First try plugging in the number 1:

p (1) = 12 − 3 ⋅ 1 + 2 = 0.

3

Wah! So “lucky”! Success on the very first try! Since p (1) = 0, by the Factor Theorem, x − 1 is a factor for p (x).

Since p (x) is quadratic (or degree 2), its other factor must be a linear (or degree-1) polynomial, i.e. of the form ax + b. To find this other factor, we could continue trying our luck with the Factor Theorem (i.e. try plugging other numbers into p (x)). But we won’t do that. It’s easier to write x2 − 3x + 2 = (x − 1) (ax + b) = ax2 + (b − a) x − b, 1

2

where a and b are unknown constants to be found. Comparing coefficients, we have a = 1 and b = −2 (no guess and check needed here). And now, we’re done: p (x) = x2 − 3x + 2 = (x − 1) (x − 2) .

Note: Above, it was actually unnecessary to write out =. From = alone, we can easily tell that a = 1 and b = −2. 2

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Example 483. Factorise p (x) = 3x2 + 5x + 2 We try plugging in the number 1:

p (1) = 3 ⋅ 12 + 5 ⋅ 1 + 2 = 10 ≠ 0.

7

Aiyah, sian, doesn’t work: Since p (1) ≠ 0, by the Factor Theorem, x − 1 is not a factor for p (x).

By the way, here’s a tip to save you time: Above we found the exact value of p (1) to be 10. But actually, we can skip finding the exact value. All we want to know is whether p (1) equals zero (or not). And by seeing that each of the three terms is obviously positive, we can tell that p (1) is also obviously positive and, in particular, not equal to zero. We next try −1:

p (−1) = 3 ⋅ (−1) + 5 ⋅ (−1) + 2 = 0. 2

3

Yay, works! Since p (−1) = 0, by the Factor Theorem, x − (−1) = x + 1 is a factor for p (x). As in the last example, to find the other factor, write

p (x) = 3x2 + 5x + 2 = (x + 1) (ax + b) .

Comparing coefficients,228 a = 3 and b = 2. So,

p (x) = 3x2 + 5x + 2 = (x + 1) (3x + 2) .

228

No guess and check needed here.

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We now try using the Factor Theorem to factorise higher-degree polynomials: Example 484. To factorise p (x) = 15x3 − 17x2 − 22x + 24, we try plugging in 1: p (1) = 15 ⋅ 13 − 17 ⋅ 12 − 22 ⋅ 1 + 24 = 0.

3

Wah! So “lucky”! Success on the very first try! Since p (1) = 0, by the Factor Theorem, x − 1 is a factor for p (x).

Since p (x) is cubic (degree 3), p (x) divided by x − 1 gives us a quadratic (degree-2) polynomial ax2 + bx + c that is now our task to find. To do so, write 15x3 − 17x2 − 22x + 24 = (x − 1) (ax2 + bx + c) .

Comparing coefficients on the x3 and constant terms, a = 15 and c = −24.

Comparing coefficients on the x2 term, −17 = −a + b = −15 + b, so that b = −2. So, ax2 + bx + c = 15x2 − 2x − 24.

We next factorise 15x2 −2x−24. To do so, we can use any of the three methods introduced so far (Compare Coefficients, quadratic formula, and Factor Theorem). Here we’ll just use the quadratic formula (which is the only one of the three methods that doesn’t involve any guesswork): The discriminant is b2 − 4ac = (−2) − 4 (15) (−24) = 4 + 1440 = 1444 > 0. Moreover, √ 1444 = 38. Hence, 2

Altogether,

15x2 − 2x − 24 = 15 (x −

2 + 38 6 4 2 − 38 ) (x − ) = 15 (x + ) (x − ) . 30 30 5 3

6 4 15x3 − 17x2 − 22x + 24 = 15 (x − 1) (x + ) (x − ) = (x − 1) (5x + 6) (3x − 4) . 5 3

By the way, when using the Factor Theorem (plus Guess and Check) method, here’s a little trick: Try numbers that are factors of the constant term. Example 485. XXX Why does this trick work? Well, suppose we are given some cubic polynomial to factorise. If the examiners are nice, then this cubic polynomial can probably be factorised as (x + a) (bx + c) (dx + e) ,

where a, c, and e are integers. In which case, ace equals the constant term and the number a (which we are trying to guess) is a factor of the constant term. 403, Contents

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Example 486. XXX Exercise 157. Factorise each polynomial. (a) 2x2 − 5x − 3

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(b) 7x2 − 19x − 6

(Answer on p. 1776.) (c) 6x2 + x − 1

(d) 2x3 − x2 − 17x

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34.7.

The Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) is another useful tool for factorising polynomials. We first describe the IVT informally with an example: Example 487. Below is part of the graph of some continuous function f ∶ R → R. We are missing the graph of f for the interval (1, 3).

Say we know that f (1) = −2 and f (3) = 2. What then can we say about the missing portion of the graph?

y

f

(3, 2)

2

1

−2

3

x

(1, −2) We have most of the graph of f , but are missing the interval (1, 3).

Since f is continuous, it must be that we can draw its entire graph without lifting our pencil. In particular, we can connect the dots (1, −2) and (3, 2) without lifting our pencil. But obviously, the only way to do so is to have our pencil “go through” every value between −2 and 2. That is (and this is what the IVT says), f must attain (or “hit”) every value between −2 and 2.

And yup, that’s all the IVT says—if f is continuous on the interval [a, b], then f must attain (or “hit”) every value between f (a) and f (b) in the interval (a, b). A bit more formally, Theorem 11. (Intermediate Value Theorem) If f is continuous on the interval [a, b], then for every y ∈ (f (a) , f (b)), there exists x ∈ (a, b) such that y = f (x).

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Proof. Omitted.229 Let’s now see how the IVT can help us factorise polynomials.

229

See e.g. O’Connor (2001).

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Example 488. To factorise p (x) = 2x2 + 9x − 5, it’s probably quicker to use the Compare Coefficients (plus Guess and Check) method or the quadratic formula.

But here, just to illustrate how the IVT can be used, we’ll start instead with the Factor Theorem (plus Guess and Check) method. We try plugging in 1: p (1) = 2 ⋅ 12 + 9 ⋅ 1 − 5 = 6 ≠ 0.

7

Aiyah, sian, doesn’t work: Since p (1) ≠ 0, by the Factor Theorem, x − 1 is not a factor for p (x).

Here we could immediately try again with the Factor Theorem. But instead, here we can first make use of the IVT: Observe that p (0) = −5. What good is this observation?

Well, since 0 is between p (0) = −5 and p (1) = 6, the IVT says there must be some a ∈ (0, 1) such that p (a) = 0. Cool! So, guided by this knowledge, let’s try the Factor Theorem again by plugging in 1/2: 1 2 1 1 p ( ) = 2 ⋅ ( ) + 9 ⋅ − 5 = 0. 2 2 2

3

Yay, works! Since p (1/2) = 0, by the Factor Theorem, x − 1/2 is a factor for p (x). 1 1 As stated earlier, if x − is a factor for p (x), then so too is 2 (x − ) = 2x − 1. So, write 2 2 2x2 + 9x − 5 = (2x − 1) (ax + b) .

Comparing coefficients, a = 1 and b = 5. Altogether, Example 489. XXX

p (x) = 2x2 + 9x − 5 = (2x − 1) (x + 5) .

Example 490. XXX Exercise 158. XXX

A158.

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34.8.

Factorising a Quartic Polynomial

For the A-Levels, you’ll often have to factorise quadratic polynomials. You may sometimes have to factorise cubic polynomials. It’s unusual that they ask you to factorise polynomials of a higher degree. And if they do, the friendly folks at the MOE will usually be nice enough to give you a little help.230 In the next example, we factorise a quartic (degree-4) polynomial using what we’ve learnt. It’s long and tedious, but conceptually, it’s not any harder than what we’ve already done: Example 491. To factorise p (x) = 6x4 + 13x3 − 29x2 − 52x + 20, we first try plugging in 1: p (1) = 6 ⋅ 14 + 13 ⋅ 13 − 29 ⋅ 12 − 52 ⋅ 1 + 20 < 0.

7

(Again, we don’t need to compute the exact value of p (1) to see that p (1) < 0.)

Aiyah, sian, doesn’t work: Since p (1) ≠ 0, by the Factor Theorem, x − 1 is not a factor for p (x). But now, observe that 0 is between p (0) = 20 and p (1) < 0. And so, the IVT says there must be some a ∈ (0, 1) such that p (a) = 0. So, let’s next try 1/2: 1 4 1 3 1 2 1 1 p ( ) = 6 ⋅ ( ) + 13 ⋅ ( ) − 29 ⋅ ( ) − 52 ⋅ ( ) + 20 < 0. 2 2 2 2 2

7

(Again, we don’t need to compute the exact value of p (1/2) to see that p (1/2) < 0.)

Aiyah, sian, still doesn’t work: Since p (1/2) ≠ 0, by the Factor Theorem, x − 1/2 is not a factor for p (x). But again, observe that 0 is between p (0) = 20 and p (1/2) < 0. And so, the IVT says that there must be some b ∈ (0, 1/2) such that p (b) = 0. So, let’s next try 1/3: 1 1 4 1 3 1 2 1 p ( ) = 6 ⋅ ( ) + 13 ⋅ ( ) − 29 ⋅ ( ) − 52 ⋅ ( ) + 20 3 3 3 3 3 6 13 29 52 2 13 29 8 15 5 = + − − + 20 = + − + = − = 0. 3 81 27 9 3 27 27 9 3 27 9

Yay, works! Since p (1/3) = 0, by the Factor Theorem, x − 1/3 is a factor for 6x4 + 13x3 − 29x2 − 52x + 20. And if x − 1/3 is a factor, then so too is 3 (x − 1/3) = 3x − 1. (Example continues on the next page ...)

230

See e.g. 9740 N2010/II/1(b) (Exercise 650).

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(... Example continued from the previous page.)

Dividing p (x) by 3x − 1 should yield some cubic (degree-3) polynomial ax3 + bx2 + cx + d that we’ll find and factorise. Write p (x) = 6x4 + 13x3 − 29x2 − 52x + 20 = (3x − 1) (ax3 + bx2 + cx + d) .

Comparing coefficients on the x4 and constant terms, 3a = 6 and −d = 20. Or a = 2 and d = −20.

Next, comparing coefficients on the x3 and x2 terms, 3b − a = 13 and 3d − c = −52. Or b = 5 and c = −8. So, ax3 + bx2 + cx + d = 2x3 + 5x2 − 8x − 20.

To factorise 2x3 + 5x2 − 8x − 20, we try plugging in 2:

p (2) = 2 ⋅ 23 + 5 ⋅ 22 − 8 ⋅ 2 − 20 = 16 + 20 − 16 − 20 = 0.

3

Yay, works! Since p (2) = 0, by the Factor Theorem, x − 2 is a factor for 2x3 + 5x2 − 8x − 20.

Dividing 2x3 + 5x2 − 8x − 20 by x − 2 should yield some quadratic polynomial ex2 + f x + g. So, write 2x3 + 5x2 − 8x − 20 = (x − 2) (ex2 + f x + g).

Comparing coefficients on the x3 and constant terms, e = 2 and −2g = −20 or g = 10. Comparing coefficients on the x2 term, f − 2e = 5 or f = 9. So, ex2 + f x + g = 2x2 + 9x + 10.

To factorise 2x2 +9x+10, we use the Compare Coefficients (plus Guess and Check) method and try (2x + 5) (x + 2) = 2x2 + 9x + 10.

3

Wah! So lucky! Success on the very first try! And now, at long last, we’re done: p (x) = 6x4 + 13x3 − 29x2 − 52x + 20 = (3x − 1) (x − 2) (2x + 5) (x + 2).

Exercise 159. Let p (x) = ax4 + bx3 − 31x2 + 3x + 3, where a and b are constants. You are told that (i) p (x) divided by x − 1 leaves a remainder of 5; and (ii) p (0.5) = 0.

(a) Find a and b.

You are now also told that (iii) p (−1/3) < 0.

(b) Factorise p (x).

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34.9.

Two Warnings

All of our examples of factorisation have so far involved a n-degree polynomial that can be fully factorised into n linear (or degree-1) factors. But this need not always be the case: Example 492. Consider the quartic polynomial x4 − 1. We can write x4 − 1 = (x2 + 1) (x + 1) (x − 1) .

Unfortunately, x2 + 1 has negative discriminant and so cannot be further factorised.231

Example 493. Consider the quartic polynomial x4 + 5x2 + 4. We can write x4 + 5x2 + 2 = (x2 + 1) (x2 + 4) .

Both x2 + 1 and x2 + 4 have negative discriminants and so cannot be further factorised.232

Example 494. The quartic polynomial x4 + x + 1 cannot be factorised at all.233

The above three examples issue two important warnings:

1. Not every n-degree polynomial can be fully factorised into n linear factors. 2. Indeed, an n-degree polynomial may not even have a single linear factor! Exercise 160. XXX

A160.

Unless we use complex numbers, which we’ll learn about in Part III. With complex numbers, we can write x2 + 1 = (x + i) (x − i) and thus x4 − 1 = (x + i) (x − i) (x + 1) (x − 1). 232 Again, with complex numbers, we can write x2 + 1 = (x + i) (x − i), x2 + 4 = (x + 2i) (x − 2i), and thus x4 + 5x2 + 2 = (x + i) (x − i) (x + 2i) (x − 2i). 233 Again, with complex numbers, it’s actually possible to factorise x4 + x + 1 into four linear factors.

231

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35.

Solving Systems of Equations

Let’s start with two warm-up questions (PSLE and O-Level style): Exercise 161. Today is Apu, Beng, and Caleb’s birthday. When Apu turned 40 some years ago, Beng was twice as old as Caleb. Today, Apu is twice as old as Beng, while Caleb turns 28. What ages do Apu and Beng turn today?234 (Answer on p. 1778.) Exercise 162. Planes A and B leave the same point at the same time. Plane A travels (precisely) northeast at a constant speed of 100 km h−1 . Plane B travels (precisely) south at a constant speed of 200 km h−1 . After (precisely) 3 hours, both planes instantly turn to start flying directly towards each other (at the same speeds as before). How many hours after making this instant turn will they collide? (Answer on p. 1778.)

✈ x

A travels northeast at 100 km h−1

B travels south at 200 km h−1

In Ch. 13, we learnt the terms solution and solution set, in the context of a single equation involving only one variable. These terms also carry the same meaning in a system of equations, which is any set of m equations in n variables:

234

Just to be clear, we use the standard western convention where one’s age on any day is an integer and increases by one on each birthday. (We do not for example use the East Asian convention.)

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Example 495. Consider this system of two equations in two variables: y =x+1 1

y = −x + 3

(x, y ∈ R).

2

and

In secondary school, we already learnt to solve such systems of equations (it’s just simple algebra):

Plug = into = to get −x + 3 = x + 1 or 2 = 2x or x = 1. Now from either = or =, we can also get y = 2. Conclude: 2

1

1

2

• “The system of equations has one solution: (x, y) = (1, 2).”

• Or more simply, “The system of equations has one solution (1, 2).” • Or, “The system of equations has solution set {(1, 2)}.”

y

y =x+1

y = −x + 3

(1, 2)

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x

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Example 496. Solve this system of two equations in two variables: x2 3 y= − 2 2 1

and

y=x

(x, y ∈ R).

2

x2 3 Plug = into = to get x = − or 0 = x2 − 2x − 3 = (x − 3) (x + 1). So, x = 3 or x = −1. 2 2 And correspondingly, y = 3 or y = −1. Conclude: 2

1

3

• “The system of equations has two solutions: (x, y) = (3, 3) , (−1, −1).”

• Or more simply, “The system of equations has two solutions: (3, 3) and (−1, −1).”

• Or, “The system of equations has solution set {(3, 3) , (−1, −1)}.”

y

y=

x2 3 − 2 2

y=x

(3, 3)

x (−1, −1)

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Example 497. Here’s a system of three equations in three variables: y = 3x − 2, 1

z = 7 − y, 2

x=y+z

(x, y, z ∈ R).

3

Plug = and = into = to get x = 3x − 2 + 7 − y = 3x + 5 − y or y = 2x + 5. 1

2

3

Plug = into = to get 2x + 5 = 3x − 2 or 7 = x. 4

4

1

Plug x = 7 into = to get y = 19. Then plug y = 19 into = to get z = −12. Conclude: 1

2

• “The system of equations has one solution: (x, y, z) = (7, 19, −12).”

• Or more simply, “The system of equations has one solution (7, 19, −12).” • Or, “The system of equations has solution set {(7, 19, −12)}.”

By the way, (7, 19, −12) is this textbook’s first example of an ordered triple. This is exactly analogous to an ordered pair, except that now there are three coordinates. As you can imagine, we also have ordered quadruples, ordered quintuples, and more generally ordered n-tuples.235 When we study vectors in Part III, we’ll learn that it’s actually possible to graph the above three equations. (Spoiler: Each of the three graphs will be a plane in 3-dimensional space.) A system of equations can have no solutions:

235

For the formal definition of an n-tuple, see Definition 263 (Appendices).

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Example 498. Consider this system of two equations in two variables: y = ln x 1

y=x 2

and

(x, y ∈ R).

Observe that the graph of = lies below that of = everywhere. Hence, no ordered pair (x, y) satisfies the above system of equations. Conclude: 1

2

• “The system of equations has no solutions.”

• Or, “The system of equations has solution set ∅.”

y

y=x y = ln x

x

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Example 499. Consider this system of two equations in two variables: y = x2 1

y = −1

(x, y ∈ R).

2

and

Observe that the graph of = lies above that of = everywhere. Hence, no ordered pair (x, y) satisfies the above system of equations. Conclude: 1

2

• “The system of equations has no solutions.”

• Or, “The system of equations has solution set ∅.”236

y

y = x2

x y = −1 Example 500. Consider this system of four equations in three variables: x = 1, 1

y = 2, 2

z = 3, 3

xy = 0 4

(x, y, z ∈ R).

There is no (x, y, z) that satisfies the above four equations. So there are zero solutions to the above system of equations. Equivalently,

• “No (x, y, z) is a solution.” • Or, “The system of equations has solution set ∅.”

At the other extreme, a system of equations can have infinitely many solutions:

236

But as we’ll learn in Part IV, if we rewrite the system of equations so that “x, y ∈ R” is replaced by “x, y ∈ C”, then we’d instead conclude:

• “The system of equations has two solutions: (x, y) = (−i, −1) , (i, −1).”

• Or more simply, “The system of equations has two solutions: (−i, −1) and (i, −1).” • Or, “The system of equations has solution set i, −1.” 416, Contents

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Example 501. Consider this system of two equations in two variables: y=x 1

2y = 2x

(x, y ∈ R).

2

and

Observe that = and = are really the same. And so, this system of equations has infinitely many solutions. Specifically, 1

2

• “Any (x, y) for which y = x is a solution.”

• Or, “The system of equations has solution set {(x, y) ∶ x = y}.” Alternatively and equivalently,

• “For any c ∈ R, (c, c) is a solution.” • Or, “The system of equations has solution set {(c, c) ∶ c ∈ R}.”

y

y=x

Also 2y = 2x

x

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Example 502. Consider this system of two equations in two variables: y = cos x 1

y=1 2

and

This system of equations has infinitely many solutions. Specifically,

(x, y ∈ R).

• “Any (x, y) for which x = 2kπ (for any k ∈ Z)and y = 0 is a solution.”

• Or, “The system of equations has solution set {(x, y) ∶ x = 2kπ, y = 0, k ∈ Z}.” Alternatively and equivalently,

• “For any k ∈ Z, (2kπ, 0) is a solution.” • Or, “The system of equations has solution set {(2kπ, 0) ∶ k ∈ Z}.”

y

(−2π, 1)

y=1

(2π, 1)

(0, 1)

x y = cos x

Example 503. Consider this system of two equations in three variables: x = yz, 1

z=0 2

(x, y, z ∈ R).

By =, any solution must have z = 0. Plug = into = to also get x = 0. There are no restrictions on what y can be. 2

2

1

So, there are infinitely many solutions. Specifically,

• “Any (x, y, z) for which x = 0 and z = 0 is a solution.” • Or, “The system of equations has solution set {(x, y, z) ∶ x = 0, z = 0}.” Alternatively and equivalently,

• “For any c ∈ R, (0, c, 0) is a solution.”

• Or, “The system of equations has solution set {(0, c, 0) ∶ c ∈ R}.”

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In the context of a system of equations, here are the formal definitions of a solution and the solution set:237 Definition 107. Given a system of equations (and/or inequalities) in n ≥ 2 variables, we call any ordered n-tuple that satisfies the system a solution (or root). And the set of all such solutions is called the solution set of that system. Exercise 163. XXX

A163. Example 504. Let a, b, c, x, y ∈ R. Suppose the equation y = ax2 + bx + c has solutions (0, 1), (2, 3), and (4, 5). Then what are a, b, and c?

Here things seem a little strange: We speak of the equation y = ax2 + bx + c and its solutions, with x and y being the variables. Yet what we really want to do is to solve the following system of equations, where a, b, and c are the variables and we’ve plugged in the given values of x and y: 1 = a ⋅ 02 + b ⋅ 0 + c = c, 1

3 = a ⋅ 22 + b ⋅ 2 + c = 4a + 2b + c, 2

5 = a ⋅ 42 + b ⋅ 4 + c = 16a + 4b + c. 3

As usual, this is just simple algebra: = − = yields 4a + 2b = 2 and = − = yields 16a + 4b = 4. 5 4 4 2 Now, = −2× = yields 8a = 0 or a = 0. Now from =, we also have b = 1. Finally, from =, we also have c = 1. Altogether, 2

1

4

3

1

5

(a, b, c) = (0, 1, 1).

Below are two similar exercises. A similar A-Level exam question is N2011/I/2 (Exercise 553). Exercise 164. Let a, b, c, x, y ∈ R.

(a) If y = ax2 + bx + c has solutions (x, y) = (1, 2) , (3, 5) , (6, 9), then what are a, b, and c? (b) If y = ax2 + bx + c has the solution (x, y) = (−1, 2) and the strict global minimum point (0, 0), then what are a, b, and c?

237

Note that these are exactly analogous to our earlier Definition 61.

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35.1.

Solving Systems of Equations with the TI84

You need to know how to use your graphing calculator to solve systems of equations. This is particularly handy when we don’t know of any method to solve the given system of equations:

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Example 505. Consider this system of equations: y = x4 − x3 − 5

and

We’ll solve this system using two methods.

y = ln x

(x > 0, y ∈ R).

Method 1. Graph both equations, then find the intersection points: 1. Enter the two equations (precise instructions omitted). 2. Press GRAPH to graph the two equations. 3. Press 2ND and TRACE to bring up the CALC (CALCULATE) menu. 4. Now press 5 to select the “intersect” function. Your TI84 now asks, “First curve?” So,

5. Press ENTER to tell the TI84 that “y1 = x4 − x3 − 5” is indeed our first curve. The TI84 now asks you, “Second curve?” This time,

6. Use the left and right arrow keys ⟨ and ⟩ to move the cursor to roughly where you think there is an intersection point. For me, I’ve moved it to (x, y) ≈ (1.702, 0.532).

7. Now hit ENTER . The TI84 will now ask you “Guess?”. 8. So hit ENTER again. After working furiously for a moment, the TI84 will inform you that the intersection point is (x, y) ≈ (1.87, 0.624).

It looks though like there might be another intersection point near x = 0. And so, repeating the above steps but now moving the cursor leftwards (precise instructions and screenshots omitted), you should find that there is another intersection point at (x, y) ≈ (0.00674, −5).

Thus, the solutions are (1.87 . . . , 0.623 . . . ) and (0.00674 . . . , −5). Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

Step 6.

Step 7.

Step 8.

(Example continues on the next page ...)

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(... Example continued from the previous page.)

Method 2. Combine the two equations into the single equation y = x4 − x3 − 5 − ln x. Graph it, then find the x-intercepts. Brief instructions: 1. Graph y = x4 − x3 − 5 − ln x. It looks like there’s an x-intercept near 2. So,

2. Use the “zero” function to find that there’s indeed an x-intercept at x ≈ 1.87. It’s not obvious, but it looks like there might be another x-intercept near the origin. So, 3. Use the “zero” function to find that there’s indeed another x-intercept at x ≈ 0.00674. With Method 2, we must plug in these two values of x into either of the original two equations to get the corresponding values of y. Then as before, we’ll conclude that the solutions are (1.87 . . . , 0.623 . . . ) and (0.00674 . . . , −5). Step 1.

Step 2.

Step 3.

Exercise 165. Use your GC to solve each system of equations. 1 √ , y = x5 − x3 + 2 (x, y ∈ R, x ≥ 0). (a) y = 1+ x

(b) (c)

y=

1 , 1 − x2

x2 + y 2 = 1,

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y = x3 + sin x y = sin x

(x, y ∈ R, x ≠ ±1).

(x, y ∈ R).

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36.

Partial Fractions Decomposition

This is yet another topic you’re supposed to have mastered in secondary school. As per p. 14 of your A-Level syllabus, you need only know how to decompose “partial fractions with cases where the denominator is no more complicated than: • (ax + b) (cx + d)

• (ax + b) (cx + d) • (ax + b) (x2 + c2 )”. 2

The lovely folks at MOE have kindly included the following on your List MF26 (p. 2): Partial fractions decomposition Non-repeated linear factors: px + q A B = + (ax + b)(cx + d ) (ax + b) (cx + d )

Repeated linear factors: px 2 + qx + r (ax + b)(cx + d )

2

=

A B C + + (ax + b) (cx + d ) (cx + d ) 2

Non-repeated quadratic factor: px 2 + qx + r (ax + b)( x + c ) 2

2

=

A Bx + C + 2 (ax + b) ( x + c 2 )

Tip for Towkays Partial fractions aren’t difficult to do—it’s just a bunch of tedious algebra. So the important thing is to go slowly and be really careful. Check and double-check that you’ve got everything exactly correct at each step of the way. This will save you time and marks, as compared to trying to do the algebra quickly and making a mistake. Let’s go through these three “categories” of partial fractions decomposition in turn:

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36.1.

Non-Repeated Linear Factors

1 Example 506. Let’s rewrite or decompose the expression 2 into partial fracx + 3x + 2 tions. To do so, first observe that x2 + 3x + 2 = (x + 1) (x + 2). So, write 1 1 = x2 + 3x + 2 (x + 1) (x + 2) =

=

A B + x+1 x+2

A (x + 2) + B (x + 1) (A + B) x + 2A + B = . (x + 1) (x + 2) (x + 1) (x + 2)

From the numerator, 1 = (A + B) x + 2A + B. Comparing coefficients, A + B = 0 and 2 2 1 2A + B = 1. Taking = − = yields A = 1 and then B = −1. Thus, 1

1 1 1 = − . x2 + 3x + 2 x + 1 x + 2

5x − 3 , observe that x2 + 3x + 2 = (x + 1) (x + 2) and Example 507. To decompose 2 x + 3x + 2 write x2

5x−3 A B A (x + 2) + B (x + 1) (A + B) x + 2A + B = + = = . + 3x + 2 x + 1 x + 2 (x + 1) (x + 2) (x + 1) (x + 2)

Comparing coefficients, A + B = 5 and 2A + B = −3. Hence, A = −8 and B = 13. Thus, x2

−8 13 5x − 3 = + . + 3x + 2 x + 1 x + 2

9x − 5 Example 508. To decompose , observe that −x2 + 5x − 6 = − (x − 3) (x − 2) 2 −x + 5x − 6 and write 9x−5 A B A (x − 2) − B (x − 3) (A − B) x−2A + 3B = + = = . + 5x − 6 − (x − 3) x − 2 − (x − 3) (x − 2) − (x − 3) (x − 2)

−x2

Comparing coefficients, A − B = 9 and −2A + 3B = −5. Hence, A = 22 and B = 13. Thus, 9x − 5 22 13 22 13 = + = + . + 5x − 6 − (x − 3) x − 2 3 − x x − 2

−x2

Exercise 166. Decompose (a) (Answer on p. 1781.)

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8 17x − 5 ; and (b) into partial fractions . x2 + x − 6 3x2 − 8x − 3

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36.2. Example 509. To decompose

=

x2 + x + 1

Repeated Linear Factors x2 + x + 1

(x + 1) (x − 1)

2,

write

B C A A (x − 1) + B (x + 1) (x − 1) + C (x + 1) + + = = 2 2 x + 1 x − 1 (x − 1)2 (x + 1) (x − 1) (x + 1) (x − 1) A (x2 − 2x + 1) + B (x2 − 1) + C (x + 1) (x + 1) (x − 1)

2

=

2

(A + B) x2 + (−2A + C) x + A − B + C (x + 1) (x − 1)

2

.

Comparing coefficients, A + B = 1, −2A + C = 1, A − B + C = 1. Summing these three equations, we get 2C = 3 or C = 3/2 and then also A = 1/4 and B = 3/4. x2 + x + 1

Thus,

(x + 1) (x − 1)

Example 510. To decompose

=

2

=

1 3 3 + + . 4 (x + 1) 4 (x − 1) 2 (x − 1)2

3x2 − x + 1

2,

(4x − 1) (x + 2)

write

B C A A (x + 2) + B (4x − 1) (x + 2) + C (4x − 1) + + = = 2 2 4x − 1 x + 2 (x + 2)2 (4x − 1) (x + 2) (4x − 1) (x + 2) 3x2 − x + 1

A (x2 + 4x + 4) + B (4x2 + 7x − 2) + C (4x − 1) (4x − 1) (x + 2)

2

2

=

(A + 4B) x2 + (4A + 7B + 4C) x + 4A − 2B − (4x − 1) (x + 2)

2

Comparing coefficients, A + 4B = 3, 4A + 7B + 4C = −1, and 4A − 2B − C = 1. 1

2

3

Rearrange = to get A = 3 − 4B. Next, = minus = yields 9B + 5C = −2 or C = − (2 + 9B) /5. 4 5 3 Plug = and = into = to get 4 (3 − 4B) − 2B + (2 + 9B) /5 = 1 or B = 19/27 and then also A = 5/27 and C = −5/3. 1

Thus,

4

3x2 − x + 1

(4x − 1) (x + 2)

2

2

=

3

5 19 5 + − . 27 (4x − 1) 27 (x + 2) 3 (x + 2)2

2x2 − x + 7 Exercise 167. Decompose 3 into partial fractions. x − x2 − x + 1

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36.3.

Example 511. To decompose

=

2x2 + x + 1 , write (5x − 1) (x2 + 1)

2x2 + x + 1 A Bx + C A (x2 + 1) + (Bx + C) (5x − 1) = + = (5x − 1) (x2 + 1) 5x − 1 x2 + 1 (5x − 1) (x2 + 1)

Ax2 + A + 5Bx2 + (5C − B) x − C (A + 5B) x2 + (5C − B) x + A − C = . (5x − 1) (x2 + 1) (5x − 1) (x2 + 1)

Comparing coefficients, A + 5B = 2, 5C − B = 1, and A − C = 1. 1

2

3

Rearrange = to get A = 2 − 5B. Rearrange = to get have C = (B + 1) /5. Plug = and = 3 into = to get 2 − 5B − (B + 1) /5 = 1 or B = 2/13 and then also A = 16/13 and C = 3/13. 1

4

2

5

4

5

2x2 + x + 1 16 2x + 3 = + . 2 (5x − 1) (x + 1) 13 (5x − 1) 13 (x2 + 1)

Thus,

Example 512. To decompose

4x2 − 3x + 2 , write (2x + 7) (x2 + 9)

4x2 − 3x + 2 A Bx + C A (x2 + 9) + (Bx + C) (2x + 7) = + = (2x + 7) (x2 + 9) 2x + 7 x2 + 9 (2x + 7) (x2 + 9)

Ax2 + 9A + 2Bx2 + (7B + 2C) x + 7C (A + 2B) x2 + (7B + 2C) x + 9A + 7C = . = (2x + 7) (x2 + 9) (2x + 7) (x2 + 9)

Comparing coefficients, A + 2B = 4, 7B + 2C = −3, and 9A + 7C = 2. 1

2

3

Rearrange = to get A = 4 − 2B. Rearrange = to get C = − (7B + 3) /2. Plug = and = 3 into = to get 9 (4 − 2B) + 7 (−7B + 3) /2 = 2 or B = 47/85 and then also A = 246/85 and C = −292/85. 1

Thus,

4

2

5

4

5

4x2 − 3x + 2 246 47x − 292 = + . (2x + 7) (x2 + 9) 85 (2x + 7) 85 (x2 + 9)

−3x2 + 5 Exercise 168. Decompose 3 into partial fractions. (Answer on p. 1781.) x − 2x2 + 4x − 8

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37. 37.1.

Solving Inequalities

Multiplying an Inequality by an Unknown Constant

Multiplication (by a constant) preserves equality. That is, given an equation a = b, we can always multiply by any constant x and the equality will still be preserved: ax = bx.

3

In contrast, multiplication (by a constant) does not generally preserve order (or inequalities): Example 513. The inequality 2 > 1 is ...

(a) Preserved if we multiply it by a positive constant like 3: 2×3>1×3

or

6 > 3.

(b) Reversed (or flipped) if we multiply it by a negative constant like −3: 2 × (−3) < 1 × (−3)

−6 < −3.

or

(c) Becomes an equality if we multiply it by zero: 2×0=1×0

or

0 = 0.

The above seems obvious. But a common mistake is to multiply an inequality by some unknown constant x and expect it to be preserved: Example 514. Let x ∈ R. Beng reasons, “We know that 8 > 5. Therefore 8x > 5x.” Beng’s reasoning is wrong.

(a) If x > 0, then yea, Beng happens to be correct and 8x > 5x. (b) If x < 0, then 8x < 5x and he’s wrong. 7 (c) If x = 0, then 8x = 5x (= 0) and he’s again wrong.

3

7

In general,

Fact 77. Let a, b, x ∈ R with a > b.

(a) If x > 0, then ax > bx. (b) If x < 0, then ax < bx.

(c) If x = 0, then ax = bx = 0.

(Positive multiplication preserves order) (Negative multiplication reverses order) (Zero multiplication creates equality)

Analogous results also hold if we replace a > b with a ≥ b.

Proof. Omitted.238 238

Although these results may seem “obvious”, to properly prove them requires various axioms and defi-

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So, in general, we may have to break our analysis into two (or three) cases, depending on whether x is positive, negative, or zero: Example 515. Suppose a, b, x ∈ R with x ≠ 0 and a 1 b > . x x

What can we say about a and b?

Beng reasons, “Multiply > by x to conclude that a > b.” 1

7

As usual, Beng is wrong. We must break our analysis down into two cases:

(a) If x > 0, then yea, he happens to be correct—we can indeed multiply > by x to conclude that a > b. 1

(b) But if x < 0, then multiplying > by x reverses the inequality, so that a < b. 1

So, the correct solution is this:

Given >, if x > 0, then a > b; but if x < 0, then a < b. 1

By the way, why did I specify that x ≠ 0?239

239

nitions (e.g. of order) that haven’t been stated in this textbook. If x = 0, then a/x and b/x would involve dividing by zero.

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37.2.

ax2 + bx + c > 0

This subchapter also doubles as yet another review of quadratic polynomials. Example 516. Solve −x2 + 3x − 1 > 0.

Since the coefficient (−1) on x2 is negative, LHS is a ∩-shaped quadratic.

And since the discriminant 32 − 4 (−1) (−1) = 5 is positive, the graph intersects the x-axis at these two points (quadratic formula): √ √ −3 ± 5 3 ∓ 5 = . −2 2 Altogether then, −x2 + 3x − 1 > 0 is true between those two roots. That is, the given inequality has this solution set: √ √ 3− 5 3+ 5 ( , ). 2 2

y x √ √ 3− 5 3+ 5 , ) x∈( 2 2 solves −x2 + 3x − 1 > 0.

y = −x2 + 3x − 1

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Example 517. Solve x2 + 3x + 1 > 0.

Since the coefficient (1) on x2 is positive, this is a ∪-shaped quadratic.

And since the discriminant 32 − 4 (1) (1) = 5 is positive, the graph intersects the x-axis at two points, given simply by the two roots of the quadratic, which are √ −3 ± 5 x= . 2

Altogether then, x2 + 3x + 1 > 0 “outside” those two roots. That is, the solution set of the given inequality is √ √ −3 − 5 −3 + 5 , ]. R∖[ 2 2

y

y = x2 + 3x + 1

√ −3 − 5 2

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x

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Example 518. Solve −x2 + 2x − 1 > 0.

Since a < 0, this is a ∩-shaped quadratic.

