581 69 44MB
English Pages 626 [753] Year 2014
Table of contents :
Preface to the Forth Edition
Contents
1. Sets
2. Relations
3. Functions
4. Mathematical Induction
5A. Basic Counting Principles
5B. Basic Counting Principles
6. Inclusionexclusion Principle
7. Recurrence Relations and Generating Functions
8. Monoids and Groups
9. Rings
10. Boolean Algebra
11. GRAPHS
12. Trees
13. Propositional Calculus
14. Matrix Algebra
15A. Arithmetic Progression
15B. Geometric Progression
16. Sequences
17. Partial Fractions
Index
Fourth Edition
SC EIE S oCIoRES
DISCRETE STRUCTURES
DISCRETE STRUCTURES For
B. Tech. and M. C.A. Courses
By
Dr SATINDER BAL GUPTA
Dr C.P. GANDHI
B.Tech., (CSE), MCA, UeCNET, Ph.D (CS)
Ph.D (Mathematics), UeCNET
Prof. Deptt. of Computer Science & Applications
Associate Professor; Deptt. of Mathematics,
Vaish College of Engineering, Rohtak,
Rayat & Bahra Institute of Engineering
Haryana.
&
BioTechnology, Kharar,
Email :[email protected]
Chandigarh.
Email :[email protected]
UNIVERSITY SCIENCE PRESS (An Imprint of Laxmi Publications Pvt. Ltd.) BANGALORE
•
CHENNAI
•
COCHIN
•
GUWAHATI
•
HYDERABAD
JALANDHAR
•
KOLKATA
•
LUCKNOW
•
MUMBAI
•
RANCHI
NEW DELHI
•
BOSTON, USA
DISCRETE STRUCTURES
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PREFACE TO THE FORTH EDITION "Discrete Structures" primarily meant for B.Tech. and M.C.A. students of various Indian Universities, has been written by taking into consideration the students' capability of solving the mathematical as well as graphical problems in a systematic and logical manner. The authors have certainly left no stone unturned in presenting the subject matter in a comprehensive and lucid way. The various topics like •
Basic Counting Principles
•
Inclusion· Exclusion Principle
•
Equivalence Relations and Partitions
•
Recurrence Relations and their Solutions by Generating Functions
•
Permutation and Symmetric Group
•
Direct Products of Groups
•
Dihedral Groups
•
Applications of Groups in Coding Theory
•
Rings
•
Applications of Boolean Algebra in Logic Circuits and Switching Designs
•
Applications of Graphs and Trees in Computer Science,
have been explained in an explanatory and methodological way. Further, the MCQs have also been provided at the end of the concerned chapters. This book gives an inkling of the capability of the authors and confirms that the authors must have been successful teachers. Errors or misprints, if any, are unintentional and regretted. Any suggestion/ recommendation for improvement of the subject matter of the book, are invited.
Dr. C.P. Gandhi Dr. Satinder Bal Gupta
(v)
CONTENTS Chapters
Pages
1. Sets
149
1 . 1. 1.2. 1.3. 1. 4. 1. 5. 1. 6. 1. 7. 1. 7. 1.8. 1.9. 1. 10. 1 . 1 1. 1. 12. 1. 13. 1. 14. 1. 15. 1. 15. 1. 15. 1. 16.
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1 Introduction . Set Formation .................. ................. ................. ................ ................. ................. 1 1 ................ ................. ................. ................ ................. Standard Notations Kinds of Sets .................................... ................. ................ ................. ................. ..... 2 Operation on Sets .............................. ................. ................ ................. ..................... 4 Algebra of Sets ................................................... ................ ................. ............... ....... 5 (a) Principle of Duality for Sets ........................ ................. ................. ................ 10 11 (b) Cardinality of a Set ...................................... ................ ................ ................. Cartesian Product of Two Sets ............................ ................. ................ .................. 12 Partitions of Sets and Venndiagrams ................ ................. ................. ................. 28 Venn diagrams ...................................................................................... .............. ..... 28 Cross Partition ..................................................................................... ............... ..... 28 Enumerable or Denumerable or Countably Infinite Set .................... .............. ..... 42 Countable Set .......................................................................................... .................. 42 Uncountable Set or Uncountably Infinite Set .................. ................. ................. 42 Fundamental Product ......................................................... ................ ............... ..... 42 (a) Minsets or Minterms .................. ................. ................ ................. .............. ..... 43 (b) Minset Normal Form .................. ................. ................ ................. .............. ..... 43 Maxsets or Maxterms ......................... ................. ................ ................. .................. 43 ....
2. Relations 5089 2.1. Introduction ....................................... ................. ................. ................. .................. 50 2.2. Ordered Pair .................................... ................. ................ ................. ............... ..... 50 2.3. Cartesian Product of Sets ................ ................ ................. ................ ............... ..... 50 2.4. Relation (or Binary Relation) ........................... ... ................. ................ .................. 50 2.5. Total Number of Relations ................................ ................ ................. .............. ..... 5 1 2.6. Domain and Range of a Relation ......................... ................. ................ .................. 5 1 2.7. Inverse Relation .................................................. ................. ................. .................. 5 1 2.8. Types of Relations ........................... ............... .. ................. ................ ............... ..... 5 1 2.9. Properties of Relations ..................... ................. ................ ................ ............... ..... 5 1 2. 10. Equivalence Relation ..................... .. ................ ................. ................ ............... ..... 53 2. 1 1. Compatible Relation ..................... .. ................. ................. ................ ............... ..... 54 2. 12. Partial Order Relation ..................... ................. ................ ................ ............... ..... 54 2. 13. Product of Sets ................................................... ................ ................. ............... ..... 54 (vii) .................................................................................................
(viii) 2. 14. 2. 16. 2. 16. 2. 17. 2. 18. 2. 19.
Ternary Relation Closure Properties of Relations ........................ ................. ........... ...... .............. ..... Composition of Relations ..................................... ................ ................. .................. Directed Graph or Digraph of a Relation .......................... ................. .............. ..... Equivalence Relations and Partitions ................................ ................ .............. ..... WarshalYs Algorithm to Find Transitive Closure .................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .
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3. Functions .............................................................................................. 90127 3.1. Function 90 3.2. Domain of a Function ......................... ................. ................ ................. .................. 91 3.3. Co·domain of a Function .................. ................. ................ ................. .............. ..... 91 3.4. Image of an Element ........................ ................ ................. ................ ............... ..... 91 3.6. Range of a Function or 1m (j) ............................... ................. ................ .................. 92 3.6. Description of a Function ..................................... ................ ................. .................. 92 3.7. (a) Everywhere Defined Function ..................... ................ ................. .............. ..... 93 3.7. (b) Graph of a Function ........................................ ................ ................. .................. 93 3.7. (c) Function as a Relation ................ ................ ................. ................ ............... ..... 94 3.8. Types of Functions .......................... ............... .. ................. ................ ............... ..... 96 3.9. Equal Functions ................................. ................ ................. ................. .................. 98 3. 10. (a) Identity Functions ........................ ................. ................ ................. .................. 98 3. 10. (b) Constant Function ..................... ................. ................. ................ ............... ..... 98 3. 1 1. Invertible (Inverse) Functions .............................................................. .................... 99 3. 12. Hashing Function .................................................................................... .................. 99 109 3. 13. Graphical Representation of One·one and Onto Functions .................... 1 14 3. 14. Composition of Functions ........................................................................ ll6 3. 16. Inverse Function ............................................... ................. .................. ll6 3. 16. Method to Find the Inverse of a Bijection .. ......................................................................................................................
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4*. Mathematical Induction .................................................................... 128137 4.1. Principle of Mathematical Induction .. 128 4.2. Working Rule ...................................................... ................ .................. 128 128 4.3. Peano's axioms ................................. ................ ................. .................. .
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SA. Basic Counting Principles .............................................................. 138166 6.1. 6.2. 6.3. 6.4.
Introduction ................. ................ ................... Basic Counting Principles ............... ................. ................ .................. Sum Rule ......................................... ................. ................. .................. Product Rule ..................................... ................ ................. .................. .......................................
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5B. Basic Counting Principles .............................................................. 142166 6.6. 6.6. 6.7. 6.8.
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Define Factorial n ............................. ................. ................ ................... Permutation ............................................................................................. Permutation with Restrictions ............................................................... Permutations When All of the Objects are Not Distinct ......................
Not meant for P,T,U, B.Tech, "Discrete Structures" (BTCS302) Course,
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(ix) 5.9. 5. 10. 5. 1 1. 5. 12. 5. 13.
Permutations with Repeated Objects ................ ................. .................. Circular Permutations ....................................... ................. ................. Combination ...................................... ................. ................ .................. Pigeonhole Principle ......................... ................. ................ .................. Extended Pigeonhole Principle
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6. Inclusionexclusion Principle ............................................................ 167179 167 6.1. InclusionBxclusion Principle .................................................................
7. Recurrence Relations and Generating Functions ........................... 180216 180 7.1. Introduction 7.2. Recurrence Relations ................. ................ ................ 180 180 7.3. Order of a Recurrence Relation ......................... ................. ................. 180 7.4. (a) Degree of Recurrence Relation ...................... ................ .................. 181 7.4. (b) Linear Recurrence Relation .. 181 7.5. Formation of Recurrence Relations . 7.6. Linear Recurrence Relation of Order n with Constant Coefficients . 183 183 7.7. Homogeneous Linear Recurrence Relation of Order n.. 7.8. Characteristic Equation . 183 7.9. Algorithm for Solving Homogeneous Linear Recurrence Relation of Order n with Constant Coefficients . 183 190 7. 10. Solution of Recurrence Relations by the Method of Substitution . 7. 1 1. Nonhomogeneous Linear Recurrence Relation of Order n with Constant Coefficient . 194 7. 12. Algorithm for Solving Nonhomogeneous Linear Recurrence Relation of Order n with Constant Coefficients . 194 7. 13. Generating Functions or Numeric Functions 205 7. 14. Generating Functions for Some Standard Sequences 207 7. 15. Solution of Recurrence Relation by the Method of Generating Functions 207 .............................................................................................. .........................................
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8. Monoids and Groups .......................................................................... 217313 8.1. Introduction 217 217 8.2. Algebraic Structure 217 8.3. Binary Operation 218 8.4. Tables of Operation 218 8.5. Properties of Binary Operations 22 1 8.6. Semigroup 225 8.7. Subsemigroup 225 8.8. Free semigroup 226 8.9. Congruent Relations and Quotient Structures 230 8.10. Homomorphism of semigroups 232 8. 1 1. Fundamental Theorem of Semigroup Homomorphism 234 8. 12. Monoid 8. 13. Submonoid 234 237 8.14. Group 8. 15. Zm (The Integers Modulo m (m 2: 1» ................... ................ ................. ................. 241 .....................
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(x) 8.16. 8. 17. 8. 17. 8.18. 8.19. 8.20. 8.20. 8.21. 8.22. 8.23. 8.24. 8.24. 8.25. 8.26. 8.27. 8.28. 8.29. 8.30. 8.30. 8.30. 8.30. 8.30. 8.30. 8.30. 8.31. 8.32. 8.33. 8.34. 8.35. 8.36. 8.37. 8.38. 8.39. 8.40. 8.41. 8.42. 8. 43. 8.44. 8.45.
Finite and Infinite Group (a) Order of a Group ......................... ................. ................ ................. .................. (b) Order of an Element ................... ................. ................ ................. .................. Subgroup .......................................... ................. ................. ................. .................. Abelian Group ....................................................................... ................. ................. (a) Cosets ............................................................................. ................. .................. (b) Coset Representative System for H in G ........................ ................ ................. Index of a Subgroup ............................................................ ................. .................. Normal Subgroup ............. ................ ................. ................ ................. ................. Quotient Group ............... ................. ................. ................ ................. .................. (a) Cyclic subGroup ................ ................ ................. ................ .................. (b) Cyclic Group ................ ................ ................. ................ ................. ................. Morphisms .................... . . ................ ................. ................. ................. .................. Kernel ! ........................... ................ ................. ................. ................. .................. Image ! ............................................. ................. ................. ................. .................. Dihedral Group ................................. ................. ................ ................. .................. Direct Product of Groups .................. ................. ................ ................. ................. Basic Terminology ............................................................... ................. .................. (a) Permutation .................................................................... ................. ................. (b) Composition of Permutations in Array Form ............... ................. ................. (c) Permutation as a Single Row .......................................... ................ ................. (d) Orbit of a Permutation ................................. ................. ................ .................. (e) Disjoint Permutations .................. ................. ................ ................. ................. (j) Cyclic Permutations .................... ................. ................. ................ .................. Multiplication of Cycles .................... ................. ................ ................ .................. Properties of Permutations ............... ................. ................ ................. ................. Even and odd permutations ............ ................. ................. ................ .................. Symmetric Group ................................................ ................ ................. ................. Alternating Group (A,) ....................................... ................. ................ .................. Applications of Groups in Coding Theory ................ ................. ................. Message .............................................................. ................. ................. .................. Word ............................... . . ................. ................ ................. ................. ................. Encoding Function ................. ................ ................. ................. .................. Weight ............................ . . ................. ................ ................. ................. ................. ................. ................ ................. ................. .................. Parity Check Code Hamming Distance ................ ................. ................. ................ .................. Minimum Distance ........................... ................. ................. ................ .................. Group Codes ...................................... ................. ................ ................. ................. More Applications of Groups ............. ................ ................ ................. ................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .
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241 241 241 242 245 252 252 255 255 261 261 261 266 266 266 276 283 287 288 288 289 289 289 289 290 293 294 297 302 304 304 304 304 305 305 305 306 306 309
9. Rings ................................................................................................... 314351 314 9.1. Ring 9.2. Commutative Ring ........................... ................ ................. ................. .................. 314 9.3. Ring with Unity ................................. ................. ................ ................. ................. 314 9.4. Finite and Infinite Ring ................... ................. ................ ................. .................. 315 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .
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(xi) 9.5. 9.5. 9.6. 9.7. 9.8. 9.9. 9.10. 9. 1 1. 9. 12. 9. 13. 9.14. 9. 14. 9. 15. 9.16. 9. 17. 9.18. 9.19. 9.20. 9.21.
(a) Ring with Zero Divisors ................................. ................ ................. ................. (b) Ring Without Zero Divisors ........................... ................ ................. ................. Ring of Integers Modulo m (m 2: 1) ..................... ................. ................ ................. Boolean Ring ..................................................... ................. ................. .................. Direct Product of Rings ..................... ................. ................ ................ .................. Morphism of Rings ........................... ................. ................. ................ .................. Subring ........................ . . . ................ ................. ................. ................. .................. Units Integral Domain ............. ................. ................ ................. ................. .................. Field .............................. . . ................. ................ ................. ................. .................. ................. ................ ................. ................. .................. Gaussian Integers Ideals ............................ . . ................. ................. ................. ................. .................. Sum of Ideals ....................................................................... ................. .................. Quotient Ring ....................................................................... ................. ................. Fundamental Theorem of Ring Homomorphism ................ ................. ................. Principal Ideal ..................................................................... ................. .................. Principal Ideal Domain (P.LD.) ......................... ................. ................ .................. Euclidean Domain Associate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .
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10. Boolean Algebra ............................................................................... 352446 10. 1. 10.2. 10.3. 10.4. 10.5. 10.6. 10.7. 10.8. 10.9. 10. 10. 10. 1 1. 10.12. 10. 13. 10. 14. 10. 15. 10. 16. 10. 17. 10. 18. 10. 19. 10.20. 10.21. 10.22.
Partially Ordered Relation Comparable Elements Non·Comparable Elements Linearly Ordered Set or Totally Ordered Set Hasse Diagrams Lattice Boolean Algebra Unary Operation Binary Operation Boolean Algebra as a Lattice Alternate Definition of Boolean Algebra Boolean Sub·algebra Atoms of a Boolean Algebra Isomorphic Boolean Algebras Representation Theorem Laws of Boolean Algebra Principle of Duality Boolean Expression or Boolean Function Literal Fundamental Product Sum· of· Products form or SOP Form Complete sum·of·Products form
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Minterm 391 BooLean Expansion Theorem ............................................................... .................. 392 Disjunctive Normal form or SumofProducts (or SOP) form ............... ................ 393 Conjunctive Normal form or Products of Sums (or POS) form ............. ................ 393 (a) Obtaining a Disjunctive Normal Form ............................................ ................. 393 (b) Obtaining a Conjunctive Normal Form ......................... ................. ................. 393 Prime Implicants ................................................................. ................. .................. 395 Karnaugh Map 395 Adjacent Fundamental Products 395 Karnaugh Map for two Variables 396 Karnaugh Map for Three Variables 396 Karnaugh Map for Four Variables 396 Looping 397 Looping Groups of Two 397 Looping Groups of Four ........................................................................................... 397 Looping Groups of Eigths ........................................................................................ 399 Karnaugh Map Method for Finding Prime Implicants and Minimal form for a Boolean Expression ................................................................................................................. 400 Basic Rectangle for Three Variable kmap ........................................... ................. 401 Applications of Boolean Algebra to Switching Circuits ........................ ................ 410 Truth Table for the Switches Connected in Parallel ........................... ................. 4 1 1 Truth Table for the Switches connected in Series ............... ................. ................ 4 1 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .
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11 GRAPHS 447524 1 1 . 1. Introduction ..................... ................. ................. ................ ................. ................. 447 11.2. Basic Terminology ............................ ................ ................. ................. .................. 447 1 l.3. Directed Graph ................................. ................. ................ ................. .................. 447 11.4. (a) Undirected Graphs ..................... ................. ................ ................. .................. 448 11.4. (b) Mixed Graph ................................ ................. ................ ................. ................. 449 11.4. (c) Finite Graph ............................ ..... ................. ................ ................. ................. 449 11.4. (d) Linear Graph .............................. ................ ................. ................. .................. 449 11.4. (e) Discrete or Null Graph ............... ................. ................ ................. .................. 449 11.5. Simple Graph .................................... ................. ................ ................. .................. 449 11.6. Complement Graph ........................... ................ ................. ................ .................. 449 11. 7. (a) Degree .......................................... ................. ................ ................. .................. 450 11. 7. (b) Indegree and Outdegree ............. ................. ................. ................ .................. 450 1 l.8. Source and Sink ................................ ................. ................ ................. ................. 450 11. 9. Even and Odd Vertex ........................ ................ ................. ................ .................. 450 11. 10. Adjacent Vertices ............................. ................. ................ ................. .................. 45 1 l l . l l. Path in a Graph 452 11. 12. Undirected Complete Graph ............. ................ ................. ................. ................. 455 11. 13. Connected Graph .............................. ................. ................ ................. .................. 456 11. 14. Disconnected Graph ......................... ................. ................ ................. .................. 456 11. 15. Connected Component ..................... ................. ................ ................. .................. 456 11. 16. Subgraph .......................................... ................. ................. ................. .................. 457 _
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(xiii) 1 1. 17. 1 1. 17. 1 1. 17. 1 1. 18. 1 1. 18. 1 1. 19. 1 1.20. 1 1.21. 1 1.22. 1 1.23. 1 1.24. 1 1.25. 1 1.26. 1 1.27. 1 1.28. 1 1.29. 1 1.30. 11.31. 1 1.32. 1 1.33. 1 1.34. 1 1.35. 1 1.36. 1 1.37. 1 1.38. 11.39. 11.40. 1 l.41. 11.42. 11.43. 11.44. 11.45. 11.46. 11.47. 11.48. 11.49. 11.50. 11.51. 11.52. 11.53. 11.54.
(a) Spanning Subgraph ....................................... ................ ................ .................. 460
(b) Complement of a Graph .............. ................. ................ ................. ................. 460
Complement of a Subgraph ................ ................ ................. ................. (a) Cut Set .......................................... ................. ................ ................. ................. ................ ................. ................. ................. (b) Cut Points or Cut Vertices Edge Connectivity ............................. ................. ................ ................. ................. Bridge (Cut Edges) ........................... ................. ................. ................ .................. Isomorphic Graphs ........................... ................. ................. ................ .................. Order and Size of Graph ................... ................ ................. ................ .................. Homeomorphic Graphs ...................................... ................. ................ .................. Weakly Connected ............................................. ................. ................. .................. Unilaterally Connected Digraph ................ ................. ................ ................. Strongly Connected Digraph ........................... .. ................. ................. ................. Disconnected Digraph ...................... ................. ................ ................. .................. Directed Complete Graph ................ ................. ................ ................. .................. Labelled Graphs ............................... ................ ................. ................. .................. Weighted Graphs ............. ................ ................. ................ ................. .................. Multiple Edges ................................. ................ ................. ................. .................. Multigraph ....................................... ................. ................. ................. .................. Traversable Multigraphs .................. ................. ................ ................. ................. Representation of Graphs ................ ................. ................ ................. .................. Other Important Graphs ................... ................ ................. ................ .................. Euler Path (or Chain) ........................ ................ ................. ................ .................. Euler Circuit (or Cycle) ..................... ................. ................ ................ .................. Euler Graph ....................................... ................. ................ ................. ................. Fleury's Algorithm ............................................ ................. ................. .................. Hamiltonian Path (or Chain) ............. ................ ................ ................. ................. Hamiltonian Circuit (Or Cycle) ............................................................................... Hamiltonian Graph .................................................................................................. Rules for Constructing Hamilton Paths (or Chains) and Hamilton Circuits (or Cycles) in a Graph .............................................................................................. Regular Graph .................................. ................ ................. ................. .................. Planar Graph .................................... ................. ................ ................. .................. Region of a Graph ................................................................. ................. ................. Properties of Planar Graphs ................................................ ................. ................. State and Prove Euler' s Theorem on Graphs ...................... ................ ................. Non Planar Graphs .............................................................. ................ .................. Properties of Non Planar Graphs ...................... ................. ................ .................. Graph Colouring ................................................ ................. ................. .................. Chromatic Number of G ................... ................. ................ ................. .................. Applications of Graph Theory ........................... ................. ................ .................. Dijkstra' s Algorithm for Shortest Path ............. ................. ................. ................. (c)
461 461 461 462 465 468 470 470 47 1 47 1 47 1 472 472 478 479 479 479 480 480 487 489 489 490 491 492 492 492 493 499 500 500 502 502 504 505 511 511 517 517
(xiv) 12. Trees .................................................................................................. 525557 12. 1. Introduction 525 12.2. Tree 525 12.3. Directed Trees ............... . . ................. ................. ................ ................. ................. 525 12.4. Ordered Trees .................................... ................. ................ ................. ................. 526 12.5. Rooted Trees ..................................... ................ ................. ................. .................. 527 12.6. Path Length of a Vertex ................... ................. ................ ................. .................. 527 12.7. Forest .............................................. .. ................ ................. ................. .................. 528 12.8. Binary Tree ....................................... ................. ................ ................. .................. 528 12.9. Basic Terminology ............................ ................ ................. ................. .................. 528 12. 10. Binary Expression Trees ................... ................ ................. ................ .................. 534 12. 1 1. Complete Binary Tree ...................... ................. ................ ................. .................. 535 12. 12. Full Binary Tree ............................... ................ ................. ................. .................. 535 12. 13. Traversing Binary Trees ................... ................ ................. ................ .................. 537 12. 14. Algorithms ......................................................... ................. ................. .................. 538 12. 15. Binary Search Trees ........................................... ................ ................. .................. 545 12. 16. Inserting into a Binary Search Tree .................. ................. ................ ................. 545 12. 17. Spanning Tree ...................................................................... ................. ................. 549 12. 18. Applications of Trees ............................................................................. .................. 550 12. 19. KruskaYs Algorithm to Find Minimum Spanning Tree ......................................... 550 ..............................................................................................................
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13*. Propositional Calculus ................................................................... 558601 13. 1. 13.2. 13.3. 13.4. 13.5. 13.6. 13.7. 13.7. 13.8. 13.9. 13. 10. 13. 1 1. 13. 12. 13. 13. 13. 14. 13. 15. 13. 16. 13. 17. 13. 18. 13. 19. 13.20. 13.21.
Basic Logic Operations ............................................................................................ Statement ......................................... ................. ................ ................. .................. Proposition ....................................... ................. ................. ................. .................. Propositional Variables ..................... ................. ................ ................ .................. Truth Table ....................................... ................. ................ ................. .................. Combination of Propositions .............................. ................. ................. ................. (a) Laws of the Algebra of Propositions ............. ................ ................. ................. (b) Variations in Conditional Statement ............ ................ ................ ................. Principle of Duality ............................................ ................. ................ .................. Logical Implication ............................................. ................. ................ .................. Logically Equivalence of Propositions ................ ................ ................. ................. Tautologies ......................................................... ................. ................. .................. Contradiction ....................................................................... ................. .................. Contingency .......................................................................... ................. ................. Functionally Complete Sets of Connectives ........................ ................ ................. Argument ............................................................................. ................. .................. Proof of Validity ................................ ................. ................ ................. ................. Quantifiers ....................................... ................. ................. ................. .................. Existential Quantifier ........................................ ................ ................. .................. Universal Quantifier .......................................... ................. ................ .................. Negation of Quantified Propositions .................. ................. ................ ................. Propositions with Multiple Quantifiers ............. ................ ................. .................
558 558 558 559 559 560 565 566 567 568 569 570 571 571 573 574 578 583 583 583 585 586
(xv) 14*. Matrix Algebra ................................................................................. 602648 14 1�Mili� � 14. 1. (b) Kinds of Matrices ........................ ................. ................. ................ .................. 602 14.2. Addition of Matrices ........................................... ................ ................. .................. 605 14.3. Multiplication of a Matrix by a Scalar ............... ................. ................ ................. 605 14.4. Properties of Matrix Addition ............................ ................. ................ .................. 605 14.5. Matrix Multiplication ......................................... ................. ................ .................. 606 14.6. Properties of Matr� Multiplication ................... ................ ................. ................. 609 14.7. Transpose of a Matr� ........................................ ................ ................. .................. 614 14.8. Properties of Transpose of a Matr� .................................... ................. ................. 614 14.9. Symmetric Matr� .................................................................................................... 615 14. 10. Skew· symmetric Matr� (or Anti·Symmetric Matr�) ............................................ 615 14. 1 1. Every Square Matrix can Uniquely be Expressed as the Sum of a Symmetric Matr� and a Skew· symmetric Matr� ................................................. 615 14.12. Orthogonal Matrix ................................................................................................... 616 14. 13. For any Two Orthogonal Matrices A and B, show that AB is an Orthogonal Matrix ................................................................................................... 616 14. 14. Adjoint of a Square Matr� .................................................................... ................. 618 14. 15. An Important Relation between a Square Matrix A and Adj A ................. 618 14. 16. Singular Matrices and Non·singular Matrices ..................................... ................. 620 14. 17. Inverse (or Reciprocal) of a Square Matr� ........................................... ................. 620 14. 18. The Inverse of a Square Matr�, if it Exists, is Unique ......................................... 620 14. 19. The Necessary and Sufficient Condition for a Square Matrix A to Possess Inverse is that I A I '" 0 (i.e., A is Non·Singular) .................................... 620 14.20. If A is Invertible, then so is Al and (Al)l = A ...................................................... 622 14.21. If A and B be Two Non·singular Square Matrices of the Same Order, then (AB)l = B1 Al 622 14.22. If A is a Non·singular Square Matrix, then so is A' and (Ajl = (Al)' .................... 622 14.23. If A and B are Two Non·singular Square Matrices of the Same Order, then adj (AB) = (adj B) (adj A) ................................................................................. 623 14.24. Solution of Simultaneous Linear Equations by Matrix Inversion Method or Matr� Method ..................................................................................................... 625 14.25. If A is Non·singular Matrix, then the System of Equations AX = B has a Unique Solution given by X = Al B ..................................................................... 626 14.26. Rank of a Matrix ............................................... ................. ................. .................. 630 14.27. To Determine the Rank of a Matr� A ............... ................ ................. ................. 63 1 14.28. Echelon Form of a Matrix ..................................................................... .................. 63 1 14.29. Rank of a Matrix by Using Echelon Form ............................................ ................. 63 1 14.30. Solution of a System of Linear Equations (Rank Method) or (Gauss Jordan Method) ........................................................................ .................. 635 ........................................... ..... ....... . ....................................................................... ....... ..................................
*
Not meant for P,T,U. B.Tech., "Discrete Structures" (BTCS402) Course,
(xvi) 14.31. Theorem: If A is a Nonsingular Matrix, then the Matrix Equation AX = B has a Unique Solution .............................................................................................. 636 14.32. Gauss Elimination Method ............... ................. ................ ................. ................. 643 15A* Arithmetic Progression 649670 15. 1. Sequence 649 15.2. Finite and Infinite Sequence ............. ................ ................ ................. ................. 649 15.3. Arithmetic Progression (AP.) 650 15.4. Important Observations 654 15.5. Sum of First n Terms of an AP 660 15.6. Arithmetic Means 667 15.7. Single AM. between any Two given Numbers 667 15.8. n AM.s between any Two given Numbers 667 _
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158*_ Geometric Progression 15.9. Geometrical Progression 15. 10. Important Observations 15. 1 1. Sum of n Terms of a G.P 15. 12. Particular Cases 15. 13. Sum to Infinity of a G.P 15. 14. Geometric Mean
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671697 67 1 67 1 681 681 687 691
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16* Sequences 16. 1. Sequence of Natural Numbers 16.2. Sequence of Squares of Natural Numbers 16.3. Sequence of Cubes of Natural Numbers _
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698704 698 698 699
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17*_ Partial Fractions 17 . 1. Introduction 17.2. Resolution of a Fraction into Partial Fractions 17.3. Method of Resolution Into Partial Fractions
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Index
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705717 705 705 705
____________________________________________________________________________________________________
Not meant for P,T,U. B.Tech., "Discrete Structures" (BTCS402) Course,
718724
SYLLABUS DETAILS The book entitled "Discrete Structures" is strictly according to the latest syllabus of various Indian universities as per details given below: Sr. N.
Name of the University
Subject
l. 2. 3. 4. 5. 6.
Anna University, Chennai Discrete Mathematics Himachal Pradesh University, Shimla Discrete Mathematics and Logic Design Kurukshetra University, Kurukshetra Discrete Structures Maharishi Discrete Structures Rohatak Dayanand University, Ponmcherry University) Puducherry Discrete Mathematics and Graph Theory Punjab Technical University, Jalandhar Foundation Computer Mathematical Punjay Technical University, Jalandhar Discrete Structures 8. Swami Vivekanand University, Discrete Structures Sironja, Sagar (MP) University of Mumbai, Mumbai Discrete Structures 10. Uttar Pradesh Technical University, Discrete Mathematical Lucknow Structures 7.
9.
(xvii)
Subject Code
(MA035) (IT 4003) (CSE205 E) (CSE203F) (MA T4l) (MCA104 (N2) BTCS402 (IT 302) (CSE305) (ECS303)
SYLLABUS ANNA UNIVERSITY, CHENNAI DISCRETE MATHEMATICS (MA 035) 1. Logic StatementsTruth TablesconnectivesNormal formsPredicate CalculusInference theory for statement calculus and Predicate Calculus.
2_ Combinatorics Review of Perm utation and combinationMathematical Inducation Pigeon hole principle Principle of inclusion and exclusionGenerating functionRecurrence relations.
3_ Groups Semi groupsMonoidsgroupspermutation groupConsetsLagranges theoremGroup homomorphismKernalRings and Fields (definitions and examples only).
4_ Lattices Partial orderingPosetsHasse diagramLatticesProperties of LatticesSub Lattices Special LatticesBoolean Algebra.
5_ Graphs Introduction to GraphsGraph terminologyRepresentation of GraphsGraph IsomorphismConnectivityEuler and Hamilton Paths
(xviii)
SYLLABUS HIMACHAL PRADESH UNIVERSITY, SHIMLA DISCRETE MATHEMATICS AND LOGIC DESIGN (IT 4003) SECTION A
Statements and Notation, Connectives; Negation; Conjunction; Disjunction; State ment Formulas and Truth Tables; Logical Capabilities of programming Languages; Conditional and Biconditional; Well formed Formulas; Tautologies; Equivalence of Formulas; Duality Law; Tautological Implications; Formulas with distinct Truth Tables; Functionally Complete sets of connectives; other connectives; Two state devices and Statement logic; Normal forms; principal disjunctive normal forms; principal conjunctive normal form; ordering and Uniqueness of Normal Forms; Completely parenthe sized Infix Notation and Polish Notation; The Theory of Interference for the statement calculus; Valid ity using Truth Table; Rules of Inference; Consistency of Premises and Indirect Method Of Proof; Auto matic Theory Proving; The Predicate Calculus; predicates; The Statement Function, variables and Quan tifiers; Predicate Formulas; Free and Bound Variables; The Universe of Discourse; Inference Theory of the Predicate Calculus, Valid Formulas and Equivalences; Some Valid Formulas over Finite Universes; Special Valid Formulas Involving Quantifiers; Theory of Inference for the Predicate Calculus; Formula Involving More Than One Quantifier. Mathematical Logic:
SECTION B
The Rules of Sumprobability, and prod uct; permutations; Combinations; Generation of permutationsIntroduction, and combinations, Discrete Information and Mutual Information. Relations and Functions: Introduction, A Relational Model for Data Bases; properties of Binary Re lations; Equivalence Relations and partitions; Partial Ordering Relations and Lattices; Chains and Antichains; A Job Scheduling problem; Functions and the Pigeonhole principle. Permutations, Combinations, and Discrete Probability:
SECTION C
Introduction, Basic Terminology, Multigraphs and Weighted Graphs, Paths and Circuits; Shortest paths in Weighted pathsplanar and circuits; and circuits, The Traveling Salesperson problem;Graphs, FactorsEulerian of Graph; Graph. Hamiltonian paths Trees and cutsets: Trees, Rooted Trees, path, Lengths in Rooted trees; prefix codes; Binary search trees; Spanning Trees and cutsets; Minimum Spanning Trees; Transport Networks.
Graphs and Planner Graphs:
SECTION D
Recurrence Relations and Recursive Algorithms: Introduction; Recurrence Relations; Linear Re currence Relations with constant coefficients; Homogeneous Solutions; Particular Solutions; total Solu tions; solution by the Method of Generating Functions; Sorting Algorithms; Matrix Multiplication Algo rithms. Groups and Rings: Introduction, Groups, Subgroups; Generators and evaluation of Powers; Co sets and Lagrange's Theorem; permutation groups and Burnside's theorem; Codes and Group codes; Isomor phisms and Automorphisms; Homomorphisms and Norma Integral Domains, and Fields; Ring Homo morphisms; Polynomial Rings and Cyclic Codes.
(xix)
SYLLABUS KURUKSHETRA UNIVERSITY, KURUKSHETRA DISCRETE STRUCTURES (CSE205 E) UNITI
Set Theory: Introduction to set theory, Set operations, Algebra of sets, combination ofsets, Duality, Finite and Infinite sets, Classes of sets, Power Sets, Multi sets, Cartesian Product, Representation of relations, Types of relation, Binary Relations, Equivalence relations and partitions, Partial ordering relations and lattices, Mathematics Induction, Principle ofInclusion and Exclusion, Propositions. Function and its types, Composition of function and relations, Cardinality and inverse relations. Functions and Pigeon Hole principles. UNITII
Propositional Calculus: Basic operations: AND, OR, NOT, Truth value of a compound
statement, propositions, tautologies, contradictions. Techniques of Counting: Rules of Sum of products, Permutations with and without repetition, Combination. Recursion and Recurrence Relation: Polynomials and their evaluation, Sequences, Introduction to AP, GP and AG series, partial fractions, linear recurrence relation with constant coefficients, Homogeneous solutions, Particular solutions, Total solution of a recurrence relation using generating functions.
UNITIII
Algebric Structures: Definition, elementary properties of algebric structures, examples
of a Monoid, Submonoid, Semigroup, Groups and rings, Homomorphism, Isomorphism and Automorphism, Subgroups and Normal subgroups, Cyclic groups, Integral domain and fields, Cosets, Lagrange's theorem, Rings, Division Ring.
UNITIV
Graphs And Trees: Introduction to graphs, Directed and Undirected graphs,
Homomorphic and Isomorphic graphs, Subgraphs, Cut points and Bridges, Multigraph and Weighted graph, Paths and circuits, Shortest path in weighted graphs, Eurelian path and circuits, Hamilton paths and circuits, Planar graphs, Euler's formula, Trees, Rooted Trees, Spanning Trees and cutsets, Binary trees and its traversals.
(xx)
SYLLABUS MAHARISHI DAYANAND UNIVERSITY, ROHATAK DISCRETE STRUCTURES (CSE203F) SECTION A
Set Theory and Propositional Calculus: Introduction to set theory, Set operations,
Algebra of sets, Duality, Finite and Infinite sets, Classes of sets, Power Sets, Multi sets, Cartesian Product, Representation of relations, Types of relation, Equivalence relations and partitions, Partial ordering relations and lattices Function and its Types: Composition offunction and relations, Cardinality and inverse relations. Introduction to propositional Calculus: Basic operations: AND, OR, NOT, Truth value of a compound statement, propositions, tautologies, contradictions.
SECTION B
Techniques of Counting and Recursion and Recurrence Relation: Permutations with and without repetition, Combination. Polynomials and their evaluation, Sequences, Introduction to AP, GP and AG series, partialfractions, linear recurrence relation with constant coefficients, Homogeneous solutions, Particular solutions, Total solution of a recurrence relation using generating functions. SECTION C
Algebric Structures: Definition and examples of a monoid, Semigroup, Groups and
rings, Homomorphism, Isomorphism and Automorphism, Subgroups and Normal subgroups, Cyclic groups, Integral domain and fields, Cosets, Lagrange's theorem.
SECTION D
Graphs and Trees: Introduction to graphs, Directed and Undirected graphs, Homomorphic and Isomorphic graphs, Subgraphs, Cut points and Bridges, Multigraph and Weighted graph, Paths and circuits, Shortest path in weighted graphs, Eurelian path and circuits, Hamilton paths and circuits, Planar graphs, Euler's formula, Trees, Spanning trees, Binary trees and its traversals.
(xxi)
SYLLABUS PONDICHERRY UNIVERSITY, PUDUCHERRY DISCRETE MATHEMATICS AND GRAPH THEORY (MA T41) UNITI
Connectives: Statement formulae, Equivalence of Statement formulae, Functionally complete set of connectives  NAND and NOR connectives, implication, Principal conjunctive and disjunctive normal forms. UNITII
Inference calculus: Derivation process  Conditional proof  Indirect method of proof
Automatic theorem proving  Predicate calculus.
UNITIII
Partial ordering: Lattices  Properties  Lattices as algebraic system  sub lattices Direct product and homomorphism  Special lattices  Complemented and Distributive lattices. UNITIV
Graphs: Applications of graphs  degree  pendant and isolated vertices  isomorphism
sub graphs  walks  paths and circuits  connected graphs  Euler graphs  operations on graphs  More on Euler graphs  Hamilton paths and circuits  complete graph.
UNITV
Trees: Properties of Trees  Pendant vertices in a Tree  Distance and Center in a Tree
 rooted and binary trees  spanning trees  Fundamental Circuits  Distance between spanning trees shortest spanning trees  Kruskal algorithm.
(xxii)
SYLLABUS PUNJAB TECHNICAL UNIVERSITY, JALANDHAR DISCRETE STRUCTURES (BTCS402) PARTA 1. Sets, Relations and Functions: Introduction, Combination of Sets, Ordered pairs,
Proofs of general identities of sets, Relations, Operations on relations, Properties of relations and Functions, Hashing Functions, Equivalence relations, Compatibility re [Chapters 1, 2, 3] [7] lations, Partial order relations. 2_ Rings and Boolean Algebra: Rings, Subrings, Morphism of rings, ideals and quo tient rings, Euclidean domains, Integral domains and fields. Boolean Algebra, direct product, Boolean subalgebra, Boolean Rings, Application of Boolean algebra (Logic Implications, Logic Gates, Karnaugh Map) [Chapters 9, 10] [8] 3_ Combinatorial Mathematics: Basic counting principles Permutations and Combina tions, Inclusion and Exclusion Principle. Recurrence relations, Generating Function, Applications. [Chapters 5A, 5B, 6, 7J [7]
PARTB
4_ Monoids and Groups: Groups, Semigroups and Monoids. Cyclic Subgroups and Submonoids, Subgroups and Cosets, Congruence Relations on Semigroups. Morphisms. Normal Subgroups. Homomorphism, Isomorphism, Dihedral Groups, Applications.
[Chapter 8] [7] 5_ Graph Theory: GraphDirected and Undirected, Eulerian Chains and cycles, Hamiltonian chains and cycles Trees, Chromatic number, Connectivity, Graph coloring, Planar and connected graphs, Isomorphism and Homomorphism. Applications.
[Chapters 11, 12] [7]
Note_ Chapters to be read for "Discrete Structures" (BTCS 402): Chapters No., 1, 2, 3, 5A, 5B, 6, 7, 8, 9, 10, 1 1, and 12 only.
(xxiii)
SYLLABUS PUNJAB TECHNICAL UNIVERSITY, JALANDHAR COMPUTER MATHEMATICAL FOUNDATION (MCAI04 (N2) SECTION A
Sets and Relations: Definition of sets, Subsets, Complement of a set, Universal set,
Intersection and Union of sets, DeMorgan's laws, Cartesian products, Equivalent sets, Countable and Uncountable sets, Minset, Partitions of sets, Relations: Basic definitions, [Chapters 1, 2] graphs of relations, Properties of relations.
SECTION B
Algebra of Logic: Propositions and logic operations, Connectives, Tautologies and Contradiction, Equivalence and implication, Principle of Mathematical Induction, Quantifiers. [Chapters 4, 13]
SECTION C
Matrix Algebra: Introduction of a Matrix, its different kinds, Matrix addition and Scalar multiplication, Multiplication of matrices, Transpose etc. Square matrices, Inverse and Rank of a square matrix, Solving simultaneous equations using Gauss eliminations, Gauss Jordan and Matrix Inversion methods. [Chapter 14] SECTION D
Graphs: A general introduction, Simple and Multigraphs, directed and Undirected
graphs, Eulerian and Hamiltonian Graphs, Shortest path algorithms, Chromatic number, Bipartite graph, Graph coloring. [Chapter 11] Note_ Chapters to be read for "Computer Mathematical Foundation (MCA 104 N2): Chapters No., 1, 2, 4, 1 1, 13 and 14 only.
(xxiv)
SYLLABUS SWAMI VIVEKANAND UNIVERSITY, SIRONJA, SAGAR (MP) DISCRETE STRUCTURES (IT 302) Un itI
Set Theory, Relation, Function, Theorem Proving Techniques : Set Theory: Definition of sets, countable and uncountable sets, Venn Diagrams, proofs of some general identities on sets Relation: Definition, types of relation, composition of relations, Pictorial representation of relation, Equivalence relation, Partial ordering relation, JobScheduling problem Function: Definition, type of functions, one to one, into and onto function, inverse function, composition of functions, recursively defined functions, pigeonhole principle. Theorem proving Techniques: Mathematical induction, Proof by contradiction. Un itII
Algebraic Structures: Definition, Properties, types: Semi Groups, Monoid, Groups, Abelian group, properties of groups, Subgroup, cyclic groups, Cosets, factor group, Permutation groups, Normal subgroup, Homomorphism and isomorphism of Groups, example and standard re sults, Rings and Fields: definition and standard results. Un itIII
Propositional Logic: Proposition, First order logic, Basic logical operation, truth tables, tau tologies, Contradictions, Algebra of Proposition, logical implications, logical equivalence, predi cates, Normal Forms, Universal and existential quantifiers. Introduction to finite state ma chine Finite state machines as models of physical system equivalence machines, Finite state machines as language recognizers Un itIV
Graph Theory: Introduction and basic terminology of graphs, Planer graphs, Multigraphs and weighted graphs, Isomorphic graphs, Paths, Cycles and connectivity, Shortest path in weighted graph, Introduction to Eulerian paths and circuits, Hamiltonian paths and circuits, Graph coloring, chromatic number, Isomorphism and Homomorphism of graphs. Un itV
Posets, Hasse Diagram and Lattices: Introduction, ordered set, Hasse diagram of partially, ordered set, isomorphic ordered set, well ordered set, properties of Lattices, bounded and complemented lattices. Combinatorics: Introduction, Permutation and combination, Binomial Theorem, Multimonial Coefficients Recurrence Relation and Generating Function: Introduc tion to Recurrence Relation and Recursive algorithms ) Linear recurrence relations with con stant coefficients, Homogeneous solutions, Particular solutions, Total solutions ) Generating functions , Solution by method of generating functions. (xxv)
SYLLABUS UNIVERSITY OF MUMBAI, MUMBAI DISCRETE STRUCTURES (CSE305) MODULEI
Set Theory: Sets, Venn diagrams, Operations on Sets; Laws of set theory, Power set and Products; Partitions of sets, The Principle of Inclusion and Exclusion. MODULEII
Logic: Propositions and logical operations, Truth tables; Equivalence, Implications;
Laws of logic, Normal Forms; Predicates and Quantifiers; Mathematical Induction. MODULEIII