And since the discriminant b2 − 4ac = 22 − 4 (−1) (−1) = 0 is zero, the graph just touches the x-axis at x=−

2 b =− = 1. 2a 2 (−1)

Altogether then, there are no values of x for which x2 + 2x + 1 > 0 is true. Equivalently, the inequality has solution set ∅.

y

x

1 y = −x2 + 2x − 1

x∈∅ (there are no values of x for which −x2 + 2x − 1 > 0)

Note that 1 does not belong to the solution set of this inequality. This is because at x = 1, we have −x2 + 2x − 1 = −12 + 2 ⋅ 1 − 1 = 0. And so, at x = 1, it is not true that −x2 + 2x − 1 > 0.

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Example 519. Solve x2 + 2x + 1 > 0.

Since a > 0, this is a ∪-shaped quadratic.

And since the discriminant b2 − 4ac = 22 − 4 (1) (1) = 0 is zero, the graph just touches the x-axis at x=−

2 b =− = −1. 2a 2 (1)

Altogether then, x2 + 2x + 1 > 0 for all values of x except −1. Equivalently, the inequality has solution set R ∖ {−1}.

y

y = x2 + 2x + 1

x ∈ R ∖ {−1} solves x2 + 2x + 1 > 0

1

x

Note that 1 does not belong to the solution set of this inequality. This is because at x = 1, we have x2 + 2x + 1 = 12 + 2 ⋅ 1 + 1 = 0. And so, at x = 1, it is not true that x2 + 2x + 1 > 0.

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Example 520. Solve −x2 + x − 1 > 0.

Since a < 0, this is a ∩-shaped quadratic.

And since the discriminant b2 − 4ac = (−1) − 4 (−1) (−1) = −3 is negative, the graph doesn’t touch the x-axis at all and is completely below the x-axis. 2

Altogether then, there are no values of x for which x2 + x + 1 > 0 holds. Equivalently, the inequality has solution set ∅.

y

x

y = −x2 + x − 1

x∈∅ (there are no values of x for which −x2 + x − 1 > 0).

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Example 521. Solve x2 + x + 1 > 0.

Since a > 0, this is a ∪-shaped quadratic.

And since the discriminant b2 − 4ac = 12 − 4 (1) (1) = −3 is negative, the graph doesn’t touch the x-axis at all and is completely above the x-axis. Altogether then, x2 + x + 1 > 0 holds for all values of x. Equivalently, the inequality has solution set R.

y

y = x2 + x + 1

x ∈ R solves x2 + x + 1 > 0

x

Here is the general solution for ax2 + bx + c > 0 (x ∈ R): Fact 78. Suppose a, b, c, x ∈ R, r1 = solution sets for the inequality in the six different cases are these:

−b −

√ b2 − 4ac −b + b2 − 4ac , and r2 = . Then the a a

ax2 + bx + c > 0

(a) b2 − 4ac > 0 (b) b2 − 4ac = 0 (c) b2 − 4ac < 0 (1) a > 0 R ∖ [r1 , r2 ] R ∖ {−b/2a}. ∅ (2) a < 0 (r2 , r1 ) ∅ ∅

Proof on the next page:

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Proof. Case 1. If a > 0, then this is a ∪-shaped quadratic.

(a) If b2 −4ac > 0, then y = ax2 +bx+c intersects the x-axis at x = r1 , r2 . Thus, ax2 +bx+c > 0 for all x “outside” those two roots—i.e. for all x except x ∈ [r1 , r2 ]. (b) If b2 − 4ac = 0, then y = ax2 + bx + c just touches the x-axis at x = −b/2a. Thus, ax2 + bx + c > 0 for all x except x = −b/2a.

(c) If b2 −4ac < 0, then y = ax2 +bx+c is completely above the x-axis and thus ax2 +bx+c > 0 for all x.

Case 1. a > 0, ∪-shaped quadratic

y

b2 − 4ac < 0

b2 − 4ac > 0

b2 − 4ac > 0

b2 − 4ac = 0

b2 − 4ac = 0

x

b2 − 4ac < 0 Case 2. a < 0, ∩-shaped quadratic

Case 2. If a < 0, then this is a ∩-shaped quadratic.

(a) If b2 −4ac > 0, then y = ax2 +bx+c intersects the x-axis at x = r1 , r2 . Thus, ax2 +bx+c > 0 for all x “between” those two roots—i.e. for all x ∈ (r2 , r1 ).240 (b) If b2 − 4ac = 0, then y = ax2 + bx + c just touches the x-axis at x = −b/2a. Thus, ax2 + bx + c > 0 for no x. (c) If b2 −4ac < 0, then y = ax2 +bx+c is completely below the x-axis and thus ax2 +bx+c > 0 for no x. Exercise 169. Solve each inequality. (a) −3x2 + x − 5 > 0. 240

Note that r2 < r1 because a < 0.

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(Answer on p. 1786.) (b) x2 − 2x − 1 > 0.

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37.3.

ax + b >0 cx + d

Example 522. 4/7 > 0 because both N = 4 and D = 7 are positive.

Example 523. −5/ − 3 > 0 because both N = −5 and D = −3 are negative. Example 524. −9/2 < 0 because N = −9 < 0 but D = 2 > 0.

Example 525. 1/ − 8 < 0 because N = 1 > 0 but D = −8 < 0.

As the above examples suggest, we have this result: Fact 79. Suppose N ∈ R and D ≠ 0. Then

N >0 D N (b) 0 or N, D < 0). N and D have different signs (i.e. N, D > 0 or N, D < 0).

Again, although these results may seem “obvious”, to properly prove them requires various axioms and definitions (e.g. of order) that haven’t been stated in this textbook.

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Example 526. Consider

1 1 < 1. Let’s first solve this inequality the wrong way: x

1. Multiply both sides by x to get 1 < x.

2. Hence, x > 1—or equivalently, the solution set for < is (1, ∞). 7 1

Again, the mistake here is to multiply both sides by the unknown quantity x and expect the inequality to be preserved. But it may not be, because x may be negative. Here’s the correct solution. First, rewrite the given inequality in Standard Form: 1 1 0 x

x−1 N 2 = > 0. x D

⇐⇒

Observe that D = 0 ⇐⇒ x = 0 and N = 0 ⇐⇒ x = 1.

Let us call the points at which either the denominator or numerator equals zero the zero points. So here, the zero points are 0 and 1. We now consider the sign of N /D, in each of the three possible cases: (a) If x < 0, then N < 0 and D < 0, so that N /D > 0. (b) If 0 < x < 1, then N < 0 and D > 0, so that N /D < 0. (c) If x > 1, then N > 0 and D > 0, so that N /D > 0.

These observations may be compactly summarised in this sign diagram: +

Conclude:

< holds 1

⇐⇒

0

> holds 2

1 ⇐⇒

Equivalently, the solution set for < is (−∞, 0) ∪ (1, ∞). 1

y

y=1

y=

+ x 1.

1 x

1

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5x + 4 > 0. −2x + 1 We note that the numerator and denominator equal zero at −4/5 and 1/2. And so, we have the sign diagram below, because

Example 527. Solve

4 (a) If x < − , then N < 0 and D > 0, so that N /D < 0. 5 4 1 (b) If − < x < , then N < 0 and D > 0, so that N /D > 0. 5 2 1 (c) If x > , then N > 0 and D > 0, so that N /D < 0. 2

4 1 Hence, − < x < . 5 2

Equivalently, the solution set is

y

y=

5x + 4 −2x + 1

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4 − 5

+

−4/5

1/2

4 1 (− , ). 5 2

1 2

x

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N x+3 = > 0. D 3x + 2 We note that the numerator and denominator equal zero at −3 and −2/3. And so, we have the sign diagram below, because

Example 528. Solve

(a) If x < −3, then N < 0 and D < 0, so that N /D > 0.

(b) If −3 < x < −2/3, then N < 0 and D > 0, so that N /D < 0. (c) If x > −2/3, then N > 0 and D > 0, so that N /D > 0.

+

Hence, x < −3 OR x > −2/3.

Equivalently, the solution set is

−3

−2/3

2 (−∞, −3) ∪ (− , ∞). 3

+

y

y= −3

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x+3 3x + 2

x

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N 4x − 1 1 = > 0. D x+2 We note that the numerator and denominator equal zero at 1/4 and −2. And so, we have the sign diagram below, because

Example 529. Solve

(a) If x < −2, then N < 0 and D < 0, so that N /D > 0.

(b) If −2 < x < 1/4, then N < 0 and D > 0, so that N /D < 0. (c) If x > 1/4, then N > 0 and D > 0, so that N /D > 0.

+

Hence, x < −2 OR x > 1/4.

Equivalently, the solution set is

−2

1/4

+

1 (−∞, −2) ∪ ( , ∞). 4

y

y=

4x − 1 x+2

−2

Exercise 170. Solve each inequality. (a)

1 4

x

−1 1 2x + 1 −3x − 18 x−1 > 0. (b) > 0. (c) > 0. (d) > 0. (e) > 0. −4 −4 −4 3x + 2 9x − 14

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3x − 2 < 3. −5x + 1 First rewrite the given inequality in Standard Form:

Example 530. Solve

3−

−15x + 3 − (3x − 2) −18x + 5 3x − 2 > 0 ⇐⇒ > 0 ⇐⇒ > 0. −5x + 1 −5x + 1 −5x + 1

Now as usual, write

1 2 −18x + 5 > 0 ⇐⇒ (−18x + 5, −5x + 1 > 0 OR − 18x + 5, −5x + 1 < 0). −5x + 1

Observe that

5 1 1 AND x < ) ⇐⇒ x < . 18 5 5 2 1 5 5 AND x > ) ⇐⇒ x > . • −18x + 5, −5x + 1 < 0 ⇐⇒ (x > 18 5 18 3x − 2 −18x + 5 1 5 Altogether then, < 3 ⇐⇒ > 0 ⇐⇒ x ∈ R ∖ [ , ]. −5x + 1 −5x + 1 5 18 • −18x + 5, −5x + 1 > 0 ⇐⇒ (x < 1

y

y=3

1 5

3x − 2 y= −5x + 1

Exercise 171. Solve each inequality. (a)

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2x + 3 < 9. −x + 7

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37.4.

The Cardinal Sin of Dividing by Zero, Revisited

Example 531. Solve xex ≥ xe (x ∈ R). 1

Beng reasons, “Divide ≥ by x to get ex ≥ e. Which is true for all x ≥ 1. Hence, the solution 1

set for ≥ is [1, ∞).” 7 1

Again, Beng’s mistake is to divide ≥ by x and assume the inequality will be preserved. 1

By the way, when given any inequality, I recommend first rewriting it in the following Standardised Form (SF), where STUFF is on the LHS of the inequality and 0 is on the RHS, like this: STUFF ⋛ 0.

Standardised Form:

(The symbol ⋛ stands for any inequality, i.e. any of >, ≥, b ∣x∣ ≥ b

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

−b < x < b. −b ≤ x ≤ b. x < −b OR x > b. x ≤ −b OR x ≥ b.

Proof. See p. 1581 (Appendices).

Example 532. ∣x∣ < 5 ⇐⇒ −5 < x < 5. The solution set is (−5, 5). Example 533. ∣x∣ ≤ 3 ⇐⇒ −3 ≤ x ≤ 3. The solution set is [−3, 3].

Example 534. ∣x∣ > 7 ⇐⇒ (x > 7 OR x < −7). The solution set is (−∞, −7) ∪ (7, ∞). Example 535. ∣x∣ ≥ 1 ⇐⇒ (x ≥ 1 OR x ≤ −1). The solution set is (−∞, −1] ∪ [1, ∞).

Here’s a more general version of the above result: Fact 81. Suppose a ∈ R, b ≥ 0. Then

(a) (b) (c) (d)

∣x − a∣ < b ∣x − a∣ ≤ b ∣x − a∣ > b ∣x − a∣ ≥ b

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

a − b < x < a + b. a − b ≤ x ≤ a + b. (x > a + b OR x < a − b). (x ≥ a + b OR x ≤ a − b).

Proof. See p. 1581 (Appendices) . Example 536. Solve ∣x − 1∣ < 5. 1

∣x − 1∣ < 5 1

⇐⇒

−5 < x − 1 < 5

⇐⇒

−4 < x < 6.

⇐⇒

−7 ≤ x ≤ −1.

Equivalently, we may say that the solution set for < is (−4, 6).

Example 537. Solve ∣x + 4∣ ≤ 3.

1

2

∣x + 4∣ ≤ 3 2

⇐⇒

−3 ≤ x + 4 ≤ 3

Equivalently, we may say that the solution set for ≤ is [−1, −7].

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Example 538. Solve ∣x − 2∣ > 7. 3

∣x − 2∣ > 7 3

⇐⇒

x − 2 > 7 OR x − 2 < −7

⇐⇒

x > 9 OR x < −5.

Equivalently, we may say that the solution set for > is (∞, −5] ∪ [9, ∞). 3

Example 539. Solve ∣x + 1∣ ≥ 1. 4

∣x + 1∣ ≥ 1 4

⇐⇒

x + 1 ≥ 1 OR x + 1 ≤ −1

⇐⇒

x ≥ 0 OR x ≤ 0

Equivalently, we may say that the solution set for ≥ is R. 4

⇐⇒

x ∈ R.

There’s actually a nice geometric interpretation of Fact 81:

(a) ∣x − a∣ < b means that the distance between x and a is less than b.

∣x − a∣ < b

( a − b

a

) a + b

x

(b) ∣x − a∣ ≤ b means that the distance between x and a is no greater than b.

∣x − a∣ ≤ b

[ a  − b

a

x

] a  + b

(c) ∣x − a∣ > b means that the distance between x and a is more than b.

∣x − a∣ > b

) a  − b

a

( a  + b

x

(d) ∣x − a∣ ≥ b means that the distance between x and a is no less than b.

∣x − a∣ ≥ b

] a  − b

a

[ a  + b

x

It will often be easier to solve inequalities with the aid of a graph:

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Example 540. Solve ∣x − 4∣ ≤ 2x. 1

We will first solve ≤ with the aid of graphs. To do so, first sketch the graphs of y = ∣x − 4∣ and y = 2x. To sketch the former, simply recall from Ch. 24.6 that the graph of y = ∣f (x)∣ is the same as that of f , but with any portion below the x-axis reflected in the x-axis. 1

y

y = 2x

y = ∣x − 4∣

P

4 3

x “Clearly”, ≤ holds ⇐⇒ x is to the right of the intersection point P . Our goal then is to find P . 1

From the graph, we see that at P , we have x − 4 < 0 and thus ∣x − 4∣ = − (x − 4) = 4 − x. And so, P is given by: 4 − x = 2x or x = 4/3. 1 4 Hence, x ≥ . Or equivalently, the solution set for ≤ is [4/3, ∞). 3

We now solve ≤ again but this time a little more rigorously and without the aid of graphs. 1

We consider two cases. If (a) x < 4, then ∣x − 4∣ ≤ 2x

⇐⇒

1

If (b) x ≥ 4, then

4 − x ≤ 2x

∣x − 4∣ ≤ 2x 1

which, of course, holds for all x ≥ 4.

⇐⇒

⇐⇒

4 ≤ 3x

x − 4 ≤ 2x

Altogether then, ≤ holds for (a) 4/3 ≤ x < 4 OR (b) x ≥ 4. 1

Or equivalently and more simply, ≤ holds for x ≥ 4/3.

⇐⇒

⇐⇒

4/3 ≤ x.

−4 ≤ x,

1

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Example 541. Solve ∣2x − 4∣ > x.

Sketching the graphs of y = ∣2x − 4∣ and y = x, we see that ∣2x − 4∣ > x ⇐⇒ x is to the left or right of the two intersection points P and Q. So, let’s find P and Q.

y

y = ∣2x − 4∣ y=x

Q

4 3

P 4

x At P , we have 2x − 4 < 0 and thus ∣2x − 4∣ = − (2x − 4) = 4 − 2x. And so, P is given by 4 − 2x = x

or

2x − 4 = x

or

4 x= . 3

At Q, we have 2x − 4 > 0 and thus ∣2x − 4∣ = 2x − 4. And so, Q is given by Conclusion: x
4 solves the inequality. In other words, the solution set is 3 4 4 (−∞, ) ∪ (4, ∞) = R ∖ [ , 4]. 3 3

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Example 542. Solve ∣3x − 4∣ ≥ 2x + 2.

Sketching the graphs of y = ∣3x − 4∣ and y = 2x + 2, we see that ∣3x − 4∣ ≥ 2x + 2 ⇐⇒ x is to the left or right of the two intersection points P and Q. So, let’s find P and Q.

y

Q

y = 2x + 2

2 5

y = ∣3x − 4∣

P

6

x

At P , we have 3x − 4 < 0 and thus ∣3x − 4∣ = − (3x − 4) = 4 − 3x. And so, P is given by 4 − 3x = 2x + 2

or

3x − 4 = 2x + 2

or

2 x= . 5

At Q, we have 3x − 4 > 0 and thus ∣3x − 4∣ = 3x − 4. And so, Q is given by Conclusion: x ≤

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x = 6.

2 or x ≥ 6 solves the inequality. In other words, the solution set is 5 2 2 (−∞, ] ∪ [6, ∞) = R ∖ ( , 6). 5 5

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Example 543. Solve ∣x + 1∣ ≥ ∣2x − 1∣.

Sketching the graphs of y = ∣x + 1∣ and y = ∣2x − 1∣, we see that ∣x + 1∣ ≥ ∣2x − 1∣ ⇐⇒ x is between the two intersection points P and Q. So, let’s find P and Q.

y

y = ∣2x − 1∣

y = ∣x + 1∣

Q

P 0

2

x

At P , we have x + 1 > 0 and 2x − 1 < 0. Thus, ∣x + 1∣ = x + 1 and ∣2x − 1∣ = 1 − 2x. And so, P is given by x + 1 = 1 − 2x

or

x + 1 = 2x − 1

or

x = 0.

At Q, we have x + 1 > 0 and 2x − 1 > 0. Thus, ∣x + 1∣ = x + 1 and ∣2x − 1∣ = 2x − 1. And so, Q is given by x = 2.

Conclusion: x ∈ [0, 2] solves the inequality. In other words, the solution set is [0, 2]. Exercise 172. Solve each inequality.

(a) ∣x − 4∣ ≤ 71. (b) ∣5 − x∣ > 13.

(c) ∣−3x + 2∣ − 4 ≥ x − 1. (d) ∣x + 6∣ > 2 ∣2x − 1∣. 448, Contents

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Fact 82. (Triangle Inequality) Suppose x, y ∈ R. Then ∣x + y∣ ≤ ∣x∣ + ∣y∣ .

Proof. We have − ∣x∣ ≤ x ≤ ∣x∣ and − ∣y∣ ≤ y ≤ ∣y∣.

Adding up, we have − (∣x∣ + ∣y∣) ≤ x + y ≤ ∣x∣ + ∣y∣. Hence, ∣x + y∣ ≤ ∣∣x∣ + ∣y∣∣ = ∣x∣ + ∣y∣.

The Triangle Inequality is so named because XXX Example 544. XXX Example 545. XXX Exercise 173. XXX

A173.

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37.6.

ax2 + bx + c >0 dx2 + ex + f

According to your syllabus, you need to solve the above inequality in those cases where both numerator and denominator are either always positive or factorisable. Let’s first look at those cases where either numerator or denominator is always positive: Fact 83. Consider the inequality

ax2 + bx + c 1 > 0. dx2 + ex + f

(a) If ax2 + bx + c is always positive, then > is equivalent to dx2 + ex + f > 0. 1

(b) If dx2 + ex + f is always positive, then > is equivalent to ax2 + bx + c > 0. 1

Proof. (a) By Fact 77(a), we can divide both sides of > by ax2 +bx+c to get 1/ (dx2 + ex + f ) > 0. And now, by Fact XXX, dx2 + ex + f > 0. 1

(b) By Fact 77(a), we can multiply both sides of > by dx2 + ex + f to get ax2 + bx + c > 0. 1

By the way, recall (Ch. 14) that a quadratic expression is always positive if and only if its coefficient on x2 is positive (so that the graph is ∪-shaped) AND the discriminant is positive (so that the graph is everywhere above the x-axis). Example 546. Consider

x2 + x + 1 > 0. 3x2 − 2x − 5

The numerator is always positive, because the coefficient on x2 is positive and the discriminant 12 − 4 (1) (1) = −3 is negative. And so by Fact 83(a), the given inequality is simply equivalent to 3x2 − 2x − 5 > 0. This is a ∪-shaped quadratic with discriminant 2 (−2) − 4 (3) (−5) = 64. Hence, 3x2 − 2x − 5 > 0 “outside” the two roots, which are √ 2 ± 64 −6 10 5 x= = , = −1, . 2⋅3 6 6 3 Hence, the given inequality’s solution set is R ∖ [−1, 5/3] or (−∞, −1) ∪ (5/3, ∞).

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Example 547. Consider

−x2 + 7x + 1 > 0. 2x2 − x + 1

The denominator is always positive, because the coefficient on x2 is positive and the 2 discriminant (−1) − 4 (2) (1) = −7 is negative. And so by Fact 83(b), the given inequality is simply equivalent to −x2 + 7x + 1 > 0. This is a ∩-shaped quadratic with discriminant 72 − 4 (−1) (1) = 53. Hence, −x2 + 7x + 1 > 0 “between” the two roots, which are √ √ −7 ± 53 7 ∓ 53 = . x= 2 ⋅ (−1) 2 √ √ 7 − 53 7 + 53 Hence, the given inequality’s solution set is ( , ). 2 2

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We next look at the case where both ax2 + bx + c and dx2 + ex + f are factorisable. Example 548. Solve

x2 + 3x + 2 > 0. 2x2 − 7x + 6

First consider the numerator N = x2 + 3x + 2. Observe that x2 + 3x + 2 = (x + 1) (x + 2). And so y = x2 + 3x + 2, which is a ∪-shaped quadratic, intersects the x-axis at x = −2, −1. Thus, N > 0 if x ∈ R ∖ [−2, −1], while N < 0 if x ∈ (−1, −2).

Next consider the denominator D = 2x2 −7x+6. Observe that 2x2 −7x+6 = (2x − 3) (x − 2). And so y = 2x2 − 7x + 6, which is a ∪-shaped quadratic, intersects the x-axis at x = 1.5, 2. Thus, D > 0 if x ∈ R ∖ [1.5, 2], while D < 0 if x ∈ (1.5, 2). The given inequality is true if N, D > 0 OR N, D < 0. We have

N, D > 0 ⇐⇒ x ∈ R ∖ [−2, −1] AND x ∈ R ∖ [1.5, 2] ⇐⇒ x ∈ R ∖ ([−2, −1] ∪ [1.5, 2]). N, D < 0 ⇐⇒ x ∈ (−1, −2) AND x ∈ (1.5, 2) ⇐⇒ x ∈ (−1, −2) ∩ (1.5, 2) = ∅.

Hence, the given inequality’s solution set is R ∖ ([−2, −1] ∪ [1.5, 2])

−x2 + 5x − 4 Example 549. Solve > 0. 3x2 − 2x − 5

or

(−∞, −2) ∪ (−1, 1.5) ∪ (2, ∞).

First consider the numerator N = −x2 +5x−4. Observe that −x2 +5x−4 = − (x − 1) (x − 4). And so y = −x2 + 5x − 4, which is a ∩-shaped quadratic, intersects the x-axis at x = 1, 4. Thus, N > 0 if x ∈ (1, 4), while N < 0 if x ∈ R ∖ [1, 4]. Next consider the denominator D = 3x2 −2x−5. Observe that 3x2 −2x−5 = (3x − 5) (x + 1). And so y = 3x2 − 2x − 5, which is a ∪-shaped quadratic, intersects the x-axis at x = −1, 5/3. Thus, D > 0 if x ∈ R ∖ [−1, 5/3], while D < 0 if x ∈ (−1, 5/3).

The given inequality is true if N, D > 0 OR N, D < 0. We have

N, D > 0 ⇐⇒ x ∈ (1, 4) AND x ∈ R ∖ [−1, 5/3] ⇐⇒ x ∈ (1, 4) ∖ [−1, 5/3] = (5/3, 4). N, D < 0 ⇐⇒ x ∈ R ∖ [1, 4] AND x ∈ (−1, 5/3) ⇐⇒ x ∈ (−1, 5/3) ∖ [1, 4] = (−1, 1).

Hence, the given inequality’s solution set is (−1, 1) ∪ (5/3, 4). Exercise 174. Solve each inequality. (a)

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x2 + 2x + 1 > 0. x2 − 3x + 2

(b)

(Answers on pp. 1787, 1788, and 1789.)

x2 − 1 > 0. x2 − 4

(c)

x2 − 3x − 18 > 0. −x2 + 9x − 14

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37.7.

Solving Inequalities by Graphical Methods

Example 550. Solve x > sin (0.5πx).

As usual, first rewrite the inequality into SF: x − sin (0.5πx) > 0.

Then graph y = x−sin (0.5πx) on your TI84. Our goal is to find the roots of this equation, i.e. the values of x for which x − sin (0.5πx) = 0. Step 1.

Step 2.

Step 3.

Step 5.

Step 6.

Step 7.

Step 4.

1. Press ON to turn on your TI84. 2. Press Y= to bring up the Y= editor. 3. Press X,T,θ,n − SIN 0 . 5 . Next press 2ND and then ∧ to enter π. Now press X,T,θ,n ) and altogether you will have entered “x − sin (0.5πx)”. 4. Now press GRAPH and the TI84 will graph y = x − sin (0.5πx).

It looks there may be some x-intercepts close to the origin. Let’s zoom in to see better.

5. Press the ZOOM button to bring up a menu of ZOOM options. 6. Press 2 to select the Zoom In option. Nothing seems to happen. But now press ENTER and the TI84 will zoom in a little for you. It looks like there are 3 horizontal intercepts. To find out what precisely they are, we’ll use the TI84’s “zero” option. 7. Press 2ND and then TRACE to bring up the CALC (CALCULATE) menu. 8. Press 2 to select the zero function. This brings you back to the graph, with a cursor flashing at the origin. Also, the TI84 prompts you with the question: “Left Bound?” (Example continues on the next page ...)

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(... Example continued from the previous page.) Step 8.

Step 9.

Step 10.

Step 11.

Step 12.

Step 13.

Recall242 that zero is another word for root. So what TI84’s zero function will do here is find the roots of the given equation (i.e. the values of x for which y = 0). Those of you accustomed to newfangled inventions like the world wide web and the wireless telephone will probably be expecting that the TI84 simply and immediately tells you what all the roots are. But alas, the TI84 is an ancient device, which means there’s plenty more work you must do to find the three roots. To find a root, you must first specify a “Left Bound” and a “Right Bound” for x. The TI84 will then check to see if there are any values of x for which y = 0 between those two bounds.

9. Using the ⟨ and ⟩ arrow keys, move the blinking cursor until it is where you want your first “Left Bound” to be. For me, I have placed it a little to the left of where I believe the leftmost horizontal intercept to be. 10. Press ENTER and you will have just entered your first “Left Bound”. TI84 now prompts you with the question: “Right Bound?”.

11. So now just repeat. Using the ⟨ and ⟩ arrow keys, move the blinking cursor until it is where you want your first “Right Bound” to be. For me, I have placed it a little to the right of where I believe the leftmost horizontal is. 12. Again press ENTER and you will have just entered your first “Right Bound”. TI84 now asks you: “Guess?” This is just asking if you want to proceed and get TI84 to work out where the horizontal intercept is. So go ahead and:

13. Press ENTER . TI84 now informs you that there is a “Zero” at “x = −1”, “y = 0” and places the blinking cursor at that point. So, x = −1 is the first root we’ve found.

To find the other two roots, “simply” repeat steps 7 through 13—two more times. You should find that the other two roots are x = 0 and x = 1. Altogether, the three roots are x = −1, 0, 1. Based on these and what the graph looks like, we conclude:

242

x > sin (0.5πx) ⇐⇒ x − sin (0.5πx) > 0 ⇐⇒ x ∈ (−1, 0) ∪ (1, ∞).

Remark 29.

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Example 551. Solve x > e + ln x.

As usual, first rewrite the inequality into SF: x − e − ln x > 0.

1. Graph y = x − e − ln x on your TI84 (precise instructions omitted).

We see that there’s clearly an x-intercept at around x ∈ (4, 5). (Note that by default, each of the little tick marks shown on your TI84 marks 1 unit.) 2. Zoom in (precise instructions omitted). Now we see that there’s probably also an x-intercept near the origin. But unfortunately, now we can no longer see the other x-intercept. To fix this: 3. Press WINDOW to bring up the WINDOW menu. We will adjust Xmin and Xmax:

4. Press 0 . We have adjusted Xmin to 0. Next, 5. Press ENTER 5 . We have adjusted Xmax to 5.

6. Now press GRAPH . We can now see the portion of the graph between x = 0 and x = 5.

7. To find the two roots, “simply” go through the steps described in the previous example—twice (precise instructions omitted). You should find that the two roots are x ≈ 0.708, 4.139. Based on these roots and what the graph looks like, we conclude that the inequality’s solution set is x ∈ (0, 0.708 . . . ) ∪ (4.139 . . . , ∞) = R+ ∖ (0.708 . . . , 4.139 . . . ). Step 1.

Step 2.

Step 5.

Step 6.

Step 3.

Step 4.

Step 7.

Exercise 175. Use a GC to solve each inequality: (a) x3 − x2 + x − 1 > ex .

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(b)

√ x > cos x.

1 > x3 + sin x. 2 1−x

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38.

Extraneous Solutions 38.1.

Example 552. Solve x = a

Let’s try these three steps:

Squaring

2 − x (x ≤ 2).

1. Square both sides: x2 = 2 − x. 2. Rearrange and factorise: x2 + x − 2 = (x − 1)(x + 2) = 0. 3. Conclude: x = 1 or x = −2. b

c

√ Observe that x = 1 does indeed solve =: 1 = 2 − 1 = 1 = 1. √ √ c a But x = −2 does not: −2 = 2 − (−2) = 4 = 2. b

1

a

3 7

So, the above steps are wrong because they produce the extraneous solution x = −2. c

It turns out that this extraneous solution was introduced in Step 1, where we applied the squaring operation. To see this more clearly, let us be more explicit about our above chain of reasoning. In particular, let us use the logical operators ⇐⇒ (“is equivalent to”) and Ô⇒ (“implies”): x= a

Ô⇒ 1

√ 2−x

x2 = 2 − x

⇐⇒ (x − 1)(x + 2) = 0 2

⇐⇒ x = 1 or x = −2. 3

b

c

We now see clearly how Step 1 differs from Steps 2 and 3. Step 1 is a “ Ô⇒ ” statement, while Steps 2 and 3 are “ ⇐⇒ ” statements. Or in plainer English, the squaring operation in Step 1 is an irreversible operation. It is generally true that:

a=b

However, the converse is false. That is:

Ô⇒

a2 = b2 .

a2 = b2

Ô⇒ /

a = b.

For example, (−1) = 12 , but −1 ≠ 1. So, squaring is an example of an irreversible operation; it produces an “ Ô⇒ ” statement and not a “ ⇐⇒ ” statement. 2

Thus, all our above chain of reasoning does is to produce the following (true) implication: b c a √ x = 2−x Ô⇒ x = 1 OR x = −2. (If you’re puzzled as to why the above implication is true, review Ch. 3.2.)

We √ must now check, on a case-by-case basis, whether each of x = 1 OR x = −2 solves a b c x = 2 − x. And when we do check, we find that x = 1 does, while x = −2 does not and is thus an extraneous solution that must be discarded. b

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The following (very) informal Theorem is our moral of the story. It describes how, when, and why extraneous solutions may arise: Theorem 12 (informal). (Extraneous Solutions Theorem) If our chain of reasoning contains only ⇐⇒ ’s, then all is well. However, if it contains even one Ô⇒ (i.e. an irreversible step), then extraneous solutions may arise and we must be careful to check for them. Note the emphasis on the word may. Extraneous solutions may arise but might not. The operation of squaring is merely one example of when extraneous solutions may be introduced. Two other examples of operations that may do likewise are those of multiplying by zero and removing logarithms:

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38.2. Example 553. To solve

Multiplying by Zero

x2 − 3x 1 a + 2 + = 0, we try these three steps:243 2 x −1 x−1

1. Multiply by x2 − 1: x2 − 3x + 2 (x2 − 1) + (x + 1) = 0.

2. Rearrange and factorise: 3x2 − 2x − 1 = (x − 1)(3x + 1) = 0. 3. Conclude: x = 1 or x = −1/3. b

c

Exercise: Verify that x = −1/3 solves =, while x = 1 does not.244 c

a

b

So, x = 1 is an extraneous solution and the above steps are wrong. Where lies the error? b

Again, to clearly detect the error, let us write out our chain of reasoning more explicitly with the aid of the logical relations Ô⇒ and ⇐⇒ : 1 a x2 − 3x + 2 + =0 x2 − 1 x−1

x2 − 3x + 2 (x2 − 1) + (x + 1) = 0

Ô⇒ 1

⇐⇒

3x2 − 2x − 1 = (x − 1)(3x + 1) = 0

2

⇐⇒

x=1

3

b

or

x = −1/3. c

Now, why is Ô⇒ an irreversible operation? The reason is that we’re multiplying by some unknown quantity x2 − 1 which might be equal to zero. 1

And multiplying by zero is an irreversible operation. In general, y = z Ô⇒ 0 ⋅ y = 0 ⋅ z,

1 = 1 Ô⇒ 0 ⋅ 1 = 0 ⋅ 1,

For example:

but but

0 ⋅ y = 0 ⋅ z Ô⇒ / y=z

0 ⋅ 2 = 0 ⋅ 3 Ô⇒ / 2 = 3.

So, our above chain of reasoning yields the following (true) implication: x2 − 3x 1 a +2+ =0 2 x −1 x−1

Ô⇒

x = 1 or x = −1/3. b

c

We must then check, case-by-case, whether each of our final solutions actually solves the c 1 b original equation. In this example, we find that x = −1/3 solves =, while x = 1 does not and is an extraneous solution that must be discarded.

The problem of multiplying by zero is thus dual to the problem of dividing by zero, which we discussed earlier in Ch. ??. Dividing by zero may cause us to lose (valid) solutions; while multiplying by zero may introduce extraneous (or invalid) solutions. This example was stolen from Manning (1970, p. 170). 2 1 1/9 + 1 3 5 3 a c a (−1/3) − 3 (−1/3) 244 Plug x = −1/3 into =: +2+ = + 2 − = − + 2 − = 0. 2 −1/3 − 1 1/9 − 1 4 4 4 (−1/3) − 1 243

In contrast, x = 1 does not solve = because x − 1 = 0, so that some terms in = are undefined. b

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a

3

a

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38.3.

Removing Logs

Example 554. To solve log x + log (x + 1) = log (2x + 2), we try these four steps: a

1. Use a Logarithm Law: log x + log (x + 1) = log (x2 + x) = log (2x + 2). a

2. Remove logs: x2 + x = 2x − 2. 3. Rearrange and factorise: x2 − x + 2 = (x + 1)(x − 2) = 0. 4. Conclude: x = −1 or x = 2. b

c

Exercise: Verify that x = 2 solves = but x = −1 does not.245 c

a

b

So, x = 1 is an extraneous solution and the above steps are wrong. Where lies the error? b

Again, to clearly detect the error, let us write out our chain of reasoning more explicitly: log x + log (x + 1) = log (2x + 2) a

⇐⇒

log x + log (x + 1) = log (x2 + x) = log (2x + 2)

⇐⇒

x2 − x + 2 = (x + 1)(x − 2) = 0

1

Ô⇒ 2

a

x2 + x = 2x − 2

3

⇐⇒

x = −1 or x = 2.

4

b

c

This time, Step 2 is the irreversible step. In general, we have log a = log b Ô⇒ a = b,

a = b Ô⇒ / log a = log b. 5

but

If a = b is non-positive, then log a or log b is undefined, so that = is necessarily false. 5

Here in this brief chapter, we have merely examined three examples of operations by which extraneous solutions may be introduced, namely squaring, multiplying by zero, and removing logarithms. These are not exhaustive and you will likely encounter more of such operations as your maths education progresses. The important thing is to remember how and why extraneous solutions arise. In particular, you should remember and understand the Extraneous Solutions Theorem.