Relations, Digraphs and Lattices: Relations, Paths and Digraphs; Properties and types of binary relations; Manipulation of relations, Closures, Warshall's algorithm; Equivalence and partial ordered relations; Posets and Hasse diagram; Lattice. MODULEIV
Functions and Pigeon Hole Principle: Definition and types of functions: Injective, Surjective and Bijective; Composition, Identity and Inverse; Pigeonhole principle. MODULEV
Generating Functions and Recurrence Relations: Series and Sequences;
Generating functions; Recurrence relations; Recursive Functions: Applications of recurrence relations e.g. , Factorial, Fibonacci, Binary search, Quick Sort etc. MODULEVI
Graphs and Subgraphs: Definitions, Paths and circuits: Eulerian and Hamiltonian;
Planer graphs, Graph coloring; Isomorphism of graphs; Subgraphs and Subgraph isomorphism. MODULEVII
Trees: Trees and weighted trees; Spanning trees and minimum spanning tree; Isomorphism of trees and sub trees; Prefix codes. MODULEVIII
Algebraic Structures: Algebraic structures with one binary operation: semigroup,
monoids and groups; Product and quotient of algebraic structures; Isomorphism, Homomorphism and Automorphism; Cyclic groups, Normal subgroups; Codes and group codes.
(xxvi)
SYLLABUS UTTAR PRADESH TECHNICAL UNIVERSITY, LUCKNOW DISCRETE MATHEMATICAL STRUCTURES (ECS303) UNITI
Set Theory: Introduction, Combination of sets, Multisets, Ordered pairs. Proofs of some general identities on sets.
Relations: Definition, Operations on relations, Properties of relations, Composite
Relations, Equality of relations, Recursive definition of relation, Order of relations. Functions: Definition, Classification offunctions, Operations on functions, Recursively defined functions. Growth of Functions. Natural Numbers: Introduction, Mathematical Induction, Variants of Induction, Induction with Nonzero Base cases. Proof Methods, Proof by counter  example, Proof by contradiction. UNITII
Algebraic Structures: Definition, Groups, Subgroups and order, Cyclic Groups, Cosets,
Lagrange's theorem, Normal Subgroups, Permutation and Symmetric groups, Group Homomorphisms, Definition and elementary properties of Rings and Fields, Integers Modulo n. UNITIII
Partial order sets: Definition, Partial order sets, Combination of partial order sets, Hasse diagram. Lattices: Definition, Properties of lattices  Bounded, Complemented, Modular and Complete lattice. Boolean Algebra: Introduction, Axioms and Theorems of Boolean algebra, Algebraic manipulation of Boolean expressions. Simplification of Boolean Functions, Karnaugh maps, Logic gates, Digital circuits and Boolean algebra. UNITIV
Propositional Logic: Proposition, well formed formula, Truth tables, Tautology, Satisfiability, Contradiction, Algebra of proposition, Theory of Inference.
Predicate Logic: First order predicate, well formed formula of predicate, quantifiers, Inference theory of predicate logic.
UNITV
Trees: Definition, Binary tree, Binary tree traversal, Binary search tree. Graphs: Definition and terminology, Representation of graphs, Multigraphs, Bipartite
graphs, Planar graphs, Isomorphism and Homeomorphism of graphs, Euler and Hamiltonian paths, Graph colouring. Recurrence Relation and Generating Function: Recursive definition of functions, Recursive algorithms, Method of solving recurrences. Combinatorics: Introduction, Counting Techniques, Pigeonhole Principle, Polya' s Counting Theory. (xxvii)
1
SETS
1 .1 . INTRODUCTION In this chapter) we will study some basic laws of algebra of sets) Venndiagrams) repre sentation of various sets) ordered and unordered partition of sets.
Set. A set is defined as a collection of distinct objects of same type or class of objects. The objects of a set are called elements or members of the set. Objects can be numbers, alphabets) names etc. e.g.,
A = {I, 2, 3, 4, 5}
A is a set of numbers containing elements 1, 2, 3, 4 and 5. Remarks. 1. A
set is usually denoted by capital letters A, B, C, D, P, Q, R, S, T etc. t or t l etc. Ql '
2. Elements of the set are defined by
p, q, r,
1 .2. SET FORMATION
PI'
r1 ,
The set can be formed by two ways : (i) Tabular form of a set (ii) Builder form of a set. (i) Tabular Form or Roster Form of a Set. If a set is defined by actually listing its members, e.g., if the set P contains elements a, b, C, d, then it is expressed as P = {a, b, c, d}.
This is called tabular form of a set. (ii) Builder Form of a Set. If a set is defined by the properties which its elements must satisfy e.g.,
P = {x : x E N, x is a multiple of 3}. R = {x : x > 1 and x < 10 and x is an odd integer}. T = {x : x is even number less than 9}.
This form is called Set builder form of a set and the method of defining the elements is called
Property Method.
1 .3. STANDARD NOTATIONS x belongs to A or x is an element of set A.
XE A x Ci A
x does not belong to set A.
x 'l A u B
=>
x 'l A and x 'l B
::::::}
x E Ac and x E Bc
::::::}
(: a E A
x 'l A n B
=>
x 'l A
=>
X E AC
=>
X E Ac u Bc
or
( ": a E A x 'l B
or
X E AC
=>
x 'l A
=>
x 'l A n B
=>
x E (A n B)'
or
X E Be x 'l B
.... (2)
A' u B' c (A n B)'
Combining (1) and (2), we get
a 'l A')
(A n B)' = A' u B'.
Hence proved.
Theorem VI. Prove Identity Laws i.e., (a) A u q, = A
(b) A n q, = q,
(c) A u U = U
Let
X E A u q,
XE A
Therefore)
X E A u q,
XE A
Hence
A u q, c A
Proof. (a) To prove A u q, = A
xE A
or
(d) A n U = A.
X E q, ( .: x 'l q" as q, is the null set) ... (1)
We know that A c A u B for any set B. For B = q" we have A c A u q, From (2) and (1), (b) To prove If x E A, then
... (2) A = A u q,.
A c A u q" A u q, c A
A n q, = q,
x 'l q,
Therefore, x E A, x 'l q, (c) To prove
=>
Hence proved. ( .: q, is null set)
A n q, = q,.
Hence proved.
AuV=V
Every set is a subset of universal set AuVcV Also,
Therefore, A u V = V.
(d) To prove We know
VcAuV Hence proved.
AnV=A
... (1)
AnVcA
So we have to show that A c A n V Let
and
XE A
XE A
XE A
XE An V
AcAnV From (1) and (2), we get
A n V = A.
XE V
( ": A c V so X E A
=>
X E V) ... (2)
Hence proved.
DISCRETE STRUCTURES
10
Theorem VII. Prove Complement Laws i.e., (b) A n A' = q, (a) A u A' = U (c) U' = q, (d) 5}, {n : n < 5}} (b) {in : n > 5}, {O}, {l, 2, 3, 4, 5}} 2 2 (e) {{n : n > ll}, {n : n < ll}}. Sol. (a) No, since 5 does not belong to any cell (b) No, since {O} is not a subset of N (c) Yes, the two cells are disjoint and their union is N.
30
DISCRETE STRUCTURES Example 5. Find all the partitions of S = {I, 2, 3j. Sol. Each partition of S contains either, 1, 2 or 3 different cells. The partition containing 1 cell is {I, 2, 3} = S The partitions containing 2 different cells are {{I}, {2, 3}}, {{2}, {I, 3}}, {{3}, {I, 2}} The partitions containing 3 different cells are {{I}, {2}, {3}} Hence, there are five different partitions of S.
Example 6. Consider the set Z of integers. Determine which of the following is a partition ofZ ? (i) P1 = {{nj " n E Z} (ii) P1 = {n : n E Z, n < OJ, P2 = {OJ, P3 = {n : n E Z and n > OJ (iii) P1 = {n " n E Z and n .'2 OJ, P2 = {n " n E Z and n '" OJ (iv) P1 = {n E Z : I n I = k, k =  1, 0, 1, 2, .... .j, P2 = {n E Z = I n I =  Ij. Sol. (i) Yes, since for different value of n, we have different sets and all the sets are
mutually disjoint. (ii) Yes, since P" P, and P3 are mutually disjoint and P 1 u P, u P3 = {n : n E Z, n < O} u {O} u {n : n E Z, n > O} = Z (iii) No, since P, and P, are not mutually disjoint as 0 E both P, and P, (iv) No, since P, = {n E Z = I n I =  I} = 3, x' > 15. Hence, we find only solutions of the equation x' + 5y = z for x = 1, 2, 3 For x = 1, we have 1 + 5y = z and the values of y, z are 1, 6; 2, I I. For x = 2, we have 4 + 5y = z and the values of y, z are 1, 9; 2, 14 For x = 3, we have 9 + 5y = z and the values of y, z are 1, 14. Hence R = {(I, 1, 6), (1, 2, 11), (2, 1, 9), (2, 2, 14), (3, 1, 14)} .
Example 9. Show, how a binary operation, say, addition (+), may be viewed as a ternary
relation.
Sol. Let the binary operation '+' may be defined as a set of ordered triples as follows:
+ = { (x, y, z) : x + y = z} Then, the relation '+' defined above is a ternary relation. For e.g., (2, 5, 7) E +, but (2, 4, 8) 'l +. Example. Let A = {l, 2, 3, ... I5}. Let R be the 4ary relation on A defined by R = {(x, y, z, t) : 4x + 3y + Z2 = t}. Write R as a set of 4tuples. Sol. Let (x, y, z, t) E R such that 4x + 3y + z, = t, Then x can assume the values 1, 2 and 3 only for all y, z, t E A. R = {(I, 1, 1, 8), (1, 1, 2, 1 1), (1, 2, 1, 1 1), (1, 2, 2, 14), (1, 3, 1, 14), Thus, (2, 1, 1, 12), (2, 1, 2, 15), (2, 2, 1, 15)}. 2.1 5. CLOSURE PROPERTIES OF RELATIONS
Consider a relation R on a given set A. Suppose the relation R does not satisfy the desired property. After adding the least number of ordered pairs to relation R, if R satisfies the desired property, then the desired property is called closure of relation R. (a) Reflexive closure. A relation RR is called reflexive closure of the relation R if R R is the smallest relation containing R having the reflexive property. For example, consider A = {7, 8, 1O}. Define a relation R by R = {(7, 8), (7, 10), (8, 8), (10, 7)}. Here R is not reflexive. We find the reflexive closure ofR. The relation R is reflexive if (x, x) E R for all x E A. :. We add the ordered pairs (7, 7), (10, 10) in R. The required reflexive closure of R is given by RR = {(7, 7), (10, 10), (7, 8), (7, 10), (8, 8), (10, 7)} (b) Symmetric closure. A relation Rs is called symmetric closure of R if Rs is the smallest relation containing R having the symmetric property. The smallest symmetric relation containing R is Rs = R U Rl For example, consider A = {4, 5, 6}. Define a relation R by R = {(4, 5), (5, 5), (5, 6)}. We find the symmetric closure of R.
DISCRETE STRUCTURES
56
Here Rl = {(5, 4), (5, 5), (6, 5)} . . The required symmetric closure of R is given by Rs = R U Rl = {(4, 5), (5, 5), (5, 6), (5, 4), (6, 5)} (c) Transitive closure. A relation R T is called transitive closure ofR if RT is the small· est relation containing R having the transitive property. For example, consider A = {4, 6, 8, 1O} and define a relation R on A by R = {(4, 4), (4, 10), (6, 6), (6, 8), (8, 1O)}. We find the transitive closure of R. Here (6, 8) E R, (8, 10) E R, but (6, 10) 'l R. So we add the ordered pair (6, 10) to R. Hence the required transitive closure of R is given by RT = {(4, 4), (4, 10), (6, 6), (6, 8), (8, 10) (6, 1O)}
(P. T. U. B. Tech. May 2007)
2.16. COMPOSITION OF RELATIONS
Let R and S be two relations from sets A to B and B to C respectively. Then a relation RoS is called composite relation from A to C where (a, c) E RoS iff we can find b E B such that (a, b) E R and (b, c) E S. The relation RoS is read as composition of R and S.
or Let A, B, C be any three sets. Let R be a relation from A to B and S be a relation from B to C. i.e., R is a subset of A x B and S is a subset of B x C. Then R and S give rise to a relation from A to C denoted by RoS and defined by a(RoS)c if for some b E B, we have aRb and bSc. Thus, RoS = {(a, c) : There exists b E B for which (a, b) E R and (b, c) E S} A = (1, 2, 3), B = (a, b, c), C = (x, y, z). e.g., Let Consider the following relation: R from A to B and S from B to C, given by R = {(I, b), (2, a), (2, c)} ; S = {(a, y), (b, x), (c, y), (c, z)} We find the composition relation RoS Draw the arrow diagram of R and S as shown Fig. 2. 1. There is an arrow from 1 to b which is followed by an arrow from b to x. Thus l(RoS)x or (1, x) E RoS. Similarly, (2, y), (2, z) E RoS. No other pairs belong to RoS. Thus, RoS = {(I, x), (2, y), (2, z)}
Fig. 2 . 1 Remark. In general RoS SoK Also, (RoS) l Sl oRl :f
=
Theorem II. Let A, B and C be three sets and R be a relation from A to B, S a relation from B to C. Then (RoSt1 = Sl oRl Proof. Let a E A, b E B, c E C. Then (a, b) E R, (b, c) E S (a, c) E S 'd a, b, C E S. Hence S is transitive. (iii) T = {(I, 1), (1, 2), (2, 2), (2, 3)}. Here (1, 2) E T, (2, 3) E T => (1, 3) E T ..
T is not transitive. (iv) The empty relation on A is always transitive. (v) The universal relation on any nonempty set A is always transitive
(d) Antisymmetric (i) R = {(I, 1), (1, 2), (1, 3), (3, 3)}. We know that a relation R is called antisymmetric relation on A iff (a, b) E R, (b, a) E R => a = b . . R is antisymmetric. (ii) S = {(I, 1), (1, 2), (2, 1), (2, 2), (3, 3)}. Here (a, b) E S, (b, a) E S => a '" b 'd a, b E S. .. S is not antisymmetric. Since (1, 2) E S, (2, 1) E S => 1 '" 2. (iii) T = {(I, 1), (1, 2), (2, 1), (1, 3), (3, 3)}. Here (1, 2) E T, (2, 1) E T => 1 ", 2. :. T is not antisymmetric.
(iv) The empty relation on A is not antisymmetric. (v) The universal relation on any nonempty set. For if, let a, b (a '" b) E A. If aRb, b R a and if R is antisymmetric, then, we must have a = b, which is a contradiction. Hence the universal relation on A is not antisymmetric.
Example lO_ Each of the following defines a relation on the set N ofpositive integers
R : x >y S : x + y = lO T : x + 4y = 10 for all x, y, E N Determine which of the relations are (a) reflexive (b) symmetric (c) transitive (d) antisymmetric SoL (a) Reflexive_ None are reflexive. For e.g., (1.1) " R, S and T R is not symmetric (b) Symmetric_ Take x = 3, y = 6, then clearly y > x, but x 1: y If (x, y) E S, then x + y = 10 => y + x = 10 => (y, x) E S. Hence S is symmetric. Also for x = 6, y = 1, x + 4y = 10 holds Le., (6, 1) E T, but (1, 6) " T i.e., T is not symmetric. (c) Transitive_ Let (x, y), (y, z) E R => x > y and y > Z => x > z :. (x, z) E R for all x, z E N R is transitive But S is not transitive. For e.g., (3, 7) E S since 3 + 7 = 10; (7, 3) E S since 7 + 3 = 10 But (3, 3) " S as 3 + 3 = 6 '" 10
RELATIONS
63
Further, if (x, y) E T, then x + 4y = 10 (y, z) E T, then y + 4z = 10 Consider x +
4z
(
10  x = x + 10  y = x + 10  
=
4
)
4x + 40  10 + x 5x + 30 '" 10 = 4 4
..
(x, z) 'l T. Hence T is not transitive. (d) Antisymmetric. If x > y and y > x, then x = y \;j x, Y E N
. . R is antisymmetric. Let (2, 8) E S => (8, 2) E S but 2 '" 8 . . S is not antisymmetric. If (x, y) E T, then x + 4y = 10 (y, x) E T, then y + 4x = 10
..
x + 4y = y + 4x 3y = 3x => x = y \;j x, Y E N
T is antisymmetric.
Example 11. (a) Find the numbers of relations from A = (a, b, c) to B = (1, 2).
(P.T.V., B.Tech. Dec. 2006) (b) Define ternary relation and give an example. (c) Define antisymmetric relation with an example. Sol. (a) We know that if a set contains m elements and B contains n elements, then
total number of relations from A to B is 2mn. Here m = 3, n = 2 required number of relations from A to B = 26 = 64. (b) A ternary relation is a set of ordered triples. In particular, if S is a set, then a subset of S x S x S is called a ternary relation on S. For example, If L is a line, then "betweenness" is a ternary relation among poin ts of L. (c) A relation R is said to be antisymmetric relation if (x, y) E R, (y, x) E R => y = x For e.g., consider A = The set of natural numbers. Define R as (x, y) E R iff x Iy. Then we know that if x divides y, y divides x then x = y for all x, y E R Hence the relation defined above is an antisymmetric relation.
Example 12. (a) Give an example to show that a reflexive relation on a set A is not necessarily symmetric. (b) Prove that a relation R on a set A is symmetric iff R = Ri . (P.T.U. M.C.A. Dec. 2005) Sol. (a) Consider A = (1, 2, 3) and define a relation R by R = {(I, 1), (2, 2), (3, 3), (1,3)}. Here (x, x) E R for all x E A. :. The relation R is reflexive. But (1, 3) E R => (3, 1) 'l R. :. The relation R is not symmetric. (b) Let R is symmetric on A. We show R = Rl Let (x, y) E R for all x, Y E A => (y, x) E R I R is symmetric By definition of inverse relation (x, y) E Rl R
C
R1
( 1)
64
DISCRETE STRUCTURES
Again let (x, y) E R1 for all x, Y E A => (y, x) E R => (x, y) E R R� c R => From (1) and (2) R = R1 Converse. Let R = Rl We show R is symmetric Let (x, y) E R => (x, y) E R1 => (y, x) E R for all x, Y E A. i.e., R is symmetric.
By definition of inverse relation I R is symmetric
(2)
I R = R1
Example 13. Let R be a relation on the set ofall lines in a plane defined by (11, 12) E R 1 1 is parallel to lz Show that R is an equivalence relation. Sol. Let A = The set of all lines in a plane and I E A. Since I is parallel to I for each I E A. :. R is reflexive. Also let (1" I,) E R for all 1" I, E A. Then by definition of R, 1, is parallel to I, => I, is also parallel to 1" It means (I" 1,) E R. . . R is symmetric. Finally, Let (1" I,) E R, (I" I,) E R for all 1" I" 13 E A. => 1, is parallel to I, Now, (1" I,) E R => I, is parallel to 13 , (I" I,) E R It means 1, is parallel to 13 , :. (1" I,) E R i.e., R is transitive. Hence we can say that the relation R is an equivalence relation. T, '" T, (T" T,) E R => T, ", T3 (T" T,) E R .. T, '" T3 => (T T,) E R. . . R is transitive. l' Hence the relation R is an equivalence relation. Example 15. Let A be a nonempty set and R be a relation on thepower set P(A) defined by (A, B) E R
Is R antisymmetric ?
Sol. Let A E P(A). Since A c A for all A E P(A) .. (A, A) E R. i. e., R is reflexive. Let (A, B) E R for all A, B E P(A) .
RELATIONS
65
Then A c B => B rz A i.e., B may not be a subset of A. Hence R is not symmetric. .. R is not an equivalence relation. But if (A, B) E R, (B, A) E R, then we have A c B, B c A => A = B. Hence R is antisymmetric.
Example 16. Consider the set Z of integers and an integer m > 1. We say that x is congruent to y modulo m, written as x = y (mod m) if x  y is divisible by m or x  y = km, for some integer k. Show that = is an equivalence relation on Z. Sol. Reflexive. For any x E Z, X = x (mod m) Since x  x = 0 is divisible by m. Hence = is reflexive Symmetry. Let x = y (mod m) => x  y is divisible by m => we can write x  y = km for some integer k.  (x  y) =  km = rm for some integer r =  k or or y  x = rm or y  x is divisible by m Hence y = x (mod m) ..
==
is symmetric.
Transitive. Let x = y (mod m), y = z (mod m)
x  y and y  z are divisible by m => x  y + y  z is also divisible by m => x  z is divisible by m => x = z (mod m) == is transitive Hence == is an equivalence relation.
Similar Problem 1. LetX = {1, 2, 3, 4, 5, 6, 7} andR = {{x, y) ,' xy is divisible by 3}. Check whether "a is congruent to b{mod 5)" is equivalence relation or not? Justify your answer. (P.T.V. B.Tech. Dec. 2007) Ans. Yes Similar Problem 2. Let R be a relation on the set Z of integers defined by a = b{mod 5)
{read as "a is congruent to b{mod 5)". Then, R is an equivalence relation on Z.
(P.T.U. B.Tech. May 2013)
Example 17. (a) Let A be the set of integers and let  be a relation on A x A defined by
(a, b)  (c, d) if a + d = b + c. Prove that  is an equivalence relation. (P.T.V. B.Tech. May 2012, May 2010) (b) Let A be a set of nonzero integers and let  be the relation on A x A defined by (a, b)  (c, d) if ad = bc. Show that  is an equivalence relation. Sol. (a) Reflexive. (a, b)  (a, b) if a + b = b + a, which is true. Hence  is reflexive. Symmetric. Let (a, b)  (c, d), then a + d = b + c or c + b = d + a and hence (c, d)  (a, b) Hence
�
is symmetric.
Transitive. Let (a, b)  (c, d) . Then a + d = b + c Let (c, d)  (e, j), then c + f= d + e Adding, a + d + c + f= b + c + d + e
a + f= b + e => (a, b)  (e, j).
=>
Hence is transitive. is equivalence relation. �
�
66
DISCRETE STRUCTURES
(b) Reflexive. (a, b)  (a, b) if ab = ba = ab, which is true is reflexive Symmetric. Let (a, b) (c, d), then ad = bc => bc = ad cb = da
a, b are non zero integers
�
�
(c, d)
�
(a, b)
a, b, c, d are nonzero integers
is symmetric Transitive. Let (a, b) (c, d) => ad = bc (c, d) (e, f) => cf= de Let Multiplying (1) and (2), we get (ad) (cf) = (bc) (de) . .
�
( 1)
�
(2)
�
cdaf = cdbe cd(af  be) = 0 af be = O af = be (a, b) (e, f)
I c, d are nonzero integers
�
is transitive Thus, is an equivalence relation. �
�
Example 18. (a) Define partial order relation with example. of b}.
(P.T.U. B.Tech. May 2006) (b) Let A be the set ofpeople and R be a relation on A defined by R = {(a, b): a is a brother
Is R equivalence relation ? Is R partial order relation ? (discard half brother, paternity brother). (c) LetR be a binary relation an A such that (a, b) E R, ifbook a costs more and contains fewer pages than book b. Is R equivalence relation ? Is R partial order relation ? Sol. (a) Partial order relation. A relation R on a set A is said to be a partial order relation on A iff it is (i) reflexive (ii) antisymmetric (iii) transitive.
Let P(A) denotes the power set of A and define a relation R as (A, B) E R A = B for all A, B E P(A) . .. R is antisymmetric. Finally, if (A, B) E R, (E, C) E R, then => A c C for all A, B, C E P(A). A c B, B c C R is transitive also. Hence the relation R is partial order relation. (b) Let a E A, then (a, a) 'l R (if a is a female, then no female can be a brother of R is not reflexive. herself). Let (a, b) E R i.e. , a is a brother of b. Then (b, a) may not belong to R (If b is a female, then b cannot be brother of a). Hence R is not symmetric. . .
RELATIONS
67
Further, if (a, b) E R and if a and b are male, then both a and b are brothers of each other. Consequently (b, a) E R, but a '" b. . . R is not antisymmetric. Transitive. If (a, b) E R and (b, c) E R. We show (a, c) E R. Case I. If (a, b) E R, then a has to be male. If (b, c) E R, then b has to be male. . . a and b are both male. Consequently (a, c) E R. Hence R is transitive. Thus, R is neither an equivalence relation nor a partial order relation. (c) Reflexive. No book costs more than itself, nor pages fewer than itself. Hence (a, a) 'l R, therefore R is not reflexive. Symmetric. If (a, b) E R, then the cost of the book 'd is more than the cost of the book 'b' . Here, (b, a) 'l R (since the cost of the book 'b' is less than the cost of the book 'd) . . R is not symmetric. Consequently R is not an equivalence relation, not a partial order relation.
Example 19. LetA = (a, b, c). Let R be a rekLtion an A defined by R = {(a, a), (a, b), (b, c), (c, a)}. Find the reflexive closure of R. Sol. Given relation R = {(a, a), (a, b), (b, c), (c, a)} is not reflexive. Adding the ordered pairs (b, b), (c, c) in R, the required reflexive closure of R is given by RR = {(a, a), (a, b), (b, c), (c, a), (b, b), (c, c)}. Example 20. Let A = (1, 2, 3, 4) and let R is defined by R = {(1, 2), (2, 3), (3, 4), (2, 1)}. Find the transitive closure of R. R = {(I, 2), (2, 3), (3, 4), (2, I)} Sol. Given Here (1, 2), (2, 1) E R. If R is transitive. Then there should be (1, 1) E R. Similarly, (1, 2) E R, (2, 3) E R. If R is transitive, there should be (1, 3) E R If (1, 3) E R, (3, 4) E R. We should also have (1, 4) E R Also (2, 1) E R, (1, 2) E R. If R is transitive, there should be (2, 2) E R. If (2, 3) E R, (3, 4) E R, we should have (2, 4) E R. Hence the required transitive closure of R is given by RT = {(I, 2), (2, 3), (3, 4), (2, 1), (1, 1), (1, 3), (1, 4), (2, 2), (2, 4)}
Example 21. Consider A = {I, 2, 3, 4}, and R, S be the relations defined by R = {(1, 2), (1, 1), (1, 3), (2, 4), (3, 2)}
Find RoS,
S = {(1, 4), (1, 3), (2, 3), (3, 1), (4, I)}.
Sol. We compute the elements of RoS. Consider the Fig. 2.6.
R
S
Fig. 2 . 6
RoS = {(I, 3), (1, 4), (1, 1), (2, 1), (3, 3)}
DISCRETE STRUCTURES
68
Example 22. Let R and S be the relations on A = (1, 2, 3, 4) A defined by R = {(1, 1), (3, 1), (3, 4), (4, 2), (4, 3)} 8 = {(1, 3), (2, 1), (3, 1), (3, 2), (4, 4)}
(a) Find the composition relation RoS (b) Find the composition relation SoR (c) Find the composition relation R2 = RoR (d) Find the composition relation R3 = RoRoR. Sol. (a) lRl and 183 => (1, 3) E R08 Also 3R1 and 1 8 3 => (3, 3) E R08 3R4 and 484 => (3, 4) E R08 => (4, 1) E R08 4R2 and 281 => (4, 1) E R08 4R3 and 38 1 => (4, 2) E R08 4R3 and 382 Thus, R08 = {(I, 3), (3, 3), (3, 4), (4, 1), (4, 2)} (b) First use 8 and then R. 183 and 3R1 => (1, 1) E 80R => (1, 4) E 80R 183 and 3R4 281 and lRl => (2, 1) E 80R => (3, 1) E 80R 381 and lRl => (4, 2) E 80R 484 and 4R2 484 and 4R3 => (4, 3) E 80R 80R = {(I, 1), (1, 4), (2, 1), (3, 1), (4, 2), (4, 3)} Thus, R = {(I, 1), (3, 1), (3, 4), (4, 2), (4, 3)} (c) Here R = {(I, 1) , (3, 1), (3, 4), (4, 2), (4, 3)} (1, 1) E R and (1, 1) E R => (1, 1) E RoR (3, 1) E R and ( 1, 1) E R => (3, 1) E RoR (3, 4) E R and (4, 2) E R => (3, 2) E RoR (3, 4) E R and (4, 3) E R => (3, 3) E RoR (4, 3) E R and (3, 4) E R => (4, 4) E RoR (4, 3) E R and (3, 1) E R => (4, 1) E RoR RoR = R' = {(I, 1) , (3, 1) , (3, 2) , (3, 3), (4, 4) , (4, I)} .. (d) Here But ( 1, 1) E (3, 1) E (3, 3) E (4, 4) E (4, 1) E ..
R3 = RoRoR = R'oR R' = {(I, 1) , (3, 1) , (3, 2) , (3, 3), (4, 4), (4, I)} R = {(I, 1) , (3, 1) , (3, 4) , (4, 2) , (4, 3)} R' and ( 1, 1) E R => ( 1, 1) E R'oR R' and ( 1, 1) E R => (3, 1) E R'oR R' and (3, 1) E R => (3, 1) E R'oR R' and (4, 2) E R => (4, 2) E R'oR R' and ( 1, 1) E R => (4, 1) E R'oR RoRoR = R'oR = {(I, 1) , (3, 1) , (4, 2) , (4, I)} .
I Using Part (c)
Example 23. Find the number of relations from A = {a, b, c} to B = {I, 2}. Sol. A contains 3 elements and B contains 2 elements. There are 3 x 2 = 6 ordered pairs
in A x B. Hence, total number of subsets of A x B is 2 6 = 64. Therefore, Total number of relations from A to B = 64 .
RELATIONS
69
Example 24. Let R be a relation on A = {I, 2, 3, 4} defined by "x is less than y". Write R as a set of ordered pairs. Find the inverse R1 of the relation R. Can R1 be described in words? Sol. R consists of the ordered pairs (x, y) where x < y. Thus R = {(I, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)} Reverse the ordered pairs of R to obtain Rl Rl = {(2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)} Thus, Rl is the relation (Ix is greater than y" Yes, Rl can be described in words, i.e. , Rl is the relation " x is greater than y". Example 25. Let R be a relation from A = (1, 2, 3, 4) to B = (x, y, z) defined by R = {(1, y), (1, z), (3, y), (4, x), (4, z)}
(a) Find the domain and range of R (b) Find the inverse relation R1 ofR. Sol. (a) The domain of R consists of the first elements of the ordered pairs of R and the
range consists of the second elements. Thus, dom (R) = {I, 3, 4} and Range (R) = {x, y, z} (b) Rl is obtained by reversing the ordered pairs in R Thus, Rl = {(y , 1), (z, 1), (y , 3), (x, 4), (z, 4)}. Example 26. Let A = {I, 2, 3, 4, 6} and R be the relation "x divides y". Write R as set of
ordered pairs. Find the inverse R1 of the relation R. Can R1 be described in words. Sol. We know that xly (read as x divides y) if there exists an integer z such that y = xz. We find those numbers in A which are divisible by 1, 2, 3, 4 and 6. Since 111, 112, 113, 114, 116, 2/2, 2/4, 2/6, 3/3, 3/6, 414, 416 Thus, R = {(I, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (4, 6)]
Reverse the ordered pairs of R to obtain R1 Thus, R1 = {(I, 1), (2, 1), (3, 1), (4, 1), (6, 1), (2, 2), (4, 2), (6, 2), (3, 3), (6, 3), (4, 4), (6, 4)] Yes, R1 can be described by the statement "x is a multiple of y".
Example 27. Let R and 5 be the relations on A = {I, 2, 3} defined by R = {(1, 1), (1, 2), (2, 3), (3, 1), (3, 3)} 5 = {(1, 2), (1, 3), (2, 1), (3, 3)}
(a) Find R n 5, R u 5 (b) Find R'. Sol. (a) R n S = {(I, 2), (3, 3)} R u S = {(I, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 3)} (b) Using the fact that A x A is the universal relation on A. Hence,
A
(1. 1) (1 . 2) (1 . 3) (2. 1) (2. 2) (2. 3) (3. 1) (3. 2) (3. 3) Fig. 2.7
70
DISCRETE STRUCTURES A x A = (1, 2, 3) x (1, 2, 3) = {(I, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} R' = A x A  R = The set of ordered pairs which are in A x A but not in R = {(I, 3), (2, 1), (2, 2), (3, 2)} Example 28. Let S = {(I, n) " n E Z}, T = (en, 1), n E Z} are relations on Z. Find
(i) ToS, (ii) SoT. Sol. (i) (n, 1) E T, (1, n) E S => (n, n) E TaS, n E Z Le., ToS = Z x Z (ii) To find SoT, here (1, n) E S, (n, 1) E T (1, 1) E SoT => .. SoT = {(I, I)}. Example 29. Consider the digraph given below Fig. 2. 8. Write relation as set of ordered pairs and check for equivalence or partial ordering. Sol. From the Fig., R = {(a, b), (b, a), (a, c), (c, c), (c, d), (b, d)} Reflexive. (a, a) 'l R . . R is not reflexive. Symmetric. (a, c) E R, but (c, a) 'l R . . R is not symmetric. Antisymmetric. (a, b) E R, (b, a) E R, but a '" b always. ..
R is not antisymmetric.
d
Transitive. (a, c) E R, (c, d) E R, but (a, d) 'l R
R is not transitive. Hence R is neither an equivalence relation nor a partial order· Fig. 2.8 ing relation. Example 30. Let A = (1, 2, 3, 4, 5, 6, 7, 8) and R be a relation on A defined by
U(x, y) E R iff y is divisible by x". Is R equivalence relation ? Is R partial order relation ? Draw its digraph. Sol. Here R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 2), (2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4,4), (4, 8), (6, 6), (6, 6), (7, 7), (8, 8) } Reflexive. (a, a) E R for all a E R R is reflexive.
Antisymmetric. R is antisymmetric. Transitive. R is transitive. Hence, R is a partial order relation. But R is not symmetric. (1, 2) E R but (2, 1) 'l R. . . R is not an equivalence relation. Digraph of R. It is shown in Fig. 2.9 Example 31. Given A = {I, 2, 3, 4} and define R = {(1, 1), (2, 2), (2, 3), (3, 2), (4, 2), (4, 4)}
Fig. 2 . 9
(a) Draw its directed graph (b) Is R (i) reflexive (ii) symmetric (iii) transitive or (iv) antisymmetric (c) Find R2 = RoR. Sol. (a) The directed graph of R is shown below (Fig. 2. 10). (b) (i) 3 E A, but (3, 3) 'l R . . R is not reflexive. (ii) (4, 2) E R, but (2, 4) 'l R . . R is not symmetric. (iii) (4, 2) E R, (2, 3) E R, but (4, 3) 'l R . . R is not transitive. (iv) (2, 3) E R, (3, 2) E R, but 2 '" 3 R is not antisymmetric.
Fig. 2 . 1 0
71
RELATIONS (c) Here R = {(I, 1), (2, 2), (2, 3), (3, 2), (4, 2) (4, 4)} R = {(I, 1), (2, 2), (2, 3), (3, 2), (4, 2), (4, 4)} (1, 1) E R and (1, 1) E R => (1, 1) E RoR
. .
(2, 2) E (2, 2) E (3, 2) E (3, 2) E (4, 2) E (4, 4) E (4, 4) E
R and (2, 2) E R R and (2, 3) E R R and (2, 3) E R R and (2, 2) E R R and (2, 3) E R R and (4, 2) E R R and (4, 4) E R R' = RoR = {(I, 1), (2, 2),
=>
=>
(2, 2) E RoR (2, 3) E RoR
(3, 3) E RoR => ( 3 , 2) E RoR => (4, 3) E RoR =>
(4, 2) E RoR (4, 4) E RoR (2, 3), (3, 2), (3, 3), (4, 2), (4, 3), (4, 4)} => =>
Example 32. Let S be a relation on X = {a, b, c, d, e, f} defined by
S = {(a, b), (b, b), (b, c), (c, f), (d, b), (e, a), (e, b), (e, f!) Draw the directed graph of S.
Sol. Write down the elements of X Draw an arrow from the letter x to the letter y if (x, y) E S. The required directed graph of S is shown in the Fig. 2. 1 1 Example 33. (a) LetA = {I, 2, 3, 4, 6} andR be the relation
d
Fig. 2 . 1 1 x I y (read as x divides y) defined by R=n � � � � � � �� � ��� � � � � � � � � � Draw the directed graph of R (b) Let R = {(1, 1), (2, 2), (5, 5), (1, 2), (1, 3), (1, 4), (2, 1), (4, 2), (3, 5), (3, 4)} be a relation (P.T.U. B.Tech Dec 2013) on the set A = {I, 2, 3, 4, 5}. Represent R as a directed graph (c) Let R be a relation defined on the set A given by R = {(1, 1), (2, 2), (2, 3), (3, 2), (4, 2), (4, 4)}. Draw its diagraph. Also Justify whether R is (i) reflexive (ii) antisymmetric (P.T.U. B.Tech Dec 2010) (iii) transitive Sol. (a) Write down the elements 1, 2, 3, 4, 6. Draw an arrow from the integer x to the integer y if x divides y. The directed graph is shown in Fig. 2. 12(a)
, ['
2 3 4 6 (a)
1 0 0 0 0
2 1 o o o
3 0 1 0 0 ( b)
4 6 1 0 1 0
i]
72
DISCRETE STRUCTURES (b) The required diagraph is shown in Fig. 2. 12(c)
Diagraph of R (d)
(c)
Fig. 2.12
(c) Given R = {(I, 1), (2, 2), (2, 3), (3, 2), (4, 2), (4, 4)}. The diagraph of R is shown in Fig. 2. 12(d) (i) R is not relflexive since (3, 2) E R (ii) R is not antisymmetric since (4, 2) E R, (2, 2) E R, but 4 '" 2 (iii) R is transitive Example 34. Find the transitive closure RT ofthe relationR an A = {I, 2, 3, 4} defined by the directed graph as shown in Fig. 2. 13.
Sol. From the directed graph, lR2, 2R2, 2R4, 4Rl, 4R3, 3R2, 3R4 :. R = {(I, 2), (2, 2), (2, 4), (4, 1), (4, 3), (3, 2), (3, 4)} Now (2, 4) E R, (4, 1) E R => (2, 1) 'l R . . We add (2, 1) in R. 3 Also (1, 2) E R, (2, 1) E R => (1, 1) 'l R . . We add (1, 1) in R (1, 2) E R, (2, 4) E R => (1, 4) 'l R . . We add (1, 4) in R (1, 4) E R, (4, 3) E R => (1, 3) 'l R . . We add (1, 3) in R (2, 1) E R, (1, 3) E R => (2, 3) 'l R . . We add (2, 3) E R (3, 2) E R, (2, 1) E R => (3, 1) 'l R . . We add (3, 1) in R (3, 1) E R, (1, 3) E R => (3, 3) 'l R . . We add (3, 3) in R (4, 1) E R, (1, 2) E R => (4, 2) 'l R . . We add (4, 2) in R (4, 1) E R, (1, 4) E R => (4, 4) 'l R . . We add (4, 4) in R . . The transitive closure of R is given by RT = {(I, 2), (2, 2), (2, 4), (4, 1), (4, 3), (3, 2), (3, 4), (2, 1), (1, 1), (1, 4), (1, 3), (2, 3), (3, 1), (3, 3), (4, 2), (4, 4)} The number of order pairs in Rr is 16
Fig. 2.13
.. RT = A x A. Example 35. (a) Let R be a relation on a set A. Give a procedure to find the symmetric
and reflexive closure ofR. (b) Using the procedure ofpart (a), find the reflexive closure ofR and symmetric closure of R where A = {I, 2, 3} and R = {(1, 1), (1, 2), (2, 3)}.
RELATIONS
73
Sol. (a) The symmetric closure ofR is R u Rl The reflexive closure ofR is R u /j.A where
/j.A is the diagonal relation
(b) The symmetric closure of R is R U Rl = {(I, 1), (1, 2), (2, 1), (2, 3), (3, 2)} Also /j.A = The diagonal relation on A = (1, 1), (2, 2), (3, 3) :.
The reflexive closure of R is R U /j.
A = {(I, 1), (1, 2), (2, 3), (2, 2), (3, 3)}
Complement of a Relation
Let R is a relation from a set A to a set B i.e., R is a subset of A x B. The complement of
R, denoted by R is defined by R
= {(a, b) : (a, b) 'l R} Example 36. Consider the sets A = {I, 2, 3, 4} and B {a, b, c}.
Let R = {(1, a), (1, b), (2, b), (2, c), (3, b), (4, b)}. Find R Sol. A x B = {(I, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c), (4, a), (4, b), (4, c)}
= A x B  R. The set of all ordered pairs in A x B but not in R = {(I, c), (2, a), (3, a), (3, c), (4, b), (4, c)}. Example 37. Let A = {a, b, c, d, e} andR be a relation onA whose corresponding diagraph :. R
is given below Fig. 2. 14. Find R .
1
Sol. From the diagraph,
R = {(a, b), (a, e), (a, d), (b, c), (b, d), (b, e), (c, c), (d, d), (e, e)}
(a, a), (a, b), (a, c), (a, d), (a, e), (b, a), (b, b), (b, c), (b, d), (b, e), (c, a), (c, b),(c, c), (c, d), Also A x A = (c, e),(d, a), (d, b), (d, c), (d, d), (d, e),(e, a), (e, b), (e, c), (e, d), (e, e)
. . R = A x A  R. The set of all ordered pairs in A x A but not in R
= {(a, a), (a, c), (b, a), (b, b), (c, a), (c, b), (c, d),
(c, e), (d, a), (d, b), (d, c), (d, e), (e, a), (e, b), (e, c), (e, d)}
I
Fig. 2.14
TEST YOUR KNOWLEDGE 2.1
SHORT ANSWER TYPE QUESTIONS
1. What is a congruent relation ? Define an equivalence relation with the help of an example, What is a relation ? Give example, If R is a relation on a finite set A having elements. What will be the number of relations on A? 5. (a) If A = (1, 2, 3, 4) and R = {(2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Examine whether R is symmetric) transitive or reflexive,
2. 3. 4.
n
74
DISCRETE STRUCTURES �
�
Let R {(I, 1),relation (1, 2), (2, 1), (2, 2), (3, 3)} be a relation on the set S {I, 2, 3} . Prove that R is an equivalence on S. Consider the relation 's on the set A (2, 3, 4, 5) . Find the inverse relation on A. 6. Consider a set A of human beings and define a relation as "is a brother of' on A. Then which of the following is true, (0 Transitive) symmetric and reflexive (ii) Transitive, but neither reflexive nor symmetric (iii) Transitive and reflexive but not symmetric (iv) Neither reflexive, nor transitive nor symmetric. 7. Examine whether the relation 'Divides' on the set of natural number is a partial order relation. Is it equivalence relation ? Prove that the relation ';:':.: on the real numbers is not an equivalence relation. Let A be a set of integers and R be a relation on A x A defined by R d) Prove that R is an equivalence relation. 8. Which of the following collections of subsets are partitions of [I, 2, .. ,6] ? (b)
=
(a)
(P. T U. B. Tech. Dec 2013)
(b)
(a)
(P. T u., B. Tech. Dec. 2009)
(b)
(PT. V., M. G.A. Dec. 2005)
=:::} a + d = b + c. (a, b) (e, (PT. U. B. Tech. May 2008, M. C.A. Dec. 2005)
(c)
9.
(a) {I, 2l. {2, 3, 4l. {4, 5, 6} (c) {2, 4, 6l. {I, 3, 5}
List all partitions of A =
� {I, 2, 3}.
(b) {I}, {2, 3, 6l. {4l. {5}
(d) {I,
4, 5l. {2, 6}.
LONG ANSWER TYPE QUESTIONS
Consider A {4, 5, 6, 7, 8}. Give an example of a relation which is (i) Reflexive) transitive but not symmetric (ii) Symmetric and antisymmetric. (iii) Antisymmetric but not reflexive (iv) Neither symmetric nor antisymmetric (v) Neither symmetric, asymmetric, nor antisymmetric. 11. Consider A (1 , 2, 3, 4) and R be a relation defined by R {(I, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 4), (4, 3), (4, 4)}. Determine whether or not R is (i) Reflexive (h) Irreflexive (iii) Symmetric (iv) Asymmetric (v�) Transitive. (v) Antisymmetric 12. Consider A {I, 2, 3}. Give an example of a relation on A which is (i) Reflexive, symmetric, transitive and antisymmetric. (ii) Neither symmetric nor antisymmetric (iii) Reflexive, symmetric and not transitive (iv) Not reflexive, not symmetric, not antisymmetric and not transitive. (v) Symmetric and transitive but not reflexive (vi) Neither reflexive nor symmetric (vii) Reflexive but not symmetric. 13. Let A (1, 2, 3, 4) and R {(2, 1), (2, 3), (3, 2), (3, 3), (2, 2), (4, 2)}. Find Reflexive closure of R Symmetric closure of R.
10.
=
�
=
(a)
�
�
(b)
RELATIONS
75
Let R and S be relations on a set A. Determine whether each of the following statements is true or not. If not true, give counter example If R and S are transitive, then R S is transitive If R and S are transitive, then R S is transitive. If R is transitive, then R1 is transitive. If R and S are reflexive, then RoS is reflexive If R is antisymmetric, then R1 is antisymmetric. 15. If A and R b)} be a relation on R. Find the transitive closure ofR. 16. Consider a set A iff = 2, and S be a relation 3, 5}. Let R be a relation defined as defined as iff :s; find (ii) SoR (iii) Is RoS SoR (i) RoS 17. Consider the following relations on the set N of positive integers: R: is greater than S: Determine which of the relations are reflexive Determine which of the relations are symmetric Determine which of the relations are antisymmetric. Determine which of the relations are transitive. Hint. See Example 18. Give examples of relations R on A 2, 3} having the stated properly R is both symmetric and antisymmetric R is neither symmetric nor antisymmetric R is transitive but R Rl is not transitive. 19. (i) Let A be a set of nonzero integers and let be the relation on A x A defined by d) if = Prove that is an equivalence relation. (ii) Show that the relation of set inclusion is not an equivalence relation. (iii) Let R be the relation on the set of positive integers defined by R is even}. Is R an equivalence relation ? (iv) Consider the relation of perpendicularity on the set L of lines in the Euclide an plane. Is an equivalence relation ? 20. Prove the following : If R and S are reflexive relations on a set A, then R S is also reflexive If R and S are symmetric relations on a set A, then R S is also symmetric If R is reflexive relation on A. Then Rl and R S are reflexive for any relation S on A Show, by a counter example, that R and S may be transitive relations on A, but R S need not be transitive. If R is any relation on A, show that R R1 is symmetric. 21. Let d}, B 2, 3} and C z}. Consider the relations R from Ato B and S from B to C defined by R 3), 3), 2)} 3), S (2. (2. z)} Find RoS. 22. Let A, B, C, D be any four sets. Suppose R is a relation from A to B, S is a relation from B to C and T is a relation from C to D. Then show that (RoS)oT Ro(SoT). 23. Show that union of two equivalence relations need not be an equivalence relation. 14.
u
(a)
II
(b) (c)
(d)
(e)
� {(a. a). (b. c). (a.
� (a. b. c)
= {I, x y
xSy
xRy
y
x+
�
x
?
x + y = 10;
y,
T � x + 4y � 10
(a)
(b) (e)
(d)
10
= {I,
(a) (b) (e)
u
(a, b)
�
ad
be.
�
c
(e,
= {(a, b): a + b
�
�
II
(a) (b) (e)
u
II
u
(d)
u
(e)
A � {a. b. c.
� {(a,
� {(I. x).
(a)
�
� {I, (b.
� {w. x. y.
(c, 1), (c,
(d.
y),
=
(P. T. U. M. G.A. Dec. 2005. 2006. May 2003)
76
DISCRETE STRUCTURES
Consider Z, the set of integers and an integer m > We say that is congruent to modulo (written as =y(mod. m). Show that this defines an equivalence relation on Z. Let R be the relation on the positive integers N defined by the equation + that is, + R Write R as a set of ordered pairs. Find : domain of R, range of R, and R1 Find the composition relation RoR. Consider a relation whose directed graph is shown in the following (Fig. Determine its inverse Rl and its complement R , Also draw the directed graphs of Rl and R , 1,
(b)
m
24.
x
y
x
� {(x, y) : x
(a)
(b)
x
3y � 12}
(ii)
(i)
3y = 12
;
(iii)
(c)
25.
26,
2 .15).
� (1, 2), (2, 3), (3, 3)
Fig,
� [1, 2, 3] ,
Let and A R. of R,Rusing composition of relation
2,15
Find the reflexive, symmetric and transitive closure
Answers
Reflexive and transitive, 7, Yes, No
5.
6,
8, 9,
10, 11. 12,
(a) (2, 2) (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4), (5, 5), (a)
(b) (iii), (b)
{{I}, {2}, {3}}, {{I}, {2, 3}}, {{2}, {I, 3}}, {{3}, {I, 2}}, {{I, 2, 3}} , (i) {(4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (4, 5), (5, 6), (4, 6)} (ii,) {(4, 4), (5, 5), (6, 6)} (ii) {(4, 4), (5, 5), (6, 6), (7, 7)}
(iv) {(4, 4), (4, 3), (3, 4), (4, 2)}
14, 15, 16,
(v) {(4, 4), (4, 5), (5, 4), (4, 6)}
R is reflexive, symmetric, transitive. R is not irreflexiv8, asymmetric, antisymmetric. R R R R R R R � Reflexive closure of R Rs Symmetric closure of R Not true, take R S True True True True, Rr The transitive closure of R RoS SoR RoS SoK But in general, RoS SoR � {(I, 1), (2, 2), (3, 3)}
(i)
� {(I, 1), (2, 2), (2, 3), (3, 2)} � {(I, 1), (2, 3), (3, 3)}
(iv) (vi)
�
(a)
(b) (a) (i)
(ii)
(iii)
(h)
� {(I, 2), (2, 1), (2, 3)}
� {(I, 1), (2, 2), (3, 3), (1, 2), (2, 3), (2, 1), (3, 2)}
(iii)
13,
(c)
and are partitions,
� {(I, 2), (2, 1), (1, 1), (3, 2)}
� {(I, 1), (2, 2), (3, 3), (2, 3)}
� {(2, 1), (2, 3), (3, 2), (3, 3), (2, 2), (4, 2), (1, 1), (4, 4)}
�
�
(v)
(vi,)
� {(I, 2)},
� {(I, 3), (1, 5), (3, 5)}
� {( 1 , 3), (1, 5), (3, 5)} =
� {(2, 1), (2, 3), (3, 2), (3, 3), (2, 2), (4, 2) (1, 2), (2, 4)} � {(2, 3)} (c) (e) (b) (d) � {(a, a) , (b, c) , (a, b), (b, b), (c, c), (c, b), (b, a)} t:
RELATIONS 17. 18. 21. 24.
77
None R and T R R RoS (a) (c) (a) (c)
S is symmetric (d) R and T. R (b)
� {(I, 1), (2, 2)} � {(I, 2)}
� fCc, x), (d, y), (d, z)} .
(a) (9, 1), (6, 2), (3, 3) (b) (1) {9, 6, 3} (c) {3, 3} .
25. (1)
� {(I, 2), (2, 1), (2, 3)}
(b)
R1
(iii) {(I, 9), (2, 6), (3, 3)}
(ii) (1, 2, 3)
� reb, a), (c, a), (b, b), (c, b), (d, c), (d,
a}+b{
d),
(a, d), (b, d)]
}+{ c
(ii)
26.
19.
RR Rs Rr
R
Directed graph of Rl d),
� [(a, a), (a, d), (b,
(b, a), (c, a), (c, b), (c, c), (d, c)]
� [(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (3, 1)] � [(1, 2), (2, 1), (2, 3), (3, 2), (3, 3)] � [(1, 2), (2, 3), (3, 3), (1, 3)]
Hints
(�)
We must show that is reflexive, symmetric, and transitive. since Hence is reflective. We have Suppose Then Accordingly, and hence d) Thus, is symmetric. Suppose d) andgives (ad) d) Then . Multiplying and and corresponding terms of the equations (de), Canceling from both sides of the equation yields and hence f). Thus is transitive. Accord ingly, is an equivalence relation. Let E A. Since R and S are reflexive, E R and E S E R nS Hence R S is reflexive. Let E R n S E R and E S But R and S are symmetric, therefore, E R and E S E RnS Hence R S is symmetric. Let E A. Since R is lreflexive, : ER and E R S for any relation S on A, Hence Rl and R S is E R reflexive then R and S are transitive. But Take R S R S is not transitive Since lE R S, E R S l R S Let E R R If E R ER Hence E R R1 , :. R R1 is symmetric, �
�
�
(c) Transitivity :
at = be,
a
..
(b) (c)
(d)
(e)
ad
(a, b)  (c,
�
(a)
ab = ba. � bc.
(a, b) (a, b) (a, b)  (c, d).
(a) Reflexivity : (b) Symmetry :
20.
Directed graph of R
II
(a, b)
a
=c}
(a, b)
(a, a)
cb � da
 (e, f). ad � be (ef) = (be) (a, b) (e, �
(a, b)
.
(a, b) (b, a)
u
u
=c}
u
cf � de C :f 0
� (a, b).
d :f 0
�
=c}
(b, a)
u
u
� (1, 2), u
� (2, 3), � {(I, 2), (2, 3)} u (2, 3) (1, 2) U (a, b)
(b, a)
(a, a)
(a, a)
(c,
(a, a)
(b, a)
II
=::}
=c}
(a, a)
(a, a)
(c,
�
=c}
(b, a)
(1, 3)
�
u
78
DISCRETE STRUCTURES
2.18. EQU IVALENCE RELATIONS AND PARTITIONS
Let S be non·empty set. By partition P of S, we mean a finite collection {Ai non·empty subsets of S such that, (i) Ai n AJ =
. If {x" x,, ...... xn} is a path in R, then any vertex other than x, and xn is called interior vertex of the path. Also for 1 < k < n, we define a Boolean matrix Wk as follows. The (i,j)th element ofWk is 1 iff there is a path from ai to a in R whose interior vertices, if any, come from the set {a" a2 , ...... an}' In other words, W n = MR' If we define Wo = Mw then we can find a sequence Wo W l ' W" ...... Wn whose first term ' is Wo = MR and the last term is Wn = MR' Each matrix Wk can be computed from Wk_1 using the following algorithm (Warshall's
algorithm) . Step I. First transfer ali I's in Wk_1 to Wk' Step II. List the locations P " P" ... in column k of Wk _ l ' where the entry is 1 and locations al ) a2 ! in row k of Wk_1 J where the entry is L Step III. Put l's at all the positions (Pi' q) ofWk (if they are not already there). ••••••
Example 1. Using Warshall's algorithm, find the transitive closure ofR defined on A = {I, 2, 3, 4} and R = (1, 1), (1, 4), (2, 1), (2, 2), (3, 3), (4, 4). Sol. If MR denotes the matrix representation ofR, then (Take Wo = MR)
[
1 1 Wo = MR = 0 o
0 1 0 0
0 0 1 0
]
1 0 0 and n = 4 1
(As MR is a 4 x 4 matrix)
We compute W4 by using Warshall's algorithm. For k = 1. In column 1 of Wo,l's are at positions 1 and 2. Hence P, = 1, P, = 2 In row 1 of Wo l's are at positions 1 and 4. ' Hence q, = 1, q, = 4. Therefore, to obtain W, we put 1 at the positions : ' {(P " q,), (P " q,), (P" q,), (P" q2 ) = (1, 1), (1, 4), (2, 1), (2, 4)}. Thus
[
1 1 W, = 0 0
0 1 0 0
0 0 1 0
1 1 0 1
]
For k = 2. In column 2 of W" l's is at positions 2. Hence P , = 2.
In row 2 of W" l's are at positions 1, 2 and 4. Hence q, = 1, q, = 2, q3 = 4. Therefore, to obtain W2 , we put Is at the positions: {(P " q,), (P " q,), (P " q, ) = (2, 1), (2, 2), (2, 4)}. Thus (using W,)
82
[
1 1 W, = 0 0
0 1 0 0
0 0 1 0
1 1 0 1
]
DISCRETE STRUCTURES
For k = 3. In column 3 ofW" ' 1' is at position 3. Hence P, = 3 In row 3 ofW" ' 1' is at position 3. Hence q, = 3 Thus, we put ' 1' at the position: {(P " q,) = (3, 3)}. Thus (using W,)
[
1 1 W3 = 0 0
0 1 0 0
0 0 1 0
1 1 0 1
]
For k = 4. In column 4 ofW3 , l's are at positions 1, 2 and 4. Hence P , = 1, P, = 2, P3 = 4. In row 4 of W3 ' 1' is at position 4. Hence q, = 4 ' Therefore, we put l's at the positions: {(P" q,), (p" q,), (P3 q,) = (1, 4), (2, 4), (4, 4)}. Thus (using W3 ). ' 1 0 0 1 1 1 0 1 W4 = 0 0 1 0 = MR' o 0 0 1
]
[
Hence, from the matrix MR', the transitive closure of R is given by W = {(I, 1), (1, 4), (2, 1), (2, 2), (2, 4), (3, 3), (4, 4)} Example 2. Let A = {I, 2, 3, 4} and letR = (O, 2), (2, 3), (3, 4), (2, I)}. Find the transitive
[� � � �
closure ofR using Warshall's algorithm.
Sol. Let MR denotes the matrix representation of R. Take Wo = Mw we have 0 1 0 0'
Wo = MR =
and n = 4 (As MR is a 4 x 4 matrix)
o 0 0 0
We compute W4 by using warshalYs algorithm. For k = 1. In column 1 of Wo ' ' 1' is at position 2. Hence P, = 2. In row 1 ofWo ' 1' is at position 2. Hence q, = 2. Therefore, to obtain W l ' we put ' l' at the ' position: {(P" q,) = (2, 2)}. Thus
[
0 1 W, = 0 o
1 1 0 0
0 1 0 0
0 0 1 0
]
For k = 2. In column 2 of W" l's are at positions 1, 2. Hence P, = 1, P, = 2.
RELATIONS
83
In row 2 of W" l's are at positions 1, 2 and 3. Hence q , = 1, q, = 2, q3 = 3 Therefore, to obtain W" we put ' 1' at the positions :
{(P " q,), (P " q,), (P " q3) (p" q,), (P2' q,), (p" q3) = (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}. '
Thus (using W,)
1 1 o 0
1 1 0 0
0 0 1 0
]
For k = 3. In column 3 of W" l's are at positions 1, 2. Hence P , = 1, P, = 2
In row 3 ofW" ' 1' is at the position 4. Hence q , = 4 Therefore, to obtain W3 we put l's at the positions: {(P" q,), (P" q,) = (1, 4), (2, 4)}. Thus (using W,)
[
1 1 W3 = 0 o
Thus
1 1 0 0
1 1 0 0
1 1 1 0
]
For k = 4. In column 4 of W3 , l's are at positions 1, 2, 3. Hence P , = 1, P2 = 2, P3 = 3 In row 4 ofW3 , ' 1' is at no position, and no new l's are added and hence MR' = W4 = W3 .
[
1 1 W4 = W3 = 0 o
1 1 0 0
1 1 0 0
l' 1 1 = MR 0 �
Thus, the transitive closure of R is given as R� = {(I, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 4)} Example 3. Let A = (1, 2, 3, 4) and let R and S be the relation on A described by
[
0 0 = 0 MR 0
0 0 1 0
0 0 0 1
] [
1 0 0 ' Ms = 0
1 0 0 0
1 1 0 1
0 0 1 0
]
0 0 0 1
Use Warshall's algorithm to compute the transitive closure of R u S.
Sol. Let MR
and
u
s
(P.T.U. B. Tech. December 2008) denotes the matrix representation or R u S, then
We compute W4 by using WarshalYs algorithm. Here n = 4 (As MR u s is a 4 x 4 matrix)
[ ]
84
DISCRETE STRUCTURES
1 o 0 o
1 0 1 1 0 0 WO = MR u S = 1 1 0 1 1 1 For k = 1. In column 1 of Wo ' '1' is at position 1. Hence P, = 1. In row 1 ofWo l's are at positions 1, 2, and 4. ' Hence q, = 1, q, = 2, q3 = 4.
[
1 o 0 o
[
1 o 0 o
Therefore, to obtain W" we put l's at the positions: {(P " q,), (P " q,), (P " q,) = (1, 1), (1, 2), (1, 4)}. Thus (using Wo) W, =
1 1 1 1
0 l' 0 0 = Wo 1 0 1 1
For k = 2. In column 2 of W" l's are at positions 1, 2, 3 and 4. Hence P , = = 3, P 4 = 4 In row 2 ofW" '1' is at position 2. Hence q, = 2 Therefore, to obtain W, , we put l's at the positions: { (P " q,), (p" q,), (P3 ' q,), (P4 ' q,) = (1, 2), (2, 2), (3, 2), (4, 2)} Thus (using W,) W, =
1 1 1 1
1, P2 = 2, P3
0 l' 0 0 = W, 1 0 1 1
For k = 3. In column 3 of W" l's are at positions 3, 4. Hence P , = 3, P, = 4
In row 3 of W" l's are at positions 2 and 3. Hence q, = 2, q, = 3 Therefore, to obtain W3 we put l's at the positions : ' { (P " q,), (P" q2) ' (P 2' q,), (P 2' q2) = (3, 2), (3, 3), (4, 2), (4, 3)}. Thus (using W2) W3 =
[
1 o 0 o
1 1 1 1
0 l' 0 0 1 0 = W2 1 1
For k = 4. In column 4 of W3 , l's are at positions 1 and 4. Hence P, = 1, P, = 4 in row 4 of W3 , l's are at to obtain W4 , positions 2, 3 and 4. Hence q, = 2, q, = 3, q3 = 4. Therefore, to obtain W4 we put l's at the positions: {(P " q,), (P" q,), (P" q,), (P " q,), (P 2 ' q,), (P" q3) = (1, 2), (1, 3), (1, 4), (4, 2), (4, 3), (4, 4)} Thus (using W,) W4 =
[ ] 1 o 0 o
1 1 1 1
1 o 1 1
1 0 = 0 M (R U S)� 1
RELATIONS
85
Thus, the transitive closure of R u S is given as (R u S) � = { ( I , 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 2), (3, 3), (4, 2), (4, 3), (4, 4)} Example 4. Let A = {I, 2, 3, 4, 5} and R=n � � � � � � � � � � � � � � � � � S = (O, 1), (2, 2), (3, 3), (4, 4), (4, 5), (5, 4), (5, 5)} Find the transitive closure of R u S by using Warshall's algorithm. Sol. Let MR and Ms denote the matrix representation of R and S respectively. Then
MR =
=
=
and
[ � �l ' [� � � l ]j � � � � � �] !] ] 1 1 0 o 0
1 1 0 0 0
0 0 1 1 1 0 0 o 1
1 1 0 o o
1 1 0 0 0
0 0 1 1 0
0 0 1 1 0
1 1 0 o o
1 1 0 0 0
0 0 1 1 0
0 0 1 1 1
Ms =
o 0 0 0 0 1 1 0 0 0 �1 �1
o 0 0 1 o 0 0 1
Iv 1 = 1 O v l = 1 etc. 1v0 = 1 OvO=O
We now compute W5 by WarshalYs algorithm. Here n = 5. (As MR u S is a 5 x 5) matrix
WO = MR u S =
1 1 0 o o
1 1 0 0 0
0 0 1 1 0
0 0 1 1 1
0 0 0 1 1
For k = 1. In column 1 of Wo' l's are at positions
1, 2. Hence P , = 1, P , = 2. In row 1 ofWo l's are at positions 1, 2. Hence q, = 1, q, = 2 ' Therefore, to obtain W" we put l's at the positions: { (P " q,), (P" q,), (p" q,), (p" q,) = (1, 1), (1, 2), (2, Thus (using W0>
W, =
[� � � � �l o 0 1 1 1 o 0 0 1 1
= Wo
1), (2, 2)}
DISCRETE STRUCTURES
86
For k = 2. In column 2 of W" l's are at positions 1, 2. Hence P , = 1, P , = 2
In row 2 ofW" l's are at positions 1, 2. Hence q , = 1, q, = 2 Therefore, to obtain W" we put l's at the positions: { (P " q ,), (P" q,), (P" q, ), (P" q,) = (1, 1), (1, 2), (2, 1), (2, 2)}. Thus (using W,)
[� � � � �l
W = = W1 ' 0 0 1 1 1 o 0 0 1 1 For k = 3. In column 3 of W" l's are at the positions 3, 4. Hence P , = 3, P, = 4. In row 3 ofW" l's are at the positions 3 and 4. Hence q , = 3, q, = 4 Therefore, to obtain W3 , we put l's at the positions: { (P " q,), (p" q,), (p" q,), (p" q,) = (3, 3), (3, 4), (4, 3), (4, 4)} Thus (using W,) W3 =
� � � � �]
= W, 0 1 1 1 0 0 1 1 For k = 4. In column 4 of W3 , l's are at positions 3, 4, 5. Hence P , = 3, P, = 4, P3 = 5. In row 4 ofW3 , l's are at the positions 3, 4 and 5. Hence q, = 3, q, = 4, q3 = 5. Therefore, to obtain W" we put l's at the positions: {(P" q,) , (P" q,), (P" q3)' (p" q ,), (p" q,), (p" q,), (P3' q ,), (P3' q,), (P3' q3) =(3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5)}. Thus (using W,)
o o
1 1 0 0 1 1 0 0 W, = 0 0 1 1 o 0 1 1 o 0 1 1
]
0 0 1 1 1
:
For k = 5. In column 5 ofW" l's are at the positions 3, 4, 5. Hence P , = 3, P, = 4, P3 = 5. In row 5 of W" l's are at the positions 3, 4, 5. Hence q , = 3, q, = 4, q3 = 5. This is similar to the case for k = 4. Hence
[
1 1 0 0 1 1 0 0 W5 = W, = 0 0 1 1 o 0 1 1 o 0 1 1
0 0 1 = M (RuS) 1 1
�
Thus, the transitive closure of R u S is given as (R u S) � = {(I, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5),
(5, 3), (5, 4), (5, 5)}
RELATIONS
87 TEST YOUR KNOWLEDGE 2.3
1. 2.
Let A � and let R � transitive closure R by using Warshall's algorithm. Let A and R be a relation on A. If
� � � � � � Wo
, 1 0 0 0 0 o 1 0 0 1
�
[� I
W2 ! = (I, 2, 3, 4).
0 0 1 0 0 1 0 0
�j [ ] 1
,�
1 0 0 1 0 1 1 0 (iii) o 1 1 0 1 0 0 1
Let A
= {a, b,
c,
MR
�
, (n) MR
1 0 1 0 0 0 0 0 1
,�
2.
MW
[
S
Ms
0 0 1 1 0 1 0 1 0 0
J
1 1 1 1 1 1 , 1 1 1
Wl �W2 �W3 �
j ()1 � �
R� � 3.
�
�
� � � � �:
A
0 1 0 0 0 0 1 0 1 0
where R� �
S.
Answers
{(I, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
[� � � � �l 1 0 0 1 0 o 1 0 0 1
u
[I j (il) i
{(I I), (1, 4), (2, 2), (3, 4), (3, 5), (4, 1), (4, 4), (5, 2), (5, 5)}
o 1 o 1 1 0 0 0 o 1
�j
II and let R and be the relations on described by 0 0 0 (iv) 1 0 0 o 1 0 o 0 1
[� � � � �l' d, e]
[
[� I
Use Warshall's algorithm to compute transitive closure of R 1.
M R
of the
Compute Wl ' W3 as in Warshall's algorithm. Also find Roo, the transitive closure of R. Let A For thes algorithm. relation R whose matrix is given, find the matrix of the transitive closure by using Warshall' , (I) MR
4.
Find the matrix
= {al l a2 ! as) a4 ! a5] MR
3.
{(I, 1), (1 2), (2, 3), (1, 3), (3, 1), (3, 2)} ,
(1, 2, 3)
0 o 1 1 0 1 0 1 0 o 1
�
[ ]
88
4.
1 0 0 1 (iii) 0 1 1 0 o 1 1 0 1 0 0 1
(iv)
u S)�
[ ] 1 1 1 1
1 1 1 1
1 1 1 1
DISCRETE STRUCTURES
1 1 1 1
� {(a, a), (a, b), (a, ) (a, d), (a, e) , (b, a), (b, b), (b, c) , (b, d), (b, e) , (c, a), (c, b), (c, c) (c, (R (c, e) , (d, a), (d, b), (d, c), (d, d), (d, e), (e, a), (e, b), (e, c), (e, d), (e, e)}
c,
d),
M U LTIPLE CHOICE QU ESTIONS ( MCQs)
1. The binary relation S
( y, x) E R. Hence R is symmetric. Transitive. If (x, y) E R, then x + y is even, say x + y = 2k If (y, z) E R, then y + Z is even, say y + Z = 2m where k, m are integers. Now, x + Z = 2k  y + 2m  y = 2k + 2m  2y = 2(k + m  y) = 2l, say l = k + m  y which is even. Hence (x, z) E R . . R is Tansitive also. Hence R is an equivalence relation. To find the number of equivalence classes, we have (x, y) E R => x + y is even. Also, y + x is even => (y, x) E R . . The number of equivalence classes is 2. (d) The number of equivalence relations = 2 4  1 = 16  1 = 15. (b) Let A and B are two sets such that A n B = x'  y' + x  y = O => => x  y = 0, => { : N ; N is one·one. Onto. Let Y E N such that {(x) = y => x' + x = y x' + x  y = O =>
x=
 1 ± F+4Y 2
Now, for Y E N, we get some values of x which are not in N.
(l Take Y = 1, we get x =  1 + J5 ;
{ : N ; N is not on to. .. (e) Oneone. Let m, n E N such that {(m) = {(n) => (m, I 10  m I ) = (n, I 10  n I ) m = n, 1 1O  m l = I lO  n l m = n, lO  m = lO  n m = n, for all m, n E N . . { : N ; N x N is one· one. Onto. Given { : N ; N x N is a function from N to N x N. Therefore co· domain {= N x N {= {{(n) : n E N} = (n, I 10  n I ) Also range
1 06
DISCRETE STRUCTURES N x N '" codomain
Le.,
f : N ; N x N is not on to function. Example 12. Let A = B = {I, 2, 3, 4, 5}. Define functions f A ; B (ifpossible) such that (P.T.U. B.Tech. Dec. 2005) (a) fis oneone and onto (b) f is neither oneone nor onto (c) f is oneone but not onto (d) f is onto but not oneone. Sol. (a) Define f : A ; B such that f(x) = x for all x E A = B f(l) = 1 , f(2) = 2, f(3) = 3 f(4) = 4, f(5) = 5
i.e., different elements of A have different images in B Fig. 3.30. .. f : A ; B is oneone. Also, we observe that for each Y E B, there exists x E A such that f(x) = y. Hence f
is on to also. A B (b) Define f : A ; B such that f(x) = 1 for all x E A, 1 E B oneone and onto Le., f(l) = 1, f(2) = 1, f(3) = 1 , f(4) = 1 , f(5) = 1 i.e., different elements of A have same images in B Fig. 3.3 1. 2 f : A ; B is not oneone. 3 Also 2, 3, 4, 5 E B have no images in A. 4 .. f : A ; B is not on to 5 (c) Since A = B, :. There is no function which is one one but not onto. A B (d) Again, as A = B, :. There is no function which is onto, but not oneone. Example 13. Consider the function f ,' N x N ; R defined by f(x, y) = (2x + 1) 2Y  1,
Fig. 3.30
Fig. 3.31
where N is the set of natural numbers including zero. Show that f is bijective.
Sol. Oneone. Let f (x" Y,) = f (x" Y,) (2x, + 1)2Y'
Le.,
=>
(2x, + 1)2Y'  1
= (2x2 + 1)2 "' , which can hold if
2X, + 1 = 2x, + 1 Xl = x2 (x" Y,) = (x" y,).
and 2Y' = 2Y' and
= (2x2 + 1)2Y'  1
if Hence, :. f is oneone. Onto. For each (x, y) E N x N, (2x + 1) 2Y  1 E R f(x, y) = (2x + 1) 2Y  1. Hence range of f = R = codomain f such that :. fis onto. Hence, f is bijective.
FUNCTIONS
1 07
Example 14. Consider the function f ,' N � N, where N is the set of natural numbers including zero defined by f(n) = n2 + 2. Check whether the function f is (i) oneone (ii) onto. Sol. Oneone. Let m, n E N such that f (m) = f (n) => I m t:  n since m, n E N :. f is oneone. Onto. Given f(n) = n' + 2, n E N m = f(n) = n' + 2
Let
For m
on R
E
=>
N, n 'l N (take m = 1, then n = ±
n' = m  2
R = ± i)
=>
n = ± �m  2
:. f is not on to. Example 15. Which of the following functions are injections, surjections, or bijections (a) f(x) =  2x (b) g(x) = x2  1. Sol. (a) We know that a function f: R � R is oneone iff f(x) = fly) => x = y for all x, y E R Let f(x) = fly)  2x =  2y x = y for all x, y E R. .. f : R � R is oneone (injection) Also, let y E R such that f(x) = y
x= y (1) 2 Now, y E R, we observe that the values of x given by (1) are in R. . . f : R � R is onto (surjection). Since f : R � R is oneone and onto. :. f is bijection also. (b) Let g(x) = gly) for all x, y E R x2  1 = y2  1 => X' = y2 x=±y g = R � R is not oneone. Also, codomain g = R and Range Rg = {g(x) : x E R} = {x'  1 : x E R} = (0, ) '" R = codomain .. g : R � R is not onto. Alternatively: Let y = g (x) = x2  1  2x = y
=>
=
=>
For Hence g is not onto
x' = l + y y E R,
3
=>
x = ± .,f1+Y
x 'l R. Take
y = 2, then x = ± � = ± i 'l R
Example 16. Consider the function f ,' N � N where N is the set of natural numbers. Defined by f(n) = n2 + n + 1. Show that fis oneone but not onto. (P.T.V. B.Tech. Dec. 2008) Sol. Given function f : N � N, where N is the set of natural numbers., is defined by f(n) = n2 + n + 1.
1 08
DISCRETE STRUCTURES We know that a function f : X ; Y is oneone if f(x,) = f(x,) => x, = x, \;j x" x, E X Also f : x ; y is onto if for each Y E y, there exists at least one x E X such that f(x) = y. Oneone_ Let n" n, E N such that f(n,) = f(n,) ::::::} n1 2 + n1 + 1 = n2 2 + n2 + 1 ::::::} n1 2  n2 2 + n1 n2 = 0 => (n,  n,) (n, + n, 1) = ° ::::::} either n1 = n2 or n1 + n2  1 = 0 If n, + n, 1 = 0, then n, = 1 n, 'l N for n, E N . . n, = n,. Hence f : N ; N is oneone. Onto_ To check whether f is onto, let m E N such that m = f(n) \;j n E N m = n' + n + 1 => n' + n + 1  m = ° 