245

Plug x = 2 into =: log 2 + log (2 + 1) = log 2 + log 3 = log 6 = log (2 ⋅ 2 + 2) = log 6. c

a

a

In contrast, x = −1 does not solve = because log −1 is not even defined. b

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Exercise 176. Suppose we are given that x ∈ [0, 2π) and sin x + cos x = 1. We attempt 1 to solve = with the steps below. Identify any errors in these steps and give the correct solution.246 (Answer on p. 1791.) 1

1. Square both sides: (sin x + cos x) = sin2 x + cos2 x + 2 sin x cos x = 12 = 1. 2. Apply the identity sin2 x + cos2 x = 1 to get 2 sin x cos x = 0. 1

2

3. Hence, sin x = 0 or cos x = 0.

4. Conclude: x = 0 or x = π or x = π/2 or x = 3π/2. 2

3

4

5

√ √ 1 Exercise 177. (Tricky.)247 Suppose we are given that x ∈ R and x2 x + x2 + x + 1 = 0. We have the following, seemingly watertight chain of reasoning: √ √ 1 x2 x + x2 + x + 1 = 0 √ (x2 + 1) ( x + 1) = 0 √ x2 + 1 = 0 or x + 1 = 0 √ N.A. or x = −1

⇐⇒ 1

⇐⇒ 2

⇐⇒ 3

⇐⇒

x = 1.

4

2

Verify that x = 1 does not solve =. Then identify any errors in the above chain of reasoning and give the correct solution. (Answer on p. 1791.) 2

1

Exercise 178. (Even trickier.)248 Suppose we are given that x ∈ R and x2 + x + 1 = 0. We have the following, seemingly watertight chain of reasoning: 1

x2 + x + 1 = 0. 1

⇐⇒ 1

⇐⇒ 2

⇐⇒ 3

⇐⇒ 4

Rearrange: x2 = −x − 1. 2

Since x = 0 doesn’t solve = anyway, we know that x ≠ 0. 1 2 3 So, we can divide = by x to get: x = −1 − . x 1 3 1 4 Now plug = into = to get: x2 + (−1 − ) + 1 = 0. x x = 1.

1

5

Verify that x = 1 solves = but not =. Then identify any errors in the above chain of reasoning and give the correct solution. (Answer on p. 1791.) 5

4

1

Stolen from Sullivan (Precalculus, 10e, 2017, p. 519), hat tip to Adapted from . 248 Stolen from . 246

.

247

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39.

O-Level Review: The Derivative 39.1.

In general, the gradient or derivative of a graph at a point is the gradient of the tangent line at that point. Informally, a tangent line (from the Latin for touching line) is a line that just touches the graph at that point.249 Example 555. Graphed below is the equation y = x2 + 1.

The red line is tangent to the graph at the point (−2, 5). It can be shown that this line’s gradient is −4. Therefore, the graph’s gradient at this point is −4.

y

y = x2 + 1

(−2, 5)

x

The blue line is tangent to the graph at the point (0, 1). It can be shown that this line’s gradient is 0. Therefore, the graph’s gradient at this point is 0.

The green line is tangent to the graph at the point (3, 10). It can be shown that this line’s gradient is 6. Therefore, the graph’s gradient at this point is 6.

249

For the formal definition of a tangent line, see Definition 211 (Part V: Calculus).

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Example 556. Consider the graph of y = x3 + 1.

The red line is tangent to the graph at the point (−2, −7). It can be shown that this line’s gradient is 12. Therefore, the graph’s gradient at this point is 12.

(−2, −7)

(0, 1)

y = x3 + 1

The blue line is tangent to the graph at the point (0, 1). It can be shown that this line’s gradient is 0. Therefore, the graph’s gradient at this point is 0.

The green line is tangent to the graph at the point (3, 28). It can be shown that this line’s gradient is 27. Therefore, the graph’s gradient at this point is 27.

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39.2.

Informally, the derivative—variously denoted as dy/dx, f ′ , and f˙—is the gradient. A little more formally, it is the function that tells us what a graph’s gradient is at each point. We’ll learn more about this in Part V. But for now, some quick examples: Example 557. Define f ∶ R → R by f (x) = 5x. The graph of f is simply the graph of the equation y = 5x. We don’t yet know how, but it’s possible to show that at every point, this graph’s gradient is equal to 5. We write this as dy = f ′ (x) = 5 dx

for all x.

So, for example, at x = −2, 0, 3, and indeed any other point, the graph’s gradient is 5.

Figure to be inserted here.

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Example 558. Define g ∶ R → R by g (x) = x2 + 1. The graph of g is simply the graph of y = x2 + 1. We don’t yet know how, but it’s possible to show that at every point, this graph’s gradient is equal to 2x: So, for example,

R dy RRRR = g ′ (−2) = 2 ⋅ (−2) = −4, RRR dx RR Rx=−2

dy = g ′ (x) = 2x. dx

R dy RRRR RRR = g ′ (0) = 2 ⋅ 0 = 0, dx RR Rx=0

R dy RRRR RRR = g ′ (3) = 2 ⋅ 3 = 6. dx RR Rx=3

That is, at each of the points x = −2, 0, or 3, the graph’s gradient is −4, 0, and 6. R dy RRRR Formally, RRR = g ′ (a) is read aloud as “the derivative of y with respect to x, evaluated dx RR Rx=a at a” or “the derivative of g at a”.

A little less formally, we can simply read it as “the gradient at a”.

Figure to be inserted here.

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Example 559. Define h ∶ R → R by h (x) = x3 + 1. The graph of h is simply the graph of y = x3 + 1. We don’t yet know how, but it’s possible to show that at every point, this graph’s gradient is equal to 3x2 : So, for example,

dy = h′ (x) = 3x2 . dx

R R R dy RRRR dy RRRR dy RRRR 2 ′ ′ 2 = h (−2) = 3 ⋅ (−2) = 12, RRR RRR = h (0) = 3 ⋅ 0 = 0, RRR = h′ (3) = 3 ⋅ 32 = 27. dx RR dx RR dx RR Rx=−2 Rx=0 Rx=3

That is, at each of the points x = −2, 0, or 3, the graph’s gradient is 12, 0, or 27.

Figure to be inserted here.

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39.3.

The Derivative as Rate of Change

dx may be interpreted as: dt • The rate of change of x with respect to t; and • The change in x resulting from a small unit change in t. Recall that

Example 560. A particle P travels along a line. Its eastward displacement x (metres) from the point O at time t (seconds) is given by x (t) = t3 − 6t2 + 9t.

1

O

2

3

4

x (metres)

• At t = 0, P starts x = 03 − 6 ⋅ 02 + 9 ⋅ 0 = 0 metres east of O. In other words, it is at O. And during t ∈ [0, 1), P travels eastwards away from O. • At t = 1, P stops and is x = 13 − 6 ⋅ 12 + 9 ⋅ 1 = 4 metres east of O. • During t ∈ (1, 3), it travels westwards, i.e. back towards O. For example, at t = 2, P is x = 23 − 6 ⋅ 22 + 9 ⋅ 2 = 2 metres east of O.

• At t = 3, P has returned to O (x = 33 − 6 ⋅ 32 + 9 ⋅ 3 = 0) and stops. • During t > 3, P keeps travelling eastwards away from O. For example, at t = 4, P is x = 43 − 6 ⋅ 42 + 9 ⋅ 4 = 4 metres east of O. And at t = 10 (not depicted in the figures), P is x = 103 − 6 ⋅ 102 + 9 ⋅ 10 = 490 metres east of O.

x (m)

5 4 3 2 1

t (s) 1

2

3

4

(Example continues on the next page ...)

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(... Example continued from the previous page.) We define velocity v (metres per second) to be the rate of change of displacement with respect to (w.r.t.) time. In other words, velocity is the (first) derivative of displacement w.r.t. time: v=

dx . dt

Using rules of differentiation that we’ll review in Ch. 39.5, it is possible to work out that P ’s (eastward) velocity is given by

11

v (t) = 3t2 − 12t + 9.

v (m s−1)

9 7 5 3 1 -1

1

2

3

4

t (s)

-3 At each instant of time, v tells us what P ’s velocity is—in other words, the rate at which x is changing per “infinitesimally small” unit of time t. If v > 0, then P is travelling eastwards. And if v < 0, then P is travelling westwards.

From the above graph, we can tell that

• During t ∈ [0, 1), P travels eastwards. • At t = 1, P stops.

• During t ∈ (1, 3), P travels westwards. • At t = 3, P stops.

• During t > 3, P travels eastwards.

(Example continues on the next page ...)

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(... Example continued from the previous page.) We define acceleration a (metres per second per second) to be the rate of change of velocity w.r.t. time. In other words, acceleration is the (first) derivative of velocity w.r.t. time and thus the second derivative of displacement w.r.t. time: v=

dv d2 x = . dt dt2

Again, it is possible to work out that P ’s (eastward) acceleration is given by

12 10 8 6 4 2 -2 -4 -6 -8 -10 -12

a (t) = 6t − 12.

a (m s−2)

1

2

3

4

t (s)

At each instant of time, a tells us what P ’s acceleration is—in other words, the rate at which v is changing per “infinitesimally small” unit of time t. If a > 0, then P ’s eastwards velocity is increasing (or equivalently, its westwards velocity is decreasing). And if a < 0, then P ’s eastwards velocity is decreasing (or equivalently, its westwards velocity is increasing) From the above graph, we can tell that

• During t ∈ [0, 2), P ’s eastwards velocity is increasing (or equivalently, its westwards velocity is decreasing). • During t > 2, P ’s eastwards velocity is decreasing (or equivalently, its westwards velocity is increasing)

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39.4.

Differentiability vs Continuity

In Ch. 16, we learnt that informally, continuity means you can draw the graph without lifting your pencil. We now introduce the concept of differentiability, which turns out to be a stronger condition than continuity. Informally, a function is differentiable if its graph is “smooth” and, in particular, doesn’t have “kinks”.250 Example 561. The function f ∶ R → R defined by f (x) = x5 − x2 − 1 is continuous everywhere, because you can draw its entire graph without lifting your pencil.

y

x f is differentiable everywhere because its graph is “smooth” and has no “kinks”.

The function f is also differentiable everywhere, because it is “smooth” everywhere and has no “kinks”. Example 562. The sine function sin is both continuous everywhere and differentiable everywhere.

y sin is differentiable everywhere because its graph is “smooth” and has no “kinks”.

x

250

We will formally define the concept of differentiability in Part V (Calculus).

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Example 563. The exponential function exp is both continuous everywhere and differentiable everywhere.

y

exp is differentiable everywhere because its graph is “smooth” and has no “kinks”.

x The above examples may leave you wondering, “So aren’t continuity and differentiability just the same thing?” Well, it turns out that every differentiable function must also be continuous.251 (This is exactly what we meant when we said that differentiability is a stronger condition than continuity.) However, the converse is false—not every continuous function is differentiable; that is, continuity does not imply differentiability. Differentiable functions are thus a subset of continuous functions. Beautiful Venn diagram drawn by an artistic genius:

All functions

Differentiable functions Continuous Functions

Let’s now look at some examples of functions that are continuous but not differentiable. The classic example is the absolute value function:

251

This assertion is formally stated and proven as Theorem 28 (Appendices).

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Example 564. The absolute value function ∣⋅∣ is continuous everywhere, because you can draw its entire graph without lifting your pencil. However, it is not differentiable everywhere, because it is not “smooth” everywhere. In particular, it has a “kink” at x = 0.

We can say though that the absolute value function is differentiable everywhere except at x = 0. Or equivalently, it is differentiable on R/ {0}.

y

∣⋅∣ is not differentiable everywhere because it has a kink at x = 0.

∣⋅∣ is differentiable on R ∖ {0}.

x

Example 565. Define the piecewise function g ∶ R → R by ⎧ ⎪ ⎪ ⎪−x, for x ≤ 0 g (x) = ⎨ ⎪ 2 ⎪ ⎪ ⎩x , for x > 0.

The function g is continuous everywhere, because you can draw its entire graph without lifting your pencil.

However, it is not differentiable everywhere because, like ∣⋅∣, it has a “kink” at x = 0. Nonetheless, again, we can say that g is differentiable everywhere except at x = 0. Or equivalently, g is differentiable on R/ {0}.

y

g g is not differentiable everywhere because it has a kink at x = 0.

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g is differentiable on R ∖ {0}.

x

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Any function that’s differentiable must also be continuous. Therefore, by the contrapositive, any function that’s non-continuous must also be non-differentiable: Example 566. The tangent function tan is not continuous everywhere. Thus, it is not differentiable everywhere either.

tan is not differentiable everywhere.

y tan is differentiable π π on (− , ). 2 2 −

x

π 2

π 2

π π It is however both (i) continuous and (ii) differentiable on the interval (− , ). This is 2 2 because this interval (i) can be drawn without lifting your pencil; and (ii) is “smooth” and has no “kinks”. Indeed, tan is both continuous and differentiable on every interval: 1 1 ((k − ) π, (k + ) π), for k ∈ Z. 2 2

Happily, most functions we’ll encounter in A-Level maths will be both continuous and differentiable. There are however exceptions, as we’ve seen here and in Ch. 16. Exercise 179. Graphed below are three functions f , g, and h. State if each is continuous everywhere and/or differentiable everywhere. If not, state the set of points on which each function is continuous or differentiable. (Answer on p. 1766.)

y

g

h

y

y

x f

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x

x

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39.5.

Rules of Differentiation

Theorem 13 below informally states several Rules of Differentiation252 that you should find familiar from secondary school. In Part V (Calculus), we will explain a little more where these rules come from. For now, you need merely “know” these rules and how to use them to “solve” differentiation problems. Theorem 13 (informal). ( Rules of Differentiation) Let c be a constant and x, y, and z be variables. Then Constant Rule

Constant Factor Rule

Power Rule

d C c=0 dx

d F dy (cy) = c dx dx

d c P c−1 x = cx dx

Sum and Difference Rules

Product Rule

d dz ± dy (y ± z) = ± dx dx dx

d dz × dy (yz) = z +y dx dx dx Sine

Cosine

dy dz d y ÷ z dx − y dx = dx z z2

d sin x = cos x dx

d cos x = − sin x dx

1 d ln x = dx x

d x x e =e dx

Quotient Rule

Natural Logarithm

Exponential

(In case you’re wondering, the Chain Rule is the topic of the next subchapter.) Here’s a common mnemonic for the Quotient Rule: Lo-D-Hi minus Hi-D-Lo, Cross over and square the Lo.

252

For a formal statement, see Proposition XXX (Appendices).

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Example 567. Example 568. Example 569. Example 570. Example 571. Example 572. Example 573. Example 574. Example 575.

d C 5 = 0. dx

(Constant Rule)

d C 500 = 0. dx

(Constant Rule)

d C (−200) = 0. dx

(Constant Rule)

d d F (5x3 ) = 5 ( x3 ) = 5 (3x2 ) = 15x2 . dx dx

(Constant Factor Rule)

d d F (500x0.3 ) = 500 ( x0.3 ) = 500 (0.3x−0.7 ) = 150x−0.7 . dx dx d d F (−200x−1 ) = −200 ( x−1 ) = −200 (−x−2 ) = 200x−2 . dx dx d 3P 2 x = 3x . dx

d 0.3 P x = 0.3x−0.7 . dx

(CFR) (CFR) (Power Rule) (Power Rule)

d −1 P −2 x = −x . dx

(Power Rule)

Example 576. Sum and Difference Rules: d (x3 + 500x0.3 ) dx d (x3 − 500x0.3 ) dx

d 3 d (500x0.3 ) = 3x2 + 150x−0.7 . x + dx dx d ± d 3 (500x0.3 ) = 3x2 − 150x−0.7 . = x − dx dx ±

=

Example 577. Product Rule and Sine:

d × (x3 sin x) = 3x2 sin x + x3 cos x. dx

Example 578. Quotient Rule and Cosine:

d x3 ÷ 3x2 cos x − x3 (− sin x) x2 = = (3 cos x + x sin x). dx cos x cos2 x cos2 x

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Example 579. Combination of various rules: d e2x d ex ex F 1 d ex ex = = dx ln x2 dx 2 ln x 2 dx ln x

d d (ex ex ) − (ex ex ) dx ln x 1 ln x dx = 2 2 (ln x)

Q

1 ln x (ex ex + ex ex ) − (ex ex ) x1 = 2 2 (ln x) P

1 e2x 2 ln x − x1 e2x = (1 − ). = 2 2 (ln x) ln x 2x ln x

R dy dy RRRR for each of the following. R Exercise 180. Find and dx dx RRRR Rx=0

(a) y = x2 .

(b) y = 3x5 − 4x2 + 7x − 2. (c) y = (x2 + 3x + 4) (3x5 − 4x2 + 7x − 2).

Exercise 181. For each of the following, find

(a) y = ex ln x.

(b) y = x2 ex ln x. sin x (c) y = . x

(d) y = tan x, given that tan x =

dy without using the chain rule. dx

sin x and sin2 x + cos2 x = 1. cos x

1 (e) y = , where z is a variable that can be expressed in terms of x. (Leave your answer z dz in terms of z and .) dx

Use (e) to solve (f), (g) and (h): (f) y = cosecx, where cosecx =

1 . sin x

1 . cos x 1 (h) y = cot x, where cot x = . tan x (g) y = sec x, where sec x =

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39.6.

The Chain Rule

You will, of course, recall the Chain Rule from secondary school. Here we will merely state it informally, as we did in secondary school:253 Theorem 14 (informal). (Chain Rule) Let x, y, and z be variables. Then dz dz dy = ⋅ . dx dy dx

One mnemonic is to think of the derivatives on the RHS as fractions—in which case, the dy’s get cancelled out and we’re left with dz/dx. (In Part V, we’ll explain why this is merely a mnemonic and why it is wrong to think of derivatives as fractions.) The Chain Rule has this informal interpretation: The change in z caused by = The change in z caused by × The change in y caused by a small unit change in x a small unit change in y a small unit change in x.

This interpretation makes (common) sense:

Example 580. Say that if I add to a cup of water 1 g of Milo, its water volume increases by 2 cm3 . And if the cup’s water volume increases by 1 cm3 , its water level rises by 0.3 cm. Then by common sense, if I add 1 g of Milo, its water level should rise by 2 × 0.3 = 0.6 cm.

Let’s now rewrite the above common-sense observations more formally:

Let x be the mass (g) of Milo in a cup of water, y be the total volume (cm3 ) of water in the cup, and z be the water level (cm) in the cup. • When x increases by 1 g, y increases by 2 cm3 . Formally,

dy = 2 cm3 g−1 . dx

• When y increases by 1 cm−3 , z increases by 0.3 cm. Formally,

dz = 0.3 cm cm−3 = 0.3 cm−2 . dy

• And so, by the chain rule, when x increases by 1 g, z increases by 2 × 0.3 = 0.6 cm. Formally,

253

dz dz dy = = 0.3 cm−2 × 2 cm3 g−1 = 0.6 cm g−1 . dx dy dx

For a formal statement, see Theorem XXX (Appendices).

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Examples of how to “use” the chain rule: Example 581. Let y = esin x . Then

Example 582. Let y =

dy desin x Ch desin x dsin x = = = esin x cos x. dx dx dsin x dx

4x − 1. Then

√ √ dy d 4x − 1 Ch d 4x − 1 d(4x − 1) 2 1 ×4= √ . = = = √ dx dx d(4x − 1) dx 2 4x − 1 4x − 1

A slightly more complicated example, where we use the Chain Rule more than once: Example 583. Let y = [sin (2x − 3) + cos (5 − 2x)] . Then 3

dy d [sin (2x − 3) + cos (5 − 2x)] = dx dx

3

d [sin (2x − 3) + cos (5 − 2x)] d[sin (2x − 3) + cos (5 − 2x)] = d[sin (2x − 3) + cos (5 − 2x)] dx

Ch

3

= 3 [sin (2x − 3) + cos (5 − 2x)] [ 2

d sin (2x − 3) d cos (5 − 2x) + ] dx dx

= 3 [sin (2x − 3) + cos (5 − 2x)] [

Ch

2

d sin (2x − 3) d (2x − 3) d cos (5 − 2x) d (5 − 2x) + ] d (2x − 3) dx d (5 − 2x) dx

= 3 [sin (2x − 3) + cos (5 − 2x)] [cos (2x − 3) × 2 + (− sin (5 − 2x)) × (−2)] 2

= 6 [sin (2x − 3) + cos (5 − 2x)] [cos (2x − 3) + sin (5 − 2x)]. 2

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R dy dy RRRR Exercise 182. For each, find and R . dx dx RRRR Rx=0 (a) y = 1 + [x − ln (x + 1)] . 2

(Answer on p. 1767.) (b) y = sin

x

2.

1 + [x − ln (x + 1)]

Exercise 183. Let F, m, v, t, and p denote force, mass, velocity, time, and momentum. Momentum is defined as the product of mass and velocity. (a) Newton’s Second Law of Motion states that the rate of change of momentum (of an object) is equal to the force applied (to that object). Write this down as an equation. (b) Acceleration a is defined as the rate of change of momentum. Explain why Newton’s Second Law simplifies to F = ma if mass is constant. (Answer on p. 1767.) Exercise 184. In Part V (Calculus), Fact 199, we will formally state and prove that d 1 1 ln x = . dx x

d exp x = exp x: dx d (a) Use the Chain Rule to write down an expression for ln (exp x). dx (b) What do you observe about the expression ln (exp x)? Use this observation to write d ln (exp x). down another expression for dx d (c) Then conclude that exp x = exp x. (Answer on p. 1768.) dx But assuming = is true, we can quite easily prove that 1

Remark 81. By the way, the following five derivatives are in List MF26 (p. 3), so no need to mug. (We’ll review the inverse trigonometric functions sin−1 , cos−1 , and tan−1 in Ch. 32. Also, we’ll restate the following as Fact 185 in Part V.) 1

sin −1 x

cos −1 x

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1− x 2 −

1 1− x 2

tan −1 x

1 1 + x2

cosec x

– cosec x cot x

sec x

sec x tan x

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39.7.

Increasing and Decreasing

Definition 108. We say that a function is • Increasing (at a certain point) if its derivative (at that point) is positive; and • Decreasing (at a certain point) if its derivative (at that point) is negative. Example 584. The function f ∶ R → R defined by f (x) = x2 is decreasing on R− and increasing on R+ . It is neither decreasing nor increasing at x = 0.

y

f

f ′ < 0 on R−

y x

f ′ > 0 on R+

g′ = 0 at x = 0

g

f′ = 0 at x = 0

g ′ > 0 on R−

g ′ < 0 on R+

x The function g ∶ R → R defined by g (x) = −x2 is increasing on R− and decreasing on R+ . It is neither increasing nor decreasing at x = 0. Example 585. The function h ∶ R → R defined by h (x) = x is increasing everywhere.

y

h′ > 0 everywhere

x h

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39.8.

Stationary and Turning Points

A stationary point (of a function) is simply any point at which that function’s derivative equals zero: Definition 109. A point x is a stationary point of a function f if f ′ (x) = 0. Example 586. Consider f ∶ R → R defined by f (x) = x2 .

The derivative of f is the function f ′ ∶ R → R defined by f ′ (x) = 2x.

We now check if f has any stationary points. By the above definition, a is a stationary point of f if and only if ⋆

f ′ (a) = 0

⇐⇒

2a = 0

⇐⇒

a = 0.

Hence, f has only one stationary point, namely 0. (We’ve labelled this point as K in the graph below.)

y f

x

K When searching for stationary points, we will often write

f ′ (a) = 0.

Because = appears so often, it has a name: the First-Order Condition (FOC).

Informally and intuitively, a turning point (of a function) is where the graph (of that function) “turns”. Formally, Definition 110. A turning point (of a function) is a point that’s both a stationary point and a strict local maximum or minimum (of that function).

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Example 587. Continue with the above example where we define f ∶ R → R by f (x) = x2 .

Above we showed that f has only one stationary point, namely 0 (labelled as K in the graph).

y f

K

x

Observe that K is also a strict local minimum of f . Since K is both a stationary point and a strict local minimum of f , by the above definition, K is a turning point of f . Remark 82. The term turning point is rarely used by mathematicians. However, it appears on your O- and A-Level syllabuses and exams.254 We shall therefore have to use it. Unfortunately, I am actually not sure what precisely the MOE or your examiners mean by turning point. In our above definition of a turning point (Definition 110), I have merely given my best guess of what I think they mean by this term.

254

The term turning point appears on your H2 Maths syllabus (pp. 5, 15) and these A-Level exam questions: N2016/I/3, N2014/I/4, N2008/I/9, N2015/I/11, N2012/I/8, N2010/I/6, N2010/II/3, N2009/1/11.

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Example 588. Consider g ∶ R → R defined by g (x) = −x2 .

Then D = (0, 0) is a stationary point the derivative (or gradient) of g at K is zero: g ′ (0) = 0.

y

x

D g

Moreover, D is also a turning point because it is both a stationary point and a strict extremum (in particular, it is a strict local maximum.) By definition, every turning point is a stationary point. However, the converse is false—that is, a stationary point need not be a turning point.

All points

Stationary points Turning points

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Example 589. Define f ∶ R → R by f (x) = x3 .

The origin (0, 0) is a stationary point of f , because the derivative at that point is zero.

However, (0, 0) is not a turning point because it is not a strict extremum.

y

x f

(0, 0) is a stationary point but not a turning point

As we’ll learn in Part V (Calculus), (0, 0) is actually an example of an inflexion point.

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Example 590. Define j ∶ R → R by

⎧ ⎪ ⎪ −81, for x ≤ −3, ⎪ ⎪ ⎪ ⎪ j (x) = ⎨0.75x5 − 3.75x4 − 5x3 + 30x2 , for − 3 < x < 5, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ for x ≥ 5. ⎩125,

y

C = (−2, 76) j

G = (5, 125)

E = (2, 44) D = (0, 0)

A = (−4, −81)

B = (−3, −81)

H = (6, 125)

x F = (4, −32)

The derivative (or gradient) of j at each of the points A, C, D, E, F , and H is zero. Hence, each of these points is a stationary point. Observe that moreover, C and E are strict maximum points, while D and F are strict local minimum points. Hence, the stationary points C, D, E, and F are also turning points. In contrast, A and H, which are not strict extrema, are not turning points. The derivative of j at each of the points B and G is not equal to zero. Indeed and as we’ll learn later on in Part V (Calculus), the derivative of j at each of those points does not even exist! Hence, neither B nor G is a stationary point. Since B and G are not stationary points, they are not turning points either. (By the way, is j continuous everywhere? Differentiable everywhere?)255

255

The function j is continuous everywhere, but not differentiable everywhere. It is differentiable everywhere except at B and G.

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Example 591. The sine function has infinitely many stationary and turning points. For every integer k, the point Ik = ((0.5 + k) π, 1) is both a stationary and a turning point.

I−2 = (−

3π , 1) 2

y

π I0 = ( , 1) 2

I2 = (

5π , 1) 2

sin

x

I−3 = (−

5π , −1) 2

π I−1 = (− , −1) 2

I1 = (

3π , −1) 2

Note that at every even integer k, the point Ik = ((0.5 + k) π, 1) is a strict local maximum. While at every odd integer l, the point Il = ((0.5 + 2l) π, 1) is a strict local minimum.

Now, here’s a subtle but important point. Every turning point is either a strict local maximum or minimum. But the converse is false—that is, a strict local maximum or minimum need not be a turning point.

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Example 592. The function i ∶ [−1, ∞) → R defined by i (x) = x has a strict local minimum at I = (−1, −1). (Indeed, I is also a strict global minimum.)

y

i

x I = (−1, −1) However, the derivative of i at I is not equal to zero. Indeed, the derivative of i at I does not even exist. Hence, I is not a stationary point and cannot be a turning point either.

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Example 593. Let A = (−5, 4), B = (1, 1), and C = (2, −3) be points. The graph G = {A, B, C} simply consists of these three isolated points. As explained earlier in Example 213, each of A, B, and C is both a strict local maximum and a strict local minimum of G.

y

A = (−5, 4)

4 2

-6

-4

-2

B = (1, 1)

2

4

x

-2 -4

C = (2, −3)

However, the graph G is nowhere-differentiable. And so, it can have neither stationary -6 points nor turning points.

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3 1 Example 594. Define f ∶ [− , ] → R by f (x) = x5 + 2x4 + x3 . 2 2

Figure to be inserted here.

Five points, A–E, are labelled and classified in the table below.

Global maximum Strict global maximum Local maximum Strict local maximum Global minimum Strict global minimum Local minimum Strict local minimum Turning point Stationary point

A B C D E 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Observe that D is a stationary point but not a turning point. (As we shall learn later in Chapter 92, D is an example of an inflexion point.) The points A and E are a minimum and a maximum point. However, neither is a turning point.

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Exercise 185. For each of the following functions, identify all turning points.

(a) f ∶ R → R defined by f (x) = x2 + 1. (b) g ∶ [−1, 1] → R “ g (x) = x2 + 1 (c) h∶R→R “ h (x) = cos x. (d) i ∶ [−1, 1] → R “ i (x) = cos x.

Exercise 186. Explain whether each of the points A–H in the graph below is a turning and/or stationary point. (Answer on p. 1768.)

y

E

A B

C

D

F

G

Assume the graph is “smooth” everywhere except at two points.

H

x

Exercise 187. Is each of the following statements true or false? If true, explain why. If false, give a counterexample from Example 594.) (Answer on p. 1769.) (a) (b) (c) (d) (e) (f)

Every Every Every Every Every Every

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maximum point or minimum point is a stationary point. maximum point or minimum point is a turning point. stationary point is a maximum point or minimum point. turning point is a maximum point or minimum point. turning point is a stationary point. stationary point is a turning point.

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40.

Conic Sections

In this chapter, we’ll study the graphs of the following eight equations. All eight are examples of conic sections.256 We first study the unit circle and ellipse centred on the origin: x2 y 2 + =1 a2 b2

x2 + y 2 = 1 We then study six types of hyperbolae: y=

1 x

y 2 x2 − =1 b2 a 2

x2 − y 2 = 1

x2 y 2 − =1 a2 b2

bx + c y= dx + e

ax2 + bx + c y= dx + e

Fun Fact The Greek word hyperbola is closely related to the English word hyperbole, which means an exaggeration or overstatement. By the way, we’ve actually already studied one example of a conic section—this was the graph of the quadratic equation y = ax2 + bx + c, which is a type of conic section called the parabola.

256

For why these are called conic sections, see Ch. 138.19 (Appendices).

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40.1.

The Ellipse x2 + y 2 = 1 (The Unit Circle)

We already discussed the circle in Chs. XXX and XXX. It turns out that the circle is an example of a conic section. And so, for completeness, here we just do a quick recap:

1

y

x2 + y 2 = 1

(0, 1) is a strict global maximum.

−1

1

A = (Ax , Ay ) Ay 1

O

x

Ax

(0, −1) is a strict global minimum.

A2x + A2y = 1

−1

Here are some characteristics of the graph of x2 + y 2 = 1:

1. Intercepts. The y-intercepts are (0, −1) and (0, 1). The x-intercepts are (−1, 0) and (1, 0).

2. Turning points. The two turning points are (0, 1) and (0, −1). (Moreover, they are, respectively, a strict global maximum and a strict global minimum.) 3. Asymptotes. There are no asymptotes.

4. Symmetry. Every line that passes through the origin is a line of symmetry. (And no other line is a line of symmetry.)

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40.2.

x2 y 2 The Ellipse 2 + 2 = 1 a b

Two transformations257 will get us from x2 + y 2 = 1 to (x/a) + (y/b) = 1: 2

2

1. First stretch x2 + y 2 = 1 horizontally, outwards from the y-axis, by a factor of a, to get 2 (x/a) + y 2 = 1.

2. Then stretch (x/a) + y 2 = 1 vertically, outwards from the x-axis, by a factor of b, to get 2 2 (x/a) + (y/b) = 1. 2

Thus, (x/a) + (y/b) = 1 is simply the unit circle stretched horizontally and vertically by factors of a and b. We call this “elongated” or “imperfect” circle an ellipse. Note that this ellipse remains centred on the origin (0, 0). 2

2

y

y 2 x 2 ( ) +( ) =1 a b

−a

(0, b) is a strict global maximum.

Line of symmetry x=0

Line of symmetry y=0

a

x

(0, −b) is a strict global minimum. 1. Intercepts. The y-intercepts are (0, −b) and 0, b. The x-intercepts are (−a, 0) and (a, 0).

2. Turning points. By observation, there are two turning points—(0, b) is a strict global maximum and (0, −b) is a strict global minimum. 3. Asymptotes. By observation, there are no asymptotes. 4. Symmetry. By observation, if a ≠ b, then there are only two lines of symmetry, namely y = 0 (the x-axis) and x = 0 (the y-axis). (Note that if a = b, then this ellipse is in fact a circle and there are again infinitely many lines of symmetry.)

257

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Exercise 188. Graph the equation below (a, b, c, d ∈ R and a, b ≠ 0). Label any turning points, asymptotes, lines of symmetry, and intercepts. (Hint in footnote.)258 (y + d) (x + c) + = 1. a2 b2 2

2

The rest of this chapter will look at six examples of hyperbolae. Our first and also the simplest example of a hyperbola is y = 1/x:

258

To find the y-intercepts, plug in x = 0. To find the x-intercepts, plug in y = 0.

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40.3.

The Hyperbola y =

1 x

All hyperbolae we’ll study will share some common features:

1. There’ll be two branches—y = 1/x has a bottom-left branch and a top-right branch. 2. There may or may not be x- and y-intercepts—y = 1/x has neither.

3. There may or may not be turning points—y = 1/x has none. 4. There’ll be two asymptotes—y = 1/x has horizontal asymptote y = 0, because as x → −∞, y → 0− and as x → ∞, y → 0+ . Also, y = 1/x has the vertical asymptote x = 0, because as x → 0− , y → −∞ and as x → 0+ , y → ∞. A rectangular hyperbola is any hyperbola whose two asymptotes are perpendicular— thus, y = 1/x is an example of a rectangular hyperbola. 5. The hyperbola’s centre is the point at which the two asymptotes intersect259 —y = 1/x has centre (0, 0).

6. There’ll be two lines of symmetry—each (i) passes through the centre; and (ii) bisects an angle formed by the two asymptotes. y = 1/x has two lines of symmetry: y = x and y = −x. Observe that indeed, each (i) passes through the centre; and (ii) bisects an angle formed by the two asymptotes.

y Line of symmetry y = −x

Horizontal asymptote y=0

y=

1 x

Centre (0, 0)

Line of symmetry y=x

x

Vertical asymptote x=0

259

For simplicity, this shall be this textbook’s definition of a hyperbola’s centre. Note though that in the usual and proper study of conic sections, the centre is instead defined as the midpoint of the line segment connecting the two foci. That the two asymptotes intersect at the centre is then a result rather than a definition. However, in H2 Maths, there is no mention of foci and so I thought it better to simply define the centre as the intersection point of the two asymptotes.

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40.4.

The Hyperbola x2 − y 2 = 1

Consider the equation x2 − y 2 = 1. Notice that no x ∈ (−1, 1) satisfy this equation. Hence, this graph contains no points for which x ∈ (−1, 1).

y

Oblique asymptote y=x

Oblique asymptote y = −x

x2 − y 2 = 1

−1

Centre (0, 0)

1

Line of symmetry y=0

x

Line of symmetry x=0

1. There are two branches—one on the left and another on the right.

2. Intercepts. The x-intercepts are (−1, 0) and (1, 0). There are no y-intercepts. 3. There are no turning points.

4. As x → −∞, y → ±x. And as x → ∞, y → ±x. So, x2 − y 2 = 1 has two oblique asymptotes y = ±x.

Since the two asymptotes y = x and y = −x are perpendicular, this is again a rectangular hyperbola. (In fact, we call this an “east-west” rectangular hyperbola.) 5. The hyperbola’s centre is (0, 0).

6. The two lines of symmetry are y = 0 (the x-axis) and x = 0 (the y-axis). Observe that each (a) passes through the centre; and (b) bisects an angle formed by the two asymptotes.

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x2 y 2 The Hyperbola 2 − 2 = 1 a b

40.5.

Consider the equation (x/a) − (y/b) = 1. Notice that again, no x ∈ (−a, a) satisfy this equation. Hence, this graph again contains no points for which x ∈ (−a, a). 2

2

Two transformations will get us from x2 − y 2 = 1 to (x/a) − (y/b) = 1: 2

2

1. Stretch x2 − y 2 = 1 horizontally, outwards from the y-axis, by a factor of a, to get 2 (x/a) − y 2 = 1. 2. Stretch (x/a) − y 2 = 1 vertically, outwards from the x-axis, by a factor of b, to get 2 2 (x/a) − (y/b) = 1. 2

Oblique asymptote

y

Oblique asymptote b y= x a

b y=− x a

Line of symmetry y=0

−a

x

a Centre (0, 0)

Line of symmetry x=0

1. There are two branches—one on the left and another on the right.

2. Intercepts. The x-intercepts are (−a, 0) and (a, 0). There are no y-intercepts. 3. There are no turning points.

4. As x → −∞, y → ±bx/a. And as x → ∞, y → ±bx/a. So, (x/a) − (y/b) = 1 has two oblique asymptotes y = ±bx/a. Since the two asymptotes y = bx/a and y = −bx/a are perpendicular, this is again a rectangular hyperbola. (This is again an “east-west” rectangular hyperbola.) 2

2

5. The hyperbola’s centre is (0, 0). 6. The two lines of symmetry are y = 0 (the x-axis) and x = 0 (the y-axis). Observe that each (a) passes through the centre; and (b) bisects an angle formed by the two asymptotes. 496, Contents

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40.6.

y 2 x2 The Hyperbola 2 − 2 = 1 b a

Take the equation from the last chapter, but switch a and b, so that we have the equation x2 /b2 − y 2 /a2 = 1. Note that this is also an east-west rectangular hyperbola, but with x-intercepts are (±b, 0) instead of (±a, 0). To get from x2 /b2 − y 2 /a2 = 1 to y 2 /b2 − x2 /a2 = 1, apply any one of these transformations: Rotate

π clockwise. 2

Rotate

Reflect in the line y = x.

π anticlockwise. 2

Notice that no y ∈ (−b, b) satisfy y 2 /b2 − x2 /a2 = 1. Hence, this graph contains no points for which y ∈ (−b, b).

y

Oblique asymptote b y=− x a

Oblique asymptote b y= x a

y 2 x2 − =1 b2 a2 b Line of symmetry y=0

−b

Centre (0, 0)

x

Line of symmetry x=0

1. There are two branches—one above and another below. 2. Intercepts. The y-intercepts are (−b, 0) and (b, 0). There are no x-intercepts. 3. The two turning points are (0, b) and (0, −b)—the former is a strict local minimum, while the latter is a strict local maximum. 4. As x → −∞, y → ±bx/a. And as x → ∞, y → ±bx/a. So, y 2 /b2 − x2 /a2 = 1 has two oblique asymptotes y = ±bx/a. Since the two asymptotes y = bx/a and y = −bx/a are perpendicular, this is again a rectangular hyperbola. (In fact, we call this a “north-south” rectangular hyperbola.) 5. The hyperbola’s centre is (0, 0). 6. The two lines of symmetry are y = 0 (the x-axis) and x = 0 (the y-axis). Each (a) passes through the centre; and (b) bisects an angle formed by the two asymptotes. 497, Contents

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40.7.

The Hyperbola y =

In the next subchapter, we’ll study the graph of: y=

ax2 + bx + c . dx + e

bx + c dx + e

But to warm up, let’s first study the simpler case where a = 0. That is, let’s first study: y=

bx + c . dx + e

We’ll assume that d ≠ 0 and cd − be ≠ 0. This is because: • If d = 0, then this is simply a linear equation; and

• If cd − be = 0, then as we’ll show below, this is simply the horizontal line y = b/d.

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Example 595. Consider y =

2x + 1 . Do the long division: x+1

2 x + 1 2x +1 2x +2 −1

y = −x + 1

Ô⇒

2x + 1 y= x+1

2x + 1 1 =2− . x+1 x+1

y=

y x = −1 (−1, 2)

(0, 1)

y =x+3 y=2

(−1/2, 0)

x

1. There are two branches—one on the top-left and another on the bottom-right.

2. Intercepts. Plug in x = 0 to get y = (2 ⋅ 0 + 1) / (0 + 1) = 1—the y-intercept is (0, 1). Plug in y = 0 to get 2x + 1 = 0 or x = −1/2—the x-intercept is (−1/2, 0).

3. There are no turning points. 4. As x → −1− , y → ∞. And as x → −1+ , y → −∞. So, y = (2x + 1) / (x + 1) has vertical asymptote x = −1. (Not coincidentally, this is the x-value for which x + 1 = 0.) As x → −∞, y → 2+ . And as x → ∞, y → 2− . So, y = (2x + 1) / (x + 1) has horizontal asymptote y = 2. (Not coincidentally, this is the quotient in the above long division.)

Since the two asymptotes y = 2 and x = −1 are perpendicular, this is again a rectangular hyperbola.

5. The hyperbola’s centre (the point at which the two asymptotes intersect) is (−1, 2). (The centre’s coordinates are simply given by the vertical and horizontal asymptotes.) 6. The two lines of symmetry are y = x + 3 and y = −x + 1. 499, Contents

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In general, given the hyperbola y = (bx + c) / (dx + e), here’s how to find the intercepts, asymptotes, and centre: • Intercepts.

Plug in x = 0 to get y = (b ⋅ 0 + c) / (d ⋅ 0 + e) = c/e and thus the y-intercept (0, c/e). (Note that if e = 0, then c/e is undefined and there is no y-intercept.)

Plug in y = 0 to get bx + c = 0 or x = −c/b. And thus, the x-intercept is (−c/b, 0). (Note that if b = 0, then −c/b is undefined and there is no x-intercept.) • Asymptotes.

The value of x for which dx + e = 0 is x = −e/d. Thus, the vertical asymptote is x = −e/d.

To find the horizontal asymptote, do the long division: b/d dx + e bx bx

Thus:

+c +be/d c − be/d

bx + c b c − bed b cd − be 1 = + = + . dx + e d dx + e d d2 x + e/d

The quotient b/d gives us the horizontal asymptote y = b/d.

Note that since we always have two perpendicular asymptotes (one horizontal and another vertical), y = (bx + c) / (dx + e) is a rectangular hyperbola.

(By the way, note that if cd−be = 0, then this hyperbola is simply the horizontal line y = b/d. This is why above we imposed the condition that cd − be ≠ 0.) • We’ve defined the centre to the point at which the two asymptotes intersect. And so, the centre is simply (−e/d, b/d). (These coordinates are simply given by the vertical and horizontal asymptotes.) • The two lines of symmetry are: y = ±x + (b + e) /d.

You need not know where the above two lines of symmetry come from (but see the proof of Fact 84 if you’re interested). Indeed, you need not even mug them. All you need remember are the following two points: The two lines of symmetry (a) Pass through the centre; and

The second point (b) implies that the two lines of symmetry may be written as y = x + α and y = −x + β, where αand β are constants you can easily find. Examples:

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Example 596. Consider y = 3.5 2x + 4 7x

7x + 3 . Do the long division: 2x + 4 +3

7x +14

y = −x + 3/2

−11

Ô⇒

7x + 3 y= 2x + 4

y=

7x + 3 7 11 = − . 2x + 4 2 2x + 4

y x = −2

(−2, 7/2)

y = x + 11/2 y = 7/2

(0, 3/4)

x

(−3/7, 0)

1. There are two branches—one on the top-left and another on the bottom-right. 2. Intercepts. Plug in x = 0 to get y = 3/4. So, the y-intercept is (0, 3/4). Plug in y = 0 to get 7x + 3 = 0 or x = −3/7. So, the x-intercept is (−3/7, 0).

3. There are no turning points. 4. Asymptotes. The value of x that makes the denominator 0 is −2—hence, the vertical asymptote is x = −2. The quotient in the long division is 7/2—hence, the horizontal asymptote is y = 7/2.

Since the two asymptotes x = −2 and y = 7/2 are perpendicular, this is again a rectangular hyperbola. 5. The hyperbola’s centre (the point at which the two asymptotes intersect) is (−2, 7/2). (These coordinates are given by the vertical and horizontal asymptotes.) 6. The two lines of symmetry may be written as y = x + α and y = −x + β and pass through the centre (−2, 7/2). Plugging in the numbers, we find that α = 11/2 and β = 3/2. Thus, the two lines of symmetry are y = x + 11/2 and y = −x + 3/2. 501, Contents

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Example 597. Consider y = −5/3 3x + 2 −5x

−5x + 1 . Do the long division: 3x + 2 +1

−5x −10/3 13/3

y=

x = −2/3

y = −x − 7/3

(0, 1/2)

y = −5/3

y =x−1

Ô⇒

5 13/3 −5x + 1 =− + . 3x + 2 3 3x + 2

y

(1/5, 0)

x

(−2/3, −5/3)

y=

−5x + 1 3x + 2

1. There are two branches—one on the top-left and another on the bottom-right. 2. Intercepts. Plug in x = 0 to get y = 1/2. So, the y-intercept is (0, 1/2). Plug in y = 0 to get −5x + 1 = 0 or x = 1/5. So, the x-intercept is (1/5, 0). 3. There are no turning points.

4. Asymptotes. The value of x that makes the denominator 0 is −2/3—hence, the vertical asymptote is x = −2/3.

The quotient in the long division is −5/3—hence, the horizontal asymptote is y = −5/3. Since the two asymptotes x = −2/3 and y = −5/3 are perpendicular, this is again a rectangular hyperbola.

5. The hyperbola’s centre (the point at which the two asymptotes intersect) is (−2/3, −5/3). (These coordinates are given by the vertical and horizontal asymptotes.) 6. The two lines of symmetry may be written as y = x + α and y = −x + β and pass through the centre (−2/3, −5/3). Plugging in the numbers, we find that α = −1 and β = −7/3. Thus, the two lines of symmetry are y = x − 1 and y = −x − 7/3.

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The following Fact summarises the features of the hyperbola y =

bx + c : dx + e

Fact 84. Let b, c, d, e ∈ R with d ≠ 0 and cd − be ≠ 0. Consider the graph of

bx + c . dx + e (a) Intercepts. If e ≠ 0, then there is one y-intercept (0, c/e). (If e = 0, then there are no y-intercepts.) And if b ≠ 0, then there is one x-intercept (−c/b, 0). (If b = 0, then there are no x-intercepts.) y=

(b) There are no turning points. (c) There is the horizontal asymptote y = b/d and the vertical asymptote x = −e/d. (The asymptotes are perpendicular and so, this is a rectangular hyperbola.)

(d) The hyperbola’s centre is (−e/d, b/d). (e) The two lines of symmetry are y = ±x + (b + e) /d.

Proof. We proved (a), (c), and (d) above. For (b) and (e), see p. 1578 (Appendices). Exam Tip for Towkays For the hyperbola y = (bx + c) / (dx + e), you should know how to find (a) the x- and y-intercepts; and (c) the horizontal and vertical asymptotes.

You’re not required to know what (d) the hyperbola’s centre is, but since this is simply the intersection of the two asymptotes (which you already know how to find), you might as well know how it, since it’ll help you sketch better graphs. You’re also not required to know how to find (e) the equations of the two lines of symmetry. But as we’ve shown in the above examples, it is not very difficult to figure out their equations. It is certainly not very difficult for you to at least sketch them. After you are done with Part V (Calculus), there is a small possibility that you are required to prove that (b) this hyperbola has no turning points. (And so you may or may not be interested in reading the proof of (b) (Appendices).) Exercise 189. Graph and describe the features of each equation. 3x + 2 . x+2 x−2 (b) y = . −2x + 1 (a) y = (c) y =

−3x + 1 . 2x + 3

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40.8.

ax2 + bx + c The Hyperbola y = dx + e

We now study the equation y = (ax2 + bx + c) / (dx + e). We’ll assume that a ≠ 0, d ≠ 0, and either c ≠ 0 or e ≠ 0. This is because • If a = 0, then this is simply the equation we studied in the last subchapter. • If d = 0, then the equation is quadratic and we’ve already studied that. • If c = 0 = e, then the equation is linear and we’ve already studied that. Example 598. Consider y = (x2 + 1) /x.

Do the long division:

Ô⇒ y=x

x x x2 +1 x2 1

y

x2 + 1 y= x

x2 + 1 1 y= =x+ . x x (−1, −2)

(0, 0) x=0

(1, 2)

y = (1 +

√ 2) x

x y = (1 −

√ 2) x

1. There are two branches—one on the bottom-left and another on the top-right.

2. Intercepts. If we plug in x = 0, then y is undefined. Thus, there are no y-intercepts. And if we plug in y = 0, then x2 + 1 = 0, an equation for which there are no (real) solutions. Thus, there are no x-intercepts.

3. There are two turning points: (−1, −2) is a strict local maximum and (1, 2) is a strict local minimum.

4. Asymptotes. The value of x that makes the denominator 0 is 0—hence, the vertical asymptote is x = 0. The quotient in the long division is x—hence, the oblique asymptote is y = x. (By the way, here for the first time, the asymptotes here are not perpendicular and so, this is a non-rectangular hyperbola.)

5. The hyperbola’s centre is (0, 0). Recall that the centre is simply the point at which the two asymptotes intersect. In this example, the intersection of the asymptotes x = 0 and y = x is (0, 0), √ 6. The two lines of symmetry are y = (1 ± 2) x.

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Here are the features of the hyperbola y = (ax2 + bx + c) / (dx + e).

• Intercepts. Plug in x = 0 to get y = c/e and thus the y-intercept (0, c/e). (Note that if e = 0, then c/e is undefined and there is no y-intercept. This was the case in the last example.) The x-intercepts are given by the values of x for which ax2 + bx + c = 0: √ −b ± b2 − 4ac , 0). ( 2a Note that if b2 − 4ac < 0, then there are no x-intercepts (this was the case in the last example). And if b2 − 4ac = 0, then there is exactly one x-intercept, namely (−b/ (2a) , 0)

• Asymptotes. The value of x that makes the denominator dx + e zero is x = −e/d and gives us the vertical asymptote x = −e/d. To find the other oblique asymptote, do the long division: ax/d + (bd − ae) /d2 dx + e ax2 +bx ax2 +aex/d (bd − ae) x/d (bd − ae) x/d Ô⇒

+c

+c + (bde − ae2 ) /d2 (cd2 + ae2 − bde) /d2 .

a bd − ae cd2 + ae2 − bde 1 y = x+ + ⋅ . d d2 d2 dx + e ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ Quotient

Remainder

The quotient gives us the oblique asymptote y = ax/d + (bd − ae) /d2 . asymptotes are not perpendicular, this hyperbola is not rectangular.)

(Since the

• The centre is the point at which the two asymptotes intersect. Its x-coordinate is given by the vertical asymptote x = −e/d. For its y-coordinate, plug x = −e/d into the equation of the oblique asymptote:

Thus, the centre is:

y=

a e bd − ae −ae + bd − ae bd − 2ae (− ) + = = . d d d2 d2 d2 (−e/d,

bd − 2ae ). d2

• You need not know how to find the equations of the lines of symmetry. You should however know how to roughly sketch them. So, just remember that they (a) pass through the centre; and (b) bisect the angles formed by the two asymptotes.

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Example 599. Consider y =

x2 + 3x + 1 . Do the long division: x+1

x2 + 3x + 1 1 y= =x+2− . x+1 x+1

1. There are two branches—one on the left and another on the right.

2. Intercepts. Plug in x = 0 to get y = 1/1 = 1. Thus, the y-intercept is (0, 1). Plug in √ y = 0 to get x2 + 3x + 1 = 0—thus, the two x-intercepts are (0.5 (−3 ± 5) , 0).

3. There are no turning points.

4. Asymptotes. The value of x that makes the denominator 0 is −1—hence, the vertical asymptote is x = −1. The quotient in the long division is x + 2—hence, the oblique asymptote is y = x + 2.

5. The centre’s x-coordinate is given by the vertical asymptote x = −1. For its ycoordinate, plug x = −1 into the oblique asymptote to get y = −1 + 2 = 1. Hence, the centre is (−1, 1). √ √ 6. The two lines of symmetry are y = (1 ± 2) x + 2 ± 2.

x2 + 3x + 1 y= x+1

y = (1 −

y =x+2

√ √ 2) x + 2 − 2

√ −3 − 5 ( , 0) 2

y = (1 + 506, Contents

x = −1

y

(−1, 1)

(0, 1) √ −3 + 5 ( , 0) 2

x

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Example 600. Consider y =

2x2 + 2x + 1 . Do the long division: −x + 1

2x2 + 2x + 1 5 5 = −2x − 4 + = −2x − 4 + . −x + 1 −x + 1 −x + 1

1. There are two branches—one on the top-left and another on the bottom-right.

2. Intercepts. Plug in x = 0 to get y = 1/1 = 1. Thus, the y-intercept is (0, 1). Plug in y = 0 to get 2x2 + 2x + 1 = 0, an equation for which there are no (real) solutions. Thus, there are no x-intercepts. √ √ 3. The two turning points are (1 ± 0.5 10, −6 ± 2 10).

4. Asymptotes. The value of x that makes the denominator 0 is 1—hence, the vertical asymptote is x = 1. The “quotient” in the long division is −2x − 4—hence, the oblique asymptote is y = −2x − 4. 5. The centre’s x-coordinate is given by the vertical asymptote x = 1. For its ycoordinate, plug x = 1 into the oblique asymptote to get y = −2 (1) − 4 = −6. Hence, the centre is (1, −6). √ √ 6. The two lines of symmetry are y = (−2 ± 5) x − 4 ± 5.

y

y = (−2 +

√ √ 5) x − 4 − 5

y=

2x2 + 2x + 1 −x + 1

(0, 1)

√ √ (1 − 10/2, −6 − 2 10) √ √ y = (−2 − 5) x − 4 + 5

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x=1

(1, −6)

x

√ √ (1 + 10/2, −6 + 2 10)

y = −2x − 4

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Given the hyperbola y=

ax2 + bx + c , dx + e

you now know how to find its intercepts, asymptotes, and centre.260 And after we’ve done Calculus (Part V), you’ll also be able to find the turning points. To repeat, you do not need to know how to find the equations of the two lines of symmetry. However, you should at least be able to roughly sketch them. Exercise 190. Graph the equations below. Label any intercepts, asymptotes, and centre. Roughly indicate or sketch any turning points and lines of symmetry. x2 + 2x + 1 . (a) y = x−4 −x2 + x − 1 (b) y = . x+1 (c) y =

260

2x2 − 2x − 1 . x+4

Fact 240 (Appendices) summarises the features of this hyperbola.

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40.9.

Eliminating the Parameter t

Above, we rewrote a single (cartesian) equation into a pair of (parametric) equations and in the process introduced the parameter t. Now we’ll go in reverse—we’ll rewrite a pair of (parametric) equations into a single (cartesian) equation and in the process eliminate the parameter t. Example 601. Consider the set S = {(x, y) ∶ x = t2 − 4t, y = t − 1, t ≥ 0}.

As usual, we interpret this set as describing the motion of a particle P in the plane, where t is time (seconds), while x and y are P ’s rightward and upward displacements (metres) 1 from the origin. With a little algebra, we can combine the two equations x = t2 − 4t and 2 y = t − 1 into a single equation and eliminate the parameter t: • First rewrite y = t − 1 as t = y + 1. 2

3

• Then plug = into = to get x = (y + 1) − 4 (y + 1) = y 2 − 2y − 3. 3

1

2

• Observe also that t ≥ 0 ⇐⇒ y ≥ 0 − 1 = −1.

Altogether, we can rewrite the set S as S = {(x, y) ∶ x = y 2 − 2y − 3, y ≥ −1}.

Note that here we have a quadratic equation. Our quadratic equations are usually in the variable x, but in this case, it is in the variable y.

The set S is the black graph below and does not include the grey portion. At t = 0, P starts at the position (x, y) = (0, −1). P never travels along the grey portion.

y

At t = 2,

(x, y) = (−4, 1) (vx, vy ) = (0, 1) (ax, ay ) = (2, 0)

During t ∈ [0, 2), P moves leftwards.

After t = 2, P moves rightwards.

√ 5, √ (x, y) = (1, 1 + 5) At t = 2 +

P always moves upwards at 1 m s−1 .

At t = 0,

(x, y) = (0, −1) (vx, vy ) = (−4, 1) (ax, ay ) = (2, 0)

(vx , vy ) = (0, 1) (ax , ay ) = (2, 0)

x

Starting point

P does not travel along this grey portion.

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(... Example continued from the previous page.) Here are P ’s velocity and acceleration, decomposed into the x- and y-directions: vx =

dx = 2t − 4, dt

vy =

dy = 1, dt

ax =

dax d2 x = 2 = 2, dt dt

ay =

day d2 y = 2 = 0. dt dt

In the y- or upwards-direction, P moves at a constant velocity of 1 m s−1 and does not accelerate. In the x- or rightwards-direction, P moves at velocity 2t − 4 m s−1 and accelerates at a constant rate of 2 m s−2 . In particular,

• When t < 2, we have 2t − 4 < 0 and so P is moving leftwards. • At t = 2, the particle is at the leftmost point of the parabola and its velocity in the x-direction is 0 m s−1 . • And when t > 2, we have 2t − 4 > 0 and so P is moving rightwards.

Example 602. Consider the set U = {(x, y) ∶ x = 2 cos t − 4, y = 3 sin t + 1, t ≥ 0}.

Again, we interpret this set as describing the motion of a particle Q in the plane. With 1 2 a little algebra, we can combine the two equations x = 2 cos t − 4 and y = 3 sin t + 1 into a single equation and eliminate the parameter t: x+4 . 2 y−1 2 • Next rewrite y = 3 sin t + 1 as sin t = . 3 • Now recall261 the trigonometric identity sin2 t + cos2 t = 1. • First rewrite x = 2 cos t − 4 as cos t = 1

Thus, we can rewrite the set U as

U = {(x, y) ∶ (

x+4 2 y−1 2 ) +( ) = 1} . 2 3

This describes an ellipse centered on (−4, 1).

Note that since x and y can be expressed in terms of trigonometric functions, we know that x and y will be periodic. That is, the particle Q will keep repeating its movement along a certain path. And so, we need not include any constraints for x and y. Note also that by eliminating the parameter t, we’ve lost one piece of information—namely Q’s starting position. (Example continues on the next page ...)

261

Fact 71(a).

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(... Example continued from the previous page.)

Centre (−4, 1)

y

Starting point

At t = 0, (x, y) = (−2, 1) (vx , vy ) = (0, 3) (ax , ay ) = (−2, 0)

x

The velocity and acceleration of Q, decomposed into the x- and y-directions, are dx vx = = −2 sin t, dt

dy dax d2 x vy = = 3 cos t, ax = = 2 = −2 cos t, dt dt dt day d2 y ay = = 2 = −3 sin t. dt dt

At t = 0, Q’s starting position and velocity are (x, y) = (2 cos 0 − 4, 3 sin 0 + 1) = (−2, 1)

and

(vx , vy ) = (−2 sin 0, 3 cos 0) = (0, 3).

So, it starts at the rightmost point of the ellipse and is moving upwards at 3 m s−1 . Thus, its direction of travel is anticlockwise around the ellipse. Every 2π s, Q completes one full revolution around the ellipse. Exercise 191. The sets A, B, and C below describe the positions (metres) of particles A, B, and C at time t (seconds), relative to the origin. For each set, (i) Rewrite the set so that the parameter t is eliminated. (ii) Sketch the graph. (iii) Describe the particle’s position and velocity as time progresses. (a) A = {(x, y) ∶ x = t − 1, y = ln (t + 1) , t ≥ 0} 1 (b) B = {(x, y) ∶ x = , y = t2 + 1, t ≥ 0}. t+1 (c) C = {(x, y) ∶ x = 2 sin t − 1, y = 3 cos2 t, t ≥ 0}.

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41.

Simple Parametric Equations

We can sometimes describe a graph (i.e. a set of points) using an equation. We can sometimes also describe a graph using parametric equations:

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Example 603. Recall that the unit circle centred on the origin is the graph of x2 +y 2 = 1, i.e. this set S = {(x, y) ∶ x2 + y 2 = 1} .

Recall262 that sin2 t + cos2 t = 1. And so, observe that by letting x = cos t and y = sin t, we have x2 + y 2 = 1. We thus have a second method for writing down the set S: S = {(x, y) ∶ x = cos t, y = sin t, t ≥ 0} .

We call the variable t a parameter (hence the name parametric equations). In words, S is the set of points such that x = cos t, y = sin t, and t ≥ 0. As t increases from 0 to 2π, we trace out, anti-clockwise, the unit circle: t = 0 Ô⇒ (x, y) = (1, 0) , √ √ t = π/4 Ô⇒ (x, y) = ( 2/2, 2/2) ,

t = π/2 Ô⇒ (x, y) = (0, 1) , √ √ t = 3π/4 Ô⇒ (x, y) = (− 2/2, 2/2) ,

Arrows indicate instantaneous direction of travel.

S = {(x, y) ∶ x + y = 1} = {(x, y) ∶ x = cos t, y = sin t, t ≥ 0} 2

At t = 1,

(x, y) ≈ (0.54, 0.84) , (vx , vy ) ≈ (−0.84, 0.54) , (ax , ay ) ≈ (−0.54, −0.84) . At t = 0,

2

\ (Example continues on the next page ...)

262

t = 3π/2 Ô⇒ (x, y) = (0, −1) , √ √ t = 7π/4 Ô⇒ (x, y) = ( 2/2, − 2/2) .

l

y

t = π Ô⇒ (x, y) = (−1, 0) , √ √ t = 5π/4 Ô⇒ (x, y) = (− 2/2, − 2/2) ,

l

(x, y) = (1, 0) , (vx , vy ) = (1, 0) , (ax , ay ) = (−1, 0) .

5π At t = , 4 √ √ 2 2 (x, y) = (− ,− ), 2 2 √ √ 2 2 (vx , vy ) = ( ,− ), 2 2 √ √ 2 2 (ax , ay ) = ( , ). 2 2

x

Fact 71.

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Example 604. Let a, b > 0. The equation x2 /a2 + y 2 /b2 = 1 describes an ellipse centred on the origin, with x-intercepts (±a, 0) and y-intercepts (0, ±b). This graph is the set: U = {(x, y) ∶ x2 /a2 + y 2 /b2 = 1}.

Observe that by letting x = a cos t and y = b sin t, we have

x2 y 2 a2 cos2 t b2 sin2 t + 2 = + = cos2 t + sin2 t = 1. 2 2 2 a b a b

We can thus use parametric equations to rewrite the set U :

U = {(x, y) ∶ x = a cos t, y = b sin t, t ≥ 0}.

We apply the same interpretation as before: t is time (seconds) and the parametric equations describe the position (metres from the origin O) of some particle R. Observe that like P , R is travelling anticlockwise.

y b

−a

x2 y 2 U = {(x, y) ∶ 2 + 2 = 1} a b = {(x, y) ∶ x = a cos t, y = b sin t, t ≥ 0}

a

x

−b

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Example 605. The equation x2 − y 2 = 1 describes the “east-west” hyperbola centred on the origin, with x-intercepts (±1, 0). Its graph is this set: W = {(x, y) ∶ x2 − y 2 = 1} .

Recall263 the trigonometric identity sec2 t − tan2 t = 1. And so, if we let x = sec t and y = tan t, then x2 − y 2 = 1. This thus gives us another way to write down the set W : W = {(x, y) ∶ x = sec t, y = tan t, t ≥ 0} .

We can again impose a similar interpretation: t is time (measured in seconds) and the parametric equations describe the motion (in metres from the origin O) of a particle A.

Arrows indicate instantaneous direction of travel.

l

At t = 0, (x, y) = (1, 0) , (vx , vy ) = (0, 1) , (ax , ay ) = (1, 0) .

x

\

During t ∈ (0.5π, 1.5π), A moves upwards along the left branch.

During t ∈ [0, 0.5π), A is moving northeast.

l

W = {(x, y) ∶ x2 − y 2 = 1} y = {(x, y) ∶ x = sec t, y = tan t, t ≥ 0}

An instant after t = 0.5π, A magically reappears “near” “bottom-left infinity”. At each instant t, A’s velocity in the x- and y-directions is given by (vx , vy ) = (

d sec t d tan t dx dy , )=( , ) = (sec t tan t, sec2 t) . dt dt dt dt

(Example continues on the next page ...)

Exercise 192. Continuing with the above example, write down A’s acceleration in the x- and y-directions at the instant t. (Answer on the next page.)

263

Fact 71(b).

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(... Example continued from the previous page.) dvx d × = (sec t tan t) = sec t tan t tan t + sec t sec2 t dt dt = sec t (tan2 t + sec2 t) = sec t (2 sec2 t − 1) ,

ax =

Thus:

ay =

(ax , ay ) = (

dvy d Ch. = (sec2 t) = 2 sec t ⋅ sec t tan t = 2 sec2 t tan t. dt dt

dvx dvy d2 x d2 y , 2)=( , ) = (sec t (2 sec2 t − 1) , 2 sec2 t tan t). 2 dt dt dt dt

Note that at t = 0.5π, 1.5π, 2.5π, , . . . , both sec t and tan t are undefined. And so, we’ll say that at these instants in time, the particle A’s position, velocity, and acceleration are simply undefined. Observe that interestingly, vy = sec2 t > 0 for all t (for which sec t is well-defined). Hence, the particle A is always moving upwards (except during the aforementioned instants in time when its velocity is undefined). At t = 0, we have:

(x, y) = (sec 0, tan 0) = (1, 0),

(vx , vy ) = (sec 0 tan 0, sec2 0) = (0, 1),

(ax , ay ) = (sec 0 (2 sec2 0 − 1) , 2 sec2 0 tan 0) = (1, 0).

So, A starts at the midpoint of the right branch of the hyperbola, is moving upwards at 1 m s−1 , and is accelerating rightwards at 1 m s−2 .

During t ∈ [0, π/2), the particle A is moving northeast. As t → π/2, it “flies off” towards the “top-right infinity” (∞, ∞) and: x, y, vx , vy , ax , ay → ∞.

An instant after t = π/2, A magically reappears “near” “bottom-left infinity” (−∞, −∞).

During t ∈ (0.5π, 1.5π), the particle travels upwards along the left branch of the hyperbola. And again, as t → 1.5π, it “flies off” towards “top-left infinity” (−∞, ∞) and we have x, vx , ax → −∞

and

Exercise 193. Continue with the above example.

y, vy , ay → ∞.

(a) What happens to particle A an instant after t = 1.5π?

(b) Describe A’s movement during t ∈ (1.5π, 2.5π).

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Part II. Sequences and Series

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[T]here is nothing as dreamy and poetic, nothing as radical, subversive, and psychedelic, as mathematics. — Paul Lockhart (2002, 2009).

[M]athematics is capable of an artistic excellence as great as that of any music, perhaps greater; not because the pleasure it gives (although very pure) is comparable, either in intensity or in the number of people who feel it, to that of music, but because it gives in absolute perfection that combination, characteristic of great art, of godlike freedom, with the sense of inevitable destiny; because, in fact, it constructs an ideal world where everything is perfect and yet true. — Bertrand Russell (1902).

Beauty is Truth, Truth Beauty.–That is all Ye know on Earth, and all ye need to know. — John Keats (1819).

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42.

Sequences

Informally, a sequence is simply an ordered list of real numbers.264 Example 606. The Fibonacci sequence is: (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . )

We call the numbers in a sequence its terms.

So the Fibonacci sequence’s 1st term is 1, 2nd is 1, 3rd is 2, etc. The first two terms of the Fibonacci sequence are fixed as 1. Each subsequent term is then simply the sum of the previous two terms. So: the the the the

3rd 4th 5th 6th

term term term term

is is is is

1+1 1+2 2+3 3+5

= 2; = 3; = 5; = 8;

the 7th term is 5 + 8 = 13; the 8th term is 8 + 13 = 21; the 9th term is 13 + 21 = 34; etc.

Letting f (n) denote the nth term, the Fibonacci sequence may be defined by: ⎧ ⎪ ⎪ ⎪1 f (n) = ⎨ ⎪ ⎪ ⎪ ⎩f (n − 2) + f (n − 1)

for n = 1, 2, for n ≥ 3.

Example 607. The sequence of square numbers is:

(1, 4, 9, 16, 25, 36, 49, 64, . . . )

This sequence’s 1st term is 1, 2nd is 4, 3rd term is 9, etc.

Letting s (n) denote the nth term, this sequence may be defined by s (n) = n2 . Example 608. The sequence of triangular numbers is: (1, 3, 6, 10, 15, 21, 28, 36, . . . )

This sequence’s 1st term is 1, 2nd is 3, 3rd term is 6, etc.

Letting t (n) denote the nth term, this sequence may be defined by t (n) = 1 + 2 + ⋅ ⋅ ⋅ + n. Remark 83. For clarity, it is wise and indeed customary to enclose a sequence in parentheses.265 And so, even though your A-Level exams do not, I will insist on doing so.

In H2 Maths, we’ll look only at sequences of real numbers. But in general, the objects in a sequence need not be real numbers and could be any objects whatsoever. 265 Note though that some writers prefer using braces {} or angle brackets ⟨⟩. 264

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The above sequences were infinite. But of course, sequences can also be finite: Example 609. The finite sequence of the first six Fibonacci numbers is: (1, 1, 2, 3, 5, 8) .

We call this a finite sequence of length 6.

Example 610. The finite sequence of the first seven square numbers is: (1, 4, 9, 16, 25, 36, 49) .

We call this a finite sequence of length 7.

Example 611. The finite sequence of the first four triangular numbers is: (1, 3, 6, 10) .

We call this a finite sequence of length 4.

Remark 84. We’ll generally be more interested in infinite sequences than finite sequences. And so, when we simply say sequence, it may be assumed that we’re talking about an infinite sequence. And when we want to talk about a finite sequence, we’ll clearly and explicitly include the word finite.

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42.1.

Sequences Are Functions

A little more formally, sequences are functions: Definition 111. A finite sequence of length k is a function with domain {1, 2, . . . , k}. Definition 112. An (infinite) sequence is a function with domain Z+ = {1, 2, 3, . . . }.

Remark 85. For H2 Maths, the objects in a sequence will always be real numbers. But in general, they can be anything whatsoever.266 We now formally rewrite our earlier examples of sequences as functions: Example 612. Formally, the Fibonacci sequence is the function f ∶ Z+ → R defined by: ⎧ ⎪ ⎪ ⎪1 f (n) = ⎨ ⎪ ⎪ ⎪ ⎩f (n − 2) + f (n − 1)

for n = 1, 2, for n ≥ 3.

Example 613. Formally, the sequence of square numbers is the function s ∶ Z+ → R defined by s (n) = n2 .

Example 614. Formally, the sequence of triangular numbers is the function t ∶ Z+ → R defined by t (n) = 1 + 2 + ⋅ ⋅ ⋅ + n.

Example 615. Formally, the finite sequence of the first six Fibonacci numbers is the function f6 ∶ {1, 2, 3, 4, 5, 6} → R defined by: ⎧ ⎪ ⎪ ⎪1 f6 (n) = ⎨ ⎪ ⎪ ⎪ ⎩f6 (n − 2) + f6 (n − 1)

for n = 1, 2,

for n = 3, 4, 5, 6.

Example 616. Formally, the finite sequence of the first seven square numbers is the function s7 ∶ {1, 2, 3, 4, 5, 6, 7} → R defined by s7 (n) = n2 .

Example 617. Formally, the finite sequence of the first four triangular numbers is the function t4 ∶ {1, 2, 3, 4} → R defined by t4 (n) = 1 + 2 + ⋅ ⋅ ⋅ + n. Remark 86. As repeatedly emphasised, the letter x we often use with functions is merely a dummy or placeholder variable that can be replaced by any another. Indeed, in the context of sequences, we’ll often prefer using n rather than x as our dummy variable.

More examples:

266

In other words, for H2 Maths, the codomain of a sequence (which is a function) will always be R. But in general, it can be any set whatsoever.

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Example 618. The function e ∶ Z+ → R defined by e (n) = 2n is the sequence of (positive) even numbers (2, 4, 6, 8, 10, 12, . . . ).

Example 619. The function g ∶ Z+ → R defined by g (n) = 2n2 − 3n + 3 is the following sequence (2, 5, 12, 23, 38, 57, 80, 107, 138, 173, . . . ).

Recall that a function need not “follow any formula” or “make any sense”. The same is true of sequences (since sequences are simply functions): Example 620. The function h ∶ {1, 2, 3, 4} → {Cow, Chicken} is defined by: h (1) = Cow,

h (2) = Cow,

h (3) = Chicken, and h (4) = Cow.

Although h does not seem to “make any sense”, it is a (perfectly) well-defined function. Indeed, the function h is also the following finite sequence of length 4: (Cow, Cow, Chicken, Cow). Although this sequence does not seem to “make any sense”, it is a (perfectly) well-defined finite sequence of length 4, simply because it is a function with domain {1, 2, 3, 4}. Example 621. The function j ∶ {1, 2, 3} → {↑, ↓, →, ←, Punch, Kick} is defined by: j (1) = ↓,

j (2) = →, and j (3) = Punch

Although j does not seem to “make any sense”, it is a (perfectly) well-defined function. Indeed, the function j is also the following finite sequence of length 3: (↓, →, Punch).