or

n=
 l ± Jl + 4(m  l)
Now for each m E N, there exist
2 n 'l N.
=
 l' ±J 4m  3 ': 

2
_
Hence f is not onto.
Example 17_ Give an explicit formula for a function from the set of integers to the set of
(P.T.V. B. Tech. May 2009) positive integers which is (a) oneone but not onto (b) onto but not oneone (c) oneone and onto ( 2X = 2Y => x = y \;j x, Y E Z :. f is oneone. If f(x) = y, then 2X = y => x = log, y Now for y E Z+, X = log, y is not an integer. Hence the function f(x) : Z ; Z+ defined by f(x) = 2x is not an onto function. Case II. Not oneone, but onto Define f : Z ; Z+ by f(x) = I x I Let f(x) = fly) => I x I = I y I => x = ± Y :. f is not oneone. Also f : Z ; Z+ is defined by f(x) = I x I Codomain of f = Z+ = set of positive integers range of f = If(x) : x E Z} = { I x I : x E Z} = Z+
FUNCTIONS
1 09
Since codomain of f = range of f . . fis onto.
Case III. Oneone and onto
Define f= Z ; Z+ by f(x) = x + 5 Let f(x) = fly) => x + 5 = y + 5 => x = y \;j x, Y E Z. . . f is oneone. f(x) = y => y = x + 5 => x = y 5 Let Now for each y E Z+ , X is an integer. Hence f is onto also. 
Case IV. Not oneone, not onto
Define f : Z ; Z+ by f(x) = x' ' Let f(x) = fly) => x' = y => x = ± Y \;j x, Y E Z f is not oneone
Also, let f(x) = y => y = x' => x = ± ..JY Now for each y E Z+, X is not an integer. Hence f is not onto.
3.13. GRAPHICAL REPRESENTATION OF ONEONE AND ONTO FUNCTIONS
When we say that a function f : A ; E is oneone, this means that there are no two distinct pairs (a" b) and (a" b) in the graph of f. Hence, each horizontal line (either below or above the xaxis) can intersect the graph offin atmost one point. When we say that a function f : A ; E is onto, this means that for every b E E, there must be at least one a E A such that (a, b) belongs to the graph of f. Hence, each horizontal line (either below or above the xaxis) must intersect the graph of fin atleast one point Further, if each horizontal line either below or above xaxis intersects the graph of fin exactly one point, we say f is both oneone and onto
Example 18. Consider the following graphs (Fig. 3. 32). Determine which of these are functions from R into R.
x
1 1
1
(b)
(a) 2
o
2 1
(e)
Fig. 3.32
110
DISCRETE STRUCTURES
Sol. (a) Recall that a set of points on a coordinate diagram is a function iff every vertical line contains exactly one point of the set. Hence the graph given in Fig. 3.32(a), (b) represents functions. But, the graph given in Fig. 3.32(c) does not represent a function. Since the vertical line drawn through the graph (Fig. 3.32) meets in two points.
Example 19. Determine which of the graphs in the followings figures (Fig. 3. 33) are functions from R to R. 2
o
2
2
2 (b)
(a) Fig. 3.33
Sol. If a vertical line is drawn, it does not contain exactly one point. Hence, both the graphs (a) and (b) do not represent any function. Example 20. Consider the functions defined as f(x) = 2', g(x) = x'I x, h(x) = x2, d(x) = x'I 
and their graphs as shown below (Fig. 3.34). Determine which of the functions are oneone. Which of them are onto ?
o
g(x) x3  x =
f(x) 2' =
(a)
(b)
o
o
(e)
d(x) x3 (d)
h(x) x'
=
=
Fig. 3.34
FUNCTIONS
111
Sol. Recall that a set of points on a coordinate diagram is a function iff every vertical line contains exactly one point of the set. Hence, the graphs shown in Fig. 3.34(a) , (b), (c) and ( fly) = k,t(x), k, E N Let Iy, x) E R => f(x) = klly), k2 E N Using (2) in (1), we have fly) = k,k2 fly)
(1) (2)
(1  k,k2) fly) = 0 1  �� = 0 I If Y E Z, then fly) E N since f : Z ; N is a function. Hence fly) # 0
=>
=>
k,k2 = 1
But k" k2 E N and (3) implies k, = 1 = k2 from (2) f(x) = klly) = fly) .. =>
x=y
Hence the relation R is antisymmetric.
To show R is transitive_
I f is oneone
112
DISCRETE STRUCTURES Let (x, y) E R (y, z) E R
=>
f(y) = k/(x) for x, Y E Z, k, E N f(z) = k,t(y) for x, y, E Z, k2 E N f(z) = k2 (k/(x» = k2k, f(x) = k2k, f(x) = k f(x) where k = k2k, E N (x, z) E R Hence R is transitive also. Therefore, the relation R is a partial order relation.
or
=>
I
TEST YOUR KNOWLEDGE 3.1
SHORT ANSWER TYPE QUESTIONS
1.
State whether or not each diagram in the given figure) defines a function from y, z}. A c} into B = {a, b,
= {x,
b e ���� (a)
2.
Consider the figure given below :
(b)
(e)
(a) Find the graph of the function and write f as a set of ordered pairs (b) Find f(S) where S = {I, 3, 5} (c) Find f 1 (T) where T = {I, 2} (d) Find f 1 (3).
FUNCTIONS 3.
1 13 =
Let 2, 3, 4}. Determine whether each of the following relations on X (set of ordered pairs)X is {I,a function from X into X. {� {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)} g � {(3, 1), (4, 2), (1, I)} � {(2, 1), (3, 4), (2, 1), (4, 4)} 3 1 then show that { (1) (b) If {(x) � 3x   +  � {(x)
(a)
(i) (ii) (iii) h
x'
x
2'
x
x
x, x < 0 x, O :S; x :s; x>l x
1
1
4,
What is the range of the function {(x) where
5.
Let f : R R be a function defined by [(x) Is toneone, onto or both ? Let A {x :  1 x I} B and f : A B is a function defined by [(x) 2 Examine whether tis bijective or not. 1 � ) is a function on R. Find the domain of f Let I(x) '\jx4 If A � {I, 2, 3}, B � { b, c} . Find the number of injections from A to B. (b) Find the number of bijective functions from A to itself when A contains 106 elements. (c) If X � {a, b} and Y � {I, 2, 3} . Find the number n of functions from Xinto Y (b) From Y into X If X has I X elements and Y has I Y I elements. Show that there are I Y I I X I functions from X into I (Hint. Use theorem I)
6. 7. 8,
=
t
Z defined by {(x) � + x for all x Z. Examine which are oneone, many one Show that the functions given by { : R > R defined by {(x) � 3x' + 5 (b) { : Q > Q defined by {(x) � 2x  3 are bijections. Which of the following functions from Z to Z are bijections ? {(x) � x' (b) {(x) � x + 2 (c {(x) � 2x + 1 (d) {(x) � x2 + x. Let f : N N is a function defined by fen) 2n + 3. Is f oneone, onto or both. (a)
10.
11.
12.
x'
E
E
(a)
(a) )
=
t
Answers 1.
(a)
No, the element b is not associated to any element of B (b) No, two elements x and are associated to c c Yes ()
z
114
DISCRETE STRUCTURES
Graph f � {(I, 3), (2, 5), (3, 5), (4, 2), (5, 3)} {3, 5} {I, 5} {4} (0 No, the element 2 is repeated twice as its first coordinate. 3. (ii) No, the element 2 does not appear as the first coordinate in any ordered pair in g. (iii) Yes, X appears as the first coordinate in the two ordered pairs, but these or deredAlthough pairs ) but2 Ethese ordered pairs are same, Neither oneone, nor onto 4. 6. Bijective i.e., oneone and onto 7. ( 2) u (2, 8. 9. Manyone Manyone. Oneone, not onto.
2.
(a)
(b)
(e)
(d)
(a)
(0, ) =
11.
(a) 6 (b)
5.
(a)
=
(b) 106 I
,

)
=
(b)
12.
3.14. COMPOSITION OF FUNCTIONS
by
Def. Let f : A ; B and g : B ; C be two functions. Then the function gof : A ; C, defined
(go!) (x) � g(f(x» for all x E A, is called composition of f and g. Thus, to find the image of 'a" under gof, we first find the image of 'a" under fand then we find the image of f(a) under g. Further, we say that gofis defined only if range R, is a subset of domain Dg' Also, gof is a functlOn from A to C. Remark f f g. g fog f f g. go! g For example, let f : {I, 2, 3} ; {a, b} is a function defined by f(l) � a, f(2) � a, f(3) � b and g : {a, b} ; {5, 6, 7} is defined by g (a) � 5, g(b) � 7 We find gof and fog. Here Range R, � {a, b}, Domain Dg � {a, b} Since Range R, c Domain D :. gofis defined and gof : {I, 2, 3} ; {5, 6, 7} g Now, (go/) (1) � g(f(l» � g(a) � 5 (go/) (2) � g(f(2» � g(a) � 5 (go/) (3) � g (f( 3» � g(b) � 7 To find fog, we observe that range Rg � {5, 6, 7} and domain D , � {I, 2, 3}. Since range Rg iZ. domam Dr .. fog is not defined. Theorem II. Let f: A ; B, g : B ; C and h : C ; D, be three functions, then fo(goh) � (fog)oh Proof. Given. f : A ; B, g : B ; C, h : C ; D => goh : B ; D, fog : A ; C => fo (goh) : A ; D, (fog)oh : A ; D Thus fo(goh) and (fog)oh are functions from A to D. We now show fo(goh) � (fog)oh Let x E A and consider (1) (fo(goh» x � f(goh)(x) � f(g(h)(x» og)(h(x» � f((g(h(x» ) (2) og)oh)(x) � Also, (f (f From (1) and (2), (fo(goh» (x) � (fog)oh) x for x E A => fo(goh) � (fog)oh.
If we view and as relations, we use the notation for the composition of and If we use and as functions, then, we use the notation for the composition of and
FUNCTIONS
115
Theorem III. Let f: A � B, g : B � C are oneone (injections), Thengof: A � C is also (P.T.V. B.Tech. Dec. 2013) oneone, not conversly. Proof. Let x, Y E A and consider (gof) (x) = (gof) (y) => g ([(x» = g([(y» f(x) = f(y) I g is oneone x =y I f is oneone .. gof is oneone. Theorem IV. Iff: A ; B, g : B ; C are onto (surjections) functions, then gof: A ; C is also onto (surjection) function, not conversly. Proof. To show gofis onto, we show that for every element in C has its preimage in A. i.e., for all Z E C there exists x E A such that (gof)(x) = z. Now Z E C and g : B ; C is onto function from B to C :. We can find y E B such that g(y) = z. Also y E B and f : A ; B is onto function :. we can find x E A such that f(x) = y (gof)(x) = g([(x» = g(y) = Z
gof is onto function from A to C.
Theorem V. Iff : A ; B, g : B ; C are oneone and onto functions or surjections, then gof: A � C is also oneone and onto (surjection). Proof. Combining theorem, II and theorem III, we get the required result. Theorem VI. Let f: A �B and IB : B �B be two functions, then IBof = f Proof. Given IB : B ; B is an identity function and f : A ; B is a function from A to B. Let x E A. Consider
(IB0f) (x) = IB ([(x» = IB(y)
I y = f(x) E B
= Y = f(x) (IB0f) (x) = f(x) IBof = f
Theorem VII. Let IA : A ; A and f: A ; B be two functions, then foIA = f Proof. Given IA : A ; A and f : A ; B :. fa IA is a function from A to B let x E A and
consider
([alp) (x) = fo(IA(X» = f(x) folA = f
3 . 1 5. INVERSE FUNCTION
Let f : A ; A and if there is a function g : A ; A such that gof = fog = lA' then g is called inverse of f and is denoted by rl i.e., g = rl If f : A ; B is a oneone and onto function, then the function defined by g : B ; A is called inverse of f Example. Consider the functions f: R ; R and g : R ; R defined by f (x) = X4 and g (x) l4 1 = X . Then for each x E R, f (x) and g (x) are inverses of one another.
116
DISCRETE STRUCTURES Sol. Here f and g are inverses of one another if Now Also Hence
(fog) (x) = Ix and (go/) (x) = Ix (fog) (x) = f (g (x» = f (X'I4) = (X'i') ' = x = Ix (go/) (x) = g (f (x» = g (x') = (X') lI' = X = Ix f = gl and gl = f
3.1 6. METHOD TO FIND THE I NVERSE OF A BIJECTION
Let f : A ; B be a bijection (one· one and onto). To find f" Step I. Take f(x) = y, y E B, x E A. Step II. Solve f(x) = y to obtain x in terms of y Step III. Replace x by r' (y) in step II, we get the required function. For example, let f : R ; R be a function defined by f(x) = x2 . Is r' exist ? If so, find r' (4), fl (O) , fl ( 1). Here r' will exist if f is one·one and onto. Oneone. Let x, y E R such that f(x) = f(y) => x' = y2 => X = ± y. If x = y, then f is one· one. Onto. Here Rf = [0, =] and co·domain f = R # Rf :. f is not onto. Hence f l (4), r' (O), r' ( 1) donot exist
onto.
Theorem VIII. Let f: A ;A is a function from A to A. Then ri exists ifffis oneone and Proof. Let r' exists, we show fis one·one and onto.
Let (a, b), (c, b) E f Then, by definition of r" (b, a) and (b, c) E r' Since r' is a function :. a = c. Hence f is one·one. Further, let b E A and since r' is a function from A to A, we can find a E A such that fl (b) = a => b = f(a). Hence fis onto. Converse. Let f is one·one and onto. We show fl exists. Consider g : A ; A is a function from A to A. We show fog = IA and gof = IA Let x E A and since f : A ; A, we can find Y E A such that f(x) = y Also, g : A ; A is a function from A to A. :. for Y E A, we can find x E A such that g(y) = x Now, (go/) (x) = g(f(x» = g(y) = x => (go/) (x) = x for all x E A => gof= IA Also, (fog)(y) = f(g(y» = f(x) = y => (fog)(y) = y for all Y E A => fog = IA Hence fl exists.
Theorem IX. A function f: A ; B has an inverse ifff is oneone and onto (bijective). Proof. Let fl exists and we show fis one·one and onto.
Let (a, b), (c, b) E ffor a, c E A, b E B. Since r' exists. :. by definition, (b, a), (b, c) E rl l But f is a function. :. We must have a = c :. f is one· one. Further, let b E B and since r' is a function from B to A, we can find a E A such that r' (b) = a => b = f(a). Hence fis onto.
FUNCTIONS
117
Converse. Let f is one·one and onto. We show r' exists. Consider g : B � A is a function from B to A. We show fog = IB , gof = IA Let x E A and since f : A � B, there exists y E B such that f(x) = y Also, g : B � A. . . we have g(y) = x Now, (go/) (x) = g(f(x» = g(y) = x, => (go/) (x) = x for all x E A =>
g� = � ( 1) Similarly, (fog) (y) = f(g(y» = f(x) = y => (fog) y = y for all y E B => (2) fog = I B Hence from (1) and (2), gof = lA' fog = IB => f1 exists. Theorem X. Iff : A � B and g : B � C are two bijections, then (gof)l = f1og1 Proof. Given f : A � B is a bijection, g : B � C is a bijection. :. gof : A � C is also a
bij ection (one·one and onto) (Theorem IV) => (gO/) l : C � A exists. Again f : A � B is a bijection => t' : B � A is also a bijection. Also, g : B � C is a bijection. => g' : C � B is also a bijection. .. r' og' is a bijection. Hence t' og' also exists. Let x E A, y E B, Z E C such that f(x) = y, g(y) = Z (go/) (x) = g(f(x» = g(y) = Z Consider => (1) (gO/) l (z) = x 1 Also, f (y) = X f(x) = y
g(y) = z .. (f'og' ) (z) From (1) and (2), (gO/) l (Z) = (f' og' ) (z) (gO/)l = r' og1
I
gl (Z) = y t' (g' (Z» = r'(y) = x
(2)
ILLUSTRATIVE EXAMPLES
Example 1. Let f : {I, 2, 3} � {a, b} be a function defined by f(1) = a, f(2) = b, f(3) = b. Let g : {a, b} � {5, 6, 7} be a function defined by g(a) = 5, g(b) = 7. Find got, fog. Sol. To find gof. Given f : {I, 2, 3} � {a, b} and g : {a, b} � {5, 6, 7}.
Rf = {a, b} = Domain Dg :. Here Range :. gof is defined (since Rf is a subset of Dg) i.e., gof : {I, 2, 3} � {5, 6, 7} Here (go/) (1) = g(f(I» = g(a) = 5 (go/) (2) = g(f(2» = g(b) = 7 (go/) (3) = g(f(3» = g(b) = 7 .. gof = {5, 7} To find fog. Given Range Rg = {5, 6, 7} Domain Df = {I, 2, 3}. :. Since Range Rg Sl Domain Df .. fog is not defined. Example 2. Let f: R � R be defined by f(x) = :x:'I and g : R � R be defined by g(x) = 3x + 1. Find (gof)(x), (fog)(x). Is gof = fog ?
DISCRETE STRUCTURES
118 f : R ; R, g : R ; R Sol. Given Here Range Rr = R, Domain Dg = R gof is defined. (since Rr is a subset of Dg) Also, Range Rg = R, Domain Dr = R fog is defined (since Rg is a subset of D r) Now, (go/) (x) = g(f(x» = g(x3) = 3x'l + 1 Also, (fog) (x) = f(g(x» = f(3x + 1) = (3x + 1)3 Since (go/)(x) = 3x'l + 1, (fog)(x) = (3x + 1)3 gof"' fog. Example 3. (a) If f(x) = x2, g(x) = x3 find (fogofog)(x). (b) Iff(x) = x + 1, g(x) = x + 3, find (fofofof)(x). (c) If f(x) = 4x  1, g(x) = x2 + 2, find (fo(fog))(1). (d) If f(x) =


1 x , g(x) = ' find (fog)(x) x 1 x+1
(e) If f(x) = x + 5, g(x) = x2, find (gof)(x), where f: R ; R, g : R ; R is given. (/) If f : R ; R and g: R ; R are two functions defined by f (x) = sinx and g(x) = x2. (P.T.U. B.Tech. May 2013) Find fog and got Is fog = got? Sol. (a) (fogofog)(x) = f(g(f(g(x» » = f(g(f(x'l» = f(g(x6» f(x) = x' ; f(x'l) = (x'l)' = x'l g(x) = x3 ; g(x6) = (X3) 6 = X' 8 (b) (fofofo/) (x) = f(f(f(f(x» » = f(f(f(x + 1» ) = f(f(x + 2» = f(x + 3) = x + 4 (fofog) ( I) = f(f(g(I» = f(f( 3» (c) I g(x) = x2 + 2, g(l) = 3 = f( l 1) = 44  1 = 43 I f(x) = 4x  1, f(3) = 1 1
I
(d)
(e) (/) Thus,
[ ) 1
(fog)(x) = f(g(x» = f __ x I
1 1 x I x I 1 = = = 1 l +xl x +1 x I x I

(go/) (x) = g(f(x» = g(x + 5) = (x + 5)' = x2 + 25 + lOx. (fog) (x) = f (g(x» = f(x') = sin x2 (go/) (x) = g (f (x» = g (sin x) = (sin x)' = sin' x gof ", fog
Example 4. Consider the functions f, g : R ; R defined by
f(x) = x' + 3x + 1, g(x) = 2x  3. Find the composition functions (ii) fog (iii) got (i) fat Sol. (i) (fo/)(x) = tIf(x) 1 = f(x' + 3x + 1) = (x' + 3x + I) ' + 3(x' + 3x + 1) + 1
I
x {(x) = __ x+l
119
FUNCTIONS = x' + 9x' + 1 + 6:0 + 2x' + 6x + 3x' + 9x + 3 + 1 = x' + 6:0 + 14x' + 15x + 5. (gof)(x) = g[f(x)] = g(x' + 3x + 1) = g(x' + 3x + 1)  3 = 2x' + 6x + 2  3 = 2x' + 6x  1. (jog)(x) = fTg(x)] = f(2x  3) = (2x  3)' + 3(2x  3) + 1 = 4x' + 9  12x + 6x  9 + 1 = 4x'  6x + 1.
(ii)
(iii)
that
Example 5. Let t, g, h be functions from N to N, where N is the set of natural numbers so f(n) = n + 1 ; g(n) = 2n and
{O ,
when n �s even h(n) = 1 , when n �s odd.
Determine fat, fog, got, goh, hog, (fog)oh. Sol. (i) fof(n) = f [f(n)] = fen + 1) = n + 2 fog(n) = f [g(n)] = f(2n) = 2n + 1 (ii) gof(n) = g[f(n)] = g(n + 1) = 2(n + 1) = 2n + 2 (iii) (iv) When n is even : goh(n) = g[h(n)] = g(O) = 0
when n is odd :
goh(n) = g[h(n)] = g(l) = 2 hog(n) = h[g(n)] = h(2n) = 1
(v) (vi) When n is even : !jog) oh(n) !jog) [hen)] [og(O) [ [g(0) ] [(0)
when
n
I 2n is even
= =I = = = !jog) oh(n) = !jog) [hen)] = [og(l) = [ [g(l)] = [(2) = 2 + 1 =3
is odd :
So, (jog)oh = 1, when n is even and (jog)oh = 3, when n is odd.
Example 6. Consider A = B = C = R and let f : A ; B and g : B ; C be defined by f(x) = x + 9 and g(y) = y 2 + 3. Find the following composition functions : (ii) (gog)(a) (i) (fof)(a) (iii) (fog)(b) (iv) (gof)(b) (vi) (fog)(4). (v) (gof)(4) Sol. (i) (jof)(a) = f[f(a)] = f(a + 9) = (a + 9) + 9 = a + 18. (ii) (gog)(a) = g[g(a)] = g(a' + 3) = (a' + 3) ' + 3 = a' + 6a' + 12. (iii) (jog)(b) = fTg(b)] = feb' + 3) = (b' + 3) + 9 = b' + 12.
1 20
DISCRETE STRUCTURES (iv) (gaf)(b) = g[f(b)] = g(b + 9) = (b + 9) ' + 3 = b ' + 18b + 84. (v) (gaf)(4) = g[f(4)] = g(13) = (13)' + 3 = 172. (vi) (fag)( 4) = fTg( 4)] = f(19) = 19 + 9 = 28.
Example 7. Let X = {a, b, c}. Define f : X ; X such that
f = {(a, b), (b, a), (e, c)} Find (i) f1 (ii) f 2 (iii) [ 3 Sol. (i) The inverse function r' = {(b, a), (a, b), (c, e)} (Fig. 3.35). (ii) The f' is faf (Fig. 3.36). (faf)(a) = f[f(a)] = f(b) = a, (faf)(b) = f[f(b)] = f(a) = b
b
Fig. 3.35
(faf)(c) = f[f(e)] = f(c) = c, f' = {(a, a), (b, b), (c, e)} . So, , (iii) The ! is fafafi.e., faf ' (Fig. 3.37). (faf) ' (a) = f[f ' (a)] = f(a) = b, (faf ')(e) = f[f'(e)] = f(e) = e �c+... a ++... b
Fig. 3.36 (faf ')(b) = f[f ' (b)] = f(b) = a
a
b
��+... c ��. c
Fig. 3.37
!' = {(a, b), (b, a), (e, e)}. 4 (iv) The f is fafafafi.e., fa! , (Fig. 3.38). (fa! ') (a) = f[f 3 (a)] = f(b) = a, (fa! ') (b) = f[f 3 (b)] = f(a) = b (fa! ') (c) = f[f 3 (c)] = f(c) = c So,
FUNCTIONS
121 3 f
a
a
a
b
b
b
c
c
f4
C
Fig. 3.38
t' = {(a, a), (b, b), (c, e)}.
So ,
Example 8. Consider the functions f, g and h as in Figs. 3. 39, 3.40 and 3.41. x,'I'"
++
\,.1+ Z,
9
Fig. 3.39
Fig. 3.40 Z, \1+ k, Z2 +jf+ k2 Z3 f,.". k3
h
Determine (i) got
Sol. (i) Consider Fig. 3.42.
Fig. 3.41
(ii) ho(gof)
(iii) (hog)of
x,'I++ \{+ Z,
Fig. 3.42. gal
(gof!(x,) = g[f(x,)] = g(y,) = Z" (gof)(x,) = g[f(x,)] = g(y,) = Z3 (gof!(x,) = g[f(x,)] = g(y,) = Z3 ' (gof)(x,) = g[f(x,)] = g(y,) = Z3 got = {(x" Z,) , (x" Z,) , (x3 ' Z3 ) ' (x" z,)}. SO, (ii) First find the composition oft with g and then with h (Fig. 3.43). ho(gof!(x,) = ho[g(f(x,» ] = hO[g(y,)] = h(Z,) = k,
DISCRETE STRUCTURES
1 22 ho(goj)(x,) = ho[g(f(x2))] = hO[g(Y3)] = h(z,) = k3 ho(goj)(x,) = ho[g(f(x,» ] = ho[g(y,)] = h(z,) = k3 ho(goj)(x,) = ho[g(f(x,))] = h[g(y,)] = h(z,) = k3
Fig. 3.43. ha(ga!).
SO, ho(goj) = {(X" k,), (X" k,), (x3 ' k,), (X" k,)}. (iii) First find the composition of g with h and then with f(Fig. 3.44).
Y2 Y3 Y4
9
h 1 ,'+ 1++ k,
\Z2
k2 +'\+ k3
Fig. 3.44. (hog) of.
So, Now,
(hog)(y,) = h(g(y,» = h(Z3 ) = k3 (hog)(y,) = h(g(y,» = h(Z,) = k" (hog)(y,) = h(g(y, » = h(z,) = k3 ' (hog)(y,) = h(g(y,» = h(z, ) = k3 hog = {(Y" k,), (Y" k,), (Y3 ' k ,), (Y" k,)}
«hog)oj)(x,) = (hog)(f(x,» = hog(y,) = k, «hog)oj)(x,) = (hog)(f(x,» = hog(y,) = k3 «hog)oj)(x,) = (hog)(f(x,» = hog(y,) = k3 «hog)oj)(x,) = (hog)(f(x,» = hog(y,) = k3 (hog)of= { (Y" k,), (Y" k,), (Y3 ' k,), (Y" k ,)} . So, Example 9. (a) Iff : A ; B and g : B ; C are oneone functions. Show that gof : A ; C is oneone. (b) If f : A ; B and g : B ; C are onto functions. Then show that gof : A ; C is onto (c) In part (b), ifgof is oneone, then f is oneone. (d) Inpart (b), ifgof is onto, then fis onto. Sol. (a) Let (goj)(x) = (goj)(y) => g(f(x» = g(f(y» => f(x) = f(y) I g is oneone I f is oneone x=y Hence gof is oneone. (b) As g is onto, therefore, for c E C, there exists b E B such that g(b) = c. As f is onto, therefore, for b E B, there exists a E A such that f(a) = b (goj) (a) = g(f(a» = g(b) = c ..
FUNCTIONS
1 23
Hence we have shown that for a E A, there exists C E C such that (go/) (a) = c. Hence gof is on to. (c) Suppose, f is not oneone, then, there exists a ", b E A such that f(a) = f(b)
..
=>
(go/)(a) = g(f(a» = g(f(b» = (go/) (b) gof is not oneone (as a ", b).
Hence, our supposition is wrong. Therefore, if gof is oneone, then fmust be oneone. (d) For a E A, consider (go/)(a) = g(f(a» E g(B) => (go/) (A) C g(B). Now, suppose, ifg is not onto, then g(B) must be properly contained in C and hence (go/) (A) is also properly contained in C. It means gofis not onto. Hence, our supposition is wrong. Therefore, if gof is onto, then g must be onto. Example 10_ Let A = {I, 2, 3j, B = (a, b c, dj, C = {7}. Define f.' A ; B, by f(1) = cJ(2) = b, f(3) = a, and define g : B ; C by g(x) = 7, for all x E B. (a) Does f1 exist ? Why ? (b) Does g l exist ? Why ? SoL (a) f1 exists iff f is oneone and onto. Here f : A ; B given by f(l) = c, f(2) = b, f(3) = a Le., different elements of A has different images in B. .. f is oneone. (Fig. 3.45) Also the element 'd' ofB has not its preimage in A. :. fis not onto. :. f1 does not exist (As we know that r' exists iff it is oneone and onto). (b) Here g : B ; C by g(x) = 7 for all x E B
Le.,
g(a) = 7, g(b) = 7, g(c) = 7, g(d) = 7 Thus, different elements of g have different images in C .. g is oneone. (Fig. 3.46) Also every element of C has its preimage in B. :. g is
onto. Hence gl exists (we know that f1 exists iff f is oneone and onto).
A
B
B
Fig. 3.46 C
Example 1 L Let (xj denotes the greatest integer function and f : R ; Z defined by
f(x) = [x]. Find f1(1).
SoL We first define the greatest integer function. For any real number, we define the greatest integer function [x] = the greatest integer less than or equal to x, x E R. [2.45] = 2, [ 2. 1] =  3, [0.32] = 0 etc. For e.g., Here f : R ; Z is defined by f(x) = [x]. Codomain f = Z, Also the range Rf = Z f is onto. .. Let x, y E R such that f(x) = fly) => [x] = [y] =>
x =y
.. f is oneone. Hence r' exists. We now find fl (l). Given f(x) = [x], X E R For O < x s 1, [x] = 1
f(x) = [x] = 1, if O < x S 1, X E R. 1 f (1) = [x : 0 < x s 1, x E R] .
DISCRETE STRUCTURES
1 24
Example 12. If f: A � B is a function from A to B, and iffl exists, then it is unique. Sol. Given fl exists => f is one·one and onto.
Let g : B � A and h : B � A be two inverses of f We show g = h. Let y E B. Since g : B � A, h : B � A. :. We can find x" x, E A such that g(y) = x" h(y) = x, Now, g(y) = x, =>
Also, =>
=>
Hence the result.
y = g' (X,) = f(x,) h(y) = x, y = hl (x,) = f(x,) f(x,) = f(x,) Xl = x2 g(y) = h(y) for y E B g = h.
Example 13. If f: R � R is given by (a) f{x) = 3x  5
I f is one·one
(b) f{x) = 5  8x
Find fl, if exists. Sol. (a) fl exists iff f : R � R is one·one and onto. Oneone. Let x, y E R such that f(x) = f(y) => 3x  5 = 3y  5 3x = 3y => x = y => :. f is one·one Onto : Let y E R such that f(x) = y => 3x  5 = y 5+y 3x = 5 + y ::::::} x = 3 
5+y
For each y E R, we have x = 3 E R. . . fis onto also. Hence fl exists. Now, f(x) = y gives x = fl (y) 
5+y fl (y) = x = 3 5+ X fl (X) = 3 5x f' (X) = 8 '


(b) Try yourself,

Example 14. Consider the function f: N x N � N defined by f (x, y) = (2x + 1) 2Y  1 where N = {O, 1, 2, 3, 4, .. .}. Show that f is bijective. Sol. Let (x, y) and (m, n) E N x N such that f (x, y) = f (m, n) => (2x + 1) 2Y  1 = (2m + 1) 2"  1 => (2x + 1) 2Y = (2m + 1) 2" which is possible only if
. . f is one·one
Let =>
=>
x = m and y = n (x, y) = (m, n)
f (a, b) = 0 => (2a + 1) 2b  1 = 0 (2a + 1) 2b = 1 = 2° = 1.20 which is possible if 2a + 1 = 1 and 2b = 2° => a = 0 and b = 0
. . fis onto. Since fis 1·1 and onto, therefore f is bijective.
=>
(a, b) = (0, 0)
FUNCTIONS
1 25 TEST YOUR KNOWLEDGE 3.2
SHORT ANSWER TYPE QUESTIONS
1. Let f and g are defined by f(x) � 2x + 1 and g(x) � x2  2 respectively. Find (b) (jog)(4) (a) gof (4) (c) (gof)(a + 2) (d) (jag)(a + 2) gof fog �f W� 2. (a) Let f: A E be a function from A to B. Show that falA � f, where IE : B B and IA : A A are the identity functions on B and A respectively. (b) hen) Let (,�g,n h 1.areThen functions Z to Z, on the set of integers, given by fen) n2 , g(n) n + find (i)from ga(joh) (ii) fa(goh) (iiI) ha(jag). 3, Let A � 2, 3}. Define f A A by f(l) � 2, f(2) � 1, f(3) � 3. Find f and t1 4, Let A � E � C � R and let f : A E, g : E C be defined by f(a) � a + 1, g(b) � b2 + 2 find (gof)(2), (jag)( 2) Let A � 2, 3, 4}, E � {a, b, c, d} andf: A E is given by (i) f� a), (2, a), (3, c , (4, d)} f� a), (2, c , (3, b), (4, d)}. Determine whether rl is a function or not. (f)
(e)
>
t
(h)
t
=
I,
>
{I,
>
5,
=
{I,
>
>
)
{(I,
(h)
{(I,
)
LONG ANSWER TYPE QUESTIONS
Let f: A B and g : B C be functions. Show that if go/is oneone, then {is oneone. Let f: A B and g : B C be functions. Show that if go/is onto) then g is onto. Let f; Z Z by f(x) � 2X2 + 7x. Is t' exist ? If not, why ? Find the formula for the inverse of g(x)  1. 10. Consider the functions [(x) 2x  3 and g(x) x2 + 3x + 5. Find a formula for the composition functions (a) gof (b) fog c f (d) f S 11. Find a formula for the inverse of[(x) 52xx  3 12. Let [(x) + b, g(x) ex + d where a, b, e and d are constants. Determine for which constants a, b, c, d it is true that fag � gal t
6. 7. 8, 9.
t
t
t
>
= x2
=
()
= ax
=
=
=
7
Answers
(a) 79 (d) 2a2 + 8a+ 5 (g) 4x + 3 2, (b) (i) (n  1) 2 + 1 3, f2 � 2, 3}, fl � f (I) No L
{I,
5,
(c) 4a2 + 20a + 23 (b) 29 (f) 2x'  3 4x2 + 4x  1 (h)  4x' (iii) (n + 1) 2  1 (ii) n2 4, (gof)(2) � 3, (jag) (2) � 5 (il) Yes (e)
x'
DISCRETE STRUCTURES
1 26
No, since {is not oneone. Also, {is not onto 9. F 2 10. (a) 4x + 5 (b) 2x' + 6x+ 7 (c) 4x9 (d) 8x 21 11. h1 (x) 75xx 3 12. (a I) d (c  l)d. 7 g l (X) =
8.
� 
�
M U LTIPLE CHOICE QU ESTIONS ( MCQs)
1. The number of functions from an m element set to an n element set is 2.
3.
4.
(a) m + n (b) mn (c) nm (d) m n. Let A and E be sets with cardinalities m and n respectively. The number of one· one mappings (injections) from A and E, when m < n, is (a) mn (b) np m (c) nem (d) None of these.
Suppose x and y are sests and I x I and I y I are their respective cardinalities. It is given that there are exactly 97 functions from x to y. From this one can conclude that (a) I x I = I y I = 97 (b) I x I = 97, I y I = (c) I x I = 97, I y I = 97 (d) None of these. Let R denote the set of real numbers. Let f : R x R ; R x R be a bijective function defined by f (x, y) = (x + y, x  y). The inverse function of f is given by
1,
1
(a) f l (x, y) =
(c) f  I (x, y) =
5.
6.
7.
8. 9.
( 1 )] [_1_,_ [; ;] x+y x
(b) f l (x, y) = (x  y, x + y)
xy
y x ,
y
(d) f l (x, y) = [2 (x  y), 2(x + y)].
Let A and E be sets with cardinalities m and n respectively. The number of one·to·one mapping from A to b where m < n is (a) mn (b) np m m (c) e (d) nem. n Let A be set of integers greater than one and smaller than thousand. Let b denote set of books in library, I E I = 999. Let f : A ; E, assigning a unique number to each book fis (a) one to one, onto (b) one to one, and not onto (c) not one to one, onto (d) not one to one, not onto. Let f(x) = x' + x and g(x) = x + then fog is : (a) x' + 3x + (b) x' + x + ' + (c) (x + I) (x + (d) None of the above. Each of the function 2n and n1cg n has growth rate ... that of any polynomial. (a) Greater than (b) Less than (d) Proportional to. (c) Equal to The number of bijective functions from A to A when A contains elements is
1 1)
� 100 1 (c)
1
1
W 100 1
(d) None.
106
FUNCTIONS 10.
1 27
Iff : N ; N is a function defined by f(n) = 2n + 3. Then (b) fis onto, but not one·one (a) fis one·one, but not onto (c) fis neither one·one, nor onto ( n = 97, m = 1, i.e., I x I = 1, I y I = 97. (c) Let f (x, y) = (x" y,) => (x + y, x  y) = (x" y,) x + Y = Xl (1)
4.
X  Y = Y,
(2)
2X = X, + Y,
Adding,
Subtracting, we get 2y = x,  Y 1 _
(x, y) 
f1 (x, y) =
5. 6.
7. 8.
9. 10.
=>
(Xl + Y Xl Y ) (; ;) 1
2
x
y

'
1
2
,x
y
 f1 _
(x" y,)
.
(b) (b) Consider the function f : A ; B defined by f(a)
= a  I for all a E A, then each element of A has
different images in B. So fis one·one. But the element 999 has not its preimage in A. Therefore, f is not onto since, if 999 E B, then there should be 1000 in A such that f(1000) = 1000  1 = 999 Hence f is one·one, but not onto. (c) fog (x) = f(g(x» = f(x + 1) = (x + 1) 2 + X + l.
A
2 '\4 3 \.H> 2 4 3
B
9 9 9 f�� 9 9 8 999
(b) (b) The number of bijective functions from f : A ; B where A and B have the same number of elements say, n is n !. (a) Oneone. Let f(n,) = f(n2) => 2n, + 3 = 2n2 + 3 f is oneone n1 = n2 • •
Onto. Let For each m E N, :. f is not on to.
f(n) = m
=>
m3 n =  'l N 2
m3
2n + 3 = m ::::::} n = 2
MATHEMATICAL INDUCTION 4. 1 . PRINCIPLE O F MATHEMATICAL INDUCTION
The process to establish the validity of a general result involving natural numbers is the principle of mathematical induction.
(P. T. U. M G.A. May 200 7)
4.2. WORKING RULE
Let no be a fixed integer. Suppose P(n) is a statement involving the natural number n and we wish to prove that P(n) is true for all n :> no' 1. Basis of Induction. P(no) is true i.e., P(n) is true for n = no' 2. Induction Step. Assume that the P(k) is true for n = k. (k :> n� Then P(k + 1) must also be true. Then P(n) is true for all n :> no ' 4.3. PEANO'S AXIOMS
Peano, a mathematician defined the natural number in the following way. According to him, the member of set N satisfying the following properties are called natural number. 1. 1 E N. 2. For all n E N, there exists a unique n' E N, such that (a) m' = n' 0). Sol. Consider pen) = 1 + 2 + 2' + 23 + ... + 2" = 2"+1  l. Basis ofInduction. For n = 0, P(o) = 1 = 2 '  1 = 1 It is true for n = 0.
(P.T.U. M.C.A. Dec. 2005)
(1)
DISCRETE STRUCTURES
1 32
Induction Step. For n = r, P(r) = 1 + 2 + 2' + 23 + . . . + 2' = 2,+1  1 is true. Adding 2'+1 to both sides, P(r + 1) = 1 + 2 + 2' + 23 + ... + 2' + 2'+1 = 2'+1  1 + 2 rt 1 = 2(2rt1)  1 = 2'+'  1 As P(r) is true, hence P(r + 1) is also true. From, (1), (2) and (3), we conclude that 1 + 2 + 22 + . . . + 2n = 2n+1 _ I , is true for n = I , 2, 3, .
(2) I Using (2)
(3)
Example 7. Prove by induction that for n :> 0 and a '" 1 ; 1 + a + a2 + ... + an =
(1) (2)
Ia
Adding a'+1 to both sides, P(r + 1) = 1 + a + a' + ... + a' + a'+1 Ia
1 a
=
1  a r+2 1 a
I
As P(r) is true, hence P(r + 1) is also true. From (1), (2) and (3), we conclude that
1 + a + a2 + . . . + an =
1  an+ 1 Ia
is true for n ;::: O.
a n+l Ia

(P.T.U. M.C.A. May 2007)
Sol. Basis ofInduction. For n = 1, 1  a 1+ 1 I  a' 1 + a' = ,  = I + a 1 a Ia It is true for n = l. Induction Step. For n = r, 1  ar + 1 P(r) = 1 + a + a' + ... + a' = is true 1  ar+ 1 = ,__ + ar + 1
1
(3) Using (2)
Hence proved.
Example 8. Show that for any integer n, 11 n+2 + 122n+1 is divisible by 133. Sol. Let P(n) = I I n+' + 12'n+1 Basis of Induction. For n = 1, P(I) = 113 + 123 = 3059 = 133 x 23 So, 133 divides P(I). Induction Step. For n = r, P(r) = 11'+' + 12,,+1 = 133 x s, say,
( 1) (2)
Now, for n = r + I, P(r + 1) = l lrt'+1 + 12' (c) +3 = 1l [133s  12 " +1 ] + 144 . 12' rt 1 I Using (2) = I I x 133s + 12,,+1 . 133 = 133[lls + 12" +1] = 133 x t, say (3) As (1), (2) and (3) all are true, hence P(n) is divisible by 133.
MATHEMATICAL INDUCTION
1 33
Example 9. Prove by induction that the sum of the cubes ofthree consecutive integers is
divisible by 9.
Sol. Let pen) = n3 + (n + 1)3 + (n + 2) 3
pen) is divisible by 9 P(l) = 1 + 8 + 27 = 36 which is divisible by 9. For n = r, per) = r3 + (r + 1) 3 + (r + 2) 3 = 9 . q, say, For n = r + 1 , per + 1) = (r + 1) 3 + (r + 2) 3 + (r + 3) 3 = r3 + (r + 1) 3 + (r + 2) 3 + [9 r' + 27r + 27] = 9q + 9(r' + 3r + 3) = 9[q + r' + 3r + 3 ] = 9.s, say, From (1), (2) and (3), we have the required result by induction. 1 n+1
1 n+2
1 2n
( 1) (2) I Using
(2)
(3) Hence proved.
13 > 24
Example lO. Prove  +  + . . . +  , for n :> 2.
Sol. For n = 2,
L.H.S. =
1
2+
1
+
7 13 = > 24 = R.H.S. 12 2+2 1
( 1)
It is true for n = 2. Now, for n = r, where r > 2 1

r+l
1
1
13
+  + . . . +  > r+2 2r 24
(2)
For n = r + I, L.H.S.
=
1

r+2 13
+
1
1
1
1
13
1
1
1
+ ... +  + + + > + r+3 2r 2r + 1 2r + 2 24 2r + 1 2r + 2 r + 1






[Using (2)]
2r + 2 + 2r + 1  2(2r + (2r + lX2r + 2)
1)
13
1
13
>=+ 24 (2r + lX2r + 2) 24 As n = r is true. Thus, the result is also true for n = r + 1. Hence, we have the result for = + 24
n :> 2 by induction.
Hence proved.
Example 11. Show that n3 + 2n is divisible by 3 for all n :> 1 by induction. Sol. Basis ofInduction. For n = 1,
P(l) = 1 3 + 2 x 1 = 3. It is divisible by 3. Induction Step. For n = r, per) = r3 + 2r = 3s, say, Assume, For n = r + I, per + 1) = (r + 1) 3 + 2(r + 1) = r3 + 1 + 3r(r + = r3 + 1 + 3 r' + 3r + 2r + 2
1)
+ 2r + 2
DISCRETE STRUCTURES
1 34 = r3 + 3 + 5r + 3r' = ,'3 + 2r + 3 + 3r + 3r' = 3s + 3 + 3r + 3r' = 3r' + 3(r + s) + 3 = 3(r' + r + s + 1) = 3t, say
( :
r3 + 2r = 3r)
It is divisible by 3. Hence, we have the required result by induction.
Example 12. Show that 2" x 2"  1 is divisible by 3 for all n :> 1 by induction. Sol. Basis ofInduction. For n = 1, 2'
X
2'

1 = 3 divisible by 3.
It is true for n = l. Induction Step. For n = r,
per) = 2' 2'
For n = r + I ,
2 '+1 . 2 '+1


1 = 3s, say
Le., 2,+1

1 = 3s
1 = 2'+1 (3s + 1)  1 = 3s . 2 "" + 2"" 1 = 3s . 2 '+1 + 3s = 3s(2'+1 + 1).
( 1) I Using (1)

It is divisible by 3. Hence, we have the required result by induction.
Example 13. Show that 1 2 + 32 + 52 + ... + {2n  1)2 = Sol. Basis of Induction. For n = 1, (2 x 1 _ 1)' = 1 = It is true of n = l.
1(2 x 1  lX2 x 1 + 1) =1 3
n(2n  1)(2n + 1) . 3
Induction Step. For n = r, l' + 3' + 5' + . . . + (2r  1)' =
r(2r  1)(2r + 1) is true. 3
( 1)
For n = r + I , per + 1) = l ' + 3' + 5' + . . . + (2r  1)' + [2(r + 1 )  1]' r(2r  lX2r + 1) r(2r  lX2r + 1) + 3(2r + 1)2 + (2r + 1)' = 3 3 (2r + 1)[2r 2  r + 6r + 3] (2r + 1)[2r 2 + 5r + 3] = ===::,''. "'3 3 2 (2r + 1)[2r + 3r + 2r + 3] (2r + lX2r + 3Xr + 1) = ��� 3 3 which proves the result for n = r + l. n(2n  lX2n + 1) l' + 3' + 5' + . . . + (2n  1) , = . So, 3 =
Example 14. Prove that n{n + 1){n + 2) is a multiple of 6. Sol. pen) = n(n + l)(n + 2)
I Using (1)
MATHEMATICAL INDUCTION
1 35
Basis of Induction. For n = 1, which is true.
P(I) = 1 . (1 + 1)(1 + 2) = 1 . 2 . 3 = 6
Induction Step. For n = r,
P(r) = r(r + 1)(r + 2) be a multiple of 6
For n = r + I, P(r + 1) = (r + 1)(r + 2)(r + 3) is also a multiple of 6. Now, P(r + 1) = (r + 1)(r + 2)(r + 3) = (r + 1)(r + 2) r + 3(r + 1)(r + 2) = r(r + l)(r + 2) + 3(2r') where (r + l)(r + 2) = even = 2r' = r(r + l)(r + 2) + 6r' which is a multiple of 6. Thus, P(r + 1) is true as P(r) is true. . . P(n) is true for all natural numbers n
i.e., n(n + l)(n + 2) is a multiple of 6.
Example 15. Prove that \In E
5 3
Hence proved.
2
N,
5
n
5 + 2 n3 + �
3
15
n is a natural number.
7 n IS. a natural number. 1 + n + 3 15 Basis of Induction. For n = 1, 1 5
Sol. Let P(n) = 
n

1 1 7 5 + 3' + 15 = 1, which is a natural number.
It is true for n = l.
Induction Step. For n = r,
(.! r5 .! r3 � r) 5
+
3
+
15
is a natural number
(1)
For n = r + I, 7 (r + 1) 1 1  (r + 1) 5 +  (r + 1) 3 + 5 3 15 1 5 1 7 (r + 1) =  (r + 5r' + lOr3 + lOr' + 5r + 1) +  (," + 3r2 + 3r + 1) + 5 3 15 
(�r5 � r3 � r)
+ + + (r' + 2r3 + 3r' + 2r) + 1 5 3 15 = (a natural number) + (a natural number) + 1 = a natural number As (P(r» is true, hence P(r + 1) is also true. Hence, by induction, P(n) is true \In E N. =

[Using (1)]
Hence proved.
DISCRETE STRUCTURES
1 36
Example 16. Prove that n{n + 1)(2n + 1) is divisible by 6. Sol. Let pen) = n(n + I)(2n + 1) is divisible by 6. Basis of Induction. For n = 1, P(l) = 1(1 + 1)(2 + 1) = 1 . 2 . 3 = 6 which is true for n = 1 Induction Step. For n = r, per) = r(r + 1)(2r + 1) be divisible by 6 For n = r + I, per + 1) = (r + l)(r + 2)(2r + 3) L.R.S. = (r + l)(r + 2)(2r + 3) = 2," + 9r' + 13r + 6 = (2r3 + 3r2 + r) + (6r' + 12r + 6) = r(r + 1)(2r + 1) + 6(r' + 2r + 1) which is divisible by 6. As (P(r» is true, hence per + 1) is also true. . . Le., n(n + 1)(2n + 1) is divisible by 6.
... (1) (2)
[From (2)] pen) is true for all natural numbers n Rence proved.
TEST YOUR KNOWLEDGE
1. If Pen) is the statement : "n(n + 1) + 1 is odd", then write P(3). 2. If Pen) is the statement "n2 + 1 is even") then write P(4). Is it true ? 3. If Pen) is the statement "ns + 2 is a multiple of 5") then show that P(4) is not true. 4. If Pen) is the statement "12n + 3 is a multiple of 5") then show that P(3) is false, whereas P(G) is true. Let Pen) be the statement "3" > n" , If Pen) is true, show that Pen + 1) is also true. 6. If P(n) is the statement : 13 + 23 + 33 + ...... + n3 = [ n(n + 1) r. Then verify that P(3), P(7) are both 2 true. 7. If Pen) is the statement : 12 + 32 + 52 + ...... (2n  1)2 _ n(4n2  1) ' then show that P(3), P(4), P(G) 3 are true. 8. Let Pen) be the statement "n2 + n is even", If P(k) is true, then show that P(k + 1) is true. Prove the following by using the principle of mathematical induction (Q. No. 916) 9. n(n + 1) + 1 is an odd number, n E N. 10. 1 + 4 + 7 + ..... + (3n  2) = n(3n2  1) ' n E N. 11. a + (a + d) + (a + 2d) + ...... + [a + (n  l)d] = "2n [2a + (n  l)d] , n E N. 1 1 l I n 12. +  +  + . . . . . . + , n E N. (2n  1) x (2n + 1) 2n + 1 1x 3 3 x 5 5 x 7 5.
:
MATHEMATICAL INDUCTION 13. 14.
15. 16. 17. 18. 19.
1.
8.
1 37
1 1 l I n ++ 1 x 4 4 x 7 7 x 10 + ...... + (3n  2)(3n + 1) = 3n + 1 ' n E N. 21 + 221 + 231 + ...... + �1 = 1  �1 , n E N. a + ar + ar2 + .. " .. + ar  a(1_rn) 1 r , n E N. 2 + 4 + 6 + ...... + 2n n(n + 1), n E N. Prove by mathematical induction that 2  24n  25 is divisible by 576. Prove by mathematical induction that (23  1) is divisible by 7 for all values of n E N. Prove by Mathematical Induction that 2 + 1 is divisible by 43 for each positive integer n, (P. T. M. May. 