Although this sequence does not seem to “make any sense”, it is a (perfectly) well-defined finite sequence of length 3, simply because it is a function with domain {1, 2, 3}. Exercise 194. Formally rewrite each sequence as a function.

(a) (1, 4, 9, 16, 25, 36, 49, 64, 81, 100).

(b) (2, 5, 8, 11, 14, 17, 20, . . . , 299). (c) (1, 8, 27, 64, 125, 216, 343).

(d) (2, 6, 6, 12, 10, 18, 14, 24, 18, 30, 22, 36, 26, 42, . . . ). (e) (5, 0, 99).

(f) (1, 2, 6, 24, 120, 720, 5 040, 40 320, . . . ). (g) (1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, . . . ).

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42.2.

Notation

Let (a1 , a2 , . . . , ak ) be a finite sequence of length k. For convenience, this sequence can also be denoted as any of the following: (an )n=1 k

or

(an )n=1,2,...,k

or

(an )n∈{1,2,...,k}

or

(an )1≤n≤k .

Example 622. Let s1 = 1, s2 = 4, s3 = 9, s4 = 16, and s5 = 25. Then we can denote the finite sequence of the first five square numbers by: (s1 , s2 , s3 , s4 , s5 ) = (sn )n=1 = (sn )n=1,2,3,4,5 = (sn )n∈{1,2,3,4,5} = (sn )1≤n≤5 . 5

Again, s and n are merely dummy or placeholder variables. And here, n in particular may also be called an index variable, because it indexes or indicates which term in the sequence we’re referring to. We could replace s or n with any other symbol, like , or ⋆. And so, we could rewrite the above example as: Example 623. Let ,1 = 1, ,2 = 4, ,3 = 9, ,4 = 16, and ,5 = 25. Then we can also denote the finite sequence of the first five square numbers by: (,1 , ,2 , ,3 , ,4 , ,5 ) = (,⋆ )⋆=1 = (,⋆ )⋆=1,2,3,4,5 = (,⋆ )⋆∈{1,2,3,4,5} = (,⋆ )1≤⋆≤5 . 5

Of course, it’s a bit strange to use symbols like , or ⋆. The point here is simply to illustrate that once again, these are mere symbols that can be replaced by any other. We’ll usually stick to using boring symbols like letters from the Latin alphabet. Next, let (a1 , a2 , . . . ) be an (infinite) sequence. For convenience, this sequence can also be denoted as any of the following: ∞

(an )n=1

or

(an )n=1,2,...

or

(an )n∈Z+

or

(an )n∈{1,2,... }

or

(an ).

Example 624. Let t1 = 1, t2 = 3, t3 = 6, t4 = 10, t5 = 15, etc. Then we can denote the (infinite) sequence of triangular numbers by: ∞

(t1 , t2 , . . . ) = (tn )n=1 = (tn )n=1,2,... = (tn )n∈Z+ = (tn )n∈{1,2,... } = (tn ) .

Again, we can replace t and n with any other symbols:

Example 625. Let ,1 = 1, ,2 = 3, ,3 = 6, ,4 = 10, ,5 = 15, etc. Then we can also denote the (infinite) sequence of triangular numbers by: ∞

(,1 , ,2 , . . . ) = (,⋆ )⋆=1 = (,⋆ )⋆=1,2,... = (,⋆ )⋆∈Z+ = (,⋆ )⋆∈{1,2,... } = (,⋆ ) .

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42.3.

Arithmetic Combinations of Sequences

In Ch. 18, we learnt to create arithmetic combinations of functions. Since sequences are functions, we can likewise create arithmetic combinations of sequences: Example 626. Let (an ) = (1, 1, 2, 3, 5, 8, 13, 21, 34, . . . ) be the Fibonacci sequence. Let (bn ) = (2, 4, 6, 8, 10, 12, 14, 16, 18, . . . ) be the sequence of even numbers. Let k = 10. Then: (an + bn ) = (3, 5, 8, 11, 15, 20, 27, 37, 52, . . . ),

(an − bn ) = (−1, −3, −4, −5, −5, −4, −1, 5, 16, . . . ),

(an bn ) = (2, 4, 12, 24, 50, 96, 182, 336, 612, . . . ), (

an 1 1 1 3 1 2 13 21 17 ) = ( , , , , , , , , , . . . ), bn 2 4 3 8 2 3 14 16 9

(kan ) = (10, 10, 20, 30, 50, 80, 130, 210, 340, . . . ),

(kbn ) = (20, 40, 60, 80, 100, 120, 140, 160, 180, . . . ).

Exercise 195. Let (cn ) be the sequence of negative odd numbers, (dn ) the sequence of cube numbers, and k = 2. Write out the first five terms of (a) (cn ); and (b) (dn ). Then write out the first five terms of each of the following. (Answer on p. 1806.) (c) (cn + dn ) (f) (

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cn ) dn

(d) (cn − bn ) (g) (kcn )

(e) (cn dn ) (h) (kdn )

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43.

Series

Definition 113. Given a finite sequence (an )n∈{1,2,...,k} , its series is the expression a1 + a2 + a3 + ⋅ ⋅ ⋅ + ak .

And the sum of this series is the number that equals the above expression. Example 627. Consider the finite sequence of the first five square numbers: (1, 4, 9, 16, 25). Its series is the expression 1 + 4 + 9 + 16 + 25, while the sum of this series is the number 55. Example 628. Consider the finite sequence of the first six even numbers: (2, 4, 6, 8, 10, 12). Its series is the expression 2 + 4 + 6 + 8 + 10 + 12, while the sum of this series is the number 42.

It may seem strange and unnecessary to distinguish between a series and its sum. Aren’t they exactly the same thing?

It turns out that expressions like a1 + a2 + a3 + ⋅ ⋅ ⋅ + ak play an important role in maths. And so, we want to reserve a special name for the expression itself, in order to distinguish it from the number that is the sum of the series. Example 629. Given the sequence (1, 3, 5, 7), we might be specifically interested in the expression 1 + 3 + 5 + 7, rather than just the number 16.

It is thus convenient to have separate names for them—we call the expression 1 + 3 + 5 + 7 the series and the number 16 the sum (of the series).

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43.1.

Convergent and Divergent Series

Definition 114. Given an (infinite) sequence (an ), its series is the expression a1 + a2 + a3 + . . .

As we saw on the previous page, every finite series has a well-defined sum—simply add up all the numbers! With an infinite series, things get a little trickier. It may sometimes be that an infinite series diverges and its limit does not exist. Example 630. Let (1, 1, 1, 1, 1, . . . ) be the (infinite) sequence that consists solely of 1s.

Its series is the expression 1 + 1 + 1 + 1 + 1 + . . .

“Clearly”, this expression is not equal to any number. And so formally, we say that this series diverges and that its limit does not exist. Also, observe that this expression “grows ever larger”. As shorthand, we write:267 1 + 1 + 1 + 1 + 1 + ⋅ ⋅ ⋅ = ∞.

Remark 87. As was the case with sequences, we’ll generally be more interested in infinite series than finite series. And so, when we simply say series, it may safely be assumed that we’re talking about an infinite series. And when we want to talk about a finite series, we’ll clearly and explicitly include the word finite. Example 631. Let (2, 4, 6, 8, 10, . . . ) be the sequence of (positive) even numbers. Its series is the expression 2 + 4 + 6 + 8 + 10 + . . .

“Clearly”, this expression is not equal to any number. And so formally, we say that this series diverges and that its limit does not exist. Also, observe that this expression “grows ever larger”. As shorthand, we write: 2 + 4 + 6 + 8 + 10 + ⋅ ⋅ ⋅ = ∞.

267

Pedantic point: this “equation” is not really an equation. Instead, it is merely shorthand for the following statement: “The expression 1 + 1 + 1 + 1 + 1 + . . . grows ever larger.”

“Grows ever larger” is, in turn, a vague and informal phrase that we clarify only in Ch. 139.1 of the Appendices. 526, Contents

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We just looked at two examples of series that diverge. We now look at examples of series that converge to some limit: Example 632. Consider the zero sequence (0, 0, 0, 0, 0, . . . ).

Its series is the expression 0 + 0 + 0 + 0 + 0 + . . .

This series converges to 0. We call 0 its limit. And as shorthand, we write: 0 + 0 + 0 + 0 + 0 + ⋅ ⋅ ⋅ = 0.

Note that what was called the sum (in the previous context of finite series) is now called the limit (in the current context of infinite series). 1 1 1 1 1 Example 633. Consider the sequence ( , , , , , . . . ). 2 4 8 16 32 1 1 1 1 1 + + + + + ... 2 4 8 16 32 As we’ll soon learn, it turns out that: Its series is the expression

1 1 1 1 1 + + + + + ⋅ ⋅ ⋅ = 1. 2 4 8 16 32

That is, this series converges to 1. (And we call 1 the limit of this series.) Here we should remark that whenever we are dealing with infinite series, we must be very careful. Here the = sign in the above equation is not the usual one. Instead, the above 1 1 1 1 1 equation is merely shorthand for “the expression + + + + + . . . converges 2 4 8 16 32 to the number 1”. In H2 Maths, we shall simply count on your intuitive and imprecise understanding of what the phrase converges to means, but you should know that it does actually have a clear and precise meaning (on this, see Ch. 139.1 (Appendices)). Example 634. Consider the sequence of reciprocals of squares (

1 1 1 1 1 + + + ⋅ ⋅ ⋅ = 1 + + + ... 12 22 32 4 9 It’s not at all obvious, but it turns out that: It series is the expression

1 1 1 , , , . . . ). 12 22 32

1 1 1 1 1 π2 + + + ⋅ ⋅ ⋅ = 1 + + + ⋅ ⋅ ⋅ = . 12 22 32 4 9 6

That is, this series converges to π/6. (And we call π/6 the limit of this series.) The problem of finding the sum of this series is among the more famous problems in the history of mathematics and is known as the Basel Problem. We will revisit this in Ch. XXX.

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Now, when does a series converge or diverge? Or equivalently, when does its limit exist? The precise definitions of these terms are beyond the scope of H2 Maths.268 You need only know—roughly and intuitively—what convergence, divergence, and limits are. It turns out that what exactly these terms mean is no simple matter. On the next page are two fun examples to give you a glimpse of the difficulties involved: Example 635. Consider Grandi’s series:269 1 − 1 + 1 − 1 + 1 − 1 + ...

Does this series converge or diverge? Or equivalently, does its limit exist? Remarkably, we can “prove” that this series is equal to 0, 1, and 1/2. • To “prove” that it equals 0, pair off the terms like so:

1 − 1 + 1 − 1 + 1 − 1 + ⋅ ⋅ ⋅ = (1 − 1) + (1 − 1) + (1 − 1) + ⋅ ⋅ ⋅ = 0 + 0 + 0 + ⋅ ⋅ ⋅ = 0. ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¸ ¹ ¹ ¹ ¶ 0

0

0

• To “prove” that it equals 1, pair off the terms after the first, like so:

1 − 1 + 1 − 1 + 1 − 1 + . . . = 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + . . . ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹¸¹ ¹ ¹ ¹ ¹ ¹ ¶ 0

0

= 1 + 0 + 0 + 0 + ⋅ ⋅ ⋅ = 1.

0

• To “prove” that it equals 1/2, let S = 1 − 1 + 1 − 1 + 1 − 1 + . . . , then “show” that 1 − S = S: 1 − S = 1 − (1 − 1 + 1 − 1 + 1 − 1 + . . . ) = 1 − 1 + 1 − 1 + 1 − 1 + ⋅ ⋅ ⋅ = S.

Since 1 − S = S, simple algebra yields 1 = 2S or S = 1/2.

We’ve just “proven” that the expression 1 − 1 + 1 − 1 + 1 − 1 + . . . equals 0, 1, and 1/2.

Well, which is it then? It turns out that the series 1 − 1 + 1 − 1 + 1 − 1 + . . . is not equal to 0, 1, or 1/2. Instead, it is divergent and its limit does not exist.270

Here is an example of a series whose convergence is unknown: Example 636. Consider the following series:

1 2 3 4 5 6 7 8 9 10 − + − + − + − + − + ... 2 3 5 7 11 13 17 19 23 29

The terms are fractions, with the numerators being the (positive) integers and the denominators being the prime numbers. This series looks “simple” enough. But remarkably, mathematicians still do not know whether it converges or diverges!271 But if you’re interested, see the brief discussion in Ch. 139.1 (Appendices). The Italian priest-mathematician Luigi Guido Grandi (1671–1742) thought he had proved that the sum of this series equalled both 0 and 1/2, and thus that “God could create the word out of nothing” (source). 270 We prove this in Example 1495 (Appendices). 271 This problem is listed as equation (8) on this Wolfram Mathworld page and as Problem E7 in Richard

268 269

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44.

Summation Notation ∑

The symbol Σ is the upper-case Greek letter sigma.272

An enlarged version of that symbol is the symbol ∑, which is read aloud as “sum”. We use the symbol ∑ to write down series more compactly, in what is called summation or sigma notation. 5

Example 637. ∑ n2 = 12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25. n=1

• The variable n is the dummy, placeholder, or index variable. (We can replace it with any other symbol without changing the meaning of the above sentence.) • The integer below ∑ is the starting point. So here, 1 tells us to start counting (the index variable n) from n = 1. • The integer above ∑ is the stopping point. So here, 5 tells us to stop at n = 5.

• The expression to the right of ∑ describes the nth term to be added up. So here, the nth term to be added up is n2 . 5

Altogether then, ∑ n2 tells us to add up the terms 12 , 22 , 32 , 42 , and 52 . n=1 3

Example 638. ∑ (2n + 3) = (2 ⋅ 1 + 3) + (2 ⋅ 2 + 3) + (2 ⋅ 3 + 3) = 5 + 7 + 9 = 21. n=1 4

Example 639. ∑ n = 1 + 2 + 3 + 4 = 10. n=1 6

Example 640. ∑ 2n = 2 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 3 + 2 ⋅ 4 + 2 ⋅ 5 + 2 ⋅ 6 = 2 + 4 + 6 + 8 + 10 + 12 = 42. n=1 7

Example 641. ∑ 2n = 21 + 22 + 23 + 24 + 25 + 26 + 27 = 2 + 4 + 8 + 16 + 32 + 64 + 128 = 254. n=1 5

Example 642. ∑ 1 = 1 + 1 + 1 + 1 + 1 = 5. Here each term to be added up is simply the n=1

5

constant 1. And so, ∑ 1 is simply the sum of five 1s. n=1

3

Example 643. ∑ (10 − 2n) = (10 − 2 ⋅ 1) + (10 − 2 ⋅ 2) + (10 − 2 ⋅ 3) = 8 + 6 + 4. n=1

272

Guy’s Unsolved Problems in Number Theory (3e, p. 316), where it is attributed to Paul Erdős. The symbol σ is the lower-case Greek letter sigma.

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4

1 1 1 1 1 = + + + . 1 2 3 4 n=1 n

Example 644. ∑ 4

Example 645. ∑

n=1

1

(n + 1)

2

=

1

(1 + 1)

2

Example 646. Let x ∈ R. Then:

+

1

(2 + 1)

2

+

1

(3 + 1)

2

+

1

(4 + 1)

2

=

1 1 1 1 + + + . 4 9 16 25

5

∑ xn−1 = x1−1 + x2−1 + x3−1 + x4−1 + x5−1 = x0 + x1 + x2 + x3 + x4 = 1 + x + x2 + x3 + x4 .

n=1

It’s nice to have 1 as the starting point and that’s what we’ll usually do. But there’s no reason why the starting point must always be 1. Examples: Example 647. Starting point 3: 5

∑ n3 = 33 + 43 + 53 = 27 + 64 + 125 = 216.

n=3

Example 648. Starting point 0: 4

n=0

√ √ √ √ √ √ √ √ n = 0 + 1 + 2 + 3 + 4 = 0 + 1 + 2 + 3 + 2.

The starting point can even be negative: Example 649. Starting point −2:

n −2 −1 0 1 2 3 1 1 1 1 3 3 = + + + + + =− − +0+ + + = . 4 4 4 4 4 4 2 4 4 2 4 4 n=−2 4 3

Exercise 196. Rewrite each series in summation notation.

(a) 1 + 2 + 6 + 24 + 120 + 720 + 5 040. (b) 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23. (c) 1/2 + 1 + 3/2 + 2 + 5/2 + 3 + 7/2. (d) 8 + 7 + 6 + 5 + 4 + 3.

Exercise 197. Redo the last exercise, but with starting point 0. (Answer on p. 1807.)

Exercise 198. Find the sum of each series. (Observe that here the dummy or index variables are not the usual n. Instead, they are i, ⋆, and x.) (Answer on p. 1807.) 4

(a) ∑ (2 − i) . i=1

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i

(b)

17

∑ (4 ⋆ +5).

⋆=16

(c)

33

∑ (x − 3).

x=31

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44.1.

Summation Notation for Infinite Series

Example 650. In the following series, the “n = 1” below the ∑ symbol indicates that it has starting point 1. ∞

1 1 1 1 1 1 1 = + + + ⋅ ⋅ ⋅ = + + + ... n 1 2 3 2 2 2 2 2 4 8 n=1 ∑

The ∞ above the ∑ symbol indicates that there is no stopping point. This is thus an infinite series.

As already mentioned and as we’ll soon learn, this series converges to 1. We may write: ∞

1 = 1. n 2 n=1 ∑

If the context makes it crystal clear what the starting and stopping points are, we will sometimes be lazy/sloppy and omit them. And so here, we may also write: ∞

1 1 = = 1. ∑ n n 2 2 n=1 ∑

Example 651. ∑ n = ∑ n = 1 + 2 + 3 + . . . n=1

“Clearly”, this series diverges. We write: ∑ n = ∑ n = ∞. n=1

1 1 1 1 1 1 1 = = + + + ⋅ ⋅ ⋅ = 1 + + + ... ∑ 2 n2 12 22 32 4 9 n=1 n

Example 652. ∑

1 1 π =∑ 2 = . 2 n 6 n=1 n

As mentioned in Example 634, this series converges to π/6: ∑ ∞

Example 653. ∑ nx2 = ∑ nx2 = 1x2 + 2x2 + 3x2 + 4x2 + . . . n=1

This infinite series has starting point 1 and each term to be added up is nx2 . ∞

Thus, ∑ nx2 is the sum of infinitely many terms, namely x2 , 2x2 , 3x2 , 4x2 ... n=1

By the way, this series diverges for all x ≠ 0. That is, for all x ≠ 0, we have: ∞

∑ nx2 = ∑ nx2 = ∞.

n=1

And “obviously”, if x = 0, then this series converges to 0: ∞

∑ n ⋅ 02 = ∑ n ⋅ 02 = 0.

n=1

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Example 654. Consider the harmonic series: ∞

1 1 1 1 1 = ∑ = + + + ... n 1 2 3 n=1 n ∑

We have: 1

2

1 1 = = 1, 1 n=1 n

1 1 1 = + = 1.5, 1 2 n=1 n

100

200

1 ∑ = 5.187 . . . , n=1 n

1 ∑ = 5.878 . . . , n=1 n

3

1 1 1 1 = + + = 1.83, 1 2 3 n=1 n ∑

104

1 000

1 ∑ = 7.485 . . . , n=1 n

300

1 = 6.282 . . . , n=1 n ∑

105

1 ∑ = 9.787 . . . , n=1 n

1 = 12.090 . . . . n=1 n ∑

Does the harmonic series converge or diverge? From the above, it’s not obvious. It turns out that it diverges. Here’s a heuristic (i.e. not 100% rigorous) proof: 1 1 1 + + + . . . —“clearly”, this series diverges. 2 2 2 As we show below, this series is “smaller than” the harmonic series.273

First, consider the series 1 +

Thus, the harmonic series must “clearly” also diverge:

=
0; (b) Exact opposite directions if u = kv for some k < 0; and (c) Different directions if u ≠ kv for any k.

Example 700. Let a = (2, 0), b = (1, 0), c = (−3, 0), and d = (1, 1). The vectors a and b point in the same direction because a = 2b.

The vectors a and c point in the exact opposite directions because c = −1.5a. The vectors a and d point in different directions because a ≠ kd for any k.

Exercise 213. Continuing with the above example, explain if b points in the same, exact opposite, or different direction from each of c and d. (Answer on p. 1814.) Remark 91. Note the special case of the zero vector 0 = (0, 0)—it does not point in the same, exact opposite, or different direction as any other vector.

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49.12.

When Are Two Vectors Parallel?

Two non-zero vectors u and v are parallel if they point in the same or the exact opposite directions and non-parallel otherwise.283 Definition 133. Two non-zero vectors u and v are parallel if u = kv for some k and non-parallel otherwise.

(So, non-parallel and point in different directions are synonyms.)

As shorthand, we write a ∥ b if a and b are parallel and a ∥/ b if they are not.

Example 701. The vectors a = (2, 0), b = (1, 0), c = (−3, 0) are parallel. And so as shorthand, we may write a ∥ b, a ∥ c, and b ∥ c.

The vector d = (1, 1) is not parallel to a, b, or c. Equivalently, d = (1, 1) points in a different direction from a, b, and c. And so as shorthand, we may write d ∥/ a, d ∥/ b, and d ∥/ c.

Remark 92. Again, note the special case of the zero vector 0 = (0, 0)—it is neither parallel nor non-parallel to any other vector.

283

Just so you know, some writers call two vectors that point in exact opposite directions anti-parallel. In contrast, we call them parallel. We will not use the term anti-parallel in this textbook.

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49.13.

Unit Vectors

Definition 134. A unit vector is any vector of length 1.

√ √ Example 702. The vectors (1, 0), (0, 1), and ( 2/2, 2/2) are unit vectors: √ 3 ∣(1, 0)∣ = 12 + 02 = 1, √ ∣(0, 1)∣ = 02 + 12 = 1, 3 ¿ √ √ √ 2 √ Á √ 2 Á 2 2 2 2 À( 2 ) + ( 2 ) = ∣( , )∣ = Á + = 1. 3 2 2 2 2 4 4 Example 703. The vectors (1, 1) and (−1, −1) are not unit vectors: √ √ ∣(1, 1)∣ = 12 + 12 = 2 ≠ 1, 3 √ √ 2 2 ∣(−1, −1)∣ = (−1) + (−1) = 2 ≠ 1. 3

ˆ , is the vector that points in the same direction, Given a vector v, its unit vector, denoted v but has length 1. Formally: Definition 135. Given a non-zero vector v, its unit vector (or the unit vector in its direction) is: v ˆ=

1 v. ∣v∣

It is easy to verify that thus defined, any vector’s unit vector has length 1: Fact 98. Given any non-zero vector, its unit vector has length 1. Proof. By Fact 97, ∣ˆ v∣ = ∣

1 1 1 v∣ = ∣ ∣ ∣v∣ = ∣v∣ = 1. ∣v∣ ∣v∣ ∣v∣

Fact 99. Let v be a vector with unit vector v ˆ. If c ∈ R, then the vector cˆ v has length ∣c∣.

Proof. See Exercise 214.

Fact 100. Let a and b be non-zero vectors. Then: (a) (b) (c) (d)

ˆ a ˆ a ˆ a ˆ a

= = = ≠

ˆ b ˆ −b ˆ ±b ˆ ±b

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

a a a a

and and and and

b b b b

point in the same direction. point in exact opposite directions. are parallel. are non-parallel.

Proof. See p. 1590 (Appendices). 567, Contents

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Exercise 214. Prove Fact 99.

Exercise 215. Let A = (1, −3), B = (2, 0), and C = (5, −1) be points. What are the unit vectors of the following six vectors? (Answer on p. 1814.) Ð→ AB,

Ð→ AC,

Ð→ BC,

Ð→ 2AB,

Ð→ 3AC,

Ð→ 4BC.

ˆ the normalised vector of u, but we shall Remark 93. Note that some writers also call u not do so.

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49.14.

The Standard Basis Vectors y

j = (0, 1)

i = (1, 0)

x

The standard basis vectors i = (1, 0) and j = (0, 1) are simply the unit vectors that point in the directions of the positive x- and y-axes. Formally: Definition 136. The standard basis vectors (in 2D space) are i = (1, 0) and j = (0, 1).

It turns out that any vector can be written as the linear combination (i.e. weighted sum) of i’s and j’s: Example 704. Let A = (−2, −1), B = (1, 2), and C = (5, −2) be points. Their position vectors can be written as linear combinations (i.e. weighted sums) of i’s and j’s:

y

⎛1⎞ ⎛0⎞ Ð→ ⎛ 1 ⎞ = i + 2j = +2 OB = ⎝2⎠ ⎝0⎠ ⎝1⎠

x

⎛1⎞ ⎛0⎞ Ð→ ⎛ −2 ⎞ OA = = −2i−j = −2 − ⎝ −1 ⎠ ⎝0⎠ ⎝1⎠

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⎛1⎞ ⎛0⎞ Ð→ ⎛ 5 ⎞ OC = = 5i−2j = 5 −2 ⎝ −2 ⎠ ⎝0⎠ ⎝1⎠

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49.15.

Any Vector Is A Linear Combination of Two Other Vectors

We just learnt that any vector can be written as the linear combination (i.e. weighted sum) of the standard basis vectors i and j. It turns out that more generally, we can do the same using any two vectors, so long as they are non-parallel: Fact 101. Let a, b, and c be vectors. If a ∥/ b, then there are α, β ∈ R such that: c = αa + βb.

Proof. The next page gives a heuristic proof. For a formal proof, see p. 1591 (Appendices). Example 705. Consider the vectors a = (1, 2) and b = (3, 4). Since a ∥/ b, by Fact 101, any vector can be expressed as the linear combination of a and b.

Consider for example the vector u = (2, 2). We will find α, β ∈ R such that u = αa + βb. To do so, first write: u=

⎛2⎞ ⎛1⎞ ⎛3⎞ = αa + βb = α +β . ⎝2⎠ ⎝2⎠ ⎝4⎠

Write out the above vector equation as the following two cartesian equations: 2 = 1α + 3β 1

and

2 = 2α + 4β. 2

Now solve this system of (two) equations: = minus 2× = yields −2β = −2 or β = 1, so that α = −1. Thus: u=

2

1

⎛2⎞ ⎛1⎞ ⎛3⎞ =− + = −a + b. ⎝2⎠ ⎝2⎠ ⎝4⎠

Exercise 216. Let a = (1, 2) and b = (3, 4). Express each of the vectors v = (3, 2) and w = (−1, 0) as the linear combination of a and b. (Answer on p. 1814.)

Exercise 217. Explain why any vector can be written as a linear combination of the vectors a = (1, 3) and b = (7, 5). Then express each of the vectors i = (1, 0), j = (0, 1), and d = (1, 1) as the linear combination of a and b. (Answer on p. 1815.) Remark 94. For a somewhat recent application of Fact 101 in the A-Level exams, see N2013/I/6(i) (Exercise 616).

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The condition that a ∥/ b is important. If a ∥ b, then Fact 101 does not apply:

Example 706. Consider the vectors a = (1, 1) and b = (2, 2). Since a ∥ b, Fact 101 does not apply. For example, we cannot express v = (1, 2) written as the linear combination of a and b. (Indeed, this can only be done for vectors that are themselves parallel to a and b.)

Example 707. The vectors c = (3, 1) and d = (−3, −1) point in exact opposite directions. Since c ∥ d, Fact 101 does not apply. For example, we cannot express v = (1, 2) written as the linear combination of c and d. (Indeed, this can only be done for vectors that are themselves parallel to c and d.)

Here is a heuristic proof-by-picture of Fact 101.

Let a, b, and v be non-zero vectors, with a ∥/ b.

Place a’s tail at v’s tail. Then also place b’s head at v’s head.

v = αa + βb a

b

βb

αa As the above figure suggests, “obviously”, we can always find real numbers α and β so that the head of αa and the tail of βb coincide. In other words, there are real numbers α and β such that v = αa + βb.

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49.16.

The Ratio Theorem

Theorem 15. Let A and B be points with position vectors a and b. Let P be the point that divides the line segment AB in the ratio λ ∶ µ. Then P ’s position vector is: p=

µa + λb . λ+µ

A

The point P has position vector: p=

λ The point A has position vector a.

P

µa + λb . λ+µ

µ The point B has position vector b. O

B

Ð→ 1 Ð→ Proof. By Fact 95, AP = p − a and AB = b − a. Ð→ Ð→ Now observe that AP points in the same direction as AB, but has λ/ (λ + µ) times the length. Thus: Ð→ 2 AP =

Putting = and = together, we have: 1

2

Ð→ p = a + AP = a +

λ λ Ð→ AB = (b − a) . λ+µ λ+µ

λ (λ + µ) a + λ (b − a) µa + λb (b − a) = = . λ+µ λ+µ λ+µ

By the way, you do not need to mug the Ratio Theorem because the following is printed on p. 4 of List MF26: The point dividing AB in the ratio λ : µ has position vector

µa + λb λ+µ

Vector product:

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Example 708. Let A = (3, 4) and B = (−1, 2) be points. Let P be the point that divides the line segment AB in the ratio 3 ∶ 2. By the Ratio Theorem: p=

2a + 3b 2 ⎛ 3 ⎞ 3 ⎛ −1 ⎞ 1 ⎛ 3 ⎞ + = . = 3+2 5 ⎝ 4 ⎠ 5 ⎝ 2 ⎠ 5 ⎝ 14 ⎠

And hence:

y 3 B = (−1, 2)

A = (3, 4)

P = (0.6, 2.8)

2

x

P = (0.6, 2.8).

Example 709. Let C = (8, 3) and D = (2, −6) be points. Let Q be the point that divides the line segment CD in the ratio 3 ∶ 7. By the Ratio Theorem: q=

7c + 3d 7 ⎛ 8 ⎞ 3 ⎛ 2 ⎞ 1 ⎛ 62 ⎞ + = = 3+7 10 ⎝ 3 ⎠ 10 ⎝ −6 ⎠ 10 ⎝ 3 ⎠

y

Q = (6.2, 0.3).

and hence

C = (8, 3) 3 Q = (6.2, 0.3)

x

7

D = (2, −6)

Exercise 218. Let A = (1, 2), B = (3, 4), C = (1, 4), D = (2, 3), E = (−1, 2), F = (3, −4) be points. Find the points P , Q, and R which divide the line segments AB, CD, and EF in the ratios 5 ∶ 6, 5 ∶ 1, and 2 ∶ 3, respectively. (Answer on p. 1815.)

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50.

Lines

Recall284 that if l is a line, then l may be written as: l = {(x, y) ∶ ax + by + c = 0} ,

where at least one of a or b is non-zero.

More simply, we can say that the line l is described by the following cartesian equation:285 ax + by + c = 0.

In this chapter, we’ll learn a second method for describing lines, namely vector equations. We’ll start by introducing the concept of a line’s direction vector:

50.1.

Direction Vector

Informally, a direction vector of a line is any vector that’s parallel to the line. Formally: Ð→ Definition 137. Given any two distinct points A and B on a line, we call the vector AB a direction vector of the line. Example 710. Consider the line l described by the cartesian equation y = 2x + 3.

It contains the points: A = (0, 3)

and

B = (−1.5, 0).

Hence, a direction vector of l is:

Ð→ AB = B − A = (0, 3) − (−1.5, 0) = (1.5, 3).

l also contains the points: C = (−1, 1)

and

D = (1, 5).

Hence, another direction vector of l is:

y

D Ð→ AB = (1.5, 3) B

A

ÐÐ→ CD = D − C = (−1, 1) − (1, 5) = (2, 4).

284 285

ÐÐ→ CD = (2, 4)

C

x

Ch. ??. Some writers also call this a scalar equation, but we shan’t do so.

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As the above example suggests, direction vectors are not unique. If v is a direction vector of a line, then so too is any vector that’s parallel to v.

But no other vector is a direction vector of the line. That is, if u ∥/ v, then u is not a direction vector of the line. (And so, although the direction vector v isn’t unique, we can say that it is unique up to non-zero scalar multiplication.) Altogether then, if a line has direction vector v, then its direction vectors are exactly those that are parallel to v. Formally: Fact 102. Let u and v be vectors. Suppose v is a line’s direction vector. Then: ⇐⇒

u is also that line’s direction vector Proof. See p. 1591 (Appendices).

u ∥ v.

Example 711. The line l described by y = 2x + 3 has direction vectors: Ð→ ⎛ 1.5 ⎞ AB = ⎝ 3 ⎠

and

ÐÐ→ Ð→ We can easily verify that CD ∥ AB:

ÐÐ→ ⎛ 2 ⎞ CD = . ⎝4⎠

ÐÐ→ ⎛ 2 ⎞ 4 ⎛ 1.5 ⎞ 4 Ð→ = AB. CD = = ⎝ 4 ⎠ 3⎝ 3 ⎠ 3

Ð→ ÐÐ→ The following are parallel to AB and CD and are thus also direction vectors of l: −5

⎛ 2 ⎞ ⎛ −10 ⎞ = , ⎝ 4 ⎠ ⎝ −20 ⎠

In contrast, the following are not: ⎛1⎞ , ⎝1⎠

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⎛2⎞ , ⎝3⎠

π

⎛ 2 ⎞ ⎛ 2π ⎞ = , ⎝ 4 ⎠ ⎝ 4π ⎠ ⎛1⎞ , ⎝0⎠

and

⎛0⎞ , ⎝1⎠

17

and

⎛ 2 ⎞ ⎛ 34 ⎞ = . ⎝ 4 ⎠ ⎝ 68 ⎠ 0=

⎛0⎞ . ⎝0⎠

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Fact 103. If a line is described by the cartesian equation ax + by + c = 0 or by = −ax − c, then it has direction vector (b, −a).

Proof. Let D = (p, q) be any point on the line. Since D is on the line, it satisfies the line’s cartesian equation—that is: ap + bq + c = 0.

Now consider the point E = (p + b, q − a). We now show that E also satisfies the line’s cartesian equation and is thus is also on the line: a (p + b) + b (q − a) + c = ap + ab + bq − ab + c = ap + bq + c = 0. 3

Since D and E are both points on the line, by Definition 137, the line has direction vector: ÐÐ→ DE = E − D = (p + b, q − a) − (p, q) = (b, −a) .

Example 712. Consider the line described by 5x − 2y + 3 = 0 or y = 2.5x + 1.5. By the above Fact, it has direction vector (−2, −5).

Since (1, 2.5) ∥ (−2, −5), by Fact 102, it also has direction vector (1, 2.5). (As we’ll see on the next page, not coincidentally, this line’s gradient is also 2.5.)

y

5x − 2y + 3 = 0 or y = 2.5x + 1.5

−2x + 1 = 0

(0, 2)

(1, 2.5)

(−2, −5)

3y − 1 = 0

(3, 0)

x

Next, the line described by 3y − 1 = 0 or y = 1/3 has direction vector (3, 0).

And the line described by −2x + 1 = 0 or x = 0.5 has direction vector (0, 2).

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The following result follows readily from Fact 103: Corollary 10. (a) A horizontal line has direction vector (1, 0).

(b) A vertical line has direction vector (0, 1).

(c) A line with gradient m has direction vector (1, m).

Proof. Suppose the line is described by ax + by + c = 0. Then by Fact 103, the line has direction vector (b, −a).

(a) If the line is horizontal, then by Fact 15, a = 0. And so, the line has direction vector (b, 0). Since (1, 0) ∥ (b, 0), by Fact 102, the line also has direction vector (1, 0). (b) Similarly, if the line is vertical, then by Fact 15, b = 0. And so, the line has direction vector (0, −a). Since (0, 1) ∥ (0, −a), by Fact 102, the line also has direction vector (0, 1).

(c) The line’s gradient is −b/a = m. But (b, −a) ∥ (−b/a, 1). And so by Fact 102, the line also has direction vector (1, m). Example 713. The horizontal line y = −1 has direction vector (1, 0). The vertical line x = 2 has direction vector (0, 1).

The oblique line y = x + 1 has gradient 1 and thus direction vector (1, 1).

y

x=2

y =x+1

(1, 1)

y = −1

(0, 1)

(1, 0)

Exercise 219. For each line, write down a direction vector. (a) 3x − y + 2 = 0. 577, Contents

x

(b) 7y = −2x − 3.

(c) y = π.

(d) x = −5.

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50.2.

Cartesian to Vector Equations

Example 714. Consider the line l = {(x, y) ∶ 3x − y + 2 = 0}. It contains exactly those points (x, y) that satisfy the cartesian equation 3x − y + 2 = 0 or y = 3x + 2. More simply, we may say that this cartesian equation describes l. It turns out that we can also describe l using a vector equation.