n l _
�
52n + n
6n
+
72n +
U.
Answers
13 is odd
2.
17 is even, No Hint
Let k2 + k 2\ where \ E N. Now (k + 1)' + (k + 1) k2 + 3k + 2 (2\ k) + 3k + 2 2(\ + k + 1). �
�
�
�
c.A.
2008)
SA
BASIC COUNTING PRINCIPLES
5 . 1 . INTRODUCTION
In this chapter, we will discuss some methods of counting which acts as 'building blocks' for all counting problems.
(P. T. U. B. Tech. Dec. 2006 ; Dec. 2005)
5.2. BASIC COUNTING PRINCIPLES
There are mainly two counting principles namely (i) Sum Rule (ii) Product Rule. These two principles form the basis of permutations and combinations and hence known
as basic counting principles. 5.3. S U M RULE
If there are two jobs such that they can be performed independently in m and n ways. Then number of ways in which either of the two jobs can be performed is m + n. 5.4. PRODUCT RULE
If there are two jobs such that one of them can be done in m ways and when it has been done, second job can be done in n ways, then the two jobs can be done in m x n ways.
ILLUSTRATIVE EXAMPLES
Example 1. In a class there are 10 boys and 8 girls. The teacher wants to select either a boy ar a girl to represent the class in a function. In how many ways the teacher can make this selection ? Sol. The teacher can select a boy in 10 ways and a girl in 8 ways.
By sum rule, the number of ways of selecting either a boy or a girl = 10 + 8 = 18. Example 2. There are 3 students for a classical, 5 for mathematical and 4 for physical
science scholarship. In how many ways, one of these scholarships be awarded ?
Sol. Classical scholarship can be awarded in 3 ways
Mathematical scholarship can be awarded in 5 ways Physical science scholarship can be awarded in 4 ways.
1 38
BASIC COUNTING PRINCIPLES
1 39
Number of ways of awarding one of three scholarship = Number of ways of awarding classical or mathematical or physical science scholarship I By Sum Rule = 3 + 5 + 4 = 12
Example 3. In a class, there are 10 boys and 8 girls. The teacher wants to select a boy and a girl to represent the class in a function. In how many ways can the teacher make this selection ? Sol. The teacher can make this selection by
(i) selecting a boy among 10 boys (ii) selecting a girl among 8 boys
The first job can be done in 10 ways and the second job can be done in 8 ways. Therefore, by product rule, the total number of ways of selecting a boy and a girl = 10 x 8 = 80. Example 4. A snack bar serves 5 different sandwiches and 3 different beverges. How
many different lunches can a person order ?
Sol. The person can place his order by
(i) asking a sandwich among 5 sandwiches (ii) asking a beverge among 3 beverges
. . By 'Product Rule', the total number of ways of placing a order for a sandwich and a beverge = 5 x 3 = 15.
Example 5. Aperson is to complete a truefalse questionaire consisting of 10 questions. How many different ways are there to answer the questionaire ? Sol. Each question can be answered in two ways (true or false). Now first question can be answered in 2 ways Second question can be answered in 2 ways
10th question can be answered in 2 ways . . By 'Product Rule' the total number of answering the questions
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2 1D Example 6. A questionaire contains 4 questions that have two possible answers and 3 questions with 5 possible answers. How many ways are there to answer the questionaire ?
Sol. There are four questions each of which can be answered in two ways each.
. . Total number of ways of answering these four questions = 2 x 2 x 2 x 2 = 16 Also the next 3 questions each of which can be answered in 5 ways . . Total number of answering these 3 questions = 5 x 5 x 5 = 125 By 'Product Rule', the total number of ways of answering the questionaire = 16 x 125 = 2000.
Example 7. A room has 6 doors. In how many ways can a man enter the room through one door and come out through a different door ?
Sol. There are 6 ways of entering the room. Therefore, a man can enter the room in 6 ways. After entering the room, the man can come out through any one of the remaining five doors. Number of leaving the room = 5. . . By 'Product rule' the required number of ways in which a man can enter a room and come out through a different room = 6 x 5 = 30.
DISCRETE STRUCTURES
1 40
Example 8. Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin (i) At any one of the 7 floors (ii) At different floors. Sol. (i) Suppose M" M2 , M3, M4, M5 are five persons. Here M , can leave the cabin at any one of the seven floors in 7 ways M2 can leave the cabin at any one of the seven floors in 7 ways Similarly, M3, M4, M5 can leave the cabin at any one of the seven floors in 7 ways . . By 'Product Rule', Total number of ways in which five persons can leave the cabin at any one of the seven floors 5
=7x7x7x7x7=7 . (ii) M , can leave the cabin at any of the seven floors in 7 ways M2 can leave the cabin at any one of the remaining 6 floors in 6 ways Similarly, M3, M4, M5 can leave the cabin in 6, 4 and 3 ways respectively. . . By Product rule, Total number of ways of leaving the cabin = 7 x 6 x 6 x 4 x 3 = 2620.
Example 9. Automobile license plates in Massachusetts usually consist of 3 digits fol
lowed by 3 letters. The first digit is never zero. How many different plates of this type could be made ? Sol. Consider the digits 0, 1, 2, ... 9. These are 10 digits. Since the first digit of the
license plate is nonzero. . . first place of the license plate can be filled up in 9 ways. Similarly, the second place of the license plate can be filled up in 10 ways and the third place in 10 ways Required number of using the digits = 9 x 10 x 10
ways.
DDD
9 10 1 0 Again, there are 26 alphabets. The first place of the license plate can be filled up in 26 Similarly, the second place can be filled up in 26 ways and third in 26 ways . . Required number of using the English alphabets = 26 x 26 x 26 By 'Product Rule', total number of making the plates = 9 x 10 x 10 x 26 x 26 x 26 = 16818400.
Example 10. For a set offive true or false questions, no student has written all correct answers and no students have given the same sequence ofanswers. Find the maximum number of students in the class for this to be possible. Sol. A truefalse question can be answered in two ways either by making it true or false ..
Number of ways of answering each of the five questions = 2 x 2 x 2 x 2 x 2 = 32. Out of these 32 ways of answering, there is only one way of answering all the five questions correctly. But no student has written all the answers correctly. Maximum number of students in the class = Number of ways except one in which all answers are correct = 32  1 = 31.
BASIC COUNTING PRINCIPLES
1 41 TEST YOUR KNOWLEDGE 5.1
1. Inmany a class 8 male choose teachers9 mathematics and 5 female teachers ways,there canarea student professorteaching ? 9 mathematics class. In how 2. There are 3 students for a classical, 5 for mathematical and 4 for physical science scholarship, In how many ways can these scholarships awarded? The flag of aThere newlyareformed countrycolours is in thetoform of three be coloured differently, six different be used. How many suchblocks, designseacharetopossible ? InII and a monthly test, the teacher decides that there will be three questions, one from each chapter Ii III of the book If there are 12 questions in chapter I, 10 in chapter II and 6 in chapter III. In how many ways can three questions be selected? 5. Find the total number of ways of answering 5 objective type questions, each questions having 4 choices. 6. How many three digit numbers can be formed without using the digits 0, 2, 3 , 4, 5, 6 ? How many numbers are there between 100 and 1000 such that 7 is in the unit place. A gentleman has 6 friends to invite. In how many ways can be send invitation cards to them, ifhe has three servants to carry the cards. 'D 'D "D
3.
4.
7. 8.
1. 13. 12 x l0 x 6 �720 9 x 10 x 1 � 90
4.
7.
Answers
6 x 5 x 4 � 120 x 5 x 4 � 60. 5 5. 4 x 4 x 4 x 4 x 4 = 4 . 6. 4 x 4 x 4 = 64 3 x 3 x 3 x 3 x 3 x 3 � 36 � 729.
2. 3 8.
3.
58
PERMUTATIONS AN D COMBINATIONS
5.5. DEFINE FACTORIAL n
The product of first n natural number is called factorial n. It is denoted by n , or � . The factorial n can also be written as
,= = 1 =1
n(n 
n
We have,
1)
,=
n(n 
1) (n  2) ,
= 1.
......... 1.
=
=
n(n  l)(n  2)(n  3)
,
and 0 '
For example : (i) Find the value of
=
1:/ .
10 ! '='::" 10':'c x 9' x8! x 8! 8 ! '':' 10 9 90. n! (ii) Determine the value of (n  1) ! n! nCn  I) ! n. Here, C""Cn::  1)""'! Cn _ I)! Here,
= = , == = 130) ,= ,=
(iii) Show that 3 ! + 4 ! '" (3 + 4) ! Here, 3,+4 (3 x 2 x + (4 x 3 x 2 x 1) 6 + 24 and 7x6x 5x4x 3x2x 7 (3 + 4) Hence 3 , + 4 , '" (3 + 4) ,
1
=
5040
5.6. PERM UTATION
A permutation is an arrangement of a number of objects in some definite order taken some or all at a time. The total number of permutations of n distinct objects taken r at a time
is denoted by n Pr or pen, r), where
1 "" r "" n.
1 42
PERMUTATIONS AND COMBINATIONS
1 43
Theorem I. Prove that the number of different permutations of n distinct objects taken r at a time, r :::; n is given by n pr =
n! ( n  r) !
= n (n  l)(n  2) ... (n  r + l).
Proof. The number of permutations of n distinct objects taken r at a time is like filling of r places with n objects. The first place can be filled in by any one of the n objects. So, this can be done in n ways n p1 = n. .. The second place can be filled in by any one of the n  1 objects because after filling first place. We are left with (n  1) objects. Thus, the first two places can be filled in n(n  1) ways. n P2 = n(n  l) Similarly, the third place can be filled in by any one of the remaining (n  2) objects. Therefore, the first three places can be filled in n(n  1) (n  2) ways. Proceeding in this way, we have the number of permutations of n different objects taken r at a time, given by n Pr = n(n  l)(n  2) ... (n  r 
= is n I.
l) n(n  lXn  2) ... (n  r + 1)(n  r) ! (n  r) !
=
n! (n  r) !
Theorem II. (a) Prove that the number ofpermutations of n things taken all at a time Proof. We know that (theorem I) np = n
n! =_ n ! =_ n! (n  n) ! O ! 1
=nI Theorem II. (b) Prove that nPr = n. n1Pr_T np = r
Proof. We know
n! (n  r) !
=n.
n (n  I) ! (n  r) ! (n  I) ! n�p� . (n  1  (r  I» ! = n . =
I Theorem. I
Example 1. Determine the value of the following (ii) 9 P3 •
4
Sol. (,) P2 • •
9
(,,) Pa
4!  (4  2) !
_
9!  (9 _ 3) ! _
=
4 x 3 x 2 ! = 12 2! 9 x 8 x 7 x 6 ! = 504 6! =
(iii) 20 P2
(iv) 52 Pr
DISCRETE STRUCTURES
1 44 ... (m) 20 P2
20 ! 20 x 19 x 18 ! = 380  (20 _ 2) ! = 18 ! 52 ! = 52 x 51 x 50 x 49 x 48 ! . 52 (w) P4 48 ! (52  4) ! _
_
=
6497400.
Example 2. Determine the value of n if (i) 4 x n P3 = n+ 1 P3 .
Sol. (i)
4x
(ii) 6 x n P3 = 3 x n+ 1 P3 .
n ! = (n + I) ! (n  3) ! (n + 1  3) ! . 4xn! (n + l) x n ! (n  3) ! ( n  2)(n  3) ! => 4(n  2) = (n + 1) 3n = 9
(ii) 6x
=> => =>
(iii)
[..
.
4n  8 = n + n = 3.
n! , = (n  r) !
]
nl (n  r) !
]
np
1
6 x n p3 = 3 x n + 1 p3
(n + I) ! ( n + 1  3) ! 3 (n + 1)(n !) = ( n  2)(n  3) ! 6(n  2) = 3 x (n + 1) 6n  3n = 12 + 3 n = 6.
n! (n  3) ! . 6 x n! (n  3) !
=
3x
=>
6n 
12 = 3n + 3 3n = 16
3 x np 4 = 7 x n1 p4
n ! = 7 x (n  I) ! T =!  I" 4) (n  4) ! (n::: 3 x n x (n  1) ! = 7 x(n "1)''! (n  4Xn  5) ! '(n ': " 5) :: ! 3x
(iii) 3 x n P4 = 7 x n 1 P4 .
3n 
3n = 7 (n  4) 7n =  28
n = 7.
=> =>
[.:
n p,
7n  28  4n =  28 3n =
Example 3. How many variahle names of 8 letters can be formed from the letters a, b, c,
d, e, f, g, h, i if no letter is repeated ? Sol. There are 9 letters and 8 are to be selected.
:. Total number of variable names of 8 letters is = 9p 8 =
9 .1 (9  8) !
9! I!
==
9 .I .
Example 4. There are 10 persons called on an interview. Each one is capahle to be selected for the job. How many permutation are there to select 4 from the 10 ? Sol. There are 10 persons and 4 are to be selected. :. Total number of permutations to select 4 persons = lO p,
PERMUTATIONS AND COMBINATIONS
1 45 10 ! (10  4) !
= ::::::: =
10 x 9 x 8 x 7 x 6 !
6!
= 5040.
Example 5. How many 6digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, if no digit is repeated. Sol. There are 8 numbers are 6 are to be selected. ..
8! 8! = (8 6) ! 2 ! 8x7x6x5x4x3x2! = 22560. 2!
Total number of 6digit numbers = 8 P6 =
=
_
5.7. PERMUTATION WITH RESTRICTIONS
The number of permutations ofn different objects taken r at a time in whichp particular objects do not occur is n P P r The number of permutations of n different objects taken r at a time in which p particu lar objects are present is n p P r _p x rpp .
Example 6. How many 6digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 if every number is to start with '30' with no digit repeated.
Sol. All the numbers begin with '30' . So, we have to choose 4digits from the remaining
7digits.
:. Total number of numbers that begins with '30' is 7p
4
=
7! (7  4) !
=
7x6x5x4x3! 3!
= 840.
Example 7. In how many ways 5 different microprocessor books and 4 different digital electronics books be arranged in a shelf so that all the four digital electronics books are together ? Sol. Consider the four digital electronics books as one unit. Thus, we have 6 units that can be arranged in 6 I ways.
For each of these arrangements, 4 digital electronics books can be arranged among themselves in 4 I ways.
:. Total number of arrangements in which all the four digital electronics books are together is
= 6 I x 4 I = 720 x 24 = 17280.
Example 8. How many permutations can be made out of the letter of word "COMPU TER" ? How many of these (i) begin with C ? (iii) begin with C and end with R ?
(ii) end with R ? (iv) C and R occupy the end places ?
Sol. There are 8 letters in the word 'COMPUTER' and all are distinct. :. The total number of permutations of these letters is 8 I = 40320.
DISCRETE STRUCTURES
1 46
(i) Permutations which begin with C. The first position can be filled in only one way i.e., C and the remaining 7 letters can be arranged in 7 I ways. . . Total number of permutations starting with C are = 1 x 7 I = 5040. (ii) Permutations which end with R. The last position can be filled in only one way i.e., R and the remaining 7 letters can be arranged in 7 I ways. . . The total number of permutations ending with R are = 7 I x 1 = 5040. (iii) Permutations begin with C and end with R. The first position can be filled in only one way i.e., C and the last place can also be filled in only one way i.e., R and the remaining 6 letters can be arranged in 6 I ways. . . The total number of permutations begin with C and end with R is = 1 x 6 I x 1 = 7 20 . (iv) Permutations is which C and R occupy end places. C and R occupy end positions in 2 I ways i.e., C, R and R, C and the remaining 6 letters can be arranged in 6 I ways. The total number of permutations in which C and R occupy end places is = 2 I x 6 I = 1440. 5.S. PERM UTATIONS WHEN ALL OF THE OBJ ECTS ARE NOT DISTINCT
Theorem III. The number ofpermutations of n objects, of which n objects are of one 1
kind and n2 objects of another kind, when all are taken at a time is
n! n] ! n2 !
Proof. Let us assume that the number of required permutations be K. Now consider a single particular permutation of these K permutations, in which n, objects of one kind is followed by n, objects of other kind. Also, assume that all n, object are distinct from all n, objects. So, number of permutations of n, objects taken all at a time is = n , Pn = n, I , Also, the number of permutations of n, objects taken all at a time is = n, Pn = n, I , By the fundamental principle of counting, these K permutations will give rise to n, I n, I permutations by arranging the objects of one kind within the places occupied by them. Therefore, K permutations will give rise to K. n, I n, I permutations. n For n distinct objects, the number of permutations is = p n = n I Therefore, K x n 1 ! n2 ! = n !
n!
This result can be generalised as follows : If n, objects are of one kind, n, objects are of second kind, n3 objects are of third kind, and so on upto n, objects are of tth type is given by
n!
[Here n, + n, + n3 + ... + n, = nl
PERMUTATIONS AND COMBINATIONS
1 47
Example 9. Determine the number ofpermutations that can be made out ofthe letters of the word 'PROGRAMMING'.
Sol. There are l l ietters in the word 'PROGRAMMING' out of which G's, M's and R's are two each. The total number of permutations is
11! 2!x2! x2!
11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 ! 2 x 1x2x 1x2!
= = :::: ,:c :: ,,:: :  : : ,: =
4989600.
Example 10. There are 4 blue, 3 red and 2 black pens in a box. These are drawn one by one. Determine all the different permutations. Sol. There are total 9 pens in the box out of which 4 are blue, 3 are red and 2 are black. The total number of permutations =
9! 4!x3!x2!
=
9x8x7x6x5x4! 4!x3x2x1x2x1
=
1080.
Example 11. How many different variable names can be formed by using the letters a, a, a, b, b, b, b, c, c, c ? Sol. There are total 10 letters out of which 3 are as, 4 are b's and there are 3 c's. Total number of permutations
10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 ! 3 ! x 4 ! x 3 ! 3 x 2 x 1x 4 ! x 3 x 2 x 1 = 10 x 3 x 4 x 7 x 5 = 4200.
=
Example 12. How many 7digits numbers can be formed using digits 1, 7, 2, 7, 6, 7, 6 ? Sol. There are total 7digits out of which 3 are 7's and 2 are 6's. ..
Total number of permutations is =
7! 3 I. x 2 I.
=
420.
5.9. PERM UTATIONS WITH REPEATED OBJECTS Theorem IV. Prove that the number of different permutations ofn distinct objects taken r at a time when every object is allowed to repeat any number of times is given by n'. Proof. Assume that with n objects we have to fill r place when repetition of objects is allowed. Therefore, the number of ways of filling the first place = n The number of ways of filling second place = n
The number of ways of filling rth place = n Thus, the total number of ways of filling r places with n elements is = n. n. n. n. ...... r times = nr,
DISCRETE STRUCTURES
1 48
Example 13. How many 4digits numbers can be formed by using the digits 2, 4, 6, 8 when repetition of digits is allowed ? Sol. We have 4digits.
So, number of ways of filling unit's place = 4. = 4. Number of ways of filling ten's place Number of ways of filling hundred's place = 4. Number of ways of filling thousand's place = 4. Therefore, the total number of 4digits numbers is = 4 x 4 x 4 x 4 = 256.
Example 14. How many 2digits even numbers can be formed by using the digits 1, 3, 4, 6, 8 when repetition of digits is allowed ? Sol. We have three even numbers and two odd number. Thus, number of ways of filling unit's place = 3. Number of ways of filling ten's place = 5. :. Total number of two digits even numbers = 3 x 5
=
15.
Example 15. In how many ways can 5 software projects be allotted to 6 final year students when all the 5 projects are not allotted to the same student ? Sol. We have 5 projects and 6 students. Each projects can be allotted in 6 ways. Thus, the number of ways of alloting 5 projects is = 6 x 6 x 6 x 6 x 6 = 65 . Number of ways in which all projects allotted to same student = 6. Therefore, total number of ways to allocate 5 projects to 6 students = 65  6 = 7770.
(P. T. U. B. Tech. Dec. 2007)
5.10. CIRCULAR PERMUTATIONS
The circular permutations are the permutations of the objects placed in a circle. Consider the letters k, I, m, n, a placed along the circle as shown in Fig. 1. Ifwe place letters linearly, there are five different permu tations i.e., k, I, m, n, 0 ; I, m, n, 0, k ; m, n, 0, k, I ; n, 0, k, I, m ; 0, k, I, m, n, but there is only one circular permutation k, I, m, n, o. Therefore, there is no starting and ending in circular permuta tion. We only consider the relative positions.
Theorem V. Prove that the number of circular permuta tions of n different objects is (n  1) I Proof. Let us consider that K be number of permutations
m n
Fig. 5.1.
required. For each such circular permutation of K, there are n corresponding linear permuta tions. As shown earlier, we start from every object of n objects in the circular permutation. Thus, for K circular permutations, we have K. n linear permutations. Therefore,
K. n = n I
or K =
� n
PERMUTATIONS AND COMBINATIONS K
or
_ _
Hence proved.
1 49
n x (n  I) ! n
or K = (n  1) I
Example 16. In how many ways can these letters a, b, c, d, e, fbe arranged in a circle ? Sol. There are 6 letters and hence the number of ways to arrange these 6 letters in a
circle is
=
(6  1) I
=
5I
=
120.
Example 17. In how many ways 10 programmers can sit on a round table to discuss the project so that project leader and a particular programmer always sit together ? Sol. There are total 10 programmers but project leader and a particular programmer always sit together. So, both become a single unit and hence there are (10  2 + 1) = 9 remains. Thus, these 9 units can be arranged on round table in (9  1) I ways. The two programmers i.e., project leader and a particular programmer can be arranged in 2 I ways. Therefore, the total number of ways in which 10 programmers can sit on a round table IS
=
(9  1) I x 2 I
=
8Ix2I
=
80640.
Example 18. Determine the number of ways in which 5 software engineers and 6 electronics engineers can be sitted at a round table so that no two software engineers can sit together.
S S
Sol. There are 6 electronics engineers that can be ar· ranged round a table in (6  1) I ways. There are 5 software
E
engineers and they are not to sit together so we have six places for software engineers and can be placed in 6 I ways as shown in Fig. 5.2. Therefore, total number of ways to arrange the engi· neers on a round table is = (6  1) I x 6 I = 5 I x 6 I
E S
S S
Fig. 5.2.
120 x 720 = 86400. Example 19. In how many ways can 5 gentlemen and =
G,
5 ladies be seated round a table so that no two ladies are (P.T.V. B. Tech. May 2009) together. Sol. Let the gentlemen be seated first leaving one seat
vacant in between each of gentlemen. This can be done in (5  1) I ways = 24 ways Now 5 ladies can be seated in 5 vacant seats, as shown in Fig. 5.S, in 5 P 5 ways = 5 I ways = 120 ways . . By fundamental principle of counting, required number of ways = 24 x 120 = 2880
S
G2 L4
L2
Fig. 5.3.
DISCRETE STRUCTURES
1 50 TEST YOUR KNOWLEDGE 5.2
SHORT ANSWER TYPE QUESTIONS 1. 3. 4. 5.
7! Compute 4 5 6 7 Compute (a) 13! 11! (b) 10! . Simplify (a) (n n!1) ! (b) (n +n !2) ! Findnif (a) nP �72, (b) nP4 �42 nP2, (c) 2nP2 � 2nP2  50. Find the number2 of distinct permutations that can be formed all the letters of each word (a) RADAR (b) UNUSUAL. !,
!,
!,
!.
2.
_
LONG ANSWER TYPE QUESTIONS
6.
7. 8. 9. 10.
11.
12. 13.
14.
15. 16.
Therecan areafour lines between A and B, and three bus lines between B and c. In how many ways manbustravel, (a) By bus from A to C by way of B (b) Roundtrip by bus from A to C by way of B (c) Roundtrip by bus from A to C by way of B ifhe does not want to use a bus line more than once. Find the number of ways that a party of seven persons can arrange themselves (a) In a row of seven chairs (b) Around a circular table. Intwohowbooks manyof ways can four books of mathematics, threeallbooks ) threesubject booksareof chemistry, sociology be arranged on a shelf so that booksofofhistory the same together ? How manyby (a) automobile licensedigits plates(b)canif the be made if each followed three different first digit is notplate zero.contains two different letters Thereonearecansixdrive roads between A and B and four roads between B and C. Find the number of ways that (a) From A to C by way of B (b) Roundtrip from A to C by way of B (c) Roundtrip from A to C by way of B without using the same road more than once. (a) Find the number of ways in which five persons can sit in a row. (b) How many ways are there if two of the persons insist on sitting next to one another. (c) Solve part (a) assuming they sit around a circular table. (d) Solve part (b) assuming they sit around a circular table. Find the number of ways in which five large books, four medium size books, and three small books can be placed on a shelf so that all books of the same size are together. (a) Find the number of permutations that can be formed from the letters of word ELEVEN. (b) How many of them begin and end with E ? (c) How many of them have the three Es together ? (d) How many begin with E and end with N? (a) In how many ways can three boys and 2 girls sit in a row ? (b) In how many ways can they sit in a row if the boys and girls are each to sit together ? (c) In how many ways can they sit in a row if just the girls are to sit together ? (a) In how many ways can you arrange the letters in the word CONGRESS ? (b) In how many ways, the two SiS not together ? How many words can be obtained by arranging the letters of the word 'UNIVERSAL' in different ways ? In how many of them (a) E, R, S occur together? (P. U. B. Tech. May (b) No two of the letters E, R, S occur together ? Find the number of different messages that can be represented by sequences of 4 dots and 6 dashes. (P.T.U. B.TechMay T.
17.
2008)
2013)
PERMUTATIONS AND COMBINATIONS
151
1. 4 � 24, 5 � 120, 6 � 720, 7 � 5040 2, (a) 156 (b) 7201 4, (a n � (b) n � (b) 3!7! � 840 5, a 2 5! 2! ! � 30 6, a 12 (b) 144 a7 (b) 6 a 468000 (b) 421200 10, a 24 (b) 576 11. a 120 (b) 24 12, 103680 13, a 120 (b) 24 14, a 120 (b) 24 15, a 20160 (b) 15120 I
I
)
9
( )
7, 9,
( ( ( ( (
) ) ) ) )
Answers
I
I
( )
3, a n (c) n � 5
9
) 8,
(c 72 41472 c 360 c 24 (d) 12 c 24 (d) 12 (c) 48 16, 362880, (a) 30240, (b) 332640,
I
I
() ()
( ) ( ) ( )
( )
(b) n2 + 3n + 2
()
Hints
6. a 4There3 are12 ways four ways B, and to go tofromgo from A to CAbytoway of B.three ways to go from B to c. Hence there are (b) There go fromroundtrip, A to C by way of B and 12 ways to return. Hence there are 12 12are 14412 ways ways toto travel c The man will traval from A to B to C to B to A. Total number of ways 4 3 2 3 72. Required number of ways 4 4 3 3 2 a Required no, � 26 25 10 8 � 468000 (b) Required no, � 26 25 8 � 421200 10, See Q, 6, 12. Required number of ways 3 5 4 3 13. a ELEVEN consists 6 letters in which three Es, 2Vs, occur, 6 x 5 x'4 ' � 120, 6 ! _ '::':' Required number of ways _ 3!_ x 2 ! ,=2 (b) The first and last place are occupied with E. Remaining places can be filled up in 4 24. c Take three Es as one word. Total number of words are 4, [EEE L,v,N] .. Required number of words 4 24. (d) The first place is occupied with E and the last placed is occupied with N. []] , L, E, V, E, lli] Required number of ways �2 ! 12. 15. a The word CONGRESS consists ofS words in which 25 occur. Total number of arranging the words �2! 20160. ()
8.
9,
( )
( )
x
x
=
=
x
x
= =
x
x
! x
x
=
x
! x
! x
! x
! x
! x
!
x 9x
x9x 9x
=
! x
!
=
()
( )
=
=
!
=
!
=
DDDDDD I I E
=
=
=
E
DISCRETE STRUCTURES
1 52
Issl C, 0, N, G, R, E (b) Consider SS as one word, then Total number of arranging the words in which two S's occcur together 7 Hence required number of arranging the words in which no two S's occur together Total number of arranging the words (number of arranging the words in which 2 S's occur together) 20160 7 15120. =
17.
�
=
!
I�
There are total 10 symbols i.e. 4 dots and 6 dashes. The total number of messages that can be represented are:
10 x 9 x 8 x 7 x 6 ! ,, = 2 10 messages.
10 !
4!x6!
4!x6! 5. 1 1 . COMBINATION
A combination is a selection of some or all, objects from a set of given objects, where order of the objects does not matter. The number of combinations of n objects, taken r at a time is represented by nc or C (n, r).
Theorem VI., The number of combinations of n different things, taken r at a time is
given by
nC = n .' , r l (n  r) 1
J
1 ::; r :::; n.
Proof. The number of permutations of n different things, taken r at a time is given by
n !, np = :(n_  r) ! ,
As there is no matter about the order of arrangement of the objects, therefore, to every combination of r things, there are r ! arrangements i.e.,
npr = r ! nCr or nc = np, = : n ! , , r ! (n  r) ! r ! ! n .e .:...:. ., _ nCr = :  ,, ) 1 :::; r ::; n. (n  r) ! r ! __
Thus,
one.
Theorem VII. Prove that the number of combinations of n things taken all at a time is Proof. We know that
nCn = (n  nn)! ! n !
�s one.
" '':c
=
n! O! n!

[ .:
=1
0 I = 1]
Theorem VIII. Prove that the number of combinations ofn things taken none at a time Proof. We know that
n! n ! ,.......,.  n! (n  O) ! O ! n ! O ! n !  1 Theorem IX. Prove that nC r = nCr> 1 :::; r :::; n. n Proof. We know that n! n! ncn  ' = n  (n  r) ! (n  r) ! (n  n + r) ! (n  r) ! Remark. If n Cr = nCm J then either m = r or m + r = n . nco
_ _

[ ' : 0 1 = 1]

_
n! r ! (n  r) !
:, , = n c " 
PERMUTATIONS AND COMBINATIONS Theorem X. Prove that numbers
Proof. Here
(n ; 1
nC
,1
(n ;
1 53 1J
where 1 :::; r :::; n and TJ n are natural (P.T.V. B. Tech. Dec
J
2010)
= n+ 1 C" Consider
+ nC' = = = =
n! n'! + ., :.,;: "r ! x (n  r) ! (r  1) ! x (n r +:: 1):! n! n! + .,�..,.�� r (r  I) ! x (n  r) ! (r I)! x (n  r� + 1)� (n� r)! n ! x (n  r + 1) + n ! x r n ! x (n  r + 1 + r) = ::'r (r  I) ! x (n  r) ! x (n  r + 1) r ! x (n  r + 1) ! n ! x (n + 1) = (n + I) ! n = + 1C ' r ! x (n  r + I)! r ! x (n + l  r) !
ILLUSTRATIVE EXAMPLES
Example 1. Determine the value offollowing
(iii) 52C4
(ii) 50C4
(i) wC6
5
(iv) 20CW'
10 x 9 x 8 x 7 x 6 ! = 10 x 3 x 7 = 210 10 ! = . (6) ! x (10  6) ! 6 ! x 4 x 3 x 2 x l 50 ! 50 x 49 x 48 x 47 x 46 x 45 ! .. (,,) 50 C4 5 = 45 ! x (50  45) ! = 45 ! x 5 x 4 x 3 x 2 x 1 = 2118760. 52 ! 52 x 51 x 50 x 49 x 48 ! = 270725 ... ('") 52 C4 . 4 x 3 x 2 x 1 x 48 ! 4 ! x (52  4) ! 20 ! 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 ! = (iv) 2DC 10 = 10 ! x (20  10) ! 10 ! x 10 ! = 184756. Sol. (i) lO C6 =
_ _
_
Example 2. Determine the value of n if
(i) n C4 = n C3 Sol. (i)
(ii) n C 2 = 10 n
nc4 = nC3
_
(iii) 2D C
n! n! = :::::c:c ,: ,, 4 ! x (n  4) ! 3 ! (n  3) ! 4 x3 ! x'(n 4) ! n ! = 4 ! x (n  4) ! =  '3 ! x (n 3) ! 3 ! x (n 3) x (n  4) ! n! 4 => n  3 = 4 => n = 7 1 = . n3
2D n + 2 = C2n  J .
DISCRETE STRUCTURES
1 54 (ii) Thus or or
=>
=>
(iii)
nC n  2 = 10
n! (n  2) ! [n  (n  2)] ! n x (n  1) x (n  2) ! (n  2) ! x 2 !
n! (n  2) ! x 2 !
= 10 or = 10
=>
= 10
n x (n  1) = 10 x 2 I = 20
n2  n  20 = 0 => n2  5n + 4n  20 = 0 (n  5) (n + 4) = 0 n =  4, 5. Since  4 is not possible, hence n = 5. 2 D Cn + = 2 D C n 2 2 1 _
Therefore, we have either
n + 2 = 2n  1 or (n + 2) + (2n  1) = 25 => =>
So
n=3 n=3 n = 3, 8.
or or
3 n = 24 n=8
I See remark below theorem IV
Example 3. How many 16bit strings are there containing exactly five O's ? Sol. A 16bit string having exactly five O's is determined if we tell which bits are O's.
This can be done in 16 C5 ways. Therefore, the total number of 16bit strings is _
16 ! 5 ! x (16  5) ! 16 x 15 x 14 x 13 x 12 x 1 1 ! 5 x 4 x 3 x 2 x l x 11! _
 1 6 C5 
=
= 4368.
Example 4. How many ways can we select a software development group of 1 project leader, 5 programmers and 6 data entry operators from a group of 5 project leaders, 20 pro grammers and 25 data entry operators ? Sol. There are 5 project leaders out of which one can be selected in 5C, ways. There are 20 programmers out of which five can be selected in 2D C5 ways. There are 25 data entry operators out of which six can be selected in 25 C6 ways. Therefore, the total number of ways to select the software development group is = 5 C, X 2 D C5 X 25 C6 = 9610 1544000.
Example 5. From 10 programmers in how many ways can 5 be selected when
(a) A particular programmer is included every time. (b) A particular programmer is not included at all. Sol. We have to select 5 programmers from the 10 programmers. So, the number of ways to select them in lO C5 =
10 ! 10 x 9 x 8 x 7 x 6 x 5 ! 5 ! x (10  5) ! = 5 x 4 x 3 x 2 x 1 x 5 ! = 252.
PERMUTATIONS AND COMBINATIONS
1 55
(a) When a particular programmer is included every time then the remaining = 5  1 4 programmers can be selected from the remaining = 10  1 = 9 programmers. This can be done in 9 C, ways =
=
9! 4 ! (9  4) !
=
'9 x 8 x 7 x 6 x 5 ! 4x3x2x 1x5!
=
126.
(b) When a particular programmer is not included at all, then the five programmers can be selected from the remaining = 10  1 = 9 programmers. This can be done in 9 C5 ways
9! 5 ! (9  5) !
= :::c ::::::: =
9 x8x7x6x5x4! 5 x4x3 x2 x 1x 4!
=
126.
5 . 1 2. PIGEONHOLE PRINCIPLE
(P. T. U. M G.A. 2005)
TheoremX. Show that ifnpigeons are assigned to m pigeonholes and m < n, then there is at least one pigeonhole that contains two or more pigeons. Proof. Let us label the n pigeons with the numbers 1 through n and the m pigeonholes with the numbers 1 through m. Now starting with pigeon 1 and Pigeonhole 1, assign each pigeon in order to the pigeonhole with the same number. So we can assign as many pigeons as possible to distinct pigeonholes, but as we know that pigeonholes are less than pigeons i.e., m < n. Thus, there remains n  m pigeons that have not yet been assigned to a pigeonhole. Hence, there is at least one pigeonhole that will be assigned a second pigeon.
Example 6. Show that ifany four numbers from 1 to 6 are chosen, then two ofthem will
add to 7.
Sol. Make three sets containing two numbers whose sum is 7.
A = {I, 6}, B = {2, 5}, C = {S, 4}. The four numbers that will be chosen assigned to the set that contains it. As there are only three sets, two numbers that are chosen is from the same set whose sum is 7.
Example 7. Show that at least two people must have their birthday in the same month if 13 people are assembled in a room. Sol. We assigned each person the month of the year on which he was born. Since there are 12 months in a year. So, according to pigeonhole principle, there must be at least two people assigned to the same month.
Example 8. Show that if any eight + ve integers are chosen, two of them will have same
remainder when divided by 7.
Sol. Take any eight +ve integers. When these are divided by 7 each have some remain·
der. Since there are eight integers and only seven distinct remainders because number 7 can generate only 7 remainders, so two +ve integers must have same remainder. 5 . 1 3. EXTENDED PIGEONHOLE PRINCIPLE
It states that if n pigeons are assigned to m pigeonholes (The number of pigeons is very large than the number of pigeonholes), then one of the pigeonholes must contain at least [(n  1)/mJ + 1 pigeons.
DISCRETE STRUCTURES
1 56
Theorem XI. Prove extended pigeonhole principle. Proof. We can prove this by the method of contradiction. Assume that each pigeon·
hole does not contain more than [(n  1)1 m] pigeons. Then, there will be at most m[(n  1)/m] '" m(n  1)/m = n  1 pigeons in all. This is in contradiction to our assumptions. Hence, for given m pigeonholes, one of these must contain at least [(n  1)/m] + 1 pigeons.
Example 9. Show that if 9 colours are used to paint 100 houses, at least 12 houses will be of the same colour. Sol. Let us assume the colours be the pigeonholes and the houses the pigeons. Now 100 pigeons are to be assigned to 9 pigeonholes. Using the extended pigeonhole principle, [(n  1)/m] + 1, where n = 100 and m = 9, we have [(100  1)/9] + 1 = 12. Thus, there are 12 houses of the same colour. ILLUSTRATIVE EXAM PLES
Example 1. How many different Bbit strings are there that end with 0111 ? Sol. An 8bit strings that end with 0 1 l l can be constructed in 4 steps i.e., ways.
By selecting 1st bit, IInd bit, IIIrd bit and IVth bit and each bit can be selected in 2 Hence, the total no. of 8bit strings that end with O l l l is = 2 . 2 . 2 . 2 = 2' .
Example 2. How many 2digits numbers greater than 40 can be formed by using the digits 1, 2, 3, 4, 6, 7 (a) When repetition is allowed (b) When repetition is not allowed. Sol. (a) When repetition is allowed We have to find the numbers greater than 40. Therefore, Ten's place can be filled up by 3 ways. Unit' s place can be filled up by 6 ways. . . The total number of 2digits numbers greater than 40 is = 3 x 6 = 18. (b) When repetition is not allowed Ten's place can be filled up by 3 ways. Unit' s place can be filled up by 6 ways. . . The total number of 2digits numbers greater than 40 is = 3 x 6 = 16.
Example 3. How many words can be constructed of three English alphabets ? (a) When repetition of alphabets is allowed (b) When repetition is not allowed.
Sol. There are 26 alphabets in English. Therefore,
(a) When repetition is allowed
First alphabet of word can be selected in 26 ways. Second alphabet of word can be selected in 26 ways. Third alphabet of word can be selected in 26 ways. Hence, total number of words of three alphabets constructed is = 26 x 26 x 26 = 17676. (b) When repetition is not allowed First alphabet of word can be selected in 26 ways.
PERMUTATIONS AND COMBINATIONS
1 57
Second alphabet of word can be selected in 25 ways. Third alphabet of word can be selected in 24 ways. Hence, the total number of words of three distinct alphabets is = 26 x 25 x 24 = 15600.
Example 4. Show that 0 ! = 1. Sol. We have n! np, = ;_...,, (n  r) !
Now, put r = n in equation (1), we have Hence 0 I = 1.
np = n
n !, (n  n) !
_ _
(1)
n!=�
O!
=>
nl 0 1 = ' = 1 n!
Example 5. There are n objects out of which r objects are to be arranged. Find the total number ofpermutations when (a) four particular objects always occur. (b) four particular objects never occur. Sol. (a) Number of ways to arrange first object = r Number of ways to arrange second object = r  1 Number of ways to arrange third object = r  2 Number of ways to arrange fourth object = r  3 Number of ways to arrange remaining n  4 objects taking r  4 at a time = n  4 p , 4 ' 
Therefore, the total number of permutation when four particular objects always occur is [Using first principle of counting] = r(r  l)(r  2)(r  3) n  4 p , _ 4 ' (b) There are four particular objects which never occur in any arrangement. Hence set aside these four particular objects. Thus, we have to find the number of arrangements of n  4 objects taking r at a time. . . The total number of arrangements is = n  4 p,.
Example 6. How many permutations can be made out ofthe letters ofthe word "Basic" ? How many of these (ii) end with C ? (i) begin with B ? (iii) B and C occupy the end places ? Sol. There are 5 letters in the word 'Basic' and all are distinct. The number of permutations of these letters = 5 I = 5 x 4 x 3 x 2 x 1 = 12 0. (i) Permutations which begin with B The first position can be filled in only one way i.e., B and the remaining 4 letters can be arranged in 4 I ways. . . Total number of permutations starting with B = 1 x 4 I = 24. (ii) Permutations which end with C The first position can be filled in only one way i.e., C and the remaining 4 letters can be arranged in 4 I ways. . . Total number of permutations ending with C = 4 I x 1 = 24.
DISCRETE STRUCTURES
1 58
(iii) Permutations in which B and C occupy end places B and C occupy end positions in 2 I ways i.e., B, C and C, B and the remaining 3 letters can be arranged in 3 I ways. . . Total number of permutations in which B and C occupy end places in = 2 I x 3 I = 12. Example 7. Show that nc, + nC, _ l = n + l C" where n :> r :> 1 and n and r are natural (P.T.V. B.Tech. Dec. 2010) numbers. Sol. Take L.R.S. of equation i.e.,
n! n! n c + n c, = , , c:;  1 r ! x (n r) ! + ;(r  1) ! x (n  r + C""7 I) ! 

n! n! + r(r  I) ! x (n  r) ! (r  I) ! x (n  r + I)(n  r) ! n ! x (n  r + 1 + r) n ! x (n  r + 1) + n ! x r = ������� = r ! x (n  r + I) ! r(r  I) ! x (n  r) ! x (n  r + 1) n ! x (n + 1) = = n + ' C' r ! x (n  r + I) ! =