To do so, first observe that l contains the point P = (0, 2). Also, it has gradient 3 and thus direction vector v = (1, 3). Since l is a straight line, it must also contain the points: P + 1v = (0, 2) + 1(1, 3) = (1, 5)

and

P − 1v = (0, 2) − 1(1, 3) = (−1, −1).

Indeed, l contains exactly those points R that can be expressed as P + λv = (0, 2) + λ(1, 3) for some real number λ. That is: l = {R ∶ R = (0, 2) + λ(1, 3) (λ ∈ R)}. 1

More simply, we may say that the vector equation = describes l. 1

Equivalently, l contains exactly those points R whose position vector r may be expressed as p + λv = (0, 2) + λ(1, 3) for some real number λ. That is: l = {R ∶ r = (0, 2) + λ(1, 3) (λ ∈ R)}. 2

Again, we may more simply say that the vector equation = describes l. 2

(By the way, = and = are subtly different—more on this in Ch. 50.3.) 1

2

As in Ch. ?? (Simple Parametric Equations), λ is a parameter. Here “λ ∈ R” says that λ takes on every value in R. And as λ varies, we get different points of the line. For example, λ = −1, λ = 0, and λ = 1 produce the following points: (−1, −1) = (0, 2)−1(1, 3),

(0, 2) = (0, 2) + 0(1, 3), and (1, 5) = (0, 2) + 1(1, 3).

l may be described by: 3x − y + 2 = 0, or 2 r = (0, 2) + λ(1, 3) (λ ∈ R).

(−1, −1) (λ = −1)

y

(0, 2) (λ = 0)

(1, 5) (λ = 1) (1, 3)

x

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(... Example continued from the previous page.)

Of course, l also contains infinitely many other points—each distinct value of λ ∈ R produces a distinct point on l. We noted in Fact 102 that a line’s direction vector is unique up to non-zero scalar multiplication. A direction vector of l is v = (1, 3), but so too is any non-zero scalar multiple of v. And so, here are three more ways to write l: l = { (x, y) ∶ r = = { (x, y) ∶ r =

= { (x, y) ∶ r =

⎛ 100 ⎞ ⎛0⎞ +λ ⎝ 300 ⎠ ⎝2⎠

⎛ −1 ⎞ ⎛ −100 ⎞ +λ ⎝ −1 ⎠ ⎝ −300 ⎠ ⎛ 1.5 ⎞ ⎛1⎞ +λ ⎝ 4.5 ⎠ ⎝5⎠

(λ ∈ R) }

(λ ∈ R) }

(λ ∈ R) }.

More generally, for any k ≠ 0 and u = k (1, 3), we may write:

l = {(x, y) ∶ r = (0, 2) + λu (λ ∈ R)} .

The foregoing discussion suggests the following general Definition of a line. Definition 138. A line is any set of points that can be written as: Ð→ {R ∶ OR = p + λv (λ ∈ R)} ,

where p and v ≠ 0 are some vectors.

The above Definition says that a line contains exactly those points R whose position vector Ð→ OR = r may be expressed as: Ð→ 2 OR = r = p + λv = (p1 , p2 ) + λ(v1 , v2 )

for some real number λ.

Equivalently, a line contains exactly those points R that may be expressed as: R = (p1 , p2 ) + λ(v1 , v2 ) 1

for some real number λ.

To repeat, here are what the vectors p and v and the number λ mean:

• p = (p1 , p2 ) is the position vector of some point on the line; • v = (v1 , v2 ) is a direction vector of the line; and • The parameter λ takes on every value in R; each distinct value produces a distinct point on the line. Note that Definition 138 is perfectly consistent with our earlier definition of a line (Definition 41). The difference is that Definition 41 “works” only in 2D space. In contrast, Definition 138 is more general—it “works” in 2D space and, as we’ll see, also in 3D space.286 286

Indeed, it also “works” in any n-dimensional space.

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And so, to write down a line’s vector equation, we need simply find any point on the line and any direction vector of the line. More examples to illustrate how this works: Example 715. The line l is described by y = 1 − x.

It contains the point (0, 1). It has gradient −1 and hence, by Corollary 10, the direction vector (1, −1). Thus, we can also describe the line l by: R = (0, 1) + λ(1, −1) 1

or

Both = and = say the same thing: 1

2

r = (0, 1) + λ(1, −1) 2

(λ ∈ R).

• = says that l contains exactly those points R that may be written as (0, 1) + λ(1, −1), for some real number λ. Ð→ 2 • = says that l contains exactly those points R whose position vector r = OR may be written as (0, 1) + λ(1, −1), for some real number λ. 1

y

The line l

(−1, 2) (λ = −1)

(1, −1)

(0, 1) (λ = 0)

l may be described by: y = 1 − x, or r = (0, 1) + λ(1, −1) (λ ∈ R).

x (1, 0) (λ = 1)

As the parameter λ takes on different values in R, we get different points of l. So for example, λ = −1, λ = 0, and λ = 1 produce the following points: (−1, 2) = (0, 1) − 1(1, −1), (0, 1) = (0, 1) + 0(1, −1), (1, 0) = (0, 1) + 1(1, −1).

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Example 716. The line l is described by y = 3.

It contains the point (0, 3). Since it’s horizontal, by Corollary 10, it has direction vector (1, 0). Hence, we can also describe the line l by: R = (0, 3) + λ(1, 0) 1

or

r = (0, 3) + λ(1, 0) 2

y

(1, 0)

The line l

(1, 3) (λ = 1)

(0, 3) (λ = 0)

(−1, 3) (λ = −1)

(λ ∈ R).

x

As λ varies, we get different points of l. So for example, λ = −1, λ = 0, and λ = 1 produce: (−1, 3) = (0, 3) − 1(1, 0), (0, 3) = (0, 3) + 0(1, 0), (1, 3) = (0, 3) + 1(1, 0).

Example 717. The line l is described by: x = −1.

It contains the point (−1, 0). Since it’s vertical, by Corollary 10, it has direction vector (0, 1) . Hence, we can also describe the line l by: R = (0, 1) + λ(0, 1) or 2 r = (0, 1) + λ(0, 1) (λ ∈ R). 1

As λ varies, we get different points of l. So for example, λ = −1, λ = 0, and λ = 1 produce: (−1, 2) = (0, 1) − 1(0, 1), (0, 1) = (0, 1) + 0(0, 1), (1, 0) = (0, 1) + 1(0, 1).

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(−1, 1) (λ = 1) (−1, 0) (λ = 0)

y

(0, 1)

x

(−1, −1) (λ = −1)

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Exercise 220. Each of the following (cartesian) equations describes a line. Rewrite each into a vector equation. Also write down the points corresponding to when your parameter takes on the values −1, 0, and 1. (Answer on p. 1816.) (a) −5x + y + 1 = 0.

(b) x − 2y − 1 = 0.

(c) y − 4 = 0.

(d) x − 4 = 0.

Exercise 221. In Definition 138 (of a line), we impose the restriction that a line’s direction vector v must be non-zero. By considering what the line becomes if v is the zero vector, explain why we impose this restriction. (Answer on p. 1816.) Remark 95. Here we repeat our earlier warning. A line {(x, y) ∶ ax + by + c = 0} is a set of points. But for the sake of convenience, we often simply say that the line may be described by the cartesian equation ax + by + c = 0. And if we’re especially lazy or sloppy, we might even say that the line is the equation ax + by + c = 0 (even though strictly speaking, this is wrong because a line is not an equation—it is a set).

Here likewise, a line {R = (x, y) ∶ r = p + λv, λ ∈ R} is a set of points. But for the sake of convenience, we will often simply say that the line may be described by the vector equation r = p + λv (λ ∈ R). And if we’re especially lazy or sloppy, we might even say that the line is the equation ax + by + c = 0 (even though again, strictly speaking, this is wrong because a line is not an equation—it is a set).

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50.3.

Pedantic Points to Test/Reinforce Your Understanding

As we’ve seen above, a line l may be described using either of the following equations: ª © 1 © R = P + λv

Point

Point

Vector

Vector

Or:

Vector

(λ ∈ R).

(λ ∈ R).

Here are three pedantic points that can serve as a useful test of your understanding: Pedantic Point #1. = is consistent with what we learnt earlier (in Ch. 49.9): 1

Point = Point + Vector.

= is also consistent with what we learnt earlier: 2

Vector = Vector + Vector.

So, both vector equations = and = are perfectly correct ways to describe the exact same line line. 1

2

The difference is that = does so “more directly” than does =. Because, to repeat: 1

2

• = says that l contains exactly those points R that may be written as P + λv, for some real number λ. Ð→ 2 • = says that l contains exactly those points R whose position vector r = OR may be written as p + λv, for some real number λ. 1

Pedantic Point #2. What would be wrong and unacceptable is the following: Vector © 3 Vector © © R = p +λv

Point

(λ ∈ R),

7

As we learnt earlier (Ch. 49.9), Vector + Vector = Vector. But the LHS of = is a Point 3 while its RHS is a Vector. Thus, = is false. 3

Similarly, the following is also wrong and unacceptable: Vector © 4 © © r = P +λv

Vector

Point

(λ ∈ R).

7

As we also learnt earlier, Point + Vector = Point. But the LHS of = is a Vector while its 4 RHS is a Point. Thus, = is false. 4

Pedantic Point #3. A line is a set of points and not a set of vectors. So, take care to note that the line l contains the points R = (x, y) and P = (p1 , p2 )—it does not contain the vectors r = (x, y) and p = (p1 , p2 ). 583, Contents

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50.4.

Vector to Cartesian Equations

Suppose a line is described by the following vector equation: r = p + λv

(λ ∈ R).

⎛v ⎞ ⎛ x ⎞ ⎛ p1 ⎞ = + λ 1 (λ ∈ R). ⎝ v2 ⎠ ⎝ y ⎠ ⎝ p2 ⎠

Or equivalently:

Then given any point (x, y) on this line, there must be some real number λ such that: x = p1 + λv1

and

y = p2 + λv2 .

We say that the line may be described by the above pair of cartesian equations. Hm ... but aren’t we supposed to be able to describe a line with just one cartesian equation? Well, if we’d like, we can do some easy algebra to eliminate the parameter λ: Example 718. The line l is described by: r = (1, 2) + λ(1, 1)

(λ ∈ R).

We can also describe l by the following pair of cartesian equations: x=1+λ⋅1 1

y y = x + 1 or r = (1, 2) + λ(1, 1)

1

y = 2 + λ ⋅ 1. 2

and

x

We can eliminate the parameter λ through sim2 1 ple algebra. = minus = yields: y−x=1

or

y = x + 1.

Example 719. The line l is described by: r = (0, 0) + λ(4, 5)

(λ ∈ R).

We can also describe l by the following pair of cartesian equations: x=0+λ⋅4 1

5 1 2 = minus × = yields: 4

and

5 y− x=0 4

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or

y = 0 + λ ⋅ 5. 2

y 5 y = x or 4 r = (0, 0) + λ(4, 5)

x

5 y = x. 4 www.EconsPhDTutor.com

Example 720. The line l is described by: r = (3, 1) + λ(0, 2)

y

(λ ∈ R).

We can also describe l by the following pair of cartesian equations: x=3+λ⋅0=3 1

x = 3 or r = (3, 1) + λ(0, 2)

y = 1 + λ ⋅ 2. 2

and

x

Observe that in this example, we cannot use algebra to eliminate λ. It turns out that this is actually a vertical line.

As λ varies, the value of x is fixed at x = 3, while y varies along with λ. And so, instead 2 of doing any algebra, we’ll simply discard =. The above pair of cartesian equations then reduces to the single equation: 1

x = 3. 1

Example 721. The line l is described by: r = (−1, 2) + λ(−1, 0)

(λ ∈ R).

We can also describe l by the following pair of cartesian equations: x = −1 + λ ⋅ (−1) 1

and

y = 2 + λ ⋅ 0 = 2. 2

y

y = 2 or r = (−1, 2) + λ(−1, 0)

x

Observe that in this example, we cannot use algebra to eliminate λ. It turns out that this is actually a horizontal line.

As λ varies, the value of y is fixed at y = 2, while x varies along with λ. And so, instead 1 of doing any algebra, we’ll simply discard =. The above pair of cartesian equations then reduces to the single equation: 2

y = 2. 2

Exercise 222. Each of the following vector equations describes a line. Rewrite each into cartesian equation form. (Answer on p. 1816.) (a) r = (−1, 3) + λ (1, −2) (λ ∈ R). (c) r = (0, −3) + λ (3, 0) (λ ∈ R).

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(b) r = (5, 6) + λ (7, 8) (λ ∈ R). (d) r = (1, 1) + λ (0, 2) (λ ∈ R).

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In general: Fact 104. Let l be the line described by r = (p1 , p2 ) + λ(v1 , v2 )

(a) If v1 , v2 ≠ 0, then l can be described by: x − p1 y − p2 = v1 v2

y=

or, rearranging:

(λ ∈ R).

v2 v2 x + p2 − p1 . v1 v1

(b) If v1 = 0, then l is vertical and can be described by x = p1 . (c) If v2 = 0, then l is horizontal and can be described by y = p2 . x = p1 + λv1 1

Proof. First, write:

Then v1 × = minus v2 × = yields: 2

1

and

y = p2 + λv2 . 2

v1 y − v2 x = v1 p2 + λv1 v2 − v2 p1 − λv1 v2 = v1 p2 − v2 p1 . v2 (x − p1 ) = v1 (y − p2 ). 3

Or:

(a) If v1 , v2 ≠ 0, then = divided by v1 v2 yields 3

(b) If v1 = 0, then = becomes x = p1 . 3

x − p1 y − p2 = . v1 v2

(c) If v2 = 0, then = becomes y = p2 . 3

Armed with Fact 104, we now revisit the last four examples. Example 722. The line r = (1, 2) + λ(1, 1) (λ ∈ R) may be described by: x−1 y−2 = 1 1

or, rearranging:

y = x + 1.

Example 723. The line r = (0, 0) + λ(4, 5) (λ ∈ R) may be described by: x−0 y−0 = 4 5

or, rearranging:

5 y = x. 4

Example 724. The line r = (3, 1) + λ(0, 2) (λ ∈ R) may be described by x = 3.

Example 725. The line r = (−1, 2) + λ(−1, 0) (λ ∈ R) may be described by y = 2. Exercise 223. Use Fact 104 to redo Exercise 222.

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51.

The Scalar Product

Definition 139. Given vectors u = (u1 , u2 ) and v = (v1 , v2 ), their scalar product, denoted u ⋅ v, is the number: Example 726. Let u = u⋅v =

u ⋅ v = u1 v1 + u2 v2 .

⎛8⎞ ⎛ −4 ⎞ ⎛2⎞ ⎛ 5 ⎞ . Then: , and x = , w= , v= ⎝7⎠ ⎝ 0 ⎠ ⎝1⎠ ⎝ −3 ⎠

⎛ 5 ⎞ ⎛2⎞ ⋅ = ⎝ −3 ⎠ ⎝ 1 ⎠

5 ⋅ 2 + (−3) ⋅ 1

= 10 − 3 = 7,

⎛ 5 ⎞ ⎛8⎞ ⋅ = ⎝ −3 ⎠ ⎝ 7 ⎠

5 ⋅ 8 + (−3) ⋅ 7

= 40 − 21 = 19,

u⋅w =

⎛ 5 ⎞ ⎛ −4 ⎞ ⋅ = 5 ⋅ (−4) + (−3) ⋅ 0 = −20 + 0 = −20, ⎝ −3 ⎠ ⎝ 0 ⎠

v⋅w =

⎛ 2 ⎞ ⎛ −4 ⎞ ⋅ = ⎝1⎠ ⎝ 0 ⎠

u⋅x =

2 ⋅ (−4) + 1 ⋅ 0

= −8 + 0 = −8.

The scalar product is itself simply a scalar (i.e. a real number). Hence the name. Remark 96. The scalar product is also called the dot product or the inner product. But it appears that your A-Level exams and syllabus do not use these terms. And so neither shall we. We will stick strictly to the term scalar product. Right now, the scalar product may seem like a totally random and useless thing, but as we’ll soon learn, it is plenty useful. Let us first learn about a few of its properties. Recall from primary school that ordinary multiplication is commutative: Example 727. 3 × 5 = 15 and 5 × 3 = 15.

Moreover, ordinary multiplication is distributive (over addition): Example 728. 3 × (5 + 11) = 3 × 5 + 3 × 11 and 18 × (7 − 31) = 18 × 7 + 18 × (−31).

It turns out that the scalar product is likewise commutative and distributive:

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Fact 105. Let a, b, and c be vectors. Then:

(a) a ⋅ b = b ⋅ a. (b) a ⋅ (b + c) = a ⋅ b + a ⋅ c.

The fact that the scalar product is both commutative and distributive is a simple consequence of the fact that multiplication is itself commutative and distributive.287 Proof. Let288 a = (a1 , a2 ), b = (b1 , b2 ), and c = (c1 , c2 ). Then:

a ⋅ b = a1 b1 + a2 b2 = b1 a1 + b2 a2 = b ⋅ a.

(a)

(b)

a ⋅ (b + c) = a1 (b1 + c1 ) + a2 (b2 + c2 ) = a1 b1 + a2 b2 + a1 c1 + a2 c2 = a ⋅ b + a ⋅ c.

Example 729. Continue to let u = (5, −3), v = (2, 1), w = (−4, 0), and x = (8, 7).

The scalar product is commutative: v ⋅ u = u ⋅ v = 7,

w ⋅ u = u ⋅ w = −20,

It is also distributive over addition: u ⋅ (v + w) =

(u + v) ⋅ w =

x ⋅ u = u ⋅ x = 19.

⎛ 5 ⎞ ⎛2−4⎞ ⋅ = −10 − 3 = −13 = 7 − 20 = u ⋅ v + u ⋅ w, ⎝ −3 ⎠ ⎝ 1 + 0 ⎠

⎛ 5 + 2 ⎞ ⎛ −4 ⎞ ⋅ = −28 + 0 = −28 = −20 − 8 = u ⋅ w + v ⋅ w. ⎝ −3 + 1 ⎠ ⎝ 0 ⎠

Here’s another “obvious” property of the scalar product:

Fact 106. Suppose a and b be vectors and c ∈ R be a scalar. Then: (ca) ⋅ b = c (a ⋅ b).

Proof. Let289 a = (a1 , a2 ) and b = (b1 , b2 ). So, ca = (ca1 , ca2 ). Thus:

(ca) ⋅ b = (ca1 ) b1 + (ca2 ) b2 = c (a1 b1 + a2 b2 ) = c (a ⋅ b) .

Exercise 224. Let v = (2, 1), w = (−4, 0), and x = (8, 7). Above we already computed v ⋅ w = −8. Now also compute the following: (Answer on p. 1817.) v ⋅ x, w ⋅ x, w ⋅ v, x ⋅ v, x ⋅ w, w ⋅ (x + v), (2v) ⋅ x, and w ⋅ (2x).

The latter is, in turn, a fact we will simply take for granted in this textbook. The proof covers only the two-dimensional case. For a more general proof, see p. 1592 (Appendices). 289 This proof covers only the two-dimensional case. For a more general proof, see p. 1592 (Appendices). 287

288

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51.1.

A Vector’s Scalar Product with Itself

It turns out that a vector’s length is the square root of the scalar product with itself: Fact 107. Suppose v be a vector. Then ∣v∣ =

√ 2 v ⋅ v and ∣v∣ = v ⋅ v.

√ Proof. By Definition 121, ∣v∣ = v12 + v22 . By Definition 139, v ⋅ v = v1 v1 + v2 v2 = v12 + v22 . √ 2 Hence, ∣v∣ = v ⋅ v and ∣v∣ = v ⋅ v. Exercise 225. Let u = (5, −3), v = (2, 1), w = (−4, 0), and x = (8, 7).

The lengths of each vector are:

∣u∣ =

∣v∣ =

∣w∣ = ∣x∣ =

√ 2 52 + (−3) √ 22 + 12 √ 2 (−4) + 02 √ 82 + 72

=

34 √ = 5 = =

4

√ 113

And the square roots of the scalar product of each vector with itself are: √ u⋅u √ v⋅v √ w⋅w √ x⋅x

√ (5, −3) ⋅ (5, −3) √ = (2, 1) ⋅ (2, 1) √ = (−4, 0) ⋅ (−4, 0) √ = (8, 7) ⋅ (8, 7) =

√ 25 + 9 √ = 4+1 √ = 16 + 0 √ = 64 + 49 =

√ 34 √ = 5 = = =

4

√ 113

Exercise 226. You are given the vectors a = (−2, 3), b = (7, 1), and c = (5, −4). Verify that the length of each vector is equal to the square root of each vector’s scalar product with itself. (Answer on p. 1817.)

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52.

The Angle Between Two Vectors v

In the figure on the right, the tails of the vectors u and v are placed at the same point. Informally, the angle between u and v is the “amount” by which we must rotate u so that it points in the same direction as v.

2π − α

Note though that we can either rotate u anticlockwise by α or clockwise by 2π − α. So, we have an ambiguity here—which is the angle between u and v? Is it α or 2π−α?

α

u To resolve this ambiguity, we will simply define the angle between two vectors so that it is always the smaller of these two angles. In other words, we’ll define the angle between two vectors so that it’s always between 0 and π.

And so, in the above figure, the angle between u and v is α (and not 2π − α).

In contrast, in the figure below, the angle between w and x is β (and not 2π − β).

x

w

β 2π − β

Previously in Ch. ??, we gave two informal definitions of angle (Definitions 82 and 84). As promised in Ch. ??, we now give our formal Definition of the angle between two vectors. Be warned that it comes seemingly outta nowhere. But don’t worry, Exercise 227 (next page) will help you understand where this Definition comes from. Definition 140. The angle between two non-zero vectors u and v is the number: cos−1

u⋅v . ∣u∣ ∣v∣

Recall290 that the range of cos−1 is [0, π]. And so, by the above Definition, the angle between two vectors is indeed always between 0 and π.

290

P. ??.

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Exercise 227. Let u and v be vectors and θ be the angle between them.

v θ

u

This Exercise will help you understand why we define θ = cos−1

u⋅v . ∣u∣ ∣v∣

(a) Write down the vector that corresponds to the third side of the above triangle. (b) Write down the lengths of the triangle’s three sides in terms of u and v. (c) The Law of Cosines (Proposition 6) states that if a triangle has sides of lengths a, b, and c and has angle C opposite the side of length c, then: c2 = a2 + b2 − 2ab cos C.

Use the Law of Cosines to write down an equation involving θ, u, and v.

(d) Use distributivity (Fact 105) to prove that (u − v) ⋅ (u − v) = u ⋅ u + v ⋅ v − 2u ⋅ v.

(e) Now take the equation you wrote down in (c), do the algebra (hint: use Fact 107), and hence show that: θ = cos−1

u⋅v . ∣u∣ ∣v∣

A simple rearrangement of Definition 140 produces the following result: Fact 108. If u and v are two non-zero vectors and θ is the angle between them, then: u ⋅ v = ∣u∣ ∣v∣ cos θ.

Example 730. Let θ be the angle between the vectors i = (1, 0) and u = (1, 1). Since i points east, while u points north-east, we know from primary school trigonometry that θ = π/4.

Let’s verify that this is consistent with Definition 140: cos−1

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i⋅u (1, 0) ⋅ (1, 1) = cos−1 ∣i∣ ∣u∣ ∣(1, 0)∣ ∣(1, 1)∣

1⋅1+0⋅1 √ = cos−1 √ 12 + 02 12 + 12 1 π = cos−1 √ = . 3 2 4

u = (1, 1) π 4

i = (1, 0)

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Example 731. Let θ be the angle between the vectors i = (1, 0) and j = (0, 1). Since i points east, while j points north, we know from primary school trigonometry that θ = π/2. Let’s verify that this is consistent with Definition 140: cos−1

(1, 0) ⋅ (0, 1) i⋅j = cos−1 ∣i∣ ∣j∣ ∣(1, 0)∣ ∣(0, 1)∣ = cos−1 √

j = (0, 1)

1⋅0+0⋅1 √ 12 + 02 02 + 12

π 2

π = cos−1 0 = . 3 2

Example 732. Consider the vectors v = (3, 2) and w = (−1, −4).

i = (1, 0) v = (3, 2)

By Definition 140: cos−1

v⋅w (3, 2) ⋅ (−1, −4) = cos−1 ∣v∣ ∣w∣ ∣(3, 2)∣ ∣(−1, −4)∣

2.404

3 × (−1) + 2 × (−4) = cos−1 √ √ 2 2 32 + 22 (−1) + (−4)

−3 − 8 −11 = cos−1 ( √ √ ) = cos−1 ( √ ) ≈ 2.404. 13 17 221

w = (−1, −4)

By the way, here’s a possible concern. We’ve defined the angle between two vectors as: cos−1

u⋅v . ∣u∣ ∣v∣

But recall291 that the domain of arccosine is [−1, 1]. So, how can we be sure that the above expression is always well-defined? In other words, how can we be sure that: −1 ≤

u⋅v ≤ 1? ∣u∣ ∣v∣

Fortunately, we can, thanks to Cauchy’s Inequality:292 Fact 109. (Cauchy’s Inequality.) Let u and v be non-zero vectors. Then:

Equivalently:

292

u⋅v ≤ 1. ∣u∣ ∣v∣

− ∣u∣ ∣v∣ ≤ u ⋅ v ≤ ∣u∣ ∣v∣

Proof. See p. 1592 (Appendices). 291

−1 ≤

or

(u ⋅ v) ≤ ∣u∣ ∣v∣ . 2

2

2

P. ??. Also known as the Cauchy-Schwarz Inequality in its more general form.

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Fact 110. Let u and v be non-zero vectors and θ be the angle between them. Then: (a) (b) (c) (d) (e)

u⋅v =1 ∣u∣ ∣v∣

u⋅v ∈ (0, 1) ∣u∣ ∣v∣ u⋅v =0 ∣u∣ ∣v∣

⇐⇒ θ = 0.

π ⇐⇒ θ ∈ (0, ). 2 π ⇐⇒ θ = . 2

And thus: (i) u ⋅ v > 0 ⇐⇒ θ is acute or zero.

u⋅v π ∈ (−1, 0) ⇐⇒ θ ∈ ( , π). ∣u∣ ∣v∣ 2 u⋅v = −1 ⇐⇒ θ = π. ∣u∣ ∣v∣

(ii) u ⋅ v = 0 ⇐⇒ θ is right.

(iii) u ⋅ v < 0 ⇐⇒ θ is obtuse or straight.

Proof. To prove this, simply apply Definition 140.

y

cos−1

−1

π

π 2

1

x

Fact 110(ii) motivates the following Definition: Definition 141. Two non-zero vectors u and v are perpendicular (or normal or orthogonal) if u ⋅ v = 0 and non-perpendicular if u ⋅ v ≠ 0.

As shorthand, we write u ⊥ v if two non-zero vectors u and v are perpendicular; and u ⊥/ v if they aren’t. Remark 97. Again, note the special case of the zero vector 0 = (0, 0)—it is neither perpendicular nor non-perpendicular to any other vector.

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Fact 111. Let u and v be non-zero vectors. Then:

u⋅v ∣u∣ ∣v∣ u⋅v (b) ∣u∣ ∣v∣ u⋅v (c) ∣u∣ ∣v∣ u⋅v (d) ∣u∣ ∣v∣ (a)

=1

⇐⇒

u and v point in the same direction.

= ±1

⇐⇒

u and v are parallel.

= −1 ∈ (−1, 1)

⇐⇒

u and v point in exact opposite directions.

⇐⇒

u and v point in different directions.

Proof. “Obviously”, (c) and (d) simply follow from (a) and (b). For the proof of (a) and (b), see p. 1594 (Appendices). More examples: Example 733. Let θ be the angle between the vectors u = (1, −3) and v = (−2, 4). Then: θ = cos−1

−2 − 12 −14 u⋅v −14 = cos−1 √ = cos−1 √ √ = cos−1 √ ≈ 2.999. √ 2 2 ∣u∣ ∣v∣ 10 20 10 2 12 + (−3) (−2) + 42

So, u and v are neither perpendicular nor parallel; instead, they point in different directions. Moreover, the angle between them is obtuse.

v = (−2, 4) θ ≈ 2.999 u = (1, −3)

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Example 734. Let θ be the angle between the vectors u = (1, −2) and v = (−2, 4). Then: θ = cos−1

u⋅v −2 − 8 −10 = cos−1 √ √ = cos−1 −1 = π. = cos−1 √ √ 2 2 ∣u∣ ∣v∣ 5 20 12 + (−2) (−2) + 42

So, u and v are parallel; more specifically, they point in exact opposite directions. Moreover, the angle between them is straight.

v = (−2, 4)

θ=π u = (1, −2)

Example 735. Let θ be the angle between the vectors u = (3, −1) and v = (1, 3). Then: θ = cos−1

3−3 π u⋅v −1 = cos−1 √ . = cos 0 = 2√ 2 ∣u∣ ∣v∣ 2 2 2 3 + (−1) 1 + 3

So, u and v are perpendicular and the angle between them is right.

v = (1, 3)

θ=

π 2 u = (3, −1)

Example 736. Let θ be the angle between the vectors u = (1, −2) and v = (2, −4). Then: θ = cos−1

u⋅v 2+8 10 = cos−1 √ = cos−1 √ √ = cos−1 1 = 0. √ 2 2 ∣u∣ ∣v∣ 5 20 12 + (−2) 22 + (−4)

So, u and v are parallel; more specifically, they point in the same direction. Moreover, the angle between them is zero.

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u = (1, −2)

θ=0

v = (2, −4)

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We can use Fact 110(i)–(iii) to quickly check if the angle between two vectors is acute, zero, or obtuse: Example 737. Even without doing any precise calculations or drawing any graphs, we can quickly see that: • The angle between the vectors (817, −2) and (39, −55) is acute, because “clearly”: 817 ⋅ 39 + (−2) ⋅ (−55) > 0.

√ √ • The vectors ( 79300, −470) and (47, 793) are perpendicular, because “clearly”: √ √ 79300 ⋅ 47 + (−470) ⋅ 793 = 0.

• If k < 0, then the angle between (67, k) and (−485, 32) is obtuse, because “clearly”: 67 ⋅ (−485) + 32k < 0.

Exercise 228. In each of the following, find the angle between u and v. Is it zero, acute, right, obtuse, or straight? Are u and v perpendicular or parallel? Do they point in the same, exact opposite, or different directions? (Answers on p. 1818.) (a) u = (2, 0) and v = (0, 17). √ (c) u = (1, 0) and v = (1, 3).

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(b) u = (5, 0) and v = (−3, 0). (d) u = (2, −3) and v = (1, 2).

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52.1.

Pythagoras’ Theorem and Triangle Inequality

Remember Pythagoras’ Theorem (Theorem 2)? Here it is again, but this time in the language of vectors: Theorem 16. (Pythagoras’ Theorem.) If u ⊥ v, then ∣u + v∣ = ∣u∣ + ∣v∣ . 2

2

2

Proof. See Exercise 229.

u+v

v

u Exercise 229. Use Facts 105 and 107 to show that ∣u + v∣ = ∣u∣ + 2u ⋅ v + ∣v∣ . Then use u ⊥ v to complete the proof of the above Theorem. (Answer on p. 1819.) 2

2

2

Remember the Triangle Inequality (Fact ??)? Here it is again, but this time in the language of vectors: Fact 112. (Triangle Inequality.) If u and v are vectors, then ∣u + v∣ ≤ ∣u∣ + ∣v∣.

Proof. See Exercise 230.

Exercise 230. In Exercise 229, we already showed that: ∣u + v∣ = ∣u∣ + 2u ⋅ v + ∣v∣ . 2

2

2

To prove Fact 112, first apply Cauchy’s Inequality (Fact 109) to the above equation; then complete the square and take square roots. (Answer on p. 1819.)

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52.2.

Direction Cosines

Definition 142. The x- and y-direction cosines of the vector v = (v1 , v2 ) are the numbers: v1 ∣v∣

and

Observe that the unit vector of v = (v1 , v2 ) is: ˆ=( v

v2 . ∣v∣

v1 v2 , ). ∣v∣ ∣v∣

And so, equivalently, v’s x- and y-direction cosines are the x- and y-coordinates of its unit vector.

We now explain why the x- and y-direction cosines are so named. Place the tail of v = (v1 , v2 ) at the origin. Let α be the angle between v and the positive x-axis. Similarly, let β be the angle between v and the positive y-axis.293

y

v

b

ˆ v

β α

a

x

ˆ = (a, b) be v’s unit vector. It has length 1 and forms the hypotenuse of two right Let v triangles. From the lower-right triangle, we have a = cos α. Similarly, from the upper-left triangle, we have b = cos β.

This explains why v’s unit vector’s x- and y-coordinates are also its x- and y-direction cosines.

293

More formally, α is the angle between v and i = (1, 0), while β is the angle between v and j = (0, 1).

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We can state and prove what was just said a bit more formally: Fact 113. Let v = (v1 , v2 ) be a non-zero vector. Let α and β be the angles between v and each of i and j. Then: cos α =

v1 ∣v∣

and

cos β =

v2 . ∣v∣

Proof. Since α is the angle between v and i, by Definition 140, we have: cos α =

v ⋅ i v1 ⋅ 1 + v2 ⋅ 0 v1 = = . ∣v∣ ∣i∣ ∣v∣ ⋅ 1 ∣v∣

Similarly, since β is the angle between v and j, we have: cos β =

v ⋅ j v1 ⋅ 0 + v2 ⋅ 1 v2 = = . ∣v∣ ∣j∣ ∣v∣ ⋅ 1 ∣v∣

Example 738. Consider the vector v = (3, 2). Its x- and y-direction cosines are: 3 v1 3 =√ =√ ∣v∣ 13 32 + 22

and

v2 2 2 =√ . =√ ∣v∣ 13 32 + 22

3 2 ˆ = ( √ , √ ). Which means, of course, that its unit vector is v 13 13

y

√ 2/ 13

v = (3, 2) ˆ v β α

√ 3/ 13

x

Let α and β be the angles it makes with the positive x- and y-axes. Then: 3 α = cos−1 √ ≈ 0.588 13

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and

2 β = cos−1 √ ≈ 0.983. 13

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Example 739. Consider the vector v = (−2, −1). Its x- and y-direction cosines are: v1 −2 −2 =√ =√ 2 2 ∣v∣ 5 (−2) + (−1)

v2 −1 −1 =√ . =√ 2 2 ∣v∣ 5 (−2) + (−1)

and

−2 −1 ˆ = ( √ , √ ). Which means, of course, that its unit vector is v 5 5

y

√ −2/ 13 ˆ v

β

x

α

v = (−2, −1)

√ −1/ 5

Let α and β be the angles it makes with the positive x- and y-axes. Then: −2 α = cos−1 √ ≈ 2.678 5

and

−1 β = cos−1 √ ≈ 2.034. 5

Exercise 231. Find each vector’s x- and y-direction cosines. Then write down its unit vector. (Answer on p. 1819.) (a) (1, 3).

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(b) (4, 2).

(c) (−1, 2).

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53.

The Angle Between Two Lines

(Below, Corollary 11 will give the “formula” for the angle between two lines. If you’re a well-trained, mindless Singaporean monkey who cares only about “knowing” the “formula” without understanding where it comes from, you can skip straight to Corollary 11.) “Obviously”, any two non-parallel lines l1 and l2 in 2D space must intersect at exactly one point. And at this intersection point, two angles are formed. Let’s call the smaller angle α, so that the larger angle is β = π − α.

Now, we have a potential ambiguity: When we talk about the angle between l1 and l2 , are we talking about the smaller angle α or the larger angle β? To resolve this ambiguity, this textbook will adopt the convention that the angle between two lines is the smaller one. So, by this convention, the angle between l1 and l2 shall be α (and not β).

l2

l1

α β =π−α

β =π−α

α

Note that by our adopted convention, the angle between two lines will always be acute (i.e. between 0 and π/2) and never obtuse. We now work towards a formal definition of the angle between two lines. Example 740. Given the lines l1 and l2 , we pick for each the direction vectors u and v. Observe that: Angle between Angle between α= = l1 and l2 u and v.

v

l2

l1 u α

That is, α is the angle between the two lines; moreover, it is also the angle between the two vectors. The above Example suggests the following “Definition” for the angle between two lines. Given two lines l1 and l2 , pick for each any direction vectors u and v. Then define: Angle between Angle between = l1 and l2 u and v.