Renee proved.
Example 8. In the 'Discrete Structures Paper' there are 8 questions. In how many ways can an examiner select five questions in all if first question is compulsory.
Sol. Since the first question is compulsory, the examiner has to select 4 questions from the remaining 7 questions. Therefore, the number of ways to select 5 questions = 7 C, =
7! 7x6x5x4! = 35 = 4 ! x (7  4) ! 4 ! x 3 x 2 x I '
Example 9. Determine the number of triangles that are formed by selecting points from a set of 15 points out of which 8 are collinear. Sol. When we take all the 15 points, the number of triangles formed is 15C3 . As 8 points lie on the same line, they do not form any triangle. Thus, sC3 triangles are lost.
The total number of triangles produced is
15 x 14 x 13 x I2 ! 8! = 3 !(8  3) ! 3 x I2 ! I5 x I4 x I3 8 x 7 x 6 = = 910  56 = 854. 3 3x2x 1
I5 ! 15 C  8C _ 3 3 3 x (15  3) ! 
8x7x6x5! 3x2x Ix5!
Example 10. How many lines can be drawn through 10 points on a circle ? Sol. As all the points on the circle are not collinear. Thus, no lines will lost. . . The total number of lines drawn through a circle = lO C, IO ! 10 x 9 x 8 ! = = = 45. 2 ! x (10  2) ! 2 x 1x 8 ! Example 11. Determine the number ofdiagonals that can be drawn byjoining the nodes
of octagon.
Sol. The number of lines that can be formed by joining 2 out of 8 points = s c, 8x7 = 28 = 2 
PERMUTATIONS AND COMBINATIONS
1 59
Out of these 28 lines, the 8 are sides of the octagon. . . The number of diagonals = 28  8 = 20.
m�ne
Example 12. In a shipment, there are 40 floppy disks of which 5 are defective. Deter
(a) in how many ways we can select five floppy disks ? (b) in how many ways we can select five nondefective floppy disks ? (e) in how many ways we can select five floppy disks containing exactly three defective floppy disks ? (d) in how many ways we can select five floppy disks containing at least 1 defective floppy disks ? Sol. (a) There are 40 floppy disks out of which we have to select 5 floppy disks. These can be done in 4 0 C5 ways i. e. J
=
40 x 39 x 38 x 37 x 36 x 35 ! 40 ! = 658008. = 5 ! (40  5) ! 5 ! x 35 !
(b) There are 40  5 = 35 nondefective floppy disks out of which we have to select 5. This can be done in 3 5C5 ways. =
35 ! 35 x 34 x 33 x 32 x 31 x 30 ! = = 324632. 5 ! (35  5) ! 5 x 4 x 3 x 2 x 1 x 30 !
(c) To select exactly three defective floppy disks out of total 5 we have 5 C3 ways and the remaining 2 floppy disks can be selected in 3 5 C, ways. Therefore, the total number of ways to select 5 floppy disks out of which exactly 3 are defective = 5 C3 x 3 5 C,
5! 5 x 4 x 3 ! 35 x 34 x 33 ! 35 ! x :,: :c: x = :: ! 3 ! (5  3) ! 2 ! x (35  2) ! 3 ! x 2 x 1 2 x 1 x  33:
= ':,,

= 5950. (d) There are five defective floppy disks out of which at least 1 must be selected. We know that the total number of ways to select 5 floppy disks out of total 40 disks = ' D C5 . Also, the number of ways to select 5 floppy disks with number one defective = 3 5 C5 ·
Therefore, the total number of ways to select 5 floppy disks out of which at least one is defective
= ' D C5  3 5 C5 = 611625. Example 13. If a finite set A has n elements. Then the power set of A i.e., P(A) has 2n elements.
Sol. By defination, p eA) is the set of subsets of A.
There are n c, subsets each consisting of one of the n elements of the given set. There are n c, subsets each consisting of any two of the n elements of the given set
Similarly, there are n C3 subsets each consisting of any three of the n elements of the given set There are n Cn subsets each consisting of all the n elements of the given set. Also, there will be one set {q,}
DISCRETE STRUCTURES
1 60 Total number of subsets = (n c , + ... + n cn> + 1 = n c o + n C + ... + n Cn 1
itself.
Remark.
=
(1 + l) n = 2n .
Number of proper subsets of A
=
2n

1, There is only one improper subset i.e.) the set
Example 14. (a) How many subsets of {I, 2, 3, ... 10j contain at least 7 elements ?
(P.T.U. B Tech Dec. 2005)
(b) If T = {I, 2, 3, 4, 5}. How many subsets of T have less than 4 elements ? (c) A set contains (2n + 1) elements. If the number of subsets of this set which contains at (P.T.U. B. Tech. December 2008) most n elements is 8192, find n. ( 2n + ' Co + 2n + ' C, + 2n + ' C, + ... + 2n + ' Cn = 8192 I n co + n c, + n c, + ... + n Cn = 2 n => 2 2n +1 = 8192 1 = 2 3 2n + 1 = 13 => 2n = 12 => n = 6 => ( = s,  s, + S3 ... + ( l) m1 sm
For 4, we have 8 neA,) + n(A.,), + n(A3) + n(A4) 8 neA, + n(A �) + n(As n A4) + n(A4 n A,) + neA, n �) + n(A n A4) 832 neA, A." �) +2 n(A." n � n A4) + n(� n A4 n A,) + neA, n A." n A24) 84 neA, A,,) A set of the form At n Ai As* A:; where each At is either A or �c, is called a fundamental product of the sets A/s. Remark 3. Consider three sets Ai B, C. Let P = A n B n Co, P , = A n B n C, 2 P 4 = A Be ee, P s = A n BC n C, Remark 1. m= = I = n A.,,) n n = n = n A." n A." n Remark 2.
II
II
P 5 = Ac n B n C, P 7 = k n � n C,
"
"
II
II
P 6 = Ac n B n Cc, P8 = k n � n �
These eight fundamental products correspond to the eight disjoint regions as shown in the Venn diagrams of sets A, B, C. (see Fig. 6.2)
I
ILLUSTRATIVE EXAMPLES
Example 1. Out of 1200 students at a college
582 took Economics 627 took English 543 took Mathematics
Fig. 6.2
INCLUSIONEXCLUSION PRINCIPLE 21 7 took both Economics and English 307 took both Economics and Mathematics 250 took both Mathematics and English 222 took all three courses. How many took none of the three ?
1 69
Sol. Let A, B, C denote the set of students studying Economics, English, Mathematics respectively. Given I A I = 582 I B I = 627 I A n B I = 2 17 I C I = 543 I B n C I = 250 I A n C I = 307 I A n B n C I = 222 The total number of students who took any of three subjects I Au B u C I = I A I + I B I + I C I  I AnB I  I B n C I  I C nA I + I A nBnC I = 582 + 627 + 543  217  307  250 + 222 = 1200 Students who took none of three subjects = (total students in the college)  (total students who took any of three subjects) = 1200  1200 = O. Example 2. 40 computer programmers interviewed for ajob. 25 knew JAVA, 28 knew ORACLE, and 7 knew neither of language. How many knew both languages ? Sol. Now,
I J I = 25 1 0 I = 28 I J u 0 I = 40  7 = 33 Computer programmers who knew both languages are I J n 0 I = I J I + I 0 I  I J u 0 I = 25 + 28  33 = 20.
Example 3. A survey of 550 television watchers produced the following information :
285 watch football games 195 watch hockey games 115 watch baseball games 45 watch football and baseball games 70 watch football and hockey games 50 watch hockey and baseball games 100 do not watch any of the three games. (a) How many people in the survey watch all three games ? (b) How many people watch exactly one of the three games ? Sol. (a) F, H, B denote the sets of watchers watching football, hockey, baseball
respectively. Given
I F I = 285 ; I H I = 195 ; I B I = 1 15 I F n B I = 45 ; I F n H I = 70 ; I H n B I = 50 I F u H u B I = 550  100 = 450 The number of people watch all three games I F n H n B I = 450  285  195  1 15 + 45 + 70 + 50 = 20.
DISCRETE STRUCTURES
1 70 (b) 20 watch all three games.
45  20 = 25 watch football and baseball but not all three. 70  20 = 50 watch football and hockey but not all three. 50  20 = 30 watch hockey and baseball but not all three. 285  25  50  20 = 190 watch only football. 195  50  30  20 = 95 watch only hockey. 115  25  30  20 = 40 watch only baseball. Number of people exactly watch one of the three games = 190 + 95 + 40 = 325. (b) Alternative. To find the number of people watch· ing, exactly one of the three games, we use Venn·diagrams (see Fig. 6.3). Required number of people watching exactly one of the three games = 190 + 95 + 40 = 325.
Fig. 6.3
Example 4. Among 100 students, 32 study Mathematics, 20 study Physics, 45 study Biology, 15 study Mathematics and Biology, 7 study Mathematics and Physics, 10 study Physics and Biology and 30 do not study any of three subjects. (a) Find the number of students studying all three subjects. (b) Find the number of students studying exactly one of the three subjects. Sol. (a) Let M, P, B denote the sets of students studying Mathematics, Physics and Biology. Given
n(M) n(MnB)
n(M n B n P) Now,
n(M n B n P)
=
=
=
=
32, n(P) = 20, n(B) = 45 15, n(MnP) = 7, n(PnB)
30
=
10
100  n(M u B u P)
100  n(M n B n P) = 100  30 = 70 . . Required number of students studying all the three subjects is given by n(M n P n B) Using inclusion€xclusion principle, we have n(M n P n B) = n(M) + n(P) + nCB)  n(M n P)  n(P n B)  nCB n M) + n(M n P n B) 70 = 32 + 20 + 45  7  10  15 + n(M n P n B) 70 = 65 + n(M n P n B) ,, n(M n P n B) = 70  65 = 5 => (b) To find the number of students studying exactly one of the subjects we use Venn·diagram (see Fig. 6.4) 5 study all three subjects 7  5 = 2 study Maths and Physics but not all three 15  5 = 10 study Maths and Biology but not all three =
Fig. 6.4
INCLUSIONEXCLUSION PRINCIPLE
1 71
10  5 = 5 study Biology and Physics but not all three 32  (10 + 2 + 5) = 15 study Maths only 20  (2 + 5 + 5) = 8 study Physics only 45  (10 + 5 + 5) = 25 study only Biology only Number of students studying exactly one of three subjects = 15 + 8 + 25 = 48. Example 5. In a survey of 300 students,
64 had taken a Mathematics course 94 had taken a English course 58 had taken a Computer course 28 had taken both a Mathematics and a Computer course 26 had taken both a English and a Mathematics course 22 had taken both a English and a Computer course 14 had taken all three courses. (a) How many students were surveyed who had taken non of the three courses ? (b) How many had taken only a Computer course ?
Sol. Let M, E, C denote the sets of students taking mathematics, English, Computer Courses respectively. Then I M I = 64 ; I E I = 94 ; I C I = 58 I M n C I = 28 ; I M n E I = 26 ; l E n C I = 22 1 M n E n C I = 14 (a) I MuEu C I = I M I + I E I + I C I  I Mn C I  I MnE I  1 EnC I + I MnEnC I = 64 + 94 + 58  28  26  22 + 14 = 154 Students who had taken none of the courses = 300  154 = 146. (b) 14 had taken all three courses. 28  14 = 14 had taken both a Mathematics and a Computer but not all three 22  14 = 8 had taken both a English and a Computer courses but not all three 58  14  8  14 = 22 had taken only Computer course. Example 6. Let A, B, C, D denote respectively, Art, Biology, Chemistry and Drama
courses. Find the number of students in a dormetory given the data. 3 Take A, B, C 12 Take A, 5 Take A and B 2 Take A, B, D 20 Take B, 7 Take A and C 2 Take B, C, D 20 Take C, 4 Take A and D 3 Take A, C, D 8 Take D, 1 6 Take B and C 4 Take B and D 2 Take A, B, C, D 3 Take C and D 71 Take none.
Sol. We first find the number of students who take at least one course. By InclusionExclusion principle. T = n(A u B u C u D) = S l  S2 + S 3  S4 (1) General Exclusion Inclusion Principle where s, = n(A) + n(B) + n(C) + n(D) = 12 + 20 + 20 + 8 = 60
DISCRETE STRUCTURES
1 72 S2 = n (A n B) +
n(B n C) + n(C n D) + n(D n A) + n(A n C) + n(B n D) = 5 + 16 + 3 + 4 + 7 + 4 = 39 S3 = n(A n B n C) + n(B n C n D) + n(A n C n D) + n(A n B n D) = 3 + 2 + 3 + 2 = 10 s, = n (A n B n C n D) = 2 From (1) T = 60  39 + 10  2 = 70  41 = 29 Hence the required number of students = Number of students who take at least one course + Number of students who take none course = 29 + 71 = 100. Example 7. Suppose that 100 of the 120 Mathematics students at a college take at least
one of the languages French, German and Russian. Also suppose 20 study French and German 65 study French, 45 study German, 25 study French and Russian 15 study German and Russian 42 study Russian, (a) Find the number of students studying all the subjects (b) Find the number of students studying taking exactly one subject. Sol. (a) Let F, G and R denote the sets of students studying French, German, Russian
respectively. Given n(F u G u R) = 100, n(F) = 65, n(G) = 45, n(R) = 42, n(F n G) = 20, n(F n R) = 25, n (G n R) = 15, we find n(F n G n R). Using InclusionExclusion principle, n(F u G u R) = n(F) + n(G) + n(R)  n(F n G)  n(G n R)  n(R n F) + n(F n G n R) => 100 = 65 + 45 + 42  20  15  25 + n(F n G n R) => 100 = 92 + n(F n G n R) => n(F n G n R) = 100  92 = 8. (b) To find the number of students taking exactly one subject, we use Venn·diagrams. (see Fig. 6.5) 8 read all three subject 25  8 = 17 read French and Russian but not German 20  8 = 12 read French and German but not Russian 15  8 = 7 read Russian and German but not French 65  (17 + 8 + 12) = 28 read French only G 45  (12 + 8 + 7) = 18 read German only 42  (17 + 8 + 7) = 10 read Russian only Required number of students who study exactly one subject = 28 + 18 + 10 = 56. Example 8. In a survey of 60 people, it was found that
Fig. 6.5
25 read Newsweek magazine 26 read Time, 26 read Fortune
INCLUSIONEXCLUSION PRINCIPLE
1 73
9 read both Newsweek and Fortune 11 read both News week and Time 8 read both Time and Fortune 3 read all three magazines (a) Find the number of students who read at least one of the three magazines (b) Find the number of students who read exactly one magazine (c) Find the number of students who read no magazine at all. Sol. Let N, F, T denote the sets of students reading ,, Newsweek, Fortune, Time magazines respectively. (a) Required number of students = n(N u T u F) = n(N) + n(T) + n(F)  n(N n T)  n(T n F)  n(F n N) + n(N n T n F) = 25 + 26 + 26  1 1  8  9 + 3 = 52. (b) To find the number of students who read exactly one magazine, we use Venndiagram. (see Fig. 6.6). Here 3 read all three magazines 11  3 = 8 read Newsweek and Time but not all three magazines 9  3 = 6 read Newsweek and Fortune but not all three magazines 8  3 = 5 read Time and Fortune but not all three magazines 25  ( 8 + 3 + 6) = 8 read only Newsweek 26  (8 + 3 + 5) = 10 read only Time 26  (6 + 3 + 5) = 12 read only Fortune Required number of students who read exactly one magazine = 8 + 10 + 12 = 30. (c) Required number of students who study no magazine = 60  n(N u T u F) = 60  52 = 8.
Fig. 6.6
Example 9. Among the first 500 positive integers :
(a) Determine the integers which are not divisible by 2, nor by 3, nor by 5. (b) Determine the integers which are exactly divisible by one of them.
Sol. Let A denotes the set of numbers of integers divisible by 2 B denotes the set of numbers of integers divisible by 3 C denotes the set of numbers of integers divisible by 5. I A I = I AnB I =
[�] [ ] [ ] 5 0
= 250 ;
500 = 83 ' 2x3

[ ] [ ] [ ]
500 I B I =  = 166 ; 3 I An C I =
500 = 50 2x5

500 500 = 33 ; = 16. I AnBn C I = 3x3x5 3x5 (a) I A u B u C I = 250 + 166 + 50  83  100  33 + 16 = 366 The integers not divisible by 2, 3 and 5 = 500  366 = 134. I BnC I =
I C I =
[�] 5 0
= 100
DISCRETE STRUCTURES
1 74 (b) The integers divisible by all the three = 16
83  16 = 67 integers are divisible by 2 and 3 but not all the three 50  16 = 34 integers are divisible by 2 and 5 but not by all the three 33  16 = 17 integers are divisible by 3 and 5 but not by all the three 250  67  34  16 = 133 integers are only divisible by 2 166  67  17  16 = 66 integers are only divisible by 3 100  34  17  16 = 33 integers are only divisible by 5 Total number of integers only divisible by 2, 3 and 5 = 133 + 33 + 66 = 232. lO.
Among the first 1000 positive integers : (a) Determine the integers which are not divisible by 5, nor by 7, nor by 9. (b) Determine the integers divisible by 5, but not by 7, not by 9.
Example
(P.T.U. B.Tech. May 2012) Sol. Let A denotes the set of numbers of integers divisible by 5 B denotes the set of number of integers divisible by 7 C denotes the set of number of integers divisible by 9. 1O 0 1000 So = 200 ; = 142 I A1 = I B I = 7 1000 1000  = 1 11 ; I AnB I = I C I = = 28 9 5x7
[ �] [ ] [ ] [ ] 
I An C I =
1000 = 22 ' 5x9
I BnC I =
[ ] [ ] [ ]
1000 = 15 7x9
1000 = 3. 5x7x9 (a) The number of integers divisible by 5, 7 and 9 I A u B u C I = 200 + 142 + 1 1 1  28  22  15 + 3 = 39l. The number of integers not divisible by 5, nor by 7, nor by 9 = Total number of integers  number of integers divisible by 5, 7 and 9 = 1000  391 = 609. (b) The integers divisible by all the three integers = 3 28  3 = 25 integers divisible by 5 and 7 but not by all the three 22  3 = 19 integers divisible by 5 and 9 but not by all the three 200  25  19  3 = 153 integers divisible by 5 but not by 7, not by 9. I An B n C I =
Example 11. How many integers between 1 and 300 (inclusive) are : (P.T.U. B.Tech. May 2010)
(a) divisible by at least one of 3, 5, 7 ? (b) divisible by 3 and 5, not by 7 ? (c) divisbile by 5 but neither by 3 nor by 7 ?
INCLUSIONEXCLUSION PRINCIPLE
1 75
Sol. Total number of integers = 300 Let A denotes the set of numbers divisible by 3 B, the set of numbers divisible by 5, C, the set of numbers divisible by 7, then,
[ 300] [300] 7[33x005 ] [3300x 7 ] [3 x3005 x 7 ]
neAl = 3 = 100 , nCB) = n(C) = n(A n B) = n(C n A) = n(A n B n C) =
[300] 5 60 =
= 42
= 20 , nCB n C) =

[ 5300x 7 ]
=8
= 14 =2
Also, n(A u B u C) = neAl + nCB) + n(C)  n(A n B)  nCB n C)  n(C n A) + n(A n B n C) = 100 + 60 + 42  20  8  14 + 2 = 162 From the Venn diagram, (a) number of integers divisible by at least 3, 5 or 7 = n(A u B u C) = 162
(b) the number of integers divisible by 3 and 5, not by 7 = n(A n B n C) = 18 (e) the number of integers divisible by 5 but not by 3, 7 = n(A n B n C ) = 28. TEST YOUR KNOWLEDGE 1.
I
(a) bottles. Of 32 people whonumber save paper or bottles (or both) for recycling, 30 save paper and 14 save Find the of people who save (i) Both paper and bottles (ii) Only paper (iii) Only bottles (b) A survey of 550 Television watchers produced the following information. 285 watch football games 195 watch hockey games 115 watch cricket 45 watch football and cricket 70 watch football and hockey 50 watch hockey and cricket 100 do not watch any of the three games. (i) How many people in the survey watch all the three games ? (ii) How many people watch exactly one of the three games ?
DISCRETE STRUCTURES
1 76 2.
3.
4.
In a class of 300 students, 60 study Mathematics course 94 study English course 58 study Discrete course 28 study both Mathematics and Discrete course 26 study both English and Mathematics course 22 study both English and Discrete course 14 study all three course, (a) Find the number of students who study none of the three courses (b) Who study only Discrete course ? A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which ofinstalled. three popular options, The survey foundairconditioning (A), radio (R) and Power windows 0N), were already 15 had airconditioning 12 had radio 11 had Power windows 5 had airconditioning and Power windows 9 had airconditioning and radio 4 had Power windows and radio 3 had all three options, Find the number of cars that had (a) Only Power windows (b) Only airconditioning Cc) Only radio (d) Radio and Power windows but not airconditioning (e) Airconditioning and radio but not Power windows if) Only one of the options (g) At least one option (h) None of the options Using InclusionExclusion principle, prove the following, Ci) nCA B) nCA) + nCB) nCA n B), for any finite sets A and B Cii) nCA B) nCA) + nCB), where A, B are finite disjoint sets. Aandsurvey wasZee conducted 550 like TV. Also,among 1000 people ofthese 595 like Metro channel, 595 like star Movies 395 of them like Metro channel and Star Movies 350 of them like Metro channel and Zee TV 400 of them like Star Movies and Zee TV 250 of them like Metro channel, Star Movies and Zee TV, Find (a) How many of them who not like Metro channel, do not like Star Movies and do not like Zee TV? (b) How many of them who like Metro channel, do not like Star Movies and do not like Zee ? Ingames a class of 60 boys, 45 boys play cards and 30 boys play carrom, How many boys play both ? How many play cards only and how many plays carroms only ? CPT. U. B.Tech. May 2008, 2010) u
u
5.
�
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do
6.
TV
INCLUSIONEXCLUSION PRINCIPLE 7.
1 77
Given that U all students in a university A Day students B mathematics majors C graduate students Also n(U)who16,are000,mathematics A 9000, nCB) 300 and n(G) 1000, Assume that the number of day students 250, 50 of which are graduate students, number of day graduate studentsmajors is 700.isDetermine the number of students who areand the total (a) evening students (b) nonmathematics majors (c) under graduate (day or evening) (d) day graduate nonmathematics majors evening graduate students if) evening graduate mathematics majors (g) evening under graduate nonmathematics majors. In a class of 80 students, 50 students know English, 55 know French and 46 know German. 37 students EnglishFind and French, 28 students know French and German, 7 students know none of theknow languages. (a How many students know all the three languages ? (b) How many students know exactly two languages ? (c How many know only one language ? Among integers 1 to 1000 a How many of them are not divisible by 3 nor by 5 nor by 7 ? (b) How many are not divisible by 5 or 7 but divisible by 3 ? It is known that in a university, 60% of teachers play tennis, 50% play bridge, 70% jog, 20% play tennis andjogbridge, play bridge and jogJustify and 30% teachers and play40%tennis and bridge. this play claim.tennis and jog. It is claimed that 20% =
=
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=
=
n(
) �
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(e)
8.
)
9.
10.
)
( )
Answers
(iii) 2 (ii) 18 (b) (,) 450 a 5 (b) 22 (b) 4 6 (/) 11 (g) 23 15, 30, 15, (b) 100 c 15000 (b) 15700 (d) 650 (g) cannot be determined with the given information. (b) 54 (c) 7 The claim is not true. (b) 229
(,) 12 146 (d) 4 a 155 a 7000 and a 12 a 457
L (a) 2, 5,
7, 8,
3, ( )
(a)
(e)
( )
6,
( )
()
(/)
( )
9,
( )
L
(a)
10.
� �
� �
Hints � �
(ii) n(P B) n(P)  n(P n B) 30  12 18 (ii,) nCB  P) nCB) n(P n B) 14  12 2
(ii) 325 c2 (h) 2. 300 ()
(e)
DISCRETE STRUCTURES
1 78 (b)
3.
20 watch all the three games.
Fig. 6.7 Number of people watch exactly one of the three games 190 + 95 + 40 325. =
=
2 5.
Fig. 6.8 (a) Number of People who like at least one of the channels n (M u S u Z) n(M) + n(S) + n(Z)  n(M n S)  n(S n Z)  n(Z n M) + n(M n S n Z) 595 + 595 + 550  395  350  400 + 250 845 Required number People 155. who do not like Metro channel, do not like Star Movies, and do ..not like Zee TV 1000of845 Given n(A) 50, nCB) 55, n(G) 46, n(A n B) 37, nCB n G) 28, n(A n G) 25. n(A u B u C) number of students knowing none of the language 7 n (u) n(A u B u G) 7 80 n(A u B u G) 7 n(A u B u G) 73 (a Number of students knowing all the three languages is given by n(A B C) Now, n(A u B u G) n(A) + nCB) + n(G)  n(An B) nCB n G)  n(G n A) + n(An B n G) n(A B C) 12. Hence, there are 12 students who know all the three languages. (b) Using Venn diagram, students knowing exactly two languages �
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8.
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=
)
�
�
�
�
�
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=
�
=:::}
II
�
�
II
�
II
II
=
n(A n B) + n (B n C) + n(A n C)  3 X n (A n B n C) 37 + 28 + 25  3
X
12
�
54.
A
B
(c) language Using Venn0 +diagram, 2 + 5 number of students knowing only one =
9.
=
7.
Let A The set of numbers between 1 to 1000 that are divisible by 3 B The set of numbers between 1 to 1000 that are divisible by 5 C The set of numbers between 1 to 1000 that are divisible by 7. =
=
=
Fig. 6.9
INCLUSIONEXCLUSION PRINCIPLE
· 1000J · 1000J n(A) � [ 1000J 3 � 333, nCB) � [ 5 � 200, n(G) � [ 7 � 142 

[1000] ] n(A n B n G) � [ 3 1000 x 5 x 7 � 9, n(A n B) � 3 x 5 � 66

[1000] n (B n G) � [1000] 5 x7 �28, n(A n G) � 3x 7 � 47 (a) The number of integers that are not divisible by 3 nor by 5 nor by 7 
�

n(A n B n C)
�
n(A u B u C)
�
U  n(A u B u C)
But, n(A u B u G) � 333 + 200 + 142  66  47  28 + 9 � 543 :. From (1), n(A n B n � 1000  543 � 457 (b) Number of integers that are not divisible by 3 but not by 5 nor by 7 n(A n B n C) n(A n B u C ) n(A  B u C ) � 333  (57 + 9 + 38) � 229. Let the total number of teachers 100 n(T) � 60, nCB) � 50, n(J) � 70 n(T n � �20, n� n J) � ®, n(T n J) � OO Number of teachers who jog and play tennis and bridge is given by n(T u B u J) � 60 + 50 + 70  20  40  30 + n(T n B n J) n(T n B n J) � 100 (90) � 10 10% of Teachers jog and play tennis and bridge. C)
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10.
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=
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Le.,
1 79
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(1)
7
RECURRENCE RELATIONS AN D GEN ERATING FUNCTIONS
7. 1 . INTRODUCTION
In this chapter) we will discuss the formation of recurrence relations and their solutions. The closed form expressions of recurrence relations and its solutions using generating functions is also discussed in this chapter. 7.2. RECURRENCE RELATIONS
(P.T.U. B.Tech. Dec.
2006)
Let S be a sequence of numbers. A recurrence relation on S is a formula that relates all, but a finite number of terms of S, to previous terms of S. For e.g the Fibonacci sequence is defined by the relation. FK = FK_2 + FK_l " K :> 2. where Fo = 1 . F, = 1 . The relation defined above is a recurrence relation and the conditions F0 = 1, F = 1 are called initial conditions. .•
1
Note.
The recurrence relations are also called difference equations.
7.3. ORDER OF A RECURRENCE RELATION
The order of a recurrence relation is the difference between the highest and the lowest subscripts of S(K). For e.g., consider the recurrence relation. FK = FK_2 + FK_l " K :> 2 Here the difference between highest and lowest subscripts of F = K  (K  2) = 2 Its order is 2. Note.
The order of a recurrence relation may or may not be defined
7.4. (a) DEGREE OF RECURRENCE RELATION
It is the highest power of S(K) occurring in the recurrence relation. For example. consider the recurrence relation. S 2 (K + 3) + 2S 2 (K + 2) + 2S(K + 1) = 0 Its order = (K + 3)  (K + 1) = 2 degree = 2(Highest power) 1 80
RECURRENCE RELATIONS AND GENERATING FUNCTIONS For example, consider S4 (K)
Its
181
+ 3S3(K  1) + 6S2(K  2) + 4S(K  3) = 0
order = K  (K  3) = 3 degree = 4.
7.4. (b) LINEAR RECURRENCE RELATION
(P.T. U. B.Tech. Dec. 2006)
A recurrence relation with degree one is called linear recurrence relation. 7.5. FORMATION OF RECURRENCE RELATIONS
We illustrate this concept in the following examples.
I
ILLUSTRATIVE EXAMPLES
Example 1. Find the order of the following recurrence relations defined by (i) T(K) = 2(T(K  1))"  KT(K  3) (ii) P(K) + 2P(K  3)  K2 = 0, (iii) S(K) = S[KI2] + 5, K ? 0 (iv) A(K)  2A(K  1)  2K = O. Sol. (i) The difference between the highest and lowest subscripts = K  (K  3) = 3
Its order = 3. (ii) The difference between the highest and lowest subscripts = K  (K  3) = 3 Its order = 3. (iii) The recurrence relation is of infinite order. Since K
 [�] becomes larger and
larger as K is made large. (iv) The difference between the highest and the lowest subscripts = K  (K  1) = 1 Its order = 1.
Example 2. Obtain the linear recurrence relation from the sequence defined by
S(K) = 5 . 2K
S(K) = 5 . 2K Changing K to K  1, we get S(K  1) = 5 . 2Kl Subtracting (1) and (2), we get S(K)  S(K  1) = 5 . 2K  5 . 2Kl
Sol. Given
... (1) ... (2)
( ) �
= 5 . 2K 1 � = . 5 . 2K = S(K) 2 2 2
=> =>
(1 �) S(K)  S(K  1) = 0
S(K)  2S(K  1) = 0, is the required linear recurrence relation.
Example 3. Obtain the recurrence relation of second order from the sequence defined by S(K) = 3K1 + 2K+ 1 + K.
DISCRETE STRUCTURES
1 82
Sol. Given S(K) = 3 K1 + 2 K+l + K Changing K to K  1 in (1), we get S(K  1)
= 3 K2 + 2 K + K  1
1
=3
3 K1
+
1
 2K+l
+
K 1
... (2)
= ..!c . 3 K1 + ..!c . 2 K +l + K _ 2 4 9
... (3)
'
Changing K to K  2 in (1), we get S(K  2)
... (1)
2 '
= 3 K3 + 2 K1 + K  2
Eliminating 3K1 from (1) and (2), we get S(K)  3S(K  1)
=
(1%)
+
2K+l
= _ ..!c . 2K+l 2
+
K  3(K  1) ... (4)
3  2K
Eliminating 3K1 from (2) and (3), we get S(K  1)  3S(K  2)
=
Gf)
2K+l
+
K  1  3(K  2)
= _ ..!c . 2K + 1 + 5  2K
... (5)
4
Eliminating 2K +l from (4) and (5), we get
S(K)  3S(K  1)  2 [S(K  1)  3 S(K  2)]
=  7 + 2K 6S(K  2) = 2K  7,
= 3  2K  2
( 5  2K)
or S(K)  5 S(K  1) + is the required recurrence relation. Its order = K  (K  2) = 2. Example 4. Obtain the recurrence relation of second order from the sequence given by S(K) = 2 . 4K  5 . ( 3jK S(K) = 2 . 4K  5 . ( 3)K
Sol. Given Changing K to K  1 in (1), we get S(K  1)
= 2.4K1  5 . =
1
_
2 '
( 3) K1
... (1)
=�. 4
4K  5. ( 3)K . ( 3)1
5 ( _ 3)K
4K + 3 '
... (2)
Changing K to K  2 in (1), we get S(K  2)
= 2 . 4K2  5 . ( 3) K2 = 2 . 4K . 42  5 . ( 3) K . ( 3)2
.!c = . . 4 K _ � . ( _ 3) K 9 8
... (3)
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
1 83
Eliminating 4K from (1) and (2) , we get S(K)  4S(K _ 1)
230 . ( we get
= _ 5 . (_ 3) K _
Eliminating 4K from (2) and (3),
_ 3) K = _
335 . (
_ 3) K
... (4)
S(K  1)  4S(K  2)
... (5)
Eliminating ( 3)K from (4) and (5), we get S(K)  4S(K  1)
or
+ 3[S(K  1)  4 S(K  2)] = 0
S(K)  S(K  1)  12S(K  2)
= 0,
is the required recurrence relation of order = 2.
7.6. LINEAR RECURRENCE RELATION OF ORDER n WITH CONSTANT COEFFICIENTS
The general linear recurrence relation of order n with constant coefficients, is given by where
S(K)
+ C,S(K  1) + C2S(K  2) + ...... + CnS(K  n) = f(K), K :> n,
Cp C2 ) ...... e n
are constants.
7.7. HOMOGENEOUS LINEAR RECURRENCE RELATION OF ORDER n A homogeneous linear recurrence
relation of order n is an equation of the form
S(K) + C, S(K  1) + C2 S(K  2) + ...... + Cn S(K  n) = O.
7.B. CHARACTERISTIC EQUATION
Consider a linear recurrence relation of order n given by
+ C,S(K  1) + C2S(K  2) + ...... + CnS(K  n) = f(K), K :> n. Then the equation n n n a + C 1 a 1 + C 2a 2 + ...... + Cn_1 a + en = 0 is called characteristic equation. S(K)
The lefthand side of this equation is known as characteristic polynomial.
7.9. ALGORITHM FOR SOLVING HOMOGENEOUS LINEAR RECURRENCE RELA TION OF ORDER n WITH CONSTANT COEFFICIENTS (P. T. U. B. Tech. Dec. 2006)
Consider the linear homogeneous recurrence relation of order n as S(K)
+ C, S(K  1) + C2S(K  2) + ...... + Cn S(K  n) = 0
... (1)
Cp C2 ) ...... en are constants. Step I. Write the characteristic equation of the equation (1), which is n n n a C 1 a 1 C 2a 2 Cn_1 a en = 0 Step II. Find the roots of the characteristic equation obtained in step 1. Case I. If the characteristic equation has n distinct roots, say, mp m2, mn, then the solution of (1) is given by C n mnK S(K) = C, m / C 2 m2K Case II. If the characteristic equation has two equal roots) say m1 = m2) then the solu tion of (1) is given by
where
+
+
+ ... +
+
•••
+
+ ...... +
DISCRETE STRUCTURES
1 84
Case III. If the characteristic equation has three equal roots, say, m1 = m2 = m3) then the solution of (1) is given by S(K) = (C, + C2K + C3K2) m/ + C4m/ + ...... + Cn mnK and so on Case IV. If the characteristic equation has imaginary roots, say, a ± i�, then solution of (1) is given by
S(K) = C, (a + i�)K + C2 (a  i�)K
Case V. If the characteristic equation has repeated imaginary roots, say, a ± i�, a ± i�, then the solution of (1) is given by
S(K) = (C, + C2K) (a + i�)K + (C3 + C4K) (a  i�)K Example 5. Solve the recurrence relation ar  3ar _ 1 + 2ar_ 2 = O. Sol. The characteristic equation is given by
s2  3s + 2 = 0 or (s  l)(s  2) = 0
Therefore, the required solution of the given homogeneous recurrence relation is ar = C 1 + C2 . 2 r .
= 3.
Example 6. (a) Solve the recurrence relation: a,  7a,_ 1 + lOa,_ 2 = 0, given that ao = 0, a1
(P.T.U. B.Tech. Dec. 2007)
(b) Solve the recurrence relation: an = 3an_1 + 4an_2) ao = 0, a1 = 5.
Sol. (a) The characteristic equation is
(P.T.U. B.Tech Dec. 2013)
s2  7s + 10 = 0 or (s  2)(s  6) = 0
Therefore, the required solution of the given homogeneous recurrence relation is ar = c1 . 2r + c2 . 5r• ... (1) Using ao = 0, (1) gives Using a, = 3, (1) gives
0 = c1 + c2
... (2)
3 = 2c, + 6c2
... (3)
Using (2) in (3), we get
3 = 2c,  6c, =  3c, c, =  1 c2 =  c, = 1
From (2) . . From (1), the required solution is
a, = _ 2r + sr. (b) Proceed yourself as in part (a).n n an = 4  (_ l) Ans. Example 7. Solve the following recurrence relations (a) tn = 6tn_1  1 1 tn _ 2 + 6tn _ 3, n :> 3, subject to to = 1, t1 = 5, t2 = 1 5. (b) tn =  3 tn _ 1  3 tn _ 2  tn _ 3, n :> 3, subject to to = 1, t1 = 2, t2 = 1. (c) an = 6an_1  12an_2 + 8an_3, ao = 3, a1 = 4, a2 = 12 (P.T. U. B.Tech Dec. 2013) Sol. (a) Given equation is tn  6tn  1 + 11 tn _ 2  6tn _ 3 = 0 Its order = n  (n  3) = 3
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
1 85
The charateristic equation is
a3  6a2 + 11a  6 = 0 Here, a = 1 satisfies (1). So a = 1 is a root of (1). By Horner's method. 1 1 6 11 6 5 6 1 1
5
The quotient is a2  5a + 6 = 0 =>
(a  3) (a  2) = 0
The required solution is
6
Remainder LiO.:O: 2
=>
a  3a  2a + 6 = 0 ::::::} a = 2) 3
tn = C, + C2 2n + C3 3n Using to = 1 in (2), 1 = C, + C 2 + C 3 5 = C, + 2C2 + 3C3 Using t, = 5 in (2), Using t2 = 15 in (2), 15 = C, + 4C2 + 9C3 (3)  (4) gives  4 =  C2  2C3 (4)  (5) gives  10 =  2C2  6C3 Multiplying (6) by 2,  8 =  2C2  4C3 Subtracting (7) and (8),  2 =  2C3 => C3 = 1 From (7),  10 =  2C 2  6 => C2 = 2 From (3), C, = 1  2  1 =  2 From (2), The required solution is tn = ( 2) + 2(2n) + 3n =  2 + 2n + 1 + 3n. (b) Given equation is tn + 3tn _ 1 + 3tn _ 2 + tn _ 3 = 0 Its order = n  (n  3) = 3 The characteristic equation is a3 + 3a2 + 3a + 1 = 0 => (a + 1)3 = 0 => a =  1,  1,  1 ..
The general solution is
tn = (C, + C2 n + C3 n2) (  l)n Using to = 1 in (1), 1 = C, Using t, =  2 in (1),  2 = (C, + C2 + C3 ) ( 1) or 2 = C, + C2 + C3 Using t2 =  1 in (1),  1 = C, + 2C2 + 4C3 Put C, = 1 in (3) and (4), we get C2 + C3 = 1 2C2 + 4C3 =  2 Solving (5) and (6), we get C3 =  2, C2 = 3 The required solution is tn = ( 1 + 3n  2n2) ( l)n.
... (2) ... (3) ... (4) ... ( 5) ... ( 6)
... (7)
... (8)
... ( 1)
... (2) ... (3) ... (4) ... (5) ... ( 6)
in part (a) . . . ( 1) Hint. The characteristic equation is a3  6a2 + 12a  8 = 0 Here, a = 2 satisfies equation (1). By Horner's method, the roots of equation ( 1) are (c) Please proceed yourself as
2, 2, 2
... (1)
Ans. an = (3  2n + n2) 2n
DISCRETE STRUCTURES
1 86
Example 8. Solve the recurrence relation for Fibonacci numbers given by
fn = fn _ 1 + fn _ 2' subject to f1 = f2 = 1. Sol. Given recurrence relation is fn  fn  fn _ 2 = 0 Its order is n  (n  2) = 2
... (1)
1
The characteristic equation is a2  a  I = 0
2
Take a = 1+ ..!5 '