This “Definition” works well in the above Example, but only because the angle between u and v happens to be acute. Unfortunately and as the next example illustrates, this “Definition” doesn’t work so well if the angle between the two chosen direction vectors is instead obtuse: 601, Contents

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Example 741. We continue with the same two lines as before. We continue to pick the direction vector u for the line l1 . But this time, we pick the direction vector w for the line l2 .

l2

l1 u

The angle between the two lines remains the acute angle α.

α

However, the angle between the two chosen direction vectors is now the obtuse angle β.

β

And so, the above “Definition” fails because: α=

Angle between Angle between ≠ = β. l1 and l2 u and w

w

In this case, the angle between the two lines, α, is actually the supplement of the angle between the chosen two direction vectors, β. That is: α = π − β.

The following Definition of the non-obtuse angle between two vectors will prove convenient: Definition 143. Let α denote the non-obtuse angle between the vectors u and v. Let β be the angle between u and v. Then: ⎧ ⎪ ⎪ ⎪β α=⎨ ⎪ ⎪ ⎪ ⎩π − β

if β is not obtuse, if β is obtuse.

Example 742. The angle between the vectors c and d is β, which is obtuse. And so, the non-obtuse angle between them is α = π − β.

e

c

α

β d

γ

f

The angle between the vectors e and f is γ, which is acute. And so, the non-obtuse angle between them is also γ. We are now ready to write down our formal Definition of the angle between two lines:

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Definition 144. Given two lines, pick for each any direction vector. We call the nonobtuse angle between these two vectors the angle between the two lines. Example 743. Given the lines l1 and l2 , we pick the direction vectors u and v. The angle between u and v is α, which is acute. And so, the non-obtuse angle between u and v is also α. Thus, by Definition 144, the angle between the two lines is α.

v

l2

l1

l3

u

w

γ

α β l4 x

Given the lines l3 and l4 , we pick the direction vectors w and x. The angle between w and x is β, which is obtuse. And so, the non-obtuse angle between w and x is γ = π − β. Thus, by Definition 144, the angle between the two lines is γ.

We just wrote down the Definition of the angle between two lines. We now work towards Corollary 11, which will give us our “formula” for the angle between two lines. u⋅v Recall that by Definition 140, the angle between u and v is: cos−1 . ∣u∣ ∣v∣ It turns out that we can get the non-obtuse angle between u and v simply by slapping ∣⋅∣ (the absolute value function) onto the numerator: Fact 114. The non-obtuse angle between two non-zero vectors u and v is: cos−1

∣u ⋅ v∣ . ∣u∣ ∣v∣

Proof. Suppose 0 ≤ θ ≤ π/2. Then u ⋅ v ≥ 0. And so, by Definition 143, the non-obtuse angle between u and v is: cos−1

u⋅v ∣u ⋅ v∣ = cos−1 . ∣u∣ ∣v∣ ∣u∣ ∣v∣

3

Suppose instead θ > π/2. Then u ⋅ v < 0. And so, by Definition 143 and the trigonometric identity π − cos−1 x = cos−1 (−x) (Fact ??), the non-obtuse angle between u and v is: 603, Contents

π − cos−1

u⋅v −u ⋅ v ∣u ⋅ v∣ = cos−1 = cos−1 . ∣u∣ ∣v∣ ∣u∣ ∣v∣ ∣u∣ ∣v∣

3

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From Fact 114, the following “formula” for the angle between two lines is immediate: Corollary 11. The angle between two lines with direction vectors u and v is: cos−1 Example 744. Two lines l1 and l2 are described by: ­ r = (1, 3) +λ (2, 1) v1

∣u ⋅ v∣ . ∣u∣ ∣v∣

y l1

(λ ∈ R),

­ r = (−1, −1) +λ (1, 3) v2

π/4 v2 = (1, 3)

(λ ∈ R).

By Corollary 11, the angle between the two lines is: cos−1

∣v1 ⋅ v2 ∣ ∣(2, 1) ⋅ (1, 3)∣ = cos−1 ∣v1 ∣ ∣v2 ∣ ∣(2, 1)∣ ∣(1, 3)∣

π/4

l2

∣5∣ 1 π = = cos−1 √ √ = cos−1 √ = . 5 10 2 4

v1 = (2, 1)

x

Example 745. Two lines l1 and l2 are described by: ³¹¹ ¹ ¹ · ¹ ¹ ¹ ¹µ r = (0, 0) + λ(−2, 3) v1

­ r = (1, 0) + λ (3, 1) v2

and

By Corollary 11, the angle between the two lines is: cos−1

∣v1 ⋅ v2 ∣ ∣(−2, 3) ⋅ (3, 1)∣ ∣−3∣ = cos−1 = cos−1 √ √ ≈ 1.305. ∣v1 ∣ ∣v2 ∣ ∣(−2, 3)∣ ∣(3, 1)∣ 13 10

l1

y

v1 = (−2, 3)

1.305

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(λ ∈ R).

l2

v2 = (3, 1)

x

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Definition 145. Two lines are (a) parallel if they have parallel direction vectors; and (b) perpendicular if they have perpendicular direction vectors. Corollary 12. Suppose θ is the angle between two lines. (a) If θ = 0, then the two lines are parallel. And (b) if θ = π/2, then they are perpendicular.

Proof. See p. 1595 (Appendices).

Example 746. Two lines l1 and l2 are described by: v1

and

By Corollary 11, the angle between l1 and l2 is: cos−1

³¹¹ ¹ ¹ ¹ ¹ ¹·¹ ¹ ¹ ¹ ¹ ¹ ¹µ r = (1, 1) + λ(−1, −1) v2

­ r = (2, −2) + λ (3, 3)

(λ ∈ R).

∣v1 ⋅ v2 ∣ ∣(3, 3) ⋅ (−1, −1)∣ ∣−6∣ = cos−1 = cos−1 √ √ = cos−1 1 = 0. ∣v1 ∣ ∣v2 ∣ ∣(3, 3)∣ ∣(−1, −1)∣ 18 2

By Corollary 12, the lines l1 and l2 are parallel.

y

y

v1 = (3, 3)

l2

v3 = (−1, 2) π/2

v2 = (−1, −1)

v4 = (6, 3)

π/2 l4

x l1

l3

x

Two lines l3 and l4 are described by: ³¹¹ ¹ ¹ · ¹ ¹ ¹ ¹µ r = (0, 3) + λ(−1, 2) v3

v4

and

By Corollary 11, the angle between l1 and l2 is: cos−1

­ r = (−1, 1) + λ (6, 3)

(λ ∈ R).

∣v3 ⋅ v4 ∣ ∣(−1, 2) ⋅ (6, 3)∣ ∣0∣ π = cos−1 = cos−1 √ √ = cos−1 0 = . ∣v3 ∣ ∣v4 ∣ ∣(−1, 2)∣ ∣(6, 3)∣ 2 5 45

By Corollary 12, the lines l3 and l4 are perpendicular. 605, Contents

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Let’s write down the “obvious” formal definition of when a line and a vector are parallel or perpendicular: Definition 146. A line and a vector are (a) parallel if the line has a direction vector that’s parallel to the given vector; and (b) perpendicular if the line has a direction vector that’s perpendicular to the given vector. Here are two “obvious” Facts you may recall from primary school: Fact 115. If two lines are: (a) Identical, then they are also parallel. (b) Distinct and parallel, then they do not intersect. (c) Distinct, then they share at most one intersection point. Proof. See p. 1596 (Appendices). Fact 116. If two lines (in 2D space) are distinct and non-parallel, then they must share exactly one intersection point. Proof. See p. 1596 (Appendices). Example 747. The lines l1 and l2 are identical. And indeed, they are also parallel. The lines l1 and l3 are distinct and parallel. And indeed they do not intersect. The lines l1 and l4 are distinct and non-parallel. And indeed they share exactly one intersection point.

l1 = l2

l3

l4

Exercise 232. Find the angle between each given pair of lines. State if they are parallel or perpendicular. (Answer on p. 1820.) (a) r = (−1, 2) + λ (−1, 1) and r = (0, 0) + λ (2, −3) (b) r = (−1, 2) + λ (1, 5) and r = (0, 0) + λ (8, 1) (c) r = (−1, 2) + λ (2, 6) and r = (0, 0) + λ (3, 2)

(λ ∈ R). “ “

Remark 98. Fact 116 applies only to 2D space. As we’ll learn later, in 3D space, two lines can be distinct, non-parallel, and yet do not intersect. (We call such lines skew lines.)

In contrast, Fact 115 applies more generally to higher dimensions, including in 3D space.

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54.

Vectors vs Scalars

We now illustrate the difference between vectors and scalars by revisiting Example ?? from Ch. ?? (Simple Parametric Equations): Example 748. A moving particle has position vector s given by: s (t) = (sx (t) , sy (t)) = (cos t, sin t)

(t ≥ 0).

The particle’s position vector s is a function of time t. For brevity of notation, we will often be lazy/sloppy and omit “(t)”. That is, we will often instead simply write: s = (sx , sy ) = (cos t, sin t)

(t ≥ 0).

At time t, the particle’s x- and y-coordinates are sx = cos t and sy = sin t, respectively. In other words, at time t, the particle is cos t m east and sin t m m north of the origin. As time t progresses from 0 to 2π seconds, we trace out, anti-clockwise, the unit circle: t = 0 Ô⇒ (sx , sy ) = (1, 0) , √ √ t = π/4 Ô⇒ (sx , sy ) = ( 2/2, 2/2) ,

t = π/2 Ô⇒ (sx , sy ) = (0, 1) , √ √ t = 3π/4 Ô⇒ (sx , sy ) = (− 2/2, 2/2) ,

y

t = 3π/2 Ô⇒ (sx , sy ) = (0, −1) , √ √ t = 7π/4 Ô⇒ (sx , sy ) = ( 2/2, − 2/2) .

At t = 1, s = (sx , sy ) ≈ (0.54 m, 0.84 m), v = (vx , vy ) ≈ (−0.84 m s−1 , 0.54 m s−1 ), v = 1 m s−1 .

l

Arrows indicate instantaneous direction of travel.

t = π Ô⇒ (sx , sy ) = (−1, 0) , √ √ t = 5π/4 Ô⇒ (sx , sy ) = (− 2/2, − 2/2) ,

At t = 0, s = (sx , sy ) ≈ (1 m, 0 m), Direction of v = (vx , vy ) ≈ (1 m s−1 , 0 m s−1 ), a and F v = 1 m s−1 . x 5π At t = , 4 √ √ 2 2 s = (sx , sy ) ≈ (− m, − m), 2 2 √ √ 2 2 −1 ms ,− m s−1 ), v = (vx , vy ) ≈ ( 2 2

l

\

(Example continues on the next page ...) 607, Contents

v = 1 m s−1 .

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(... Example continued from the previous page.)

We next define the particle’s velocity vector v (t) to be the first derivative of s with respect to time t: v (t) = (vx (t) , vy (t)) =

dsx dsy d cos t d sin t ds =( , )=( , ) = (− sin t, cos t) . dt dt dt dt dt

So, at time t, the particle is travelling eastwards at vx = − sin t m s−1 (or equivalently, westwards at sin t m s−1 ) and northwards at vy = cos t m s−1 . The magnitude of the particle’s velocity vector is denoted v and is called its speed: √ v = ∣v∣ = vx2 + vy2 .

Recalling294 the trigonometric identity sin2 t + cos2 t = 1, we have: √ √ √ 2 2 v = ∣v∣ = vx + vy = sin2 t + cos2 t = 1 = 1.

Aha! So, interestingly, the particle travels at the constant speed of 1 m s−1 . That is, at every instant in time t, it is moving 1 m s−1 in its direction of travel.

We now prove that the particle always moves in a direction tangent to the circle. In other words, its direction of travel is always perpendicular to its position vector. To do so, we need simply prove that v ⋅ s = 0 for all t:

v ⋅ s = (− sin t, cos t) ⋅ (cos t, sin t) = − sin t cos t + cos t sin t = 0.

3

Velocity is a vector—it has both magnitude and direction. In contrast, speed is a scalar—it has only magnitude. (Example continues on the next page ...)

294

Fact ??(a).

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(... Example continued from the previous page.) Similarly, the particle’s acceleration vector is defined as the first derivative of the velocity vector (or, equivalently, the second derivative of the position vector): dvx dvy d (− sin t) d cos t dv d2 s = 2 =( , )=( , ) = (− cos t, − sin t) . a (t) = (ax (t) , ay (t)) = dt dt dt dt dt dt

So, at time t, the particle is accelerating eastwards at ax = − cos t m s−2 and northwards at ay = − sin t m s−2 . Or equivalently, it is accelerating westwards at cos t m s−2 and southwards at sin t m s−2 . (Note that m s−2 is abbreviation for metre per second per second.) The magnitude of the particle’s acceleration vector is denoted a: √ √ √ 2 2 2 2 a = ∣a∣ = ax + ay = (− cos t) + (− sin t) = 1 = 1.

Aha! So, interestingly, the particle accelerates at the constant rate of 1 m s−2 . That is, at every instant in time t, it is accelerating 1 m s−2 in its direction of acceleration. (Note that for velocity, we gave its magnitude the special name of speed. But in contrast, the magnitude of acceleration has no special name. We simply call it the magnitude of acceleration.)

Above we proved that the particle’s direction of movement is always tangent to the circle. Here we can similarly prove that its direction of acceleration is always towards the centre of the circle. (Or equivalently, the acceleration vector points in the exactly opposite direction as the position vector.) To prove this, we need simply observe that the acceleration vector a = (− cos t, − sin t) and the position vector s = (cos t, sin t) point in exact opposite directions.295 Suppose the particle’s mass is m = 1 kg. Recall from physics Newton’s Second Law:296 F = ma

­ ­ ³¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ µ Force = Mass × Acceleration. Vector

or

Scalar

Vector

Note that mass is a scalar quantity. Hence, force, being a product of a scalar and a vector, is itself a vector quantity. The force vector points in the same direction as the acceleration vector (i.e. towards the centre of the circle). Moreover, it has constant magnitude: ∣F∣ = ∣ma∣ = ∣m∣ ∣a∣ = (1 kg) × (1 m s−2 ) = 1 kg m s−2 = 1 N,

where N is for newton (the SI unit for force and which is equal to kg m s−2 ).

Physicists call such a force (which results in circular movement) a centripetal force.

Note though that it would be wrong to write a = −s. This is because acceleration is measured in m s−2 , while position is measured in m. 296 See e.g. Exercise 183. 295

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55.

The Projection and Rejection Vectors

Let a and b be vectors.

a

The projection of a on b, denoted projb a, is the vector that is:297 • Parallel to b; and • Perpendicular to a − projb a (the vector depicted in blue).

a − projb a

θ

b

projb a

Let us work towards a formal definition of projb a.

ˆ for some k ≠ 0. First, since projb a ∥ b, by Definition 133, we must have projb a = k b

Next, the length of projb a is ∣k∣. But what is k?

We observe that in the above figure, a right triangle is formed. This right triangle’s hypotenuse corresponds to the vector a, while its base corresponds to the vector projb a. Let θ be the angle between these two vectors. Then by our right-triangle definition of cosine, we have: cos θ =

∣projb a∣ 1 ∣k∣ “Adjacent′′ = = . “Hypotenuse′′ ∣a∣ ∣a∣

But by Definition 140, we also have: Putting together = and =, we have: 1

2

ˆ ∣k∣ a ⋅ b = ∣a∣ ∣a∣

cos θ = or

ˆ a⋅b 2 a⋅b = . ∣a∣ ∣b∣ ∣a∣ ˆ ∣k∣ = a ⋅ b.

The above discussion motivates the following definition of the projection vector: Definition 147. Let a and b be vectors. Then the projection of a on b, denoted projb a, is the following vector: ˆ b ˆ projb a = (a ⋅ b)

(or equivalently, projb a =

a⋅b ∣b∣

2

b).

For convenience, let’s call the “blue vector” the rejection vector and denote it rejb a: Definition 148. Let a and b be non-zero vectors. Then the rejection of a on b is the vector denoted rejb a and defined by: rejb a = a − projb a.

297

These two properties are to hold so long as projb a and a − projb a are non-zero.

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The following result is “obvious” from our above two Definitions: ˆ b. ˆ Fact 117. Let a and b be vectors, and projb a = (a ⋅ b)

ˆ > 0, then proj a is a positive scalar multiple of b. (a) If a ⋅ b b ˆ < 0, then proj a is a negative scalar multiple of b. (b) If a ⋅ b b ˆ = 0, then proj a = 0 and rej a = a. (c) If a ⋅ b b

b

Here’s what the above result says geometrically. Let θ be the angle between a and b. Then: (a) If θ is acute, then projb a points in the same direction as b. (b) If θ is obtuse, then projb a points in the exact opposite direction as b. (c) If θ is right (i.e. if a ⊥ b), then projb a = 0 and rejb a = a.

(a)

a

(b)

a rejb a = a − projb a

θ

θ

b

projb a

projb a

b

(c)

a rejb a = a θ = π/2

projb a = 0

b

ˆ We now formally prove Above we already argued that the length of projb a must be ∣a ⋅ b∣. that this is so: Fact 118. Let a and b be vectors. Then:

ˆ . ∣projb a∣ = ∣a ⋅ b∣

Proof. By Definition 147 and Fact 97:

ˆ b∣ ˆ = ∣a ⋅ b∣ ˆ ∣b∣ ˆ = ∣a ⋅ b∣ ˆ ⋅ 1 = ∣a ⋅ b∣ ˆ . ∣projb a∣ = ∣(a ⋅ b) 611, Contents

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Example 749. Let u = (3, 2) and v = (1, 1) be vectors and projv u be the projection of u on v. Then: ˆ∣ = ∣ ∣projv u∣ = ∣u ⋅ v

5 (3, 2) ⋅ (1, 1) 3 + 2 ∣= √ = √ . ∣(1, 1)∣ 2 2

Now consider w = (−2, −2)—it points in the exact opposite direction from v and has twice the length. It turns out that the projection of u on w, projw u, is identical to projv u.

projv u

u = (3, 2)

v = (1, 1)

projw u u = (3, 2) We can easily verify that ∣projw u∣ = ∣projv u∣: ˆ =∣ ∣projw u∣ = ∣u ⋅ w∣

w = (−2, −2)

(3, 2) ⋅ (−2, −2) ∣ ∣(−2, −2)∣

−6 − 4 −10 5 =∣ √ ∣=∣√ ∣= √ . 8 8 2

3

As the above example suggests, if v and w are parallel, then the projections of any vector u on v and w are identical. Formally: Fact 119. Let u, v, and w be vectors. If v ∥ w, then: projv u = projw u.

ˆ = ±w. ˆ And so: Proof. If v ∥ w, then by Fact 100, v

ˆ) v ˆ = [u ⋅ (±w)] ˆ (±w) ˆ = (u ⋅ w) ˆ w ˆ = projw u. projv u = (u ⋅ v

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Example 750. Let a = (−6, 1) and b = (2, 0) be vectors and projb a be the projection of a on b. Then: ˆ = ∣ (−6, 1) ⋅ (2, 0) ∣ = ∣ −12 + 0 ∣ = ∣ −12 ∣ = 6. ∣projb a∣ = ∣a ⋅ b∣ ∣(2, 0)∣ 2 2

projb a

a = (−6, 1)

b = (2, 0)

Now consider c = (3, 0)—it points in the same direction as b, but is half again as long. By Fact 119, the projection of a on c is the same as that of a on b. That is: projc a = projb a.

We can easily verify that ∣projc a∣ = ∣projb a∣ = 6: ∣projc a∣ = ∣a ⋅ cˆ∣ = ∣

projc a

−18 + 0 −18 (−6, 1) ⋅ (3, 0) ∣=∣ ∣=∣ ∣ = 6. ∣(3, 0)∣ 3 3

a = (−6, 1)

c = (3, 0)

Fact 119 can help us simplify some calculations:

√ √ Example 751. Let u = (5, −7) and v = (51 347, 68 347). What is ∣projv u∣ (i.e. the length of the projection of u onto v)?

Here it seems that the calculations will be pretty tedious. But observe that v is a multiple of w = (3, 4). Thus, v ∥ w. And so, by Fact 119: projv u = projw u.

Hence, instead of computing ∣projv u∣, we can simply compute ∣projw u∣: ˆ =∣ ∣projv u∣ = ∣projw u∣ = ∣u ⋅ w∣

(5, −7) ⋅ (3, 4) 15 ⋅ −28 −13 u⋅w ∣=∣ ∣ = ∣√ ∣=∣ ∣ = 2.6. ∣w∣ ∣(3, 4)∣ 5 32 + 42

Exercise 233. Find the lengths of the projections of: (a) (1, 0) on (33, 33);

and

(b) (33, 33) on (1, 0).

(c) Hence conclude if the following statement is true or false:

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“Given any vectors a and b, ∣projb a∣ = ∣proja b∣.”

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56.

Collinearity

Definition 149. Two or more points are collinear if some line contains all of them. Example 752. The points A and B are collinear.

Ð→ AB

Ð→ Ð→ r = OA + λAB (λ ∈ R)

B

A

“Obviously”, any two points must be collinear. Indeed, given any two points, there is a unique line that contains both of them: Fact 120. Suppose A and B are distinct points. Then the unique line that contains both A and B is described by: Ð→ Ð→ r = OA + λAB

(λ ∈ R).

Proof. First, plug in λ = 0 and λ = 1 to verify that the given line contains A and B.

Next, this line is unique because any line that contains both A and B must have direction Ð→ vector AB and must thus be described by: Ð→ Ð→ r = OA + λAB

(λ ∈ R) .

In contrast, three distinct points can be collinear but will not generally be: Example 753. The points A, B, and C are collinear:

A, B, and C are collinear.

A

Ð→ AB

B

C

Ð→ r = a + λAB (λ ∈ R)

In contrast, the points D, E, and F are not:

D, E, and F are not collinear.

D

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ÐÐ→ DE

F

E

ÐÐ→ r = d + λDE (λ ∈ R)

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Here is one possible procedure for checking whether three points are collinear: 1. First use Fact 120 to write down the unique line that contains two of the three points. 2. Then check whether this line also contains the third point. Example 754. Let A = (1, 2), B = (4, 5), and C = (7, 8) be points.

To check if they are collinear:

1. First write down the unique line that contains both A and B: Ð→ Ð→ r = OA + λAB = (1, 2) + λ(3, 3)

(λ ∈ R).

ˆ such that: 2. If this line also contains C, then there exists λ ⎛ 7 ⎞ ⎛ 1 ⎞ ˆ⎛ 3 ⎞ +λ = C= ⎝3⎠ ⎝8⎠ ⎝2⎠

or

1 ˆ 7 = 1 + 3λ, 2 ˆ 8 = 2 + 3λ.

ˆ = 2 solves the above vector equation (or system of two equations). As you can verify, λ Hence, our line also contains C. Thus, A, B, and C are collinear.

Example 755. Let D = (1, 0), E = (0, 1), and F = (0, 0) be points. To check if they are collinear: 1. First write down a line that contains both D and E: ÐÐ→ ÐÐ→ r = OD + λDE = (1, 0) + λ(−1, 1)

(λ ∈ R).

ˆ such that: 2. If this line also contains F , then there exists λ ⎛ 0 ⎞ ⎛ 1 ⎞ ˆ ⎛ −1 ⎞ F= = +λ ⎝0⎠ ⎝0⎠ ⎝ 1 ⎠

or

1 ˆ 0 = 1 − 1λ, 2 ˆ 0 = 0 + 1λ.

ˆ = 1. But this contradicts =. This contradiction means that there is no From =, we have λ solution to the above vector equation (or system of two equations). 1

2

Hence, our line does not contain F . Thus, D, E, and F are not collinear.

Exercise 234. In each of the following, three points A, B, and C are given. Determine if they are collinear. (a) A = (3, 1), B = (1, 6), and C = (0, −1). (b) A = (1, 2), B = (0, 0), and C = (3, 6).

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57.

The Vector Product

Definition 150. Let u = (u1 , u2 ) and v = (v1 , v2 ) be vectors. Their vector product, denoted u × v, is the number: u × v = u1 v2 − u2 v1 .

Remark 99. The vector product is also called the cross product. But your A-Level exams and syllabus do not seem to use this term and so neither shall we. We will stick strictly to the term vector product. Example 756. Let u = (5, −3), v = (2, 1), w = (−4, 0), and x = (8, 7). Then: u×v =

⎛2⎞ ⎛ 5 ⎞ = 5 ⋅ 1 − (−3) ⋅ 2 = 5 + 6 = 11, × ⎝1⎠ ⎝ −3 ⎠

u×x =

⎛ 5 ⎞ ⎛8⎞ × = 5 ⋅ 7 − (−3) ⋅ 8 = 35 + 24 = 59, ⎝ −3 ⎠ ⎝7⎠

u×w =

v×w =

⎛ 5 ⎞ ⎛ −4 ⎞ × = 5 ⋅ 0 − (−3) ⋅ (−4) = 0 − 12 = −12, ⎝ −3 ⎠ ⎝ 0 ⎠ ⎛2⎞ ⎛ −4 ⎞ × = 2 ⋅ 0 − 1 ⋅ (−4) = 0 + 4 = 4. ⎝1⎠ ⎝ 0 ⎠

We now discuss three properties of the vector product.

Recall that ordinary multiplication and the scalar product are both distributive (over addition) and commutative. It turns out that the vector product is also distributive: Example 757. Continuing with the above example: u × (v + w) =

⎛ 5 ⎞ ⎝ −3 ⎠

×

⎛ 2−4 ⎞ = 5 ⋅ 1 − (−3) ⋅ (−2) ⎝ 1+0 ⎠

= −1 = 11 + (−12) = u × v + u × w.

(u + v) × w =

⎛ 5+2 ⎞ ⎛ −4 ⎞ × = 7 ⋅ 0 − (−2) ⋅ (−4) ⎝ −3 + 1 ⎠ ⎝ 0 ⎠

= −8 = −12 + 4 = u × w + v × w.

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However, the vector product is not commutative. Instead, it is anti-commutative: a × b = −b × a.

Example 758. Let u = (5, −3), v = (2, 1), and w = (−4, 0). We already showed that: u × v = 11

and

We now show that v × u = −11 and w × u = 12:

w × u = −12.

v × u = (2, 1) × (5, −3) = 2 ⋅ (−3) − 1 ⋅ 5 = −6 − 5 = −11.

w × u = (−4, 0) × (5, −3) (−4) ⋅ (−3) − 0 ⋅ 5 = 12 − 0 = 12.

The third property is that a vector’s vector product with itself is 0: Example 759. Continuing with the above example, we have:

u × u = (5, −3) × (5, −3) = 5 ⋅ (−3) − (−3) ⋅ 5 = −15 + 15 = 0. v × v = (2, 1) × (2, 1) 2 ⋅ 1 − 1 ⋅ 2 = 2 − 2 = 0.

In summary:

w × w = (−4, 0) × (−4, 0) (−4) ⋅ 0 − 0 ⋅ (−4) = 0 − 0 = 0.

Fact 121. Let a, b, and c be vectors. Then:

(a) a × (b + c) = a × b + a × c (b) a × b = −b × a (c) a × a = 0

(Distributivity over addition) (Anti-commutativity) (Self vector product equals zero)

Proof. See Exercise 235.

Exercise 235. Let a = (a1 , a2 ), b = (b1 , b2 ), and c = (c1 , c2 ) be vectors. Prove the following (Fact 121): (Answer on p. 1822.) (a) a × (b + c) = a × b + a × c

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(b) a × b = −b × a

(c) a × a = 0

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From Fact 121(c), we have the following result: Corollary 13. If a ∥ b, then a × b = 0.

Proof. If a ∥ b, then there exists c ≠ 0 such that ca = cb. Thus:

a × b = a × (ca) = c (a × a) = c ⋅ 0 = 0.

Example 760. Let a = (1, 2) and b = (−2, −4). Since a ∥ b, by Corollary 13, a × b = 0.

The converse of Corollary 13 is also true, but is harder to prove:

Fact 122. Let a and b be non-zero vectors. If a × b = 0, then a ∥ b.

Proof. See p. 1601 (Appendices).

Example 761. Let a = (3, −1) and b = (2, k), where k is some unknown constant. We are now told that a × b = 0. What is k? Well, by Fact 122, a × b = 0 implies that a ∥ b. That is, b is a multiple of a.

Hence, 2/3 = k/ (−1). And so, k = −2/3.

The following result is simply Corollary 13 and Fact 122 combined: Corollary 14. Suppose a and b are non-zero vectors. Then: a×b=0

⇐⇒

a ∥ b.

Here’s another “obvious” property of the vector product:

Fact 123. Suppose a and b are vectors and c ∈ R is a scalar. Then: (ca) × b = c (a × b).

Proof. Let a = (a1 , a2 ) and b = (b1 , b2 ). Then ca = (ca1 , ca2 ) and:

(ca) × b = (ca1 ) ⋅ b2 − (ca2 ) b1 = c (a1 b2 − a2 b1 ) = c (a × b) .

Exercise 236. Let a = (1, −2), b = (3, 0), and c = (4, 1). Compute a × b, a × c, b × c, b × a, c × a, c × b, and a × (b + c). (Answer on p. 1822.)

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57.1.

The Angle between Two Vectors Using the Vector Product

Recall298 that with the scalar product, we had a ⋅ b = ∣a∣ ∣b∣ cos θ.

It turns out that with the vector product, we have a very similar result: Fact 124. Let θ be the angle between the vectors a and b. Then:

Proof. See Exercise 237.

∣a × b∣ = ∣a∣ ∣b∣ sin θ.

Exercise 237. Let θ be the angle between the vectors a = (a1 , a2 ) and b = (b1 , b2 ).

(a) Express ∣a∣, ∣b∣, ∣a × b∣, and cos θ in terms of a1 , a2 , b1 , and b2 . (You do not need to expand the squared terms.) (b) Since θ ∈ [0, π], what can you say about the sign of sin θ? (That is, is sin θ positive, negative, non-positive, or non-negative?) (c) Now use a trigonometric identity to express sin θ in terms of cos θ. (Hint: You should find that there are two possibilities. Use what you found in (b) to explain why you can discard one of these possibilities.)

(d) Plug the expression you wrote down for cos θ in (a) into what you found in (c).

(e) Prove299 that (a21 + a22 ) (b21 + b22 )−(a1 b1 + a2 b2 ) = (a1 b2 − a2 b1 ) . (Hint: Simply expand the terms and do the algebra.) 2

2

(f) Use (a) and (d) to express ∣a∣ ∣b∣ sin θ in terms of a1 , a2 , b1 , and b2 . Then use (e) to prove that ∣a × b∣ = ∣a∣ ∣b∣ sin θ. (Answer on p. 1822.)

Fact 124 immediately yields the following result :

Corollary 15. If θ ∈ [0, π] is the angle between the vectors u and v, then: θ = sin−1

∣u × v∣ . ∣u∣ ∣v∣

Corollary 15 is thus the sine or vector product analogue of Definition 140. Note though that we won’t be using Corollary 15 to compute the angle between two vectors. This is because, as we’ll see shortly, it’s generally easier to compute the scalar product than the vector product. And so, it’s easier to just use Definition 140.

298 299

Fact 108. By the way, this is simply an instance of Lagrange’s Identity.

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57.2.

The Length of the Rejection Vector

The vector product will mostly be useful only when we look at 3D space. Nonetheless, even in 2D space, it has the following use: Fact 125. Let a and b be vectors. Then: ˆ . ∣rejb a∣ = ∣a × b∣

a rejb a = a − projb a

θ projb a

b

Fact 125 is thus the vector product analogue of Fact 118. Proof. By the right-triangle definition of sine:300 sin θ =

Opposite ∣rejb a∣ = . Hypotenuse ∣a∣

ˆ = 1 and Fact 124: Below, we first rearrange, then use ∣b∣

ˆ sin θ = ∣a × b∣ ˆ . ∣rejb a∣ = ∣a∣ sin θ = ∣a∣ ∣b∣

3

As we’ll see next, Fact 125 will help us compute the distance between a point and the line.

300

For a proof that makes no mention of the sine function, see p. 1600 (Appendices).

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58.

The Foot of the Perpendicular From a Point to a Line

Definition 151. Let A be a point that isn’t on the line l. The foot Ð→ of the perpendicular from A to l is the point B on l such that AB ⊥ l.

Ð→ AB

A

B

P

l

v

Ð→ projv P A

B

A

l

Suppose the line l has direction vector v. Pick any point P on Ð→ Ð→ l. Consider projv P A—the projection of P A on v. Ð→ Observe that B = P + projv P A is a foot of the perpendicular from A to l. Moreover, it is unique.301 Formally:

Ð→ Fact 126. Suppose l is the line described by r = OP + λv (λ ∈ R) and A is a point that isn’t on l. Then the unique foot of the perpendicular from A to l is the following point: Proof. See Exercise 239.

Ð→ P + projv P A.

Example 762. Let A = (1, 2) be a point and l be the line described by: Ð→ r = OP + λv = (0, 1) + λ(9, 1)

Ð→ Compute P A = (1, 2) − (0, 1) = (1, 1) and:

(λ ∈ R).

Ð→ 5 (1, 1) ⋅ (9, 1) ⎛ 9 ⎞ 9 + 1 projv P A = proj(9,1) (1, 1) = = (9, 1) = (9, 1). 92 + 12 ⎝ 1 ⎠ 82 41

By Fact 126, the foot of the perpendicular from A to l is: Ð→ 5 1 B = P + projv P A = (0, 1) + (9, 1) = (45, 46) . 41 41

By the way, let’s call this Method 1 or the Formula Method for finding the foot of the perpendicular (so named because we simply plug in the formula given in Fact 126). Below, we’ll also two more methods for finding the same.

A = (1, 2)

B P = (0, 1)

l

v = (9, 1)

Exercise 238. Find the feet of the perpendiculars from the points A = (−1, 0) and Ð→ B = (3, 2) to the line described by r = OP + λv = (2, −3) + λ (5, 1) (λ ∈ R). (Answer on p. 1823.)

301

Hence justifying the use of the definite article in Definition 151.

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Exercise 239. This Exercise guides you through a proof of Fact 126. Let l be the line Ð→ Ð→ described by r = OP + λv (λ ∈ R), A be a point that isn’t on l, and B = P + projv P A. To prove that B is a foot of the perpendicular from A to l, we must prove that (a) B is Ð→ on l; and (b) AB ⊥ l :

(a) To prove that B is on l, follow these two steps: Ð→ (i) Explain why projv P A can be written as a scalar multiple of v. That is, explain Ð→ why projv P A = λv for some λ ∈ R. (ii) Hence explain why B satisfies l’s vector equation and is thus on l. Ð→ (b) Next, to prove that AB ⊥ l, follow these steps: Ð→ Ð→ (i) Show that AB = −rejv P A. (Hint: Definition 148.) Ð→ (ii) Explain why rejv P A ⊥ v. Ð→ (iii) Hence explain why AB ⊥ l.

(c) To prove that B is the unique foot of the perpendicular from A to l, let C ≠ B be a Ð→ point on l—we will show that AC ⊥/ l and hence that C cannot also be a foot of the perpendicular from A to l: Ð→ Ð→ (i) First explain why AB ⋅ BC = 0. Ð→ Ð→ Ð→ (ii) Now prove that AC ⊥/ BC and hence that AC ⊥/ l. (Answer on p. 1823.)

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58.1.

The Distance Between a Point and a Line

We define the distance between a point and a line as the minimum distance between them: Definition 152. Let A be a point and l be a line. Suppose B is the point on l that’s Ð→ closest to A. Then the distance between A and l is ∣AB∣. Remark 100. In the trivial case where A is on l, the point on l that’s closest to A is A Ð→ itself. And so, the distance between A and l is ∣AA∣ = ∣0∣ = 0 (as we’d expect).

Fact 127 states what you may find intuitively “obvious”: If B is the foot of the perpendicular from a point A to a line l, then B is also the point on the line that’s closest to A.

A The line l

B, the foot of the perpendicular, is also the closest point.

Fact 127. If B is the foot of the perpendicular from a point A to a line l, then B is also the point on l that’s closest to A.

A

Proof. Let C ≠ B be any other point on l. Observe that ABC forms a right triangle with hypotenuse AC.

By Pythagoras’ Theorem, the leg AB must be shorter than the hypotenuse AC. That is, ∣AB∣ < ∣AC∣. We’ve just shown that B is closer to A than any other point on q. Thus, B is the point on l that’s closest to A.

l B

C

From the above Definition and Fact, the following Corollary is immediate: Corollary 16. Suppose l is a line, A is a point, and B is the foot of the perpendicular Ð→ from A to l. Then the distance between A and l is ∣AB∣.

Example 763. As in Example 762, let A = (1, 2) be a point and l be the line described by Ð→ r = OP +λv = (0, 1)+λ (9, 1) (λ ∈ R). Previously, using Method 1 (Formula Method), 1 we already found that the foot of the perpendicular from A to l is B = (45, 46). 41

Ð→ 1 1 4 AB = B − A = (45, 46) − (1, 2) (4, −36) = (1, −9) 41 41 41 Ð→ 4√ And so, by Corollary 16, the distance between A and l is ∣AB∣ = 82. 41 We will next introduce two more methods for finding B and hence the distance between ˜ (9, 1) be the foot of the A and l. In each method, we begin by letting B = (0, 1) + λ ˜ is some unknown to be found. perpendicular from A to l, where λ

Now:

(Example continues on the next page ...)

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(... Example continued from the previous page.)

Ð→ Method 2 (Perpendicular Method). First write down AB:

Ð→ ˜ (9, 1) − (1, 2) = (9λ ˜ − 1, λ ˜ − 1). AB = B − A = (0, 1) + λ

Ð→ Ð→ Ð→ Since B is the foot of the perpendicular, we have AB ⊥ l or AB ⊥ v or AB ⋅ (9, 1) = 0: Ð→ ˜ − 1, λ ˜ − 1) ⋅ (9, 1) = 9 (9λ ˜ − 1) + (λ ˜ − 1) = 82λ ˜ − 10. 0 = AB ⋅ (9, 1) = (9λ

˜ = 10/82 = 5/41 and so B = (0, 1) + λ ˜ (9, 1) = (0, 1) + 5 (9, 1) = 1 (45, 46). Rearranging, λ 41 41 Lovely—this is the same as what we found in Method 1. And now, if we’d like, we can Ð→ calculate ∣AB∣ (the distance between A and l) as we did before. Ð→ Method 3 (Calculus Method). Let R be a generic point on l. Then AR = (9λ − 1, λ − 1) and the distance between A and R is: √ √ Ð→ 2 2 ∣AR∣ = (9λ − 1) + (λ − 1) = 82λ2 − 20λ + 2. Ð→ ˜ Since B is the point on l that’s closest to A, the value of λ that minimises ∣AR∣ must be λ. Ð→ √ Our goal then is to determine when ∣AR∣ = 82λ2 − 20λ + 2 is minimised—or equivalently, when 82λ2 − 20λ + 2 is minimised.

To do so, we’ll use calculus. We’ll learn how more about how this calculus thing works in Part V, but for now we’ll rely on what you may (or may not) remember from secondary school. First differentiate 82λ2 − 20λ + 2 with respect to λ: d (82λ2 − 20λ + 2) = 164λ − 20. dλ

And so, by the First Order Condition (FOC), we have: (164λ − 20) ∣λ=λ˜ = 0

or

˜ − 20 = 0 164λ

or

˜ = 20 = 5 . λ 164 41

Lovely—this is the same as what we found in Method 2. And now, if we’d like, we can Ð→ find B and ∣AB∣ as we did before. By the way, instead of explicitly using calculus, an alternative is to use what we learnt earlier in Part I. Recall302 that quadratic expressions are minimised at “−b/2a”. Hence: ˜ = “ − b ” = − −20 = 5 . λ 2a 2 ⋅ 82 41

This alternative is probably quicker (provided of course you can recall the “−b/2a” thing).

302

Fact 27 (whose proof, by the way, actually uses calculus).

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The following is a “formula” for the distance between a point and a line. Ð→ Corollary 17. Suppose A is a point, l is the line described by r = OP + λv (λ ∈ R), and d is the distance between A and l. Then: Proof. See Exercise 240.

Ð→ ˆ ∣. d = ∣P A × v

Example 764. We continue with the point A = (1, 2) and the line l described by r = Ð→ OP + λv = (0, 1) + λ (9, 1) (λ ∈ R). By Corollary 17, the distance between A and l is:

Ð→ (9, 1) 1⋅1−1⋅9 8 ˆ ∣ = ∣(1, 1) × √ ∣P A × v ∣=∣ √ ∣= √ =4 82 82 92 + 12

2 . 41

Happily, this coincides with what we found before.

Ð→ Exercise 240. Let A be a point, l be the line described by r = OP + λv (λ ∈ R), and d be the distance between A and l. (Answer on p. 1823.)

(a) Prove Corollary 17 in the trivial case where the point A is on the line l.

In the rest of this exercise (or proof), we will suppose that A is not on l. Let B be the foot of the perpendicular from A to l. Ð→ (b) What is the relationship between d and AB? Ð→ Ð→ (c) Express AB in terms of rejv P A. (Hint: Refer to the figure on p. 621.)

(d) Now use a result from the previous chapter to complete the proof of Corollary 17. And now a brand new example where we illustrate all three methods:

Ð→ Example 765. Let A = (−1, 0) be a point, l be the line described by r = OP + λv = (3, 2) + λ(5, 1) (λ ∈ R), and B be the foot of the perpendicular from A to l. Ð→ Method 1 (Formula Method). First compute P A = (−1, 0) − (3, 2) = (−4, −2) and:

So:

Ð→ (−4, −2) ⋅ (5, 1) ⎛ 5 ⎞ −20 − 2 ⎛ 5 ⎞ −11 ⎛ 5 ⎞ projv P A = proj(5,1) (−4, −2) = = = . 52 + 12 26 ⎝ 1 ⎠ 13 ⎝ 1 ⎠ ⎝1⎠

And:

Ð→ −11 1 B = P + projv P A = (3, 2) + (5, 1) = (−16, 15). 13 13

√ Ð→ Ð→ (5, 1) −4 − (−10) 6 3 26 ˆ ∣ = ∣(−4, −2) × √ AB = ∣P A × v ∣=∣ √ ∣= √ = . 13 26 26 52 + 12

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(... Example continued from the previous page.)

˜ (5, 1). Write: Method 2 (Perpendicular Method). Let B = (3, 2) + λ Ð→ ˜ (5, 1) − (−1, 0) = (5λ ˜ + 4, λ ˜ + 2). AB = B − A = (3, 2) + λ

Ð→ Ð→ Since AB ⊥ l, we have AB ⊥ v or:

Ð→ ˜ + 4, λ ˜ + 2) ⋅ (5, 1) = 5 (5λ ˜ + 4) + (λ ˜ + 2) = 26λ ˜ + 22. 0 = AB ⋅ (5, 1) = (5λ

˜ = −22/26 = −11/13 and so: Rearranging, λ

˜ (5, 1) = (3, 2) − 11 (5, 1) = 1 (−16, 15) . B = (3, 2) + λ 13 13

And now, the distance between A and l is:

√ Ð→ 1 3 3 26 1 ∣(−1, 5)∣ = . ∣AB∣ = ∣B − A∣ = ∣ (−16, 15) − (−1, 0)∣ = ∣ (−3, 15)∣ = 13 13 13 13

Lovely—these coincide with what we found using Method 1.

Method 3 (Calculus Method). Let R be a generic point on l. Then: Ð→ AR = (5λ + 4, λ + 2)

Differentiate: FOC:

and

Ð→ ∣AR∣ =

√ √ 2 2 (5λ + 4) + (λ + 2) = 26λ2 + 44λ + 20.

d (26λ2 + 44λ + 20) = 52λ + 44. dλ

(52λ + 44) ∣λ=λ˜ = 0

or

Alternatively, we could simply have used “−b/2a”:

˜ = − 44 = 11 . λ 52 13

˜ = − 44 = − 44 = − 11 . λ 2 ⋅ 26 52 13

Ð→ And now, we can find B and ∣AB∣ as we did in Method 2.

Exercise 241. In each of the following, a point A and line l are given. Let B be the foot of the perpendicular from A to l. Find B and also the distance between A and l. The point A The line l (a) (7, 3) r = (8, 3) + λ (9, 3) (b) (8, 0) Contains the points (4, 4) and (6, 11) (c) (8, 5) r = (8, 4) + λ (5, 6)

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Answer on p. 1824. 1825. 1826.

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So far, we’ve looked only at two-dimensional (2D) space (or the cartesian plane). In the remainder of Part III, we’ll look instead at three-dimensional (3D) space. I will also often make use of Paul Seeburger’s CalcPlot3D.303 Whenever you see a tiny version of this icon:

click/touch it304 and you’ll be brought to the relevant 3D graph, where you can pan, zoom, rotate, etc., so as to get a better sense of what the 3D graph looks like.

I’ve examined dozens of 3D graphing software and all things considered (user-friendliness, accessibility, features, etc.), this is the best 3D graphing web app I’ve found so far. Please let me know if you know of any other better software/app. (I was gonna use GeoGebra, but it had too many critical flaws.) 304 Note that you’ll be routed through TinyURL.com first. The reason is that the CalcPlot3D links are often thousands of characters long and were confusing my computer. 303

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59.

Three-Dimensional (3D) Space

In 2D space, we had ordered pairs. In 3D space, we’ll instead have ordered triples:305 Definition 153. Given an ordered triple (a, b, c), we call a its first or x-coordinate, b its second or y-coordinate, and c its third or z-coordinate.

Example 766. The ordered triple (Cow, Chicken, Dog) has x-coordinate Cow, ycoordinate Chicken, and z-coordinate Dog. As with ordered pairs, the order of the coordinates matters. So for example: (Cow, Chicken, Dog) ≠ (Cow, Dog, Chicken) ≠ (Chicken, Cow, Dog) .

In contrast, with a set of three elements, order doesn’t matter:

{Cow, Chicken, Dog} = {Cow, Dog, Chicken} = {Chicken, Cow, Dog} .

Example 767. The ordered triple (2, 5, −π) has x-coordinate 2, y-coordinate 5, and z-coordinate −π. Again, order matters, so that for example: (2, 5, −π) ≠ (2, −π, 5) ≠ (5, 2, −π) .

Again, in contrast, with a set of three elements, order doesn’t matter: {2, 5, −π} = {2, −π, 5} = {5, 2, −π} .

In 2D space, a point was simply any ordered pair of real numbers. Now in 3D space: Definition 154. A point is any ordered triple of real numbers. Example 768. The ordered triple (Cow, Chicken, Dog) is not a point because at least one of its coordinates is not a real number. (Indeed, all three aren’t.) Example 769. The ordered triple (2, 5, −π) is a point because all three of its coordinates are real numbers.

305

For the formal definition of an ordered triple (and n-tuple), see Definition 263 (Appendices).

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In 2D space (the cartesian plane), we could depict points (ordered pairs of real numbers) by drawing on a piece of paper. The x-axis went right and the y-axis up.

y a2

A = (a1 , a2 , a3 )

z

a3

O = (0, 0, 0)

A = (2, 1)

x

x a1

y

In 3D space, we can again depict points (now ordered triples of real numbers) by drawing on a piece of paper. Again, the x-axis goes right and the y-axis up. But now, we also have the z-axis, which “comes out of the paper towards your face” and is perpendicular to both the x- and y-axes.

We say that this coordinate system follows the right-hand rule. To see why, have the palm of your right hand face you. Fold your ring and pinky fingers. Have your thumb point right, your index finger up, and your middle finger towards your face. Then these three fingers correspond to the x-, y-, and z-axes. (Try it!) (If instead the z-axis “goes into the paper away from your face”, then our coordinate system would instead follow the left-hand rule. Can you explain why?)

In 2D space, the origin was the point O = (0, 0) (Definition 37) and was where the x- and y-axes intersected. And the generic point A = (a1 , a2 ) was a1 units to the right and a2 units above the origin. Analogously, in 3D space:

Definition 155. The origin is the point O = (0, 0, 0).

In 3D space, the origin is where the x-, y-, and z-axes intersect. And relative to the origin, the generic point A = (a1 , a2 , a3 ) is a1 units right, a2 units up, and a3 units “out (towards your face)”.

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59.1.

Graphs (in 3D)

In 2D space, a graph was any set of points, where points were ordered pairs of real numbers (Definition 38). And the graph of an equation was the set of points (x, y) for which the equation was true (Definition 39).

Likewise, in 3D space, a graph remains any set of points, the only difference being that points are now ordered triples of real numbers. And: Definition 156. The graph of an equation (or system of equations) is the set of points (x, y, z) for which the equation (or system of equations) is true.

Shortly, we’ll learn about the equations (and systems of equations) used to describe planes (and lines). For now, here are two quick examples: Example 770. Consider the equation:

The plane q

x + y + z = 1.

B = (0, 1, 0)

Specifically, q contains exactly those points (x, y, z) that satisfy:

x A = (1, 0, 0)

C = (0, 0, 1)

x + y + z = 1.

So for example, it contains the points (1, 0, 0), (0, 1, 0), and (0, 0, 1). A little more formally, the plane q is a set:

y

z

q = {(x, y, z) ∶ x ∈ R, y ∈ R, z ∈ R, x + y + z = 1}.

In words, q is the set of ordered triples (x, y, z) such that x, y, and z are real numbers satisfy x + y + z = 1. As with ordered pairs, we will generally be looking only at ordered triples of real numbers, i.e. points. And so, we shall be a little lazy/sloppy and not bother mentioning that x, y, and z are real numbers. That is, we’ll usually more simply write: q = {(x, y, z) ∶ x + y + z = 1} .

In words, q is the set of points (x, y, z) such that x, y, and z satisfy x + y + z = 1.

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Example 771. Consider the following system of (two) equations:

Or equivalently and more simply:

x=y

and x = y = z.

y = z.

It turns out that this system of (two) equations describes a line l in 3D space. (We’ll learn more about this in Ch. 63.) The line l contains exactly those points that can be written as (λ, λ, λ), for some real number λ. , So, for example, it contains the points (1, 1, 1), O = (0, 0, 0), and (−1, −1, −1).

y

The line l

(1, 1, 1)

x z (−1, −1, −1) A little more formally, the line l is a set: l = {(x, y, z) ∶ x = y = z}.

In words, l is the set of ordered triples (x, y, z) such that x, y, and z are real numbers that satisfy x = y = z. As per , above, we can also write:

l = {(λ, λ, λ) ∶ λ ∈ R} = {λ (1, 1, 1) ∶ λ ∈ R} .

In words, l is the set of points that can be written as (λ, λ, λ) or λ (1, 1, 1) for some real number λ.

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60.

Vectors (in 3D)

We now give the basic definitions and results concerning vectors in 3D space. Everything we learnt about vectors in 2D space finds its analogy in 3D space. (Indeed, we’ll simply reproduce verbatim many of the definitions and results from before). Most of the time, the analogy is obvious. We will therefore go fairly briskly in this chapter. Definition 157. Given the points A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ), the vector from A Ð→ to B is AB = (b1 − a1 , b2 − a2 , b3 − a3 ). Example 772. The vector from the point A = (1, 5, 0) to the point B = (−2, 6, 3) is: ⎛ −3 ⎞ Ð→ ⎟ AB = (−3, 1, 3) = ⎜ ⎜ 1 ⎟ = u. ⎝ 3 ⎠

Observe that there are, again, at least four ways to denote a single vector.

y B = (−2, 6, 3) Ð→ AB = (−3, 1, 3) (The vector from A to B)

z

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A = (1, 5, 0)

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Definition 120. A scalar is any real number. We may again contrast vectors with scalars: vectors are two-dimensional objects, while scalars are one-dimensional. Ð→ Definition 122. Given a point A, its position vector is the vector OA .

Ð→ And so, the point A = (a1 , a2 , a3 ) has position vector OA = a = (a1 , a2 , a3 ).

Ð→ Example 773. The point A = (1, 5, 0) has position vector OA = a = (1, 5, 0).

Once again, do not confuse a point (a zero-dimensional object) with a vector (a twodimensional object). Definition 123. The zero vector, denoted 0, is the origin’s position vector.

Ð→ And so, the zero vector (in 3D space) is 0 = OO = (0, 0, 0).

Definition 124. Suppose a moving object starts at point A and ends at point B. Then Ð→ we call AB its displacement vector.

And so, if a moving object starts at A = (a1 , a2 , a3 ) and ends at B = (b1 , b2 , b3 ), then its Ð→ displacement vector is AB = (b1 − a1 , b2 − a2 , b3 − a3 ). Exercise 242. Let A = (2, 5, 8) and B = (0, 1, 1) be points.

(a) What is the vector from A to B?

(b) What are the position vectors of A and B? (c) If a particle starts at A, travels to B, then travels back to A and stops there, then what is its displacement vector? (Answer on p. 1827.)

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60.1.

The Magnitude or Length of a Vector

√ In 2D space, the length of a vector u = (u1 , u2 ) was defined as ∣u∣ = u21 + u22 (Def. 121). Our Definition of a vector’s length in 3D space is very much analogous:

Definition 158. Given the vector u = (u1 , u2 , u3 ), its magnitude or length, denoted ∣u∣, is the number: √ ∣u∣ = u21 + u22 + u23 .

Example 774. If u = (1, 2, 3), then the length of u is ∣u∣ =

y

B = (0, a2 , a3 )

a2

D

√ 14.

It is natural to define the length of the Ð→ vector OA to be the length of the line segment OA, i.e. ∣OA∣.

A = (a1 , a2 , a3 )

a1 a3

12 + 22 + 32 =

To see why the above definition makes sense, pick any point A = (a1 , a2 , a3 ).

x

z

O = (0, 0, 0)

Our goal then is to find ∣OA∣. To do so, we first consider the point B = (0, a2 , a3 ).

Observe that the line segment OB is the hypotenuse of the right triangle ODB. And so, by Pythagoras’ Theorem: √ ∣OB∣ = a22 + a23 .

Now, observe that the line segment OA is the hypotenuse of the right triangle OBA. Moreover, ∣BA∣ = a1 . And so, again by Pythagoras’ Theorem: √ √ √ √ 2 2 2 2 2 2 ∣OA∣ = ∣BA∣ + ∣OB∣ = a1 + ( a2 + a3 ) = a21 + a22 + a23 . This completes our explanation of why the above Definition makes sense.

As before, the length of every vector must be non-negative. Moreover, a vector has zero length if and only if it is the zero vector: Fact 93. Suppose v is a vector. Then ∣v∣ ≥ 0. Moreover, ∣v∣ = 0 ⇐⇒ v = 0.

Exercise 243. Let A = (2, 5, 8) and B = (0, 1, 1) be points. What is the length of the vector from A to B? (Answer on p. 1827.)

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60.2.

Sums and Differences of Points and Vectors

As before:

1. 2. 3. 4.

Point Point Point Point

+ − + −

Point Point Vector Vector

= = = =

Undefined. Vector. Point. Point.

Ð→ Definition 125. Given two points A and B, the difference B − A is the vector AB.

And so, given the points A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ), their difference is: Ð→ B − A = AB = (b1 − a1 , b2 − a2 , b3 − a3 ) .

Example 775. Given the points A = (1, 5, 0) and B = (−2, 6, 3), their difference is the Ð→ vector B − A = AB = (−2 − 1, 6 − 5, 3 − 0) = (−3, 1, 3).

y

B = (−2, 6, 3) Ð→ B − A = AB = (−3, 1, 3) (The difference between the points A and B)

z

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Definition 159. Given a point A = (a1 , a2 , a3 ) and a vector v = (v1 , v2 , v3 ), their sum A + v is the following point: A + v = (a1 + v1 , a2 + v2 , a3 + v3 ).

Example 776. Given the point A = (1, 5, 0) and the vector v = (−3, 1, 3), their sum is the point A + v = (1 − 3, 5 + 1, 0 + 3) = (−2, 6, 3).

(The sum of a point A and a vector v)

y

A + v = (−2, 6, 3)

v = (−3, 1, 3)

z

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Definition 160. Given a point B = (b1 , b2 , b3 ) and a vector v = (v1 , v2 , v3 ), their difference B − v is the following point: B − v = (b1 − v1 , b2 − v2 , b3 − v3 ) .

Example 777. Given the point B = (−2, 6, 3) and the vector v = (−3, 1, 3), their difference is the point B − v = (−2 − (−3) , 6 − 1, 3 − 3) = (1, 5, 0).

y

B = (−2, 6, 3) v = (−3, 1, 3)

z

B − v = (1, 5, 0) (The difference between a point B and a vector v)

x

Exercise 244. Let A = (1, 2, 3), B = (−1, 0, 7), and C = (5, −2, 3) be points. What are (a) A + B; (b) A − B; (c) A + (B + C); and (d) A + (B − C)? (Answer on p. 1827.)

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60.3.

Sum, Additive Inverse, and Difference of Vectors 1. Vector + Vector = Vector. 2. − Vector = Vector. 3. Vector − Vector = Vector.

As before:

Definition 161. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors. Then their sum, denoted u + v, is the vector u + v = (u1 + v1 , u2 + v2 , u3 + v3 ).

Example 778. Given the vectors u = (1, 2, 3) and v = (−1, 0, 1), their sum is the vector u + v = (1 − 1, 2 + 0, 3 + 1) = (0, 2, 4).

v = (−1, 0, 1)

u + v = (0, 2, 4) (The sum of two vectors)

y

u = (1, 2, 3)

Place the tail of v at the head of u. Then the sum u + v is the vector from the tail of u to the head of v.

x

z

Definition 162. The additive inverse of u = (u1 , u2 , u3 ) is the vector: −u = (−u1 , −u2 , −u3 ).

Example 779. The additive inverse of u = (1, 2, 3) is the vector −u = (−1, −2, −3).

y

u = (1, 2, 3)

Flip u in the exact opposite direction to get its additive inverse −u.

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Definition 130. Given two vectors u and v, the difference u − v is the sum of u and the additive inverse of v. That is: u − v = u + (−v).

Fact 128. If u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are vectors, then: u − v = (u1 − v1 , u2 − v2 , u3 − v3 ) .

Proof. By Definition 162, −v = (−v1 , −v2 , −v3 ). And so by Definition 130, u − v = u + (−v) = (u1 − v1 , u2 − v2 , u3 − v3 ) .

Example 780. Given the vectors u = (1, 2, 3) and v = (−1, 0, 1), their difference is the vector u − v = (1 − (−1) , 2 − 0, 3 − 1) = (2, 2, 2).

v = (−1, 0, 1)

u = (1, 2, 3)

Place the heads of u and v at the same point. Then the difference u − v is the vector from the tail of u to that of v.

z

y

u − v = (2, 2, 2) (The difference of two vectors)

x

Exercise 245. Let u = (1, 2, 3), v = (−1, 0, 7), and w = (5, −2, 3) be vectors. What are (a) u + v; (b) u − v; (c) u + (v + w); and (d) u + (v − w)? (Answer on p. 1827.)

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Ð→ Ð→ Ð→ Fact 95. If A and B are points, then OB − OA = AB.

Proof. Let A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ). Then: Ð→ AB = (b1 − a1 , b2 − a2 , b3 − a3 ),

Ð→ Ð→ OA = (a1 , a2 , a3 ), and OB = (b1 , b2 , b3 ).

Ð→ Ð→ Ð→ Ð→ Ð→ By Fact 128, OB − OA = (b1 − a1 , b2 − a2 , b3 − a3 ). Thus, AB = OB − OA.

Example 781. Let A = (1, 5, 0) and B = (−2, 6, 3) be points. Ð→ Ð→ Ð→ Then AB = B − A = (−2, 6, 3) − (1, 5, 0) = (−3, 1, 3), OA = (1, 5, 0), and OB = (−2, 6, 3). Ð→ Ð→ Ð→ And indeed, OB − OA = (−2, 6, 3) − (1, 5, 0) = (−3, 1, 3) = AB. 3

y

B = (−2, 6, 3)

Ð→ Ð→ Ð→ OB − OA = AB = (−3, 1, 3)

A = (1, 5, 0)

Ð→ OB = (−2, 6, 3)

z

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Ð→ OA = (1, 5, 0)

x

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Ð→ Ð→ Ð→ Ð→ Ð→ Ð→ Fact 96. If A, B, and C are points, then AB − AC = CB and AB + BC = AC.

Proof. Let A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ), and C = (c1 , c2 , c3 ). Then:

And:

Ð→ AB = (b1 − a1 , b2 − a2 , b3 − a3 ), Ð→ AC = (c1 − a1 , c2 − a2 , c3 − a3 ), Ð→ CB = (b1 − c1 , b2 − c2 , b3 − c3 ),

Ð→ Ð→ AB − AC = (b1 − a1 , b2 − a2 , b3 − a3 ) − (c1 − a1 , c2 − a2 , c3 − a3 ) Ð→ = (b1 − c1 , b2 − c2 , b3 − c3 ) = CB. 3

Ð→ Ð→ Ð→ Ð→ Ð→ Observing that −CB = BC and rearranging, we also have AB + BC = AC. 3

Example 782. Let A = (1, 5, 0), B = (−2, 6, 3), and C = (4, −2, 1) be points. Ð→ Ð→ Then AB = B − A = (−2, 6, 3) − (1, 5, 0) = (−3, 1, 3), AC = C − A = (4, −2, 1) − (1, 5, 0) = Ð→ (3, −7, 1), and BC = C − B = (4, −2, 1) − (−2, 6, 3) = (6, −8, −2). Ð→ Ð→ Ð→ Ð→ And indeed, AB − AC = (−3, 1, 3) − (3, −7, 1) = (−6, 8, 2) = −BC = CB. 3 Ð→ Ð→ Ð→ Also, AB + BC = (−3, 1, 3) + (6, −8, −2) = (3, −7, 1) = AC. 3

B = (−2, 6, 3)

Ð→ AB = (−3, 1, 3)

Ð→ BC = (6, −8, −2)

y

A = (1, 5, 0) Ð→ AC = (3, −7, 1)

x z C = (4, −2, 1)

Ð→ Exercise 246. Let A = (5, −1, 0), B = (3, 6, −5), and C = (2, 2, 3) be points. Find AB, Ð→ Ð→ Ð→ Ð→ Ð→ Ð→ Ð→ Ð→ AC, and BC; and show that AB − AC = CB and AB + BC = AC. (Answer on p. 1827.)

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60.4.

Scalar Multiplication and When Two Vectors Are Parallel

Definition 163. Let v = (v1 , v2 , v3 ) be a vector and c ∈ R be a scalar. Then: cv = (cv1 , cv2 , cv3 ).

Fact 97. If v is a vector and c ∈ R, then ∣cv∣ = ∣c∣ ∣v∣.

Example 783. Let v = (−1, 0, 1) be a vector. Then 2v = (−2, 0, 2) and −3v = (3, 0, −3).

−3v = (3, 0, −3)

y

2v = (−2, 0, 2)

x

z

v = (−1, 0, 1)

Now: And so by Fact 97, we have:

√ √ 2 ∣v∣ = (−1) + 02 + 12 = 2

√ ∣2v∣ = 2 2

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and

√ ∣−3v∣ = 3 2.

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Definition 132. Two non-zero vectors u and v are said to point in:

(a) The same direction if u = kv for some k > 0; (b) Exact opposite directions if u = kv for some k < 0; and (c) Different directions if u ≠ kv for any k.

Definition 133. Two non-zero vectors u and v are parallel if u = kv for some k and non-parallel otherwise. Example 784. Let u = (1, 0, 1). Then u points in:

• The same direction as v = (3, 0, 3) because v = 3u.

• The exact opposite direction as w = (−2, 0, −2) because w = −2u. • A different direction from x = (5, 1, 0) because x ≠ ku for any k.

y

x = (5, 1, 0)

w = (−2, 0, −2) u = (1, 0, 1)

x

v = (3, 0, 3)

z

So, u is parallel to both v and w, but not to x. As shorthand, we may write: u ∥ v, w

and

u ∥/ x.

Remark 101. Again, note the special case of the zero vector 0 = (0, 0, 0). It does not point in the same, exact opposite, or different direction as any other vector. Also, it is neither parallel nor non-parallel to any other vector. Exercise 247. Continue to let u = (1, 0, 1), v = (3, 0, 3), w = (−2, 0, −2), and x = (5, 1, 0). State if each of the following pairs of vectors point in the same, exact opposite, or different directions; and also if they are parallel. (Answer on p. 1827.) (a) v and w.

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(b) v and x.

(c) w and x.

(d) u and 0.

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60.5.

Unit Vectors

The following definitions and results about unit vectors are exactly the same as before: Definition 134. A unit vector is any vector of length 1. Definition 135. Given a non-zero vector v, its unit vector (or the unit vector in its direction) is: v ˆ=

1 v. ∣v∣

Fact 98. Given any non-zero vector, its unit vector has length 1. Fact 99. Let v be a vector with unit vector v ˆ. If c ∈ R, then the vector cˆ v has length ∣c∣. Fact 100. Let a and b be non-zero vectors. Then: (a) (b) (c) (d)

ˆ a ˆ a ˆ a ˆ a

= = = ≠

ˆ b ˆ −b ˆ ±b ˆ ±b

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

a a a a

and and and and

b b b b

point in the same direction. point in exact opposite directions. are parallel. are non-parallel.

Exercise 248. Find the length and unit vector of each vector. (a) a = (1, 2, 3). (d) 2a.

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(b) b = (4, 5, 6). (e) 3b.

(c) a − b. (f) −4 (a − b).

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60.6.

The Standard Basis Vectors

Recall306 that in 2D space, the (two) standard basis vectors were the unit vectors that point in the directions of the positive x- and y-axes: i = (1, 0)

j = (0, 1).

and

Analogously, in 3D space, the (three) standard basis vectors are the three unit vectors that point in the directions of the positive x-, y-, and z-axes: Definition 164. The standard basis vectors are: i = (1, 0, 0),

k = (0, 0, 1)

z

j = (0, 1, 0),

and

y

j = (0, 1, 0)

k = (0, 0, 1).

x

i = (1, 0, 0)

Not surprisingly, every vector can be written as the linear combination of i, j, and k:307 Example 785. Let u = (7, 5, 3). Then u = 7i + 5j + 3k.

Exercise 249. Write each of the vectors v = (9, 0, −1) and w = (−7, 3, 5) as a linear combination of the standard basis vectors. (Answer on p. 1828.) 306 307

Ch. 49.14. You may also recall (Fact 101) that in 2D space, every vector can be written as the linear combination of two non-parallel vectors. It turns out that there is an analogous result in 3D space. For this result, we first define what it means for three (or more) vectors to be linearly independent: Definition 165. Three (or more) non-zero vectors are linearly independent if the first vector cannot be written as a linear combination of the other two vectors. We then have the following Fact (proof omitted). This Fact is definitely out of the H2 Maths syllabus and isn’t something you need worry about. Fact 129. Every vector can be written as the linear combination of three linearly independent vectors.

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60.7.

The Ratio Theorem

The Ratio Theorem is exactly the same as before and now reproduced: Theorem 15. Let A and B be points with position vectors a and b. Let P be the point that divides the line segment AB in the ratio λ ∶ µ. Then P ’s position vector is: p=

µa + λb . λ+µ

A

The point P has position vector: p=

λ The point A has position vector a.

P

µa + λb . λ+µ

µ The point B has position vector b. O

B

Exercise 250. Let A = (1, 2, 3) on B = (4, 5, 6) be points. Find the point that divides the line segment AB in the ratio 2 ∶ 3. (Answer on p. 1828.)

Let’s end this chapter with three more exercises: Exercise 251. Fill in the blanks.

(a) Informally, a vector is an “arrow” with two properties: ____ and ____. (b) A point and a vector are entirely different objects and should not be confused. Nonetheless, each can be described by ____. (c) Let A = (a1 , a2 , a3 ) be a point and a = (a1 , a2 , a3 ) be a vector. We say that a is A’s _____. (d) The vector a = (a1 , a2 , a3 ) carries us from the ____ to the point A = (a1 , a2 , a3 ). Exercise 252. Let A = (a1 , a2 , a3 ) be a point. Write down the vector from the origin to A in every possible way. (Answers on p. 1828.)

Ð→ Exercise 253. Let A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ) be points. What are A+B, A+ OB, Ð→ Ð→ Ð→ Ð→ Ð→ Ð→ OA + OB, OA − OB, OA − BA? (Answers on p. 1828.) 646, Contents

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61.

The Scalar Product (in 3D)

In 2D space, we defined the scalar product of u = (u1 , u2 ) and v = (v1 , v2 ) to be: u ⋅ v = u1 v1 + u2 v2 .

We define the scalar product in 3D space analogously: Definition 166. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors. Their scalar product u ⋅ v is the number: u ⋅ v = u1 v1 + u2 v2 + u3 v3 .

Example 786. Let u = (5, −3, 1), v = (2, 1, −2), and w = (0, −4, 3). Then: ⎛ 5 ⎞ ⎛ 2 ⎟ ⎜ u⋅v =⎜ ⎜ −3 ⎟ ⋅ ⎜ 1 ⎝ 1 ⎠ ⎝ −2

⎛ 5 ⎞ ⎛ 0 ⎟ ⎜ u⋅w =⎜ ⎜ −3 ⎟ ⋅ ⎜ −4 ⎝ 1 ⎠ ⎝ 3 ⎛ 2 ⎞ ⎛ 0 ⎟ ⎜ v⋅w =⎜ ⎜ 1 ⎟ ⋅ ⎜ −4 ⎝ −2 ⎠ ⎝ 3

⎞ ⎟ = 10 − 3 − 2 = 5. ⎟ ⎠

⎞ ⎟ = 0 + 12 + 3 = 15. ⎟ ⎠

⎞ ⎟ = 0 − 4 − 6 = −10. ⎟ ⎠

Recall308 that in 2D space, the scalar product was both commutative and distributive over addition. The same remains true of the scalar product in 3D space: Fact 130. Let a, b, and c be vectors. Then:

(a) a ⋅ b = b ⋅ a. (b) a ⋅ (b + c) = a ⋅ b + a ⋅ c.

Proof. Let309 a = (a1 , a2 , a3 ), b = (b1 , b2 , b3 ), and c = (c1 , c2 , c3 ). Then: (a)

(b)

308 309

a ⋅ b = a1 b1 + a2 b2 + a3 b3 = b1 a1 + b2 a2 + b3 a3 = b ⋅ a.

a ⋅ (b + c) = a1 (b1 + c1 ) + a2 (b2 + c2 ) + a3 (b3 + c3 ) = a1 b1 + a2 b2 + a3 b3 + a1 c1 + a2 c2 + a3 c3 = a ⋅ b + a ⋅ c.

Fact 105. Our proof here covers only the 3D case. For a more general proof, see p. 1592 (Appendices).

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Example 787. Continue to let u = (5, −3, 1), v = (2, 1, −2), and w = (0, −4, 3).

To illustrate commutativity, we can easily verify that: v⋅u=u⋅v=5

and

w ⋅ u = u ⋅ w = 5.

And to illustrate distributivity, we compute the following: ⎛ 5 ⎞ ⎛ 2+0 ⎟ ⎜ u ⋅ (v + w) = ⎜ ⎜ −3 ⎟ ⋅ ⎜ 1 − 4 ⎝ 1 ⎠ ⎝ −2 + 3 ⎛ 5+2 (u + v) ⋅ w = ⎜ ⎜ −3 + 1 ⎝ 1−2

⎞ ⎟ = 10 + 9 + 1 = 20. ⎟ ⎠

⎞ ⎛ 0 ⎞ ⎟ ⋅ ⎜ −4 ⎟ = 0 + 8 − 3 = 5. ⎟ ⎜ ⎟ ⎠ ⎝ 3 ⎠

And again, a vector’s length is the square root of its scalar product with itself: Fact 107. Suppose v be a vector. Then ∣v∣ =

√ 2 v ⋅ v and ∣v∣ = v ⋅ v.

√ Proof. By Definition 158, ∣v∣ = v12 + v22 + v32 . By Definition 139, v ⋅ v = v1 v1 + v2 v2 + v3 v3 = √ 2 v12 + v22 + v32 . Hence, ∣v∣ = v ⋅ v and ∣v∣ = v ⋅ v. Exercise 254. Compute (1, 2, 3)⋅(4, 5, 6) and (−2, 4, −6)⋅(1, −2, 3). (Answer on p. 1829.)

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61.1.

The Angle between Two Vectors

Place the tails of two vectors at the same point. Then as before, the angle between these two vectors is, informally, simply the (smaller) “amount” by which we must rotate one of the two vectors so that both point in the same direction. Example 788. As shown in the figure below, a, b, c, and d are vectors. The angle between a and b is α, while that between c and d is β.

y a α b c

d

β

x

z (Observe that here it so happens that α is acute, while β is obtuse.) And formally, we’ll use the exact same Definition as before: Definition 140. The angle between two non-zero vectors u and v is the number: cos−1

u⋅v . ∣u∣ ∣v∣

And as before, a simple rearrangement of Definition 140 yields: Fact 108. If u and v are two non-zero vectors and θ is the angle between them, then: u ⋅ v = ∣u∣ ∣v∣ cos θ.

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