=>
1 ±'1 ± ..!5 ..J1+4 a=2 2
� = 1  ..!5 . Clearly, a + � = 1 , a� =  l 2
The solution of (1) is given as fn. = C an + C2 Rf'n Given f, = C, a + C2 � = 1
... (2)
1
... (3)
f2 = C, a2 + C2�2 = 1
... (4)
Solving (3) and (4), we get � C2
2 2 3 1 2 2 2 2 _ _ _ � � a a a� a � C = _ f3(1  �) af3(�  a) a� ( : a + � = 1)  a""' f3("" � a') 2 1 1 a � 1  ..!5  1..!5 ..!5 �  a� a(a  1) C2 =  ='=l a+�=l a�(�  a) af3(�  a) 1 = � 1 a �  ..!5 n n 11 +..!5 1 ..!5 . d soIutlOn. .fn = J5 . 2 1f ) IS the requlre 2 1 Example 9. Solve the recurrence relation tn = 4(tn _1  tn _ 2 ), subject to initial conditions for n = 0, 1. Sol. Given equation is tn  4tn + 4tn _ 2 = 0 Its order = n  (n  2) = 2 The characteristic equation is a2  4a + 4 = 0 => (a  2)2 = 0 => a = 2, 2 The solution is given by tn = (C, + C2n)2n Using tn = 1 for n = 0 in (1), we get 1 = C, 1 Using tn = 1 for n = 1 in (1), we get 1 = (1 + C�2 => C2 =  "2 1
( )f[
tn = 1
J [
J
1
( �n) 2n = (2  n) 2n
tn = 1
1
,
is the required solution.
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
1 87
Example 10. Given that white tiger population of Orissa is 30 at time n = 0 and 32 at time n = 1. Also the increase from time n  1 to time n is twice the increase from time n  2 to time n 1. Find the recurrence relation and solve it for growth rate of tiger.
{
Sol. Given initial conditions are
30, 32, tn = 3tn_1  2tn_2 ,
I
According to given,
n� n� n>
0 1 1
Increase from time Increase from time
n  1 to time n is tn  tn _ I " n  2 to time n  1 is tn 1  tn _ 2'
tn  tn _ 1 = 2(tn _1  tn _ ) tn = 3tn _ 1  2tn _ 2 ::::::} For n > 1, the recurrence relation is tn  3tn _ 1 + 2tn _ 2 = 0 The characteristic equation is
a2  3a + 2 = 0 => a2  2a  a + 2 = 0 => (a  1) (a  2) = 0 => a = 1, 2 . . The solution is tn = C, + C2 2 n For n = 0, 30 = C, + C2 For n = 1, 32 = C, + 2C2 Solving (1) and (2), we get C, = 28, C2 = 2 . . The required solution is tn = 28 + 2n + 1 .
Example
... (1)
... (2)
11. Find the solution of the homogeneous recurrence relation
YK  YK 1  YK 2 = o. Sol. The characteristic equation is s2  S  1 = 0
s=
l ± ..j1+4
2
l±
.J5  2
Therefore, the required solution of the given homogeneous recurrence relation is
r
r
1 J5 1 .J5 C = C, + h \ 2 2. Example 12. Find the solution of the homogeneous recurrence relation ar 4 + 2ar 3 + 3ar 2 + 2ar 1 + ar = o. Sol. The characteristic equation is S4 + 2s3 + 3s2 + 2s + 1 = 0 S4 + s2 + 1 + 2s3 + 2s + 2s2 = 0 I (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca => (S2 + s + 1)2 = 0
[
+
or by
+
[
+
+
s =  l ± iJ3  l ± i J3 2 2
Therefore) the required solution of the given homogeneous recurrence relation is given
1 88
DISCRETE STRUCTURES Example 13.
Find the solution of the homogeneous recurrence relation YK + 4 + 4YK + 3
+ 8yK + 2 + 8yK + 1 + 4yK = O.
Sol. The characteristic equation is 84 + 483 + 882 + 88 + 4 = 0 84 + 482 + 4 + 483 + 88 + 482 = 0 I (a + b + c)2 = a2 + b 2 + c2 + 2ab + 2bc + 2ca (82 + 28 + 2)2 = 0
s =  l ± i)  l ± i
or
Therefore) the required solution is given by
YK = (C, + C2 K)( l + ilK + (C3 + C4 K)( 1  ilK Example 14. Solve T(K) = 7T(K  1)  lOT(K  2) where T(O) = 4, TO) = 1 7. Sol. Given equation is T(K)  7T(K  1) + 10T(K  2) = 0 Its order = K  (K  2) = 2 The characteristic equation is a2  7a + 10 = 0 a2  5a  2a + 10 = 0 => (a  5) (a  2) = 0 a = 5) 2 ..
The required solution is
Given
=4 =
T(O)
= C,5K + C22K 4 = C, + C2 (1) gives 1 7 = 5C, + 2 C 2
T(l) 1 7 . . Multiplying the equation (2) by 2 and subtracting from
8  1 7 = 2 C,  5C, =>  3C, =  9 C, = 3 4 = 3 + C2 => C2 = 1
=>
... (1) ... (2)
T(K) . . (1) gives
... (3) (3), we get
From (2), From (1), we have
= 3 . 5K + 2K, is the required solution. Example 15. Solve the recurrence relation (P.T.V. B.Tech. May 2006) (a) a,  2a,_1 + a,_2 = 0 given that ao = 1, a1 = 2. (b) an = 6 an_1  8 an_2, ao = 1, a1 = 10. (P.T.V. B.Tech. Dec. 2012) (c) an =  3 an_1 + 10 an_2, n :> 2, given ao = 1, a1 = 4. (P.T.V. B.Tech. Dec. 2009) Sol. (a) Given recurrence relation can be written as 8(r)  2 8(r  1) + 8(r  2) = 0 ... (1) where 8(0) = 1 , 8(1) = 2 Its order = r  (r  2) = 2 The characteristic equation is a2  2a + 1 = 0 (a  1)2 = 0 => a = 1, 1 T(K)
The solution to (1) is
Given
8(r) = (C, + C2r) . l' 8(0) = 1 (2) gives 1 = C, (2) gives 2 = C, + 2 C2 8( 1) = 2 1 2 C2 = 2  1 = 1 => C2 = "2
... (2)
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
r=
The required solution of (1) is S( )
(b) Given equation is an  6 an_1 +
ao = a1 = 10
where 1) (1) can be written as Its
..
1 89
( + � r) . l' 1
8an_2 = 0
 6S(n  1) + 8S(n  2) = 0, S(O) = 1 , S(l) = 10 = n  (n  2) = 2 The characteristic equation is a2  6a + 8 = 0 a2  4a  2a + 8 = 0 => a(a  4)  2(a  4) = 0 (a  2) (a  4) = 0 => a = 2, 4
... (1)
S( n) order
The solution of (1) is given by
S(n) = C,.2n + C2. 4n Given S(O) = 1 . . (2) gives 1 = C, + C2 S(l) = 10 . . (2) gives 1 0 = 2C, + 4C2 Multiplying (3) by 2 and subtracting from (4), we get  8 =  2C2 => C2 = 4 From (3), C, + 4 = 1 => C, =  3
... (2) ... (3) ... (4)
The required solution is
S(n) =  3 . 2n + 4n+1 (c) Given recurrence relation is an + 3an 1  lOan _ 2 = 0 Its order = n  (n  2) = 2 The characteristic equation is
a2 + 3a  10 = 0 => a2 + 5a  2a  10 = 0 (a + 5) (a  2) = 0 => a = 2,  5 => The general solution is an = C, 2n + C2 ( 5) n Using ao = 1 , (1) gives 1 = C, + C2 Using a, =  4, (1) gives  4 = 2C,  5C2 6 1 Solving (2) and (3), we get C 2 = "7 ' C, = "7
(%)
3 9 C = 6  = 2 2 1
(K) = � . ( 5)K + �2 , is the required solution. 2 Example 2_ Solve the following recurrence relation T(K)  7T(K  1) + lOT(K  2) = 6 + 8K (2) gives
where
. .
... (2)
S
T(O) = 1, TO) = 2. Sol. The given equation is
(P.T.U. B.Tech. Dec.
2005)
 7T(K  1) + 10T(K  2) = 6 + SK ... (1) Its order = K  (K  2) = 2 The characteristic equation is a2  7a + 10 = 0 a2  5a  2a + 10 = 0 => a(a  5)  2(a  5) = 0 (a  5) (a  2) = 0 => a = 5, 2 The homogeneous solution of (1) is given by Th (K) = C , . 5 K + C2 . 2 K To find the particular solution of (1), we observe that R.H.S. of (1) is a linear equation of the form 6 + SK. Therefore, assume the particular solution of (1) as Tp(K) = do + d,K Tp(K  1) = do + d,(K  1) Here Tp(K  2) = do + d, (K  2) Using all these in (1), we get do + d,K  7(do + d,(K  1» + 10 (do + d,(K  2» = 6 + SK (do  7do + 7d, + 10do  20d,) + (d,  7d, + 10d,) K = 6 + SK => 4do  13d, + 4d,K = 6 + SK ... (2) T(K)
DISCRETE STRUCTURES
1 96
Equating the coefficients of constant term, in (2), we get
4do  13d, = 6 Equating the coefficient of K in (2), we get 4d, = 8 => d, = 2 . . From (3), 4do  26 = 6 => 4do = 32 do = 8
Tp(K) = 8 + 2K Hence the general solution of (1) is given by T(K) = Th(K) + Tp(K) = C, . 5K + C2 . 2 K + 8 + 2 K Given T(O) = 1 gives 1 = C, + C2 + 8 T(l) = 2 (4) gives 2 = 5C, + 2 C2 + 10 Multiplying (5) by 2 and subtracting from (6), we get => C, = 2 o =  3C , + 6 From (5), 1 = 2 + C2 + 8 => C2 =  9 . . (4) reduces to T(K) = 2 . 5K  9 . 2K + 8 + 2K, is the required solution.
... (3)
... (4) ... (5)
... (6)
Example 3. Find the general solution of the following recurrence relation S(K)  3S(K  1)  4S(K  2) = 4K (P.T.U. B.Tech. May 2010)
Sol. The given equation is
S(K)  3S(K  1)  4S(K  2) = 4K ... (1) Its order = K  (K  2) = 2 The characteristic equation is a2  3a  4 = 0 a2  4a + a  4 = 0 => a(a  4) + (a  4) = 0 (a + 1) (a  4) = 0 => a =  1, 4 The homogeneous solution is given by Sh(K) = C , ( l)K + C24K ... (2) Particular solution. Corresponding to the term 4K, [R.H.S. of (1)], we assume the general form of the solution as do 4K but due to occurence of this term in equation (2), we multiply this by suitable power ofK so that none of the term will occur in equation (2). Hence, we multiply by K. Hence the particular solution of (1) becomes as Sp(K) = doK 4K Sp(K  1) = do(K 1) 4KI Here Sp(K  2) = do(K  2) 4K2 Using all these values in (1), we get doK 4K  3do(K  1) 4KI  4do(K  2) 4K2 = 4K Dividing by 4K2, we get
16 Kdo  12do(K  1)  4do(K  2) = 16 do(16K  12K + 12  4K + 8) = 16 20do = 16 => do = 4/5
S P(K) = � K 4K 5
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
1 97
The general solution of (1) is given by Example 4. Find the particular solution of the recurrence relation ar + 2  3ar + + 2ar = Z",
... (1)
1
where Z is some constant.
Sol. The general form of solution is = A . Z'
Now putting this solution on L.R.S. of equation (1), we get
ar + 2  3ar + 1 + 2ar = AZH 2 3AZH 1 + 2AZ' = (Z2  3Z + 2)AZ' Equating equation (2) with R.R.S. of equation (1), we get (Z2  3Z + 2)A = 1 1 A = Z2 1 2 (Z 1XZ 2) (Z '" 1, Z '" 2)  3Z + Z' Therefore) the particular solution is a,(P) (Z  lXZ 2) . 
or
_
... (2)
_

Example 5. Find the particular solution of the recurrence relation
a, + 2  5a, +
1
+
6a, = 5'.
. .. (1)
Sol. Let us assume the general form of the solution a,(p) = A . 5'.
Now to find the value of A, put this solution on L.R.S. of the equation (1), then this becomes ar + 2  5ar + 1 + 6ar = A' 5H 2 5 . A5H 1 + 6 . A5' 
= 25A . 5'  25A . 5' + 6A . 5' = 6A . 5'
... (2)
Equating equation (2) to R.R.S. of equation (1), we get
6A = 1
=>
A = 61
1
Therefore) the particular solution of the difference equation ar(p) = '6' 5r.
Example 6. Find the homogeneous solution and particular solution of the recurrence
relation
ar + 2  4ar = r2 + r  1.
Sol. The characteristic equation is given by =>
a2  4 = 0 (a  2) (a + 2) = 0
The homogeneous solution of the recurrence relation is given by
a,(h) = C,(2? + C2 ( 2?
To find the particular solution) let us assume the general form of the solution is
a,(p) = A, r2 + A2 r + A3.
. .. (1)
1 98
DISCRETE STRUCTURES
(1), we get ad 2  4ar = A, (r + 2)2 + A2 (r + 2) + A3  4A, r2  4A2 r  4A3 =  3A, r2 + (4A,  3A)r + (4A, + 2A2  3A) Equating equation (2) with R.R.S. of equation (1), we get  3A, = 1 4A,  3A2 = 1 4A, + 2A2  3A3 =  1
Putting this solution in L.R.S. of equation
... (2)
After solving these three equations) we get
17 27 � E. Therefore, the particular solution is a,(P) =  C 3 9r27 A1
=  �3 ) A2 = �9 ) A3 _
_ _
_
Example 7. Find the homogeneous solution and particular solution of the recurrence
relation
ar + 2  2ar + 1 + ar = 3r + 5.
... (1)
Sol. The characteristic equation is =>
a2  2a + 1 = 0 (a 1)2 = O
=>
a = l, l
The homogeneous solution of the recurrence relation is given by
a,(h) = C, + C2 r
3r
... (2)
Particular solution. Corresponding to the term + 5, we assume the general form of the solution as + but due to occurrence of these terms in equation we multiply this by suitable power of so that none of the term will occur in equation Thus multiply by
A, r A2 r
(2).
(2),
r2.
Hence, the general form of the solution becomes
ar(p) = Al r3 + A2 r2
(1), we get 3 a, + 2  2ad + a, = A, (r + 2) + A2 (r + 2)2  2A, (r + 1)3  2A2 (r + 1)2 + A, r3 + A2 r2 = A, (r3 + S + 6r2 + 12r) + A2(r2 + 4 + 4r)  2A, (r3 + 1 + 3r2 + 3 r)  2A2(r2 + 1 + 2r) + A, r3 + A2 r2 = (12A, + 4A2  6A,  4A)r + (SA, + 4A2  2A,  2A) ... (3) = (6A,)r + (6A, + 2A) Equating equation (3) with R.R.S. of equation (1), we get 1 A, = "2 6A, = 3 6A, + 2A2 = 5 1 Therefore, the particular solution is ar(p) = 2' T3 + r2 .
Putting this solution in L.R.S. of equation I
. .
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
1 99
Example 8. Find the particular solution of the recurrence relation
ar + 2 + ar + 1 + ar r . 2r. =
Sol. Let us assume the general form of the solution a,(p) = (Ao + A, r) . 2'
. .. (1)
(1), we get d 2 ad 2 + ad 1 + a, = 2 [Ao + A, (r + 2)] + 2d 1 [Ao + A, (r + 1)] + 2' (Ao + A, r) = 4. 2' (Ao + A,r + 2A,) + 2 . 2' (Ao + A, r + A,) + 2' (Ao + A, r) ... (2) = r. 2'(7A,) + 2' (7Ao + 10A,) Equating equation (2) with R.H.S. of equation (1), we get 7A, = 1 A, = 71 10 7Ao + 10A, = 0 Ao = 49  0+ Therefore, the particular solution is ar(p) = 2' 4 Now, put this solution in the L.H.S. of equation
( � � r).
Example 9. Find the homogeneous and particular solution of the recurrence relation
a,  4a,_ 1 + 4a, _ 2 (r + 1) . 2'. Sol. The characteristic equation is a2  4a + 4 = 0 (a  2)2 = 0 => a = 2, 2 =>
. .. (1)
=
The homogeneous solution of the difference equation is given by
a,(h) = (C, + C2r) . 2'
... (2)
To find the particular solution) let us assume the general form of the solution is
= 2' (A,r + Ao), but due to occurrence of there terms in equation (2), we multiply this by suit· able power of r so that none of the terms will occur in equation (2). Thus multiply by r2 . Hence, the general form of the solution becomes = 2' (A,r + Ao) . r2 Putting this solution in L.H.S. of equation (1), we get a,  4a,_ 1 + 4a, _ 2 = 2' . (A,r + Ao) . r2  4 . 2'  1 [A, (r  1) + Ao] . (r  1)2 + 4 . 2'  2 [A, (r  2) + Ao] . (r  2)2 = 2' . (A,r + Ao) . r 2  2(r2 + 1  2r) . 2' (A,r  A, + Ao) + (r2 + 4  4r) . 2' . (A,r  2A, + Ao) ... (3) = r . 2' (6A,) + 2' ( 6A, + 2Ao) Equating equation (3) with R.H.S. of equation (1), we get 6A, = 1  6A, + 2Ao = 1
A1 = .!.6
( )
ar(p) = r2. 2' % + 1 +1 Example 10. Solve S(n)  6S(n  1) + 9S(n  2) 3n Sol. Given equation is S(n)  6S(n  1) + 9S(n  2) = 3.3n Its order = n  (n  2) = 2 Therefore, the particular solution is
=
(P.T.V.
B. Tech. May 2010)
... (1)
DISCRETE STRUCTURES
200
The characteristic equation is a2  6a + 9 = 0 (a  3)2 = 0 => a = 3, 3 . . The homogeneous solution of (1) is given by Sh (n) = (C, + C2n) 3n ... (2) n 1 + Corresponding to the term 3 [R.H.S. of (1)], we assume that general form of the solution as do 3n+1 , but due to occurence of this term in equation (2), we multiply this by suitable power of n so that none of the term will occur in equation (2). Thus multiply by n2• Hence the particular solution of (1) becomes as Sp(n) = do n23n Here Sp(n  1) = do(n  1)2 3n1 Sp (n  2) = do(n  2)2 3n2 Using all these in (1), we get don23n  6do(n  1)2 3n1 + 9do (n  2)2 3n2 = 3 . 3n Dividing by 3n2, we get 9don2  18 do(n  1)2 + 9do(n  2)2 = 27 2 => do[9n  18n2  18 + 36n + 9n2 + 36  36n] = 27 =>
1
do [54] = 27 => do = '2 Sp(n) = � n2 . 3n 2 The general solution is given by
S(n) = Sh(n) + Sp(n) = (C, + C2n) 3n + '21 n2 . 3n. Example 11. Solve the following recurrence relation Q(J)  Q(J  1)  12Q(J  2) = ( 3y + 6 . 4J Sol. Given equation is
Q(J)  Q(J  1)  12Q(J  2) = ( 3)" + 6 . 4J ... (1) Its order = J  (J  2) = 2 The characteristic equation is a2  a  12 = 0 a2  4a + 3a  12 = 0 => a(a  4) + 3(a  4) = 0 (a  4) (a + 3) = 0 => a = 4,  3 The homogeneous solution of (1) given by ... (1) Qh (J) = C,4J + C2( 3)J To find the particular solution of (1), we observe that the R.H.S. of (1) is a combination of the terms ( 3)J and4J. But these terms also occur in the homogeneous solution (1). Therefore we assume the particular solution of (1) as Qp(J) = doJ( 3)" + d,J 4J Here Qp (J  1) = do (J  1) ( 3)J1 + d,(J  1) 4J1 Qp(J  2) = do(J  2) ( 3)J2 + d,(J  2) 4J2 Using all these values in (1), we get doJ ( 3)J + d,J 4J  [do(J  1) ( 3)J1 + d,(J  1) 4J1] _ 12 [do(J  2) ( 3)"2 + d,(J  2) 4J2] = ( 3)J + 6 . 4J
RECURRENCE RELATIONS AND GENERATING FUNCTIONS or or
201
do ( S)"2 [9J + 3(J  1)  12(J  2)] + d, 4J2 [16J  4(J  1)  12(J  2)] = ( S)" + 6 . 4J do( S)"2 ( 21) + d, 4J2 (28) = ( 3)" + 6 . 4J 21 do = 1 => d  � Equating the coefficients of ( 3)J, we get 9 7 96 = 24 Equating the coefficient of 4J, we get => d1 = 28 7 0
The general solution of
=
(1) is given by
4 Q(J) = Qh(J) + Qp(J) = C,4J + Ck 3)"  � J( 3)" + 27 J 4J. 7 Example 12. Solve S(K)  4S(K  1) + 4S(K  2) = 3K + 2K where S(O) = 1, S(1) = 1. Sol. Given equation is S(K)  4S(K  1) + 4S(K  2) = 3K + 2K ... (1) Its order = K  (K  2) = 2 The characteristic equation is given by a2  4a + 4 = 0 (a  2)2 = 0 => a = 2, 2 The homogeneous solution of (1) is given by Sh(K) = (C, + C2K). 2K = C,. 2K + C2K . 2K To find the particular solution of (1), we observe that the R.H.S. of (1) contains the terms 3K and 2 K But the terms 2K and K2 K also occur in Sh(K). Hence we divide the particular solution in two parts.
Particular solution corresponding to the term 3K is
S� (K) = do + d, K S�(K  1) = do + d, (K  1), S� (K  2) = do + d, (K  2) Using all these values in (1), we get do + d,K  4 (do + d,(K  1) + 4(do + d,(K  2» = 3K => do (1  4 + 4) + d, [K  4(K  1) + 4(K  2)] = 3K => do + d, (K  4K + 4 + 4K  8) = 3K do + d, (K  4) = 3K => Equating the coefficient of constant term, we get do  4d, = 0 Equating the coefficient of K, we get d, = 3 From (2), S�(K) = 12 + 3K Particular solution corresponding to the term 2K Let S�(K) = d2K2 . 2K .. S�(K  1) = d2(K  1)2 . 2K  1 S�(K  2) = d2(K  2)2 . 2K  2
... (2) ... (3) ... (4)
DISCRETE STRUCTURES
202
(1), we get 2 2 K d2K .2  4 d2(K  1) 2Kl + 4 d2(K  2)2 . 2K2 = 2K d2 • 2K 2(4K2  8(K  1)2 + 4(K  2),,) = 2K 2 2 2 2 => d2 • 2K (4K  8K  8 + 16K + 4K + 16  16K) = 2K d22K2(8) = 2K => 8d2 = 1 4
Using all these values in
. . . (5)
(1) becomes Sp(K) = S� (K) + S�(K) = 12 + 3K + K2 . 2Kl Hence the general solution of (1) is given by S(K) = Sh(K) + Sp(K) = C, . 2K + C2K . 2K + 12 + 3K + K2 . 2K  l Given S(O) = 1, (7) gives C, =  11 1 = C, + 12 => S(l) = 1 (7) gives 1 = 2C, + 2C2 + 12 + 3 + 1 The particular solution of
... (6) I
Using
(4) and (5)
S(K) =  11 . 2 K + '!...2 . K . 2 K + 12 + 3K + K2 . 2 Kl = 12 + 3K + 2K  '(K2 + 7K  22). Example 13. Solve S(K) = rS(K  1) + a, where S(O) = 0, where r, a > 0, r ", 1. Sol. Given equation is S(K)  rS(K  1) = a The characteristic equation is given by a  r = O ::::::} a=r Sh(K) = C,rK Also, let Sp(K) = do Sp(K  1) = do Using all these in (1), we get do  rdo = a
. . . (7)
From (7),
do(1  r) = a
=>
a Sp (K) = 1r
a d0 = _ _ 1r
... (1)
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
203
The general solution is given by
. . . (2) Given 8 (0) =
0
(2) gives O = C1 +
a

1r
=>
C1 = 
a

1r
a K a 1r . r + 1r a(l  rK) . . = the requlre . d soIutlOn. 1  r IS Example 14. Solve the recurrence relation a, + 5a,_1 + 6a,_2 = f(r) r = O, 1,5 f(r) = 6, otherwise where given that (P.T.U. B.Tech. Dec. 2006) ao = a1 = 0. From
(2),
8(K) = 

{a,
Sol. Given equation is 8 ( r) + 5 8 (r  1) + 6 8 ( r  2) = f(r) Its order = r  (r  2) = 2
Sa + 6 = 0 2a + 6 = 0 (a + 3) (a + 2) = 0 The homogeneous solution of (1) is given by 8 h ( r) = C , ( 3)' + C2 (  2)' The characteristic equation is a2 + a2 + 3a +
... (1)
=>
=>
a(a + 3) + 2(a + 3) = 0 a =  3,  2
To find the particular solution, Case I. When r = 0, 1, 5, then from given, we have f(r) = O. Consequently, 8p(r) = 0
(1) is given by 8(r) = 8h(r) + 8p(r) = C, ( 3)' + C2 ( 2)' Case II. When r '" 0, 1, 5, then we are given f(r) = 6 . . Let 8p(r) = do 8p(r  1) = do => 8p(r  2) = do Using all these in (1), we get do + 5do + 6do = 6 12do = 6 => do = 1/2 => 8p(r) = "21 Hence the general solution of
The required general solution is given by
204
DISCRETE STRUCTURES Example 15. Solve S(K)  4S(K  1) + 3S(K  2) = K2 Sol. Given equation is S(K)  4S(K  1) + 3S(K  2) = K2
(P.T.U. B.Tech. May 2012) . . . (1)
Its order = K  (K  2) = 2 The characteristic equation is a2  4a + 3 = 0 a2  3a  a + 3 = 0 => a(a  3)  (a  3) = 0 (a  3) (a  1) = 0
=>
a = 1, 3
The homogeneous solution of (1) is given by
K
K
K
Sh(K) = C , . 1 + C2 3 = C , + C 2 3 . . . (2) Let the particular solution of (1) is given by Sp(K) = K(Ao + A, K + A2K2) (We have multiplied by K since (2) contains the constant term) 3 Sp(K) = AoK + A , K2 + A2K => .. Sp(K  1) = Ao (K  1) + A , (K 1 ) 2 + A2 (K  1) 3 Sp(K  2) = Ao (K  2) + A, (K  2) 2 + A2 (K  2) 3 Using all these in (1), we get 3 3 AoK + A, K2 + A2K  4[Ao (K  1) + A , (K  1) 2 + A2 (K  1) ] + 3[Ao (K  2) + A, (K  2) 2 + A2 (K  2)3] = K2 Ao [K  4(K  1) + 3(K  2)] + A, [K2  4K2 + SK  4 + 3K2  12K + 12] => 3 3 3 + A2 [K  4K + 12K2  12K + 4 + 3K  lSK2 + 36K  24] = K2  2Ao + A, [ 4K + S] + A2 [ 6K2 + 24K  20] = K2 Equating the coefficient of powers of K, we get
_
1
Coefficient of K2 ,
A2 =  6
Coefficient of K,  4A, + 24A2 = 0  4A ,  4 = 0 A, =  1 =>
Coefficient of constant term,  2Ao + SA ,  20A2 = 0 2Ao =
_ 4S + 20
S p(K) = 
6
2S 6
20  2Ao  S +  = 0 6 =>
3 7... K  K2  .!. K 6
Ao =
_ 14 = _ 7... 3
6
3
Hence the general solution is given by S(K) = Sh(K) + Sp(K) = C , + C2 . 3
K3 K  7 K  K2  3
6
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
L
2.
I
205
TEST YOUR KNOWLEDGE 7.2
Solve the following recurrence relations (,) 8(K)  28(K  1) = 3.2K (ii) 8(K) + 58(K  1) + 68(K  2) = 3K' (P.T. U. B.Tech. Dec. 2007) (iii) 8(K)  28(K  1) + 8(K  2) = 2, 8(0) = 25, 8(1) = 16 (iv) 8(K)  8(K  1)  68(K  2) =  30, 8(0) = 20, 8(1) =  5 (v) 8(K)  58(K  1) = 5K, 8(0) = 3 (vi) 8(K)  58(K  1) + 6 8(K  2) = 2, 8(0) = 1, 8(1) =  1. Solve the following recurrence relations r 8 2 2 (i) ar 2  2ar 1 + ar :::: r . 2 (ii) a 3 + aK 2  aK 1  12aK :::: 2K + 5 K n n (iv) an 4an_1 :::: 7(4) , n ;::': 1 (iii) an  4an_1 :::: 7(5) , n 1 (P.T. U. B.Tech. May 2013) +
+
+
;::.:
+
+
Answers
1 2 17 (i) 8(K) = Cj 2K + 3K 2K (ii) 8(K) = Cj ( 2)K + C, ( 3)K + 2115 88 + 24 K + '4 K (iv) 8(K) = 11(  2)K + 4 . 3K + 5 (iii) 8(K) = K'  10K + 25 (v) 8(K) = (3 + K)5K (vi) 8(K) =  2 . 3K + 2 . 2K + 1. 2. (i) a, = Cj + C,r + 2'(r'  Pr + 20) 1 K 17 (ii) aK = C13K + (C2 + C3K) ( 2)K K92 + 27 54
1.
_ _
7.13. GENERATING FUNCTIONS OR NUMERIC FUNCTIONS Let
be a sequence defined for n :> 0, then the infinite sum G(S, z) where G(S, z) = So + SjZ + S2z2 + S3z3 + ...... = L S nzn , is called generating function or numeric function of the n=O �
ILLUSTRATIVE EXAMPLES
Example 1. Consider a sequence defined by 8(n) = 2", n :> O. Obtain its generating
function.
(P.T.V. B.Tech. May
2010)
Sol. By definition, the required generating function is given by �
G(S,
�
�
z) = L S(n) zn = L 2nzn = L (2z)n n=O n=O n=O 3 2 = 1 + 2z + (2z) + (2z) + ...... =
=
1 1  2z
S
a
Sum of infinite G.P. is .  
where
a=
�
First term,
lr '
r = common ratio
DISCRETE STRUCTURES
206 Example
2. If S(n) = ban, n :> 0, obtain its generating function.
(P.T.V. B.Tech. May 2012)
Sol. By definition, the generating function is given by �
�
�
z) = L S(n) zn = L ban. zn = b L (az)n n=O n=O n=O 3 2 = b (1 + az + (az) + (az) + ...... =) 1 b _ b . 1 az _ 1 az ' Example 3. If S(n) = n, n :> 0, obtain its generating function. G(S,
DO
S
�_ a_ 1r
Sol. By definition, the generating function is given by �
Example
�
z) = L S(n) zn = L nzn n=O n=O 2 = 0 + 1.z + 2z + 3z3 + 4Z4 + ...... = = z(l + 2z + 3z2 + 4z3 + ...... =) z . = z . (1  z)2 = (1 Z)2
G(S,
(1
 x)2 =
1
+ 2x + 3x2 + ...
4. If S(n) = �, find its generating function.
n!
Sol. By definition, the generating function is given by G(S,
Example
G/'"S
zn = 1 +Z +Z2 + Z3 + ..... . = L n=O n ! O ! I! 2 ! 3 ! Z2 Z3 = 1 + z +  +  + ...... = = e2 . 2! 3!
z) = L n=O
=
5. Obtain partial fraction decompositions for the expression
z) = 1  611z 29z + 30z2 Sol. Consider
and identi' y the sequence having the expression asgeneratingfunction. I. G(S
, z) = =
Let
S(n) zn
6 "::": 29z 6  29z = "':2 2 z;;''6z  5z +""'l l 11z + 30z "'3"'06  29z 6  29z = c:::cc,,6z(5z  1)  (5z  1) (6z  1) (5z  1)
A B 6:..: 29:..:=Z _ _ _ _ =_ + ::= :: ,:(6z  1) (5z  1) 6z  1 5z  1
... (1)
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
207
Put
(1) by (6z  1) (5z  1), we get 6  29z = A(5z  1) + B(6z  1) 5z  1 = 0 in (2), we get 6  2 = B
1
Put
6z  1 = 0 in (2), we get
1
Multiplying
: (� ) 6  2: (% ) =A
... (2) 1
=>
B
=>
A=
=
7
7 7 1 + 5z 1 1 = 1 , (6z 6 1) 29z  1  5z (5z  1) = 6z  1 1  6z G(S, z) = 7.6n  1.5n b Then G(S, z) =  = 7.6n  5n 1  az
From ( )
7.14. GENERATING FUNCTIONS FOR SOME STANDARD SEQUENCES Sequence Sen) Sen) = an, n � Sen) = n
Generating functions G(S, z)
0
1 G(S, z) = 1  az
+ 1, n � 0
Sen) = (n
G(S, z) =
+ 1) bn , n � 0
Sen) = n, n � O Sen) = nan, n �
G(S, z) =
nCr. 0
1 Z
(1  z)
G(S, z) =
0
G(S, z) = e< G(S, z) =
2 2
(1  bz)
0
1 Sen) =  ' n � n ., S(n) :::; r :::; n ::::
G(S , z) =
1 (1  z)
2
az (1  az)
2
(1 + z)n
7.15. SOLUTION OF RECURRENCE RELATION BY THE METHOD OF GENERATING FUNCTIONS Recurrence relation can also be solved by using the generating functions. Let the recur rence relation is
+
1 + zn
2 + ...... +
C S(n  K) = fen), n :> K Sen) C , Sen  ) C 2 S(n  ) k Multiplying both sides by and summing up from n = K to where K = order of the given recurrence relation, we get =,
208
DISCRETE STRUCTURES �
L S(n) zn + C, L n=K n=K �
S(n  1)
zn + C2 L n=K
S(n  2)
zn + ... �
+ CK L n=K �
Writing each term in the form
L n=O
S(n)zn ,
zn = L f(n)zn n=K �
S(n  K)
we will get the solution in terms of G(S,
z).
Using the results of 7.13, we can find the required solution. The following examples illustrate the concepts more clearly_
Example 6. Solve the recurrence relation S(K)  3S(K  1)  2 = 0, K? 1, where S(O) = 1, by using generating functions. ... (1) Sol. Given equation is S(K)  3S(K  1) = 2
zK, we get S(K) zK  3S(K  1) zK = 2zK Summing up for all K ? 1 , we get, (order = K  (K  1) = 1) Multiplying (1) by
�
L K=l
�
S(K)
zK  3 L K=l
S(K  1)
�
G(S,
Consider
z) = L K=O
S(K)
�
zK = 2 L zK K=l
zK = S(O) + S(l) z + S(2) z2 + ..... .
�
= S(O) + L K=l �
L K=l
=>
S(K)
... (2)
S(K)
zK
zK = G(S, z)  S(O)
For the second term in (2), we have
L K=l
S(K  1)
For the third term in
zK = z L K=l
S(K  1)
zKl = z L K=O
(2), we have
2 3 � L... zK = z + z + z + ...... 00 =
K=l
Using all these in (2), we get G(S, z)  S(O)  3 zG(S, z)
=
2z
1z

z _ 1z
_
S(K)
zK = zG(S, z) I
Changing K to K
+1
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
2z
(1  3z) G(S, z) = = G(S , z) =
1z
209
I
+1
2z + 1 z 1z 1+z
=
z+l 1z ... (3)
( 1  z ) (1  3z)

B A l+z ' =  + 1  3z 1z (1  z) (1  3z)
Consider
Multiplying (4) by (1  z) (1  3z), we get
... (4)
1 + z = A(l  3z) + B(l  z) Put
1  3z = O in (5), we get
Put
1  z = 0 in (5), we get
l+
i
( i) 
= B l
1 + 1 = A(l  3)
S(O) =1
=>
=>
... (5) B=2
A=1
1 2 From (3) and (4), we get G(S , z) =  + 1 z 1  3z Hence the required solution of the given recurrence relation is S(K) =  (l)K + 2.3K, K :> O.
7. Find the generating function from the recurrence relation given by S(K)  6S(K  1) + 5S(K  2) = 0 where S(O) = 1, S(1) = 2.
Example
(P.T.V. B.Tech. Dec. 2010)
Sol. Given recurrence relation is
... (1)
S(K)  6S(K  1) + 5S(K  2) = 0 Its order = K  (K  2) = 2 Multiplying (1) both sides by zK and summing up from K = 2 to �
L
K=2
�
S(K) zK  6
L
K=2
S(K  1) zK + 5
�
Consider
G(S, z) =
L
�o
�
L
Also
�
L
K=2
L
K =2
�
[
�
S(K  l) ZK' = Z
= z 8(0) +
�
L
K =2
S(K) zK
S(K) zK
S(K) zK = G(S, z)  1  2z
S(K  l) ZK = Z
we get
S(K  2) zK = 0
S(K) zK = S(O) + S(l) z +
= 1 + 2z +
K =2
=,
�
[�
S(K) Z K
,
) [�
S(K) Z K  8(0»
= z(G(S, z)  1) = zG(S, z)  z
)
=z
I Changing K to K + 1
S(K) Z K
)
 S(O»
210
DISCRETE STRUCTURES �
Also
L
K=2
�
L
S(K  2) zK = z2
�
L
S(K  2) zK2 = z2
K=2 K=O = z2 G(S, z) Using all these in (2), we get [G(S, z)  1  2z]  6 [z G(S, z)  z] + 5[z2 G(S, z)] = 0 (1  6z + 5z2) G(S, z)  1  2z + 6z = 0 => 1  4z
G(S, z) =
1  6z + 5z
S(K) zK
I
Changing K to K + 2
2
is the required generating function. Example
8. Find the generating function from the recurrence relation S(n  2) = S(n  1) + S(n) where S(O) = 1, S(1) = 1, n :> O.
(P.T.U. B.Tech. May 2012)
Sol. Given recurrence relation is S(n) + S( n  1)  S(n  2) = 0 Its order = n  (n  2) = 2 Multiplying (1) both sides by zn and summing up all the terms from �
L
S( n) zn +
n =2
�
L
S(n  1) zn 
n =2 �
Consider
G(S, z) =
L
L
�
Also
L
n =2
n =2
n =O
�
L
n =2
S ( n  1 ) zn = Z
=z �
L
n =2
S(n  2) zn = 0
n =2
�
L
n =2
=,
... (1) we get ... (2)
S(n) zn
S(n) zn
S(n) zn = G(S, z)  1  z �
L
[ [�
n =2
S ( n  1 ) zn1 = Z
= z S(O) +
Also
L
S(n) zn = S(O) + S(l) z +
= 1 +z+ �
�
n = 2 to
S(n  2) zn = z2
L
n =2
t,
�
L
n =l
]
S( n) zn
I
Changing
n to n +
1
S(n ) z n  S(O)
]
S(n) zn  S(O) = z[G(S, z)  1] = zG(S, z)  z S ( n  2) zn2 = z2
= z2 G(S, z)
Using all these terms in (2) , we get
[G(S, z)  1  z] + [z(G(S, z)  1)]  z2 G(S, z) = 0 => ( 1 + z  z2) G(S, z) = 1 + z + z
L
n =O
S( n) zn
I
Changing
n to n + 2
21 1
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
2z + 1 '=::"":"0": 1 + z  z2 )
G(S, z) =
is the required generating function. Example 9. Use generating functions to solve the recurrence relation. (P.T.U. B.Tech May 2010) ak = ak_1 + 2ak_2 + 2k where ao = 4, a 1 = 12. Sol. Given recurrence relation can be written as ak  ak_ 1  2ak_2 = 2k Multiplying both sides of (1) by zk and summing up from � L..J DO
Consider
k=2
ak
Z
k
� L..J DO
G(S, z) =
L
k=2
ak z
k
�
k=2
L
k=O
akl ak z
k
k 2 � L..J ak  2 Z
=
ao + a,z + L
Z
k
�
k=2
k=2
= G(S, z)  4  12z
= Z
[
aD
+
f
k�
�
DO
ak z
k
 aD
DO
= L..J
k=2
ak z
k
. . . (1)
k = 2 to k =
2k
Z
k=O
we get . . . (2)
= 4 + 12z +
ak z
,
k
J [ f 4J = Z
=
k

Further
L akz k
k=2
. . . (3)
(Changing k to
k+
1)
= zG(S, z)  4z . . . (4) (Changing k to
k + 2) . . . (5)
Also
= 1 + 2z + = 1 + 2z +
L 2k Zk
 1  2z = 1 + 2z +
L (2z)'
3 = 1 + 2z + (1 + 2z + (2z)2 + (2z) + . . .
k=2 k=O
L 2k Zk
k=O
1 1  2z
= 1 + 2z + 
. . . (6)
Using (3), (4), (5) and (6) in (2), we get
Is�
1 G(S, z)  4  12z  (zG(S, z)  4z)  2(z2 G(S, z» = 1 + 2z + 1  2z
1 1 (1  z  2z2) G(S, z) = 1 + 2z +  + 4 + 12z  4z = 5 + 10z + =>
(2z  1) (z + 1) G(S, z) =
1  2z 5(1 + 2z) (1  2z ) (1 _ 2z )
1  2z
)
=
=
a
_ _
1 r
212
DISCRETE STRUCTURES
5(1  4z 2 ) + 1 (2z _ 1)2 (Z + 1)
,
6  20Z2 (z + 1) (2z  1)
' '' = ,,:G (S z) =  ,:,,
. . . (7)
Consider
6  20Z 2 = A + B + C ":' _ 1)72 1 (2z''z+1 2z_ 1)72 (Z + 1)'(:": 2z'::'Multiplying (8) by (z + 1) (2z  1) 2 , we get 6  20z2 = A(2z  1)2 + B(z + 1)(2z  1) + c(z + 1) 14 Put z + 1 = 0 in (9), we get 14 = 9A => A = _
. . . (8) . . . (9)
9
Put 2z  1 =
0 in (9), we get
=>
3
1 = "2 C
2
=>
C = "3
Equating the coefficient of constant terms in (9), we get
14
2
9
3
6 = A  B + C =   B + 
14
2
9
3
B =  +   6 = G(S, z) =
From (7)
 14
14 + 6  54 9
=
62 9
1 62 1 2 1 "9 ' Z + 1  "9 ' 2z  1 + "3 ' (2z _ 1)2
Hence) the required solution is
ak =
2k + �3 . (k + 1) 2k  149 ( _1)k + 62 9
1 , then S(n) = an , n :> 0 G(S, z ) = 1  az 1 , then S(n ) = (n + l)bn , n :> 0 If G(S, z ) = (1 bz)2 Example 10. Find a closed form expression (generating function) for If
Fibonacci sequence
the terms of
(P.T.V. B.Tech. May 2005)
Sol. Consider the Fibonacci sequence F(K), given by F(K) = F(K where
 1) + F(K  2), K :> 2
F(O) = 1 , F(l) = 1
Its order = K  (K Multiplying by
L K=2 Consider
 2) = 2
zK and summing up all the terms from K = 2 to
F(K)
zK  L K=2
G(F, Z) =
F(K
L K=O
 1) zK  L K=2
F(K)
F(K
=,
we get
 2) zK = 0
zK = F(O) + F(l) z + L F(K)zK K=2
... (1)
RECURRENCE RELATIONS AND GENERATING FUNCTIONS
213
�
=1
+z+ L
K =2
F(K) ZK
�
L
K=2
F(K) ZK = G(F, Z)  1  Z
�
L
Also
K=2
F(K  1) zK = z
L
[
K =2
F(K 2) zK = z2
L
l F(K  1) zK = z
= Z F(O) +
�
L
F(K) zK
] {�;)
K =l
F(K) Z K  F(O) =
I
F(K  2) zK2 = z2
K=2 2 = z G(F, z)
]
�
L
K=O
F(K) zK
+1
F(K) Z K  1
= z (G(F, z)  1) = zG(F, z)  z �
Changing K to K
I
Changing K to K
+ 2.
Using all these values in (1), we get [G(F, z)  1  z]  [zG(F, z)  z]  z2 G(F, z) = 0 (1  z  z2) G(F, z)  1  z z = 0
+
(1  z  z2) G(F, z) = 1 G(F, z) =
1
1z z
2
is the required generating function (or closed form expression) for the Fibonacci sequence.
I
TEST YOUR KNOWLEDGE 7.3
SHORT ANSWER TYPE QUESTIONS
L 2. 3.
4. 5. 6.
Find the generating function for the sequence Sn :::: 21"1, n ;::': o. If Sn :::: arl, n ;::': 0, then find the generating function G(S, z). Find the generating function for the sequence Sn :::: n, n o. Write the generating function for the sequence U(n) :::: 2.3/"1, n � o. Write the generating function for the sequence V(n) :::: 5, n :::: o. What is the generating function for the sequence T(n) :::: 7( l)nP ? ;::.:
LONG ANSWER TYPE QUESTIONS
7. 8.
If 8(n) is a sequence defined by 8(n) 2.3" + 5 + 7.( 1)", find the generating function for 8(n). Find the generating function for the sequence (ii) V(n) 2" [3 + 2( 1)"] . (i) V(n) 2"+1 + 5" If 8(n) 28(n  1) + 38(n  2), n 2 where 8(0) 3, 8(1) l. (a) Find the generating function (b) Find the sequence which satisfies the given recurrence relation. =
=
9.
(P.T. U. B.Tech. Dec. 2013)
=
2, Consider one of the guests, say, Mr. X . , By definition, the number of handshakes made by remaining (n  1) guests among themselves is Hn l' Now, the guest Mr. X will shake· hands with (n  1) guests, gives (n  1) additional handshakes.
(b)
_
_
_
Hence
Hn
= Hn
1
+ (n  1), n :> 2, H,
= O.
8
MONOIDS AN D GROUPS
8.1 . INTRODUCTION In the present chapter) we introduce the concept of algebraic system) binary operations and groups. The study of cyclic groups) normal groups) group homomorphism etc. help us in understanding various applications of computer science. Groups play an important role in coding theory.
8.2. ALGEBRAIC STRUCTURE If there exists a system such that it consists of a nonempty set and one or more opera tions on that set, then that system is called an algebraic system. It is generally denoted by (A) oP p oP2 ) ... ) oP n») where A is a nonempty set and oP p oP2 ) ... ) 0Pn are operations on A. An algebraic system is also called an algebraic structure because the operations on the set A define a structure on the elements of A.
8.3. BINARY OPERATION Consider a nonempty set A and a function f such that f : A x A ; A, then f is called a binary operation on A whose domain is the set of ordered pairs of elements of A. If * is a binary operation on A) then it may be written as a * b. A binary operation can be denoted by any of the symbols
+)
)

*) 8\ A) 0) v) /\ etc.
The value of the binary operation is denoted by placing the operator between the two operands.
e.g.,
(i) The operation of addition is a binary operation on the set of natural numbers.
(ii) The operation of subtraction is a binary operation on set of integers. But) the operation of subtraction is not a binary operation on the set of natural numbers because the subtraction of two natural numbers may or may not be a natural number. (iii) The operation of multiplication is a binary operation on the set of natural numbers) set of integers and set of complex numbers. (iv) The operation of set union is a binary operation on the set of subsets of a universal set. Similarly) the operation of set intersection is a binary operation on the set of subsets of a universal set.
217
218
DISCRETE STRUCTURES
8.4. TABLES OF OPERATION Consider a nonempty finite set A = {ap described by means of table as shown below:
*
a1
a2
a1
a1 *a1
a1 * a2
a2
a2*a1
a2* a2
a2 ) a3 ) ... , an}' A binary operation * on A can be a3
a
n
as * as
a3 a
n
a *a n
The empty cell in the ph row and kth column represent the element a; *ak •
n
ILLUSTRATIVE EXAMPLES
Example 1. Consider the set A = {I, 2, 3} and a binary operation * on the set A defined by a * b = 2a + 2b. Represent operation * as a table on A. Sol. The table of operation is shown below: (Table S.I) Table 8. 1
1 4 6 8
*
1 2 3
2 6 8 10
3 8 10 12
8.5. PROPERTIES OF BINARY OPERATIONS There are many properties of the binary operations which are as follows : 1. Closure Property. Consider a non·empty set A and a binary operation * on A. Then A is closed under the operation *, if a * b E A, where a and b are elements of A. For example, the operation of addition on the set of integers is a closed operation. i.e., if a, b E Z , then a + b E Z \;j a, b E Z.
Example 2. Consider the set A = { i, D, I}. Determine whether A is closed under (i) addition (ii) multiplication. Sol. (i) The sum of the elements is ( 1) + ( 1) =  2 and 1 + 1 = 2 does not belong to A.
Hence A is not closed under addition.
(ii) The multiplication of every two elements of the set are 1 *0=0;
1*1 = 1;
0*1=0;
0*1 = 0;
1 * 1 =  1 ;
1 *0 = 0;
1*1=1
Since, each multiplication belongs to A hence A is closed under multiplication.
MONOIDSAND GROUPS
219
Example 3. Consider the set A = {I, 3, 5, 7, 9, .. .}, the set of odd +ve integers. Determine whether A is closed under (i) addition (ii) multiplication. Sol. (i) The set A is not closed under addition because the addition of two odd numbers produces an even number which does not belong to A.
(ii) The set A is closed under the operation multiplication because the multiplication of two odd numbers produces an odd number. So) for every a, b E A, we have a * b E A. 2. Associative Property. Consider a non·empty set A and a binary operation * on A. Then the operation * on A is associative, if for every a, b, c, E A, we have (a * b) * c = a * (b * c).
Example 4. (a) Consider the binary operation * on Q, the set of rational numbers, defined by a * b = a + b  ab \;j a, b E Q. Determine whether * is associative. (b) Consider the binary operation * on the set N ofpositive integers defined by a * b = ab Determine whether * is associative? Sol. (a) Let us assume some elements a, b, C E Q, then by definition (a * b) * c = (a + b  ab) * c = (a + b  ab) + c  (a + b  ab)c = a + b  ab + c  ca  be + abc = a + b + c  ab  ac  be + abc. Similarly, we have
Therefore, Hence
a * (b * c) = a + b + c  ab  ac  be + abc (a * b) * c = a * (b * c).
* is associative.
(b) * will be associative if a * (b * c) = (a * b) * c \;j a, b, c E N a = 2, b = 2, c = 3 and consider Take a * (b * c) = 2 * (2 * 3) = 2 * 2 3 = 2 * 8 = 28 = 256 and (a * b) * c = (2 * 2) * 3 = 2 2 * 3 = 4 * 3 = 43 = 64 a * (b * c) '" (a * b) * c Hence * is nonassociative. 3. Commutative Property. Consider a non·empty set A and a binary operation * on A. Then the operation * on A is commutative, if for every a, b E A, we have a * b = b * a. by
Example 5. (a) Consider the binary operation * on Q, the set ofrational numbers, defined
a * b = a2 + b2 \;j a, b E Q. Determine whether * is commutative. (b) Consider S = {a, b, c, d} and be a binary operation on S defined by as shown in the following table. *
220 *
a
a b
a b
d
d
c
and
Table 8.2
c
DISCRETE STRUCTURES
b
c
d
b a b a
a a a
c
d b a a
Determine (i) whether * is associative ? (ii) whether * is commutative? Sol. (a) Let us assume some elements a, b E Q, then by definition a * b = a2 + b2 = b 2 + a2 = b * a Hence * is commutative. (b) (i) Let a, b, e E S and consider b * (e * c) = b * a = b (b * c) * e = a * e = e b * (e * c) '" (b * c) * e Thus) * is nonassociative (ii) b * e = a and e * b = b ::::::} b * c * c * b * is noncommutative
Example 6. Consider the binary operation * and Q, the set of rational numbers defined by ab a * b =  \;j a, b E Q. 2 Determine whether * is (i) associative (ii) commutative. Sol. (i) Let a, b E Q, then we have ab ba a*b===b*a 2 2 Hence * is commutative. (ii) Let a, b, C E Q, then by definition we have ab ' e abc ab 2(a * b) * e = '2 * e = =4 2 abc be abc 2 Similarly, a * (b * c) = a * '2 =  = 4 2 a * (b * c) = a * (b * c) Therefore) Hence) * is associative. 4. Identity. Consider a non· empty set A and a binary operation * on A. Then the opera· tion * has an identity property if there exists an element, e, in A such that a * e (right identity) = e * a(left identity) = a \;j a E A. Theorem I. Prove that e/ = e where e/ is a right identity and e is a left identity ofa binary operation. Proof. We know that e/, is a right identity.
( )
()
/
'
/
'
. .. (1)
MONOIDSAND GROUPS
221
et is a left identity. el l! * e/ = e/ . .. (2) From (1) and (2), we have e/ = e/,. Thus, we can say that if e is a right identity of a binary operation, then e is also a left
Also, we know that Hence,
identity.
Example 7. Consider the binary operation * on 1+, the set ofpositive integers defined by ab a * b = 2' Determine the identity for the binary operation *, if exists. Sol. Let us assume that e be a +ve integer number, then e * a = a, a E 1+ ea  = a => e = 2 2
... (1)
Similarly,
ae =a
2
Form (1) and
or
e=2
... (2)
(2) for e = 2, we have e * a = a * e = a
2 is the identity element for *. 5. Inverse. Consider a nonempty set A and a binary operation * on A. Then operation
Therefore,
* has the inverse property if for each a E A, there exists an element b in A such that a * b (right inverse) = b * a (left inverse) = e, where b is called an inverse of a. 6. Idempotent. Consider a non· empty set A and a binary operation * on A. Then the operation * has the idempotent property, if for each a E A, we have a * a = a V a E A. 7. Distributivity. Consider a non·empty set A and two binary operations * and + on A. Then the operation * distributes over +, if for every a, b, C E A, we have [Left distributivity] a * (b + c) = (a * b) + (a * c) and (b + c) * a = (b * a) + (c * a) [Right distributivity] 8. Cancellation. Consider a non·empty set A and a binary operation * on A. Then the operation * has the cancellation property, if for every a, b, C E A, we have a * b = a * c => b = c [Left cancellation]
and
[Right cancellation]
(P.T. U. B.Tech. Dec.
8.6. SEMIGROUP
2009,
May 2008)
Let us consider, an algebraic system (A, *) , where * is a binary operation on A. Then, the system (A, *) is said to be a semi· group if it satisfies the following properties : 1. The operation * is a closed operation on set A.
2. The operation * is an associative operation. Example 8. Consider an algebraic system (A, *), where A = {I, 3, 5, 7, 9, ...}, the set ofall positive odd integers and is a binary operation means multiplication. Determine whether (A, *) is a semigroup. *
222
DISCRETE STRUCTURES
Sol. Closure property. The operation * is a closed operation because multiplication of two +ve odd integers is a +ve odd number. Associative property. The operation * is an associative operation on set A. Since for b, C E A, we have (a * b) * c = a * (b * c) Hence, the algebraic system (A, *) is a semigroup.
every a,
Example 9. Consider the algebraic system ({D, i), *), where * is a multiplication opera tion. Determine whether ({D, i), *) is a semigroup. Sol. Closure property. The operation * is a closed operation on the given set since 0 * 0 = 0 ; 0 * 1 = 0 ; 1 * 0 = 0 ; 1 * 1 = 1. Associative property. The operation * is associative since we have (a * b) * c = a * (b * c) \;j a, b, c Since, the algebraic system is closed and associative. Hence, it is a semigroup.
Example 10. Let S be a semigroup with an identity element e and if b and b' are inverses of an element a E S, then b = b' i.e., inverse are unique, if they exist. Sol. Given b is an inverse of a, therefore, we have a*b=e=b*a Also, b' is an inverse of a, therefore, we have a * b' = e = b' * a b * (a * bi = b * e = b Consider ... (1) and (b * a) * b' = e * b' = b' ... (2) Now, S is a semigroup, associativity holds in S i.e., b * (a * bi = (b * a) * b' b = b'. I Using (1) and (2) e=} Example 11. Let N be the set of positive integers and let * be the binary operation of least common multiple (L. c.lYI) on N. Find (a) 4 * 6, 3 * 5, 9 * i8, i * 6 (b) Is (N, *) a semigroup (c) Is N commutative (d) Find the identity element ofN (e) Which elements of N have inverses ? Sol. (a) Let x, Y E N and x * y = L.C.M. of x and y .. 4 * 6 = L.C.M. of 4 and 6 = 12 3 * 5 = L.C.M. of 3 and 5 = 15 9 * 18 = L.C.M. of 9 and 18 = 18 1 * 6 = L.C.M. of 1 and 6 = 6 (b) We know that the operation of L.C.M. is associative, i.e., a * (b * c) = (a * b) * c \;j a, b, c E N . . N is a semigroup under *. (c) Also for a, b E N, a * b = L.C.M. of a and b = L.C.M. of b and a = b * a N is commutative also.
MONOIDSAND GROUPS (d) For a E
223
N, consider
a * 1 = L.C.M. of a and 1 = a 1 * a = L.C.M. of 1 and a = a a*l=a=l*a
.. i.e., 1 is the identity element of N. (e) Consider a * b = 1 i.e., L.C.M. of a and b is 1, which is possible iff a = 1 and b = 1. i.e.) the only element which has an inverse is 1 and it is its own inverse. Example 12. Consider the set Q of rational numbers and let * be the operation on Q defined by a * b = a + b  ab 1 (a) Find 3 * 4, 2 * ( 5), 7 * '2 (b) Is (Q, *) a semigroup ? (c) Is Q commutative ? (d) Find the identity element of Q. (e) Which elements of Q have inverses and what are they ? Sol. Given a * b = a + b  ab for a, b E Q (a) 3 * 4 = 3 + 4  12 =  5 2 * ( 5) = 2 + ( 5)  ( 10) = 2  5 + 10 = 7
1
1 7
7 *  = 7 +    = 4.
2 2 2 (b) Q will be a semigroup if it holds associativity under Consider a * (b * c) = a * (b + c  be)
* for a, b, C E
= a + (b + c  be)  a(b + c  be) = a + b + c  be  ab  ac + abc (a * b) * c = (a + b  ab) * c = a + b  ab + c  (a + b  ab) c = a + b + c  ab  ac  be + abc = a + b + c  be  ab  ac + abc From (1) and (2), a * (b * c) = (a * b) * c Hence) (Q) *) is a semigroup. (c) For a, b E Q Consider
a * b = a + b  ab = b + a  ba = b * a . . Q is commutative. (d) Let e is the identity element of Q, therefore, for a E a + e  ae = a e  ea = 0 e(l  a) = 0 e = O if a i" l
The identity of Q is O.
Q, we have
Q.
... (1)
... (2)
DISCRETE STRUCTURES
224 (e) If x is the inverse of a E Q, then a * x = 0 a + x  ax = O a + x(l  a) = 0 a = x(a  1) x= Thus
a has an inverse
(identity)
a ) a :t 1 a1

a
_ _
a1
.
Example 13. Consider a nonempty set S with the operation a * b (a) Is the operation associative ? (b) Is the operation commutative ? (c) Show that the right cancellation law holds. (d) Does the left cancellation law hold ? Sol. (a) For a, b, c E S, Consider and
=a
a * (b * c) = a * b = a (a * b) * c = c * a = a
* is associative. (b) For a '" b E S, Consider
a * b = a and b * a = b a * b ", b * a
* is not commutative. (c) For a, b, C E Q, Consider
a*c=b*c a=b
I
Using given a
*b=a
Right cancellation laws hold.
(d) The left cancellation law does not hold. For example, suppose b '" c, then a*b=a*c b = c) a contradiction => Hence) the result.
Example 14. Let (A, *) be semigroup. Show that for a, b, c in A, if a * c = c * a and b * c = c * b, then (a * b) * c = c * (a * b). Sol. Take L.R.S., we have [": * is associative] (a * b) * c = a * (b * c) = a * (c * b) [ .: b * c = c * b] = (a * c) * b [": * is associative] = (c * a) * b [ .: a * c = c * a] = c * (a * b) [": * is associative] (a * b) * c = c * (a * b).
MONOIDSAND GROUPS
225
8.7. SUBSEMIGROUP Let A be a non· empty subset of a semi group S. Then A is called a subsemigroup of S if A is itself a semigroup with respect of the same operation on S. Since the elements of A are also elements of S, the associative law automatically holds for the elements of A. Therefore, A is a subsemigroup of S iff A closed under the operation of S. Example 15. Let A and B are the sets of even and odd positive intergers. Then (A, x) and (E, x) are subsemigroups of(N, +). Is (A, +) subsemigroup of(N, +) ? Is (E, +) subsemigroup of (N, +) ? Sol. Given A = {set of even positive integers} B = {set of odd positive integers} Also N is a semigroup under addtion and multiplication. Let m, n E A ::::::} m x n E A i.e.) A is closed under multiplication. Therefore) A is a subsemigroup of N
under multiplication.
Similarly) B is also closed under multiplication. Therefore) B is also a subsemigroup of
N under multiplication.
Further) sum of two even positive integers is also a positive even integer. Therefore, A is also closed under addition. Hence A is a subsemigroup of (N, +). But sum of two positive odd integers is not always an odd integer (e.g., B is not closed under
+.
Hence B is not a subsemigroup of (N,
5 + 3 = 8 (even»
+).
For example,
Consider a semigroup (N, +), where N is the set of all natural numbers and + is an addition operation. The algebraic system (E, +) is a subsemigroup of (N, +), where E is a set of all even natural number.
8.8. FREE SEMIGROUP Consider a nonempty set A. A word For example, if we take A = {a, b, c}.
w on A is a finite sequence of its elements.
abcabca, aabbcc, baccaaaa, etc. are words on A. Also, the length of a word w, denoted by l(w), is the number of elements in w. Then, the elements
Consider the following words
u 1 = abcabca, u2 = aabbcc, u3 = baccaaaa l(u,) = the length of the word u , = number of elements in u 1 =7 Similarly, l(u) = 6, l(u) = 8 We now define concatenation of two or more words on the set A. By concatenation of words u 1 and u2 ) written as u 1 * u2 or u 1 u2 ) we mean the word obtained by writing down the elements of u , followed by the elements of u2 • Thus, u , * u2 = u,u2 = (abcabca) (aabbcc) = abcabca3 b 2c2 Here
226
DISCRETE STRUCTURES Here
Also,
=>
l(u , U) = length of the word u , u2 = number of elements in the word u 1 u2 = 13 l(u ,) = 7 and l(u) = 6 l(u,) + l(u) = 7 + 6 = 13 l(u , u) = l(u,) + l(u)
Thus, Let F denotes the collection of all words on a given set A under the operation of concat enation. Then) for up u2 E F, we have proved that u 1 U2 E F.
Also, for up u2' u3 E F, the number of elements in the words u,(U2 uj and (u , u) u3 will be same. Hence, we can say that F is a semigroup under the operation of concatenation and this semigroup is called free semigroup on A and the elements of A are called generators of F.
8.9. CONGRUENT RELATIONS AND QUOTIENT STRUCTURES
R is an equivalence relation on a given set S and let a E S. Then, [a] = Equivalence class of a in S = {x E S: (a, x) E R} Thus, [a] denotes the set of all elements of S to which a is related under R. Further, the collection of all equivalence classes of elements of S under an equivalence relation R is de Suppose
noted by SIR. Thus,
SIR = ([a] : a E S) where SIR (read as the quotient space of S by R) is called the
quotient set.
Further, suppose that the equivalence relation R on S has the following property:
If aRa', bRb' => abRa'b' \;j a, b E S Then R is called a congruence relation on S. We now show that SIR forms a classes, defined by
semigroup
under the operation
*
on the equivalence
[a] * [b] = [a * b] or [a] [b] = lab] Since SIR is closed under the operation * on the equivalence classes (by definition), it remains to show SIR is associative Consider ([a] * [b]) * [c] = [a * b] x [c] Also
= [a * b * c] [a] * ([b] * [c]) = [a] * [b * c] = [a * b * c] [a] * ([b] * [c]) = ([a] * [b]) * [c]
Thus, Hence, associativity holds in SIR. Thus, SIR forms a semigroup under the operation * on the equivalence classes. This, semigroup SIR is called quotient of S by R.
Example 16. Let F be a free semigroup on a set A and R be an equivalence relation on F defined by "uRu' if l(u) = l(u )" Show that R is a congruence relation on F. Sol. Let uRu', vRv'. then l(u) = l(u') = m, say and l(v) = l(v') = n, say
227
MONOIDSAND GROUPS l(uv) = l(u) + l(v) = m + n l(u'v') = l(u') + l(v') = m + n => uvRu'v'
Consider and
Thus R is a congruence relation of F. Example 17. Let Z be the set of integers and m > 1 be any positive integer. Consider a relation "congruence modulo m" defined by a = b (mod m) if ml a  b V a, b E Z (read as "a is congruent to b modulo m if m divides (a  b)'). Show that is a congruence relation on Z. Sol. We first show that "congruence modulo m" is an equivalence relation on Z. We know that for any integer a E Z , the difference a  a = 0 is divisible by m where m > 1 '=:0'
is any positive integer.
Hence a == a(mod m») which proves that '=' is reflexive Also, if a = b (mod m), then m/(a  b) V a, b E Z
=> =>
ml(a  b) = b  a => m/(b  a) b = a(mod m)
which proves that '=' is symmetric. Further, if a = b (mod m), b = c (mod m), we show a = c (mod m) V a, b, c E Z Now a = b (mod m) => m/(a  b) V, a, b E Z b oo c (mod m) => m/(b  c) V, b, c E Z which further implies that m/(a  b) + (b  c) = a  c => m/(a  c) => a = c(mod m) Hence the relation '=' is an equivalence relation on Z. Finally, we show that the equivalence relation '=' is a congruence relation on Z. Consider a = c (mod m) and b = d(mod m), we show ab = cd (mod m). Now a = c (mod m) => m/(a  c) => mlb(a  c) Also, b oo d (mod m) => m/(b  d) => mlc(b  d) V a, b, c E Z Hence mlb(a  c) and mlc(b  d) V a, b, c, d E Z mlb(a  c) + c(b  d) = ba  bc + cb  cd =>
mlba  cd = ab  cd => ab = cd (mod m).
mlab  cd
Hence the result.
Example 18. Let (J, +) be a semigroup and R is an equivalence relation on J defined by aRb iff a = b (MOD 3). Sol. If a and b yield the same remainder when divided by 3, then we have 3 divides a  b i.e., 31a  b. Now, if a = b (MOD 3) and c = d (MOD 3), then 3 divides a  b and 3 divides c  d. Thus, we can write a  b = 3m ... (1) and c  d = 3n ... (2) I al b => b = at for some t Here m and n are some integers of I. Adding (1) and (2), we have (a  b) + (c  d) = 3m + 3n or (a + c)  (b + d) = 3(m + n) => 3 divides (a + c)  (b + d)
228
DISCRETE STRUCTURES a + c = (b + d) (MOD 3)
or
Thus) the relation is a congruence relation.
Example 19. Consider the set A = {a, b}. Let (A*, ) is the semigroup generated by A, also let R is a relation on A defined by aR� iff a and � have the same number of a's. Show whether the relation R is a congruence on (A*, .). Sol. First of all we will show that R is an equivalence relation. So, for that we will check
reflexive) symmetric and transitive properties of the relation R.
Reflexive.
reflexive.
aRa for any a
E
A* since a has same number of a's as itself. Thus, R is
Symmetric. If a and � have same number of a's, then aR� or we can say �Ra. Thus, R
is symmetric.
Transitive. If aR�, it means a and � have same number of a's. If �Ry, it means � and y have same number of a's. It implies a and y have same number of a's i.e., aRy. Thus, R is transitive.
Hence, R is an equivalence relation. To show that R is a congruence relation, let us assume that aRa, and �R�" It means a and a, have same number of a's and � and �, have same number of a's. We know that the number of a's in a . � is the sum of number of a's in a and the number of a's in �. From the above discussion, we can say that the number of a's in a .� is same as in aI" � ' l Hence (a . �) R (a, . �,) which shows that R is a congruence relation.
Example 20. Consider the semigroup (I, +), where + is an addition operation. Let f(x) = x2  2x 3 and also let R is a relation on I defined by aRb ifff(a) = f(b). Show whether R is a congruence relation. Sol. We first show that the relation R is an equivalence relation on the set 1. (i) f(a) = f(a) => aRa i.e., R is symmetric (ii) aRb => f(a) = f(b) => f(b) = f(a) i.e., bRa. Hence R is symmetric (iii) If aRb, bRc, thenf(b) = f(b) andf(b) = f(c) => f(a) = f(c) => aRc i.e., R is transitive
To check whether R is congruence relation or not, we will try to find two pair of num bers aRb and cRd but (a + b) R (c + d), if possible. Then we will say R is not a congruence relation. Thus, we have and
2RO Le., f(2) = f(O) =  3 f( 2) = f(4) = 5  2R4 i.e., But (2 + ( 2» R (0 + 4) Le., 0 R 4 As f(O) =  3 and f(4) = 5 Hence, R is not a congruence relation.
Example 21. Let (S,*) be a commutative semigroup. Show that ifx * x = x and y * y = y, then (x * y) * (x * y) = x * y. Sol. Take L.H.S. (x * y) * (x * y) = (x * y) * (y * x) (8, *) is a commutative semigroup] =x*y*y *x=x*y *x [ .; y * y = y] . [ ; Commutative semigroup] =x*x*y =x*y [ .; x * x = x] = (x * y) * (x * y) = x * y.
MONOIDSAND GROUPS
229
Example 22. Let ({x, y), .) be a semigroup where x . x = y show that (i) x . y = y . x (ii) y . y = y. x . y = y. x Sol. (i) To show that We have X . X . X = X . X . x ::::::} x . y = y . x
[ .: x . x = y]
Hence proved.
(ii) To show that We know that the set
Case. I
Let Consider
Case. II
Let
Consider
y . y =y (x, y) is closed under the operation. Therefore, we have two cases x. y = x) x . y = y x.y=x y . y = y . (x . x) = (y . x) . x [ ": . is associative] = (x . y) . x [ .: x . y = y . x] =x.X [ .: x . y = x] =y x.y=y y . y = (x . x) . y [ .: x . x = y] = x . (x . y) [ ": . is associative] =x.y [ .: x . y = y] =y
Hence proved.
Example 23. Let (A, *) be a semigroup. Further more, for every a and b in A, if a # b then a * b # b * a. (a) Show that for every a in A, a * a = a (b) Show that for every a, b in A, a * b * a = a (c) Show that for every a, b, c in A, a * b * c = a * c. Sol. (a) We know that A is a semigroup. .. (a * b) * c = a * (b * c) Now putting b = a and c = a) we have (a * a) * a = a * (a * a) Since A is not commutative semigroup. i.e.) a * b t: b * a) H�� a*a=a ... (1) (b) Let us assume that b E A, then we have b*b=b Multiplying both sides by a, we get a * b * b = a * b or (a * b) * b = a * b [ ": * is associative] Hence) a*b=a ... (2) * is associative] a * b * a = (a * b) * a So, a * b = a from (2)] =a*a =a a * a = a from (1)] a * b * c = (a * b) * c (c) We know that [ ": * is associative] =a*c [ .: a * b = a from (2)]
230
DISCRETE STRUCTURES
8.10. HOMOMORPHISM OF SEMIGROUPS Let (8. *) and (8'. *') be any two semi·groups. Then. a function ffrom 8 to 8' is called a
semigroup homomorphism (or a homomorphism) if f(a * b) = f(a) *' f(b) \j a. b E 8 In addition) if f is oneone and onto) then f is called an isomorphism of S and 8' and we write S
==
8'. Also) S and 8' are said to isomorphic subgroups.
Example 24. Consider G = {I, 3, 7, 9} under multiplication modulo iO and r Z4 ; G be a function defined by f(O) = i, f(1) = 3, f(2) = 9, f(3) = 7 Show that f is a semigroup homomorphism. Also prove that Z4 G. Sol. By definition Z4 = (O. 1. 2. 3. +4) G = {I . 3. 7. 9. x l O} We first show that (Z4 ' +4) and (G. x lO) are both semigroups under the addition modulo 4 and multiplication modulo 10 respectively. The addition modulo table (Table 2a) for Z4 and the multiplication modulo table (Table 2b) '"
for G are shown below.
Table 2 +4
0
1
2
3
X ,O
1
3
7
9
0 1 2 3
0 1 2 3
1 2 3 0
2 3 0 1
3 0 1 2
1 3 7 9
1 3 7 9
3 9 1 7
7 1 9 3
9 7 3 1
(a)
(b)
From Table 2a. we observe that each element in the interior of the table is also on element of Z4' It means Z4 is closed under +4" Also) for a,
b, C E Z4) we know a +4(b +4 c) = (a +4 b) +4 c is true i.e., associativity holds in Z4' Hence) (Z4 ) +4) is a semigroup Similarly) we can prove that (G) x l O) is a semigroup. To show f is a semigroup homomorphism. We need to show f(a +4 b) = f(a) x , O f(b) \j a. b E Z4 We first find (a +4 b) Now) the only elements in Z4 are 0) 1) 2 or 3. Therefore) the value of a +4 b can be either o or 1 or 2 or 3 (since Z4 is closed under +4)' Consider the following possibilities: (1) If the value of a +4 b is 0 (2) If the value of a +4 b is 1 (3) If the value of a +4 b is 2 (4) If the value of a +4 b is 3 We discuss the case (1) when the value of a +4 b is O. From Table 2. we observe that a and b can take the following possible values.
MONOIDSAND GROUPS
231
b = 0 so that a +4 b = 0 => [(a +4 b) = [(0) = 1 b = 3 so that a +4 b = 1 + 3 = 4 = 0 => [(a +4 b) = [(0) = 1 b = 2 so that a +4 b = 2 + 2 = 4 = 0 => [(a +4 b) = [(0) = 1 b = 1 so that a +4 b = 3 + 1 = 4 = 0 => [(a +4 b) = [(0) = 1 We summarise the above obtained values in the following Table 3.
(i) a = 0, (ii) a = 1, (iii) a = 2, (iv) a = 3,
Table 3. Calculation off(a +4 b) a
0 1 2 3
f(a +4 b)
a +4 b
b
0 3 2 1
1 1 1 1
0 0 0 0
Similarly, the various values of [(a) x lO [(b) are calculated in Table 4. Table 4. Calculation of [(a) x,O [(b) a
0 1 2 3
f(a)
b
f(b)
1 3 9 7
0 3 2 1
1 7 9 3
f(a) xlO f(b)
l X10 1 :::: 1
3 9 7
x,O x,O
x,O
7 21 1 9 81 1 3 21 1 =
=
=
=
=
=
On comparing Table 3 and Table 4, we observe that the last column of these two tables are identitcaL It means [(a +4 b) = [(a) x,O [(b) \;j a, b E Z4 Hence, fis a semigroup homomrphism
To show fis 11 and onto: Here [ : Z4 ; G such that [(0) = 1 , [(1) = 3 , [(2) = 9, [(3) = 7. S.l)
Consider the following (Fig.
Z4
G Fig.
8.1
Since different elements of Z4 have different images in G, therefore, fis oneone. Also, each element of G has its preimage in Z4' Therefore, fis onto also. Thus, we have shown that f is homomorphism, oneone and onto. Thus, S 4
==
G.
232
DISCRETE STRUCTURES Example 25. Let M be the set of 2 x 2 matrices over integers of the types
(� �)
s. t.
ad  bc '" O. Define f: M ; M such that f(A) = det. A \;j A E M. Then f is a semigroup homomorphism on (M, x), but not on (M, +). Sol. We know that (M, +) and (M, x) are both semigroups. Let A, B E M and consider f(A + B) = det. (A + B) '" det . A + det . B '" f(A) + f(B) Thus, f is a not a semigroup homomorphism on (M, +). Now, consider
f(AB) = det. (AB) = (det. A) (det. B) = f(A) f(B)
Thus, f is a semigroup homomorphism on (M, x).
Example 26. Let  be a congruence relation on a semigroup S. Let q, : S ; SI be the natural mapping from "S to the quotient space of S by  " (denoted by SI) defined by q,(a) = fa] then q, is a semigroup homomorphism. Sol. Let a, b E S and consider q,(ab) = lab] = [a][b] = q,(a)q,(b) thus) q, is a semigroup homomorphism.
Remark. We know that SIR denotes the set of all equivalence classes of elements of S under an equivalence relation R. Also, if R is a congruence relation on S, then SIR forms a semigroup under the operation on the equivalence classes.
8.11 FUN DAMENTAL THEOREM OF SEMIGROUP HOMOMORPHISM
Theorem. Let f: S ; S' be a semigroup homomorphism. Let  be a relation on S defined by "a  b iff(a) = f(b)" then (i) ,..., is a congruence relation on S (ii) SI is isomorphic to f(S) Proof. (i) We know that a relation R on a given set S is said to be a congruence relation on S if R is an equivalence relation on S and satisfies the following. If aRa', bRb' => abRa'b' \;j a, b E S We first show that R is an equivalence relation on S. (a) Reflexive. Since f(a) = f(a) => a  a \;j a E S �
is reflexive
(b) Symmetric. Let a  b �
=>
f(a) = f(b)
=>
f(b) = f(a)
is symmetric (c) Transitive. Let a  b => f(a) = f(b) \;j a, b E S Let b  c => f(b) = f(c) \;j b, c E S It implies that f(a) = f(b) = f(c) f(a) = f(c) => a  c \;j a, c E S is transitive . .
�
=>
b  a \;j a, b E
S
233
MONOIDSAND GROUPS Hence)
� is an equivalence relation on S.
We next show that R is a congruence relation on Let a � a' and b � b'. Then f(a) = f(ai and f(b) = f(bi Consider f(ab) = f(a) f(b)
S. I
As f is a homomorphism
= f(ai f(bi = f(a'bi ab � a'b'
. . � is a congruence relation on S (ii) To show S/� is isomorphic to f(S). Define q, : S/� ; f(S) by q,([a]) = f(a) We show q, is welldefined) homomorphism) oneone and onto. (a) q, is welldefined. Consider [aJ = [bJ => a � b => f(a) = f(b) => q,([a]) = q,([b]) . . q, is well·defined (b) q, is homomorphism. Consider q,([aJ [b]) = q,([ab]) = f(ab) = f(a) f(b) I fis a semi·group homomorphism = q,([a]) q,([b]) . . q, is homomorphism (c) q, is oneone. Let q,([a]) = q,([b]) => f(a) = f(b) a�b [aJ = [bJ . . q, is oneone (d) q, is onto. Let Y E f(S) => 3 a E S such that y = f(a) Now q,([a]) = f(a) = y Th us q, is onto Hence, q, : S/� ; f(S) is an isomomorphism Example 27. Let F be a free semigroup on A = {a, b}. Define f = F ; Z such that f(u) = l(u) where l(u) denotes the length of the word 'u' in F. Let a � b if f(a) = f(b). Show that f is a homomorphism. Also, FI� is isomorphic to N. Sol. Suppose up u2 E F and consider f(u , u) = l(u , u) = l(u,) l(u) = f(u,) f(u) :. f is a homomorphism Also, for u E F, f(F) = l(u) = a natural number .. f(F) = N By fundamental theorem of semigroup homomorphism, F/� is isomorphic to f(F) = N Thus F/� is isomorphic to N
234
DISCRETE STRUCTURES Example 28. Let M be the set ofall 2 x 2 type matrices over integers of the type
(� �)
and let f: M ; Z be a function defined by f(A) = det. A Let A � B if f(A) = f(B). Then MI� is isomorphic to Z. Sol. We know that (M, x) is a semi·group under the multiplication of matrices. Also (Z, ) is a semi·group Let A, B E M and consider f(AB) = det. (AB) = (det. A) (det. B) = f(A) f(B) Thus, f is a semi·group homomorphism from M to Z. By fundamental theorem of semi· group homomorphism, M/� is isomorphic to f(Z). But f(Z) = (b E Z: There exists A E M for which f(A) = b) = (b E Z: det. A = b) =Z l\V� is isomorphic to Z 8.12. MONOID
(P.T.U. B.Tech., Dec. 2013, May 2013, Dec. 2012, May 2008)
Let us consider an algebraic system (A) 0») where 0 is a binary operation on A. Then the system (A, 0) is said to be a monoid if it satisfies the following properties. (i) The operation 0 is a closed operation on set A. (ii) The operation 0 is an associative operation. (iii) There exists an identity element w.r.t. the operation o. Examples. (N, x), (Z, +), (Q, +) are monoids
Example 29. Consider an algebraic system (L +), where the set 1 = {O, 1, 2, 3, 4, .. .} the set of natural numbers including zero and + is an addition operation. Determine whether (L +) is a monoid. Sol. Closure property. The operation + is closed since sum of two natural numbers is a natural number.
Associative property. The operation + is an associative property since we have (a + b) + c = a + (b + c) \;j a, b, c E I. Identity. There exists an identity element in set I w.r.t. the operation +. The element 0 is an identity element w.r.t. the operation +. Since) the operation + is a closed) associative and there exists an identity. Hence, the algebraic system (I, +) is a monoid. Example 30. Let S be a finite set and F(s) be the collection of all functions f : S ; S under the operation ofcomposition offunctions. Show that F(s) is a semigroup. Is F(s) a monoid? Sol. Let t. g, h E F(s), then we know that composition of functions is associative i.e., fo(goh) = (fog)oh \;j f, g, h E F(s). Hence, F(s) is a semigroup. Also the identify function is an identify element of F(s). F(s) is a monoid. 8.13. SUBMONOID Let us consider a monoid (M, 0), also let 8 c M. Then (M, 0), if and only if it satisfies the following properties.
(8, 0) is called a submonoid of
MONOIDSAND GROUPS
235
(i) S is closed under the operation o. (ii) S is associative under the operation o. (iii) There exists an identity element e E S. e.g., Let us consider) a monoid (M) *»)where * is a binary operation and M is a set of all inte gers. Then (M ' *) is a submonoid of (M, *), where M, is defined as l i M, = {a I i is from 0 to n, a positive integer and a E M}.
Examples 31. Since N e Z and Z c Q, N is a submonoid of Z and Z is a sub monoid of Q. Theorem II. Let [M; *i be a monoid and K is a nonempty subset of M. Then K is a submonoid of M iff (i) a, b E K => a * b E K i.e. K is closed under * (ii) For each a E K, there exists e E K such that a * e = e * a = a. Proof. Let K is a submonoid of M. Then, by defination, K must be closed under *. Also for each a E K, there exists e E K such that a * e = e * a. Hence (i) and (ii) hold Converse. Let (i) and (ii) hold From (i), a, b E K => a * b E K i.e., K is closed under * Also) if a) b) C E K and K c M. :. a, b, C E M. But M is a monoid .. a * (b * c) = (a * b) * c \;j a, b, c E K i.e., the associativity holds for each element of K. From (ii). K has an identity element K is a monoid. But K c M
K is a submonoid of M under *.
L
2. 3.
4. 5.
I
TEST YOUR KNOWLEDGE 8.1
Let * be the operation on the set R of real numbers defined by a * b :::: a + b + 2ab (a) Find 2 * 3, 3 * ( 5), 7 * (1/2) (b) Is (R, *) a semigroup ? Is it commutative ? (c) Find the identity element (d) Which elements have inverses and what are they ? Let S be a semigroup with identity e and let b and b' be inverses of a. Show that b :::: b' i.e., inverses are uniques, if they exist. Prove that for any commutative monoid (M *), the set of idempotent elements of M form a submonoid. If a, b are elements of a monoid M and a * b :::: b * a. Show that (a * b) * (a * b) = (a * a) * (b * b) Let S :::: Q x Q, the set of ordered pairs of rational numbers, with the operation * defined by (a, b) * (x, y) = (ax, ay + b) (a) Find (3, 4) * (1, 2) and ( 1, 3) * (5, 2) (b) Is S a semigroup ? Is it commutative ? (c) Find the identity element of S (d) Which elements, if any, have inverses and what are they ? ;
236 6.
DISCRETE STRUCTURES
Let S :::: N x N, the set of ordered pairs of positive integers with the operation * defined by (a, b) * (e, d) (ad + be, bd) (a) Find (3, 4) * (1, 5) and (2, 1) * (4, 7) (b) Is S a semigroup ? Is S commutative ? Let A be a nonempty set with the operation * defined by a * b :::: a and assume A has more than one element. Then (a) Is A a semigroups ? (b) Is A commutative ? (c) Does A have an identity element ? (d) Which elements, if any have inverses and what are they ? Let A be a nonempty set with the operation * defined by a * b :::: a, and assume A has more than one element. (a) Is A a semigroup? (b) Is A commutative? (c) Does A have an identity element? (d) Which elements, if any, have inverses and what are they ? Let A :::: {a, b}. Find the number of operations on A , and exhibit one which is neither associative nor commutative. Find a set A of three numbers which is closed under: (a) multiplication, (b) addition. Let S be an infinite set. Let A be the collection of finite subsets of S and let B be the collection of infinite subset of S. (a) Is A closed under : (i) union, (ii) intersection, (iii) complements ? (b) Is B closed under : (i) union, (ii) intersection, (iii) complements ? =
7.
8.
9.
10. 1L
Answers 1. 5.
6. 7. 8. 9.
29 17,  32, ""2 (b) Yes, Yes a (e) zero (d) If a :f ...!.. , then a has an inverse which is 2 (1 + 2a) (a) (3, 10), ( 5, 1) (b) Yes, No (e) (1, 0) (d) The element (a, b) has an inverse if a :f 0 and its inverse is (�,  %) (a) (19, 20), (18, 7) (b) Yes, Yes (a) Yes (b) No (c) No (d) It is meaningless to talk about inverses when no identity element exists. (a) Yes (b) No (c) No (d) If identity element does not exist, then it is meaningless to talk about inverses. 16, as there are two choices a or b for each of the four products aa, ab, ba and bb (Fig.). Further, From Fig. 8.1 ab :f ba. Also a b [ (aa)b =bb = a a b a But a(ab) = aa = a (a)
*
10. 11. L
(a) {(a) (i)
(aa) b " a(ab) OJ (b) (ii) (iii)
1, 1, No such set exists Yes Yes No
Given
b
(b) (i)
Hints
Yes (ii) No (iii) No.
(a) a * b :::: a + b + 2ab . . 2 * 3 = 2 + 3 + 12 = (b) (R, *) a * (b * e) = (a * b) * e \j a, b, e E R
17 is semigroup iff we can show
b
a
Fig. 8.1
237
MONOIDSAND GROUPS
Consider a * (b * c) :::: a * (b + c + 2bc) = a + (b + c + 2bc) + 2a(b + c + 2bc) ... (1) :::: a + b + c + 2bc + 2ab + 2ac + 2abc Also (a * b) * c = (a + b + 2ab) * c = a + b + 2ab + c + 2(a + b + 2ab) c ... (2) :::: a + b + 2ab + c + 2ac + 2bc + 2abc From (1) and (2). we have a * (b * c) = (a * b) * c (R, *) is a semigroup Further, (R, *) is commutative iff we can show a * b :::: b * a a, b R Now a * b :::: a + b + 2ab :::: b + a + 2ba :::: b * a. (c) Let e is an identity element of R, then for each a R, we have a * e :::: a a + e + 2ae = a e(l + 2a) = 0 1 1 + 2a ," O e=O (d) Let b R is the inverse of a. Then we must have e :::: 0, the identity a * b =O a + b + 2ab = 0 a + b(l + 2a) = 0 b(l + 2a) =  a b =  all + 2a. 1 + 2a ," 0 Since S is a semigroup with identity e and b and b' are inverses of a ... (1) b*a=e=a*b b' * a :::: e :::: a * b' ... (2) Now S is a semigroup under *. Therefore associativity holds among the elements of S under *. b * (a * b') = (b * a) * b' Le., b * e :::: e * b' I Using (1) and (2) b :::: b' Let S be the set of idempotent elements i.e., S :::: [a M: a * a :::: a]. We show S is a submonoid. Let a, b S, then a * a :::: a, b b :::: b V
E
�
E
=>
E
2.
=:::}
3.
E
:f
(a * b) * (a * b) = a * (b * a) * b = a * (a * b) * b = (a * a) * (b * b) =a*b S is closed under *.
Consider
4.
E
I Associativity
I Commutative
Also e * e :::: e e S i.e., S has an identity element. Therefore, S is a submonoid. (a * b) * (a * b) = a * (b * (a * b» I Associativity = a * ((b * a) * b) I Associativity = a * ((a * b) * b) I a * b :::: b * a Associativity = a * (a * (b * b» = (a * a) * (b * b) I Associativity
8.14. GROUP
::::}
E
(P.T. U. B.Tech. Dec. 2013, May 2008, 2006)
Let us consider an algebraic system (G, *), where * is a binary operation on G. Then the system (G. *) is said to be a group if it satisfies the following properties:
238
DISCRETE STRUCTURES
(i) The operation * is a closed operation. (ii) The operation * is an associative operation. (iii) There exists an identity element w.r.t. the operation *. (iv) For every a E G) there exists an element a I E G such that a I * a = a * a I = e For example, the algebraic system (I, +), where I is the set of all integers and + is an addition operation) is a group. The element 0 is the identity element w.r.t. the operation +. The inverse of every element a E I is  a E I. Examples (i) The sets (Q, +) (R, +) and (C, +) are groups under addition. (ii) The sets R* (set of nonzero reals), Q* (set of nonzero rationals) and C* (set of non zero complex numbers) are groups under multiplication.
ILLUSTRATIVE EXAMPLES
Example 1. Determine whether the algebraic system (Q, +) is a group where Q is the set of all rational numbers and + is an addition operation. Sol. Closure Property. The set Q is closed under operation +, since the addition of two rational numbers is a rational number. Associative Property. The operation + is associative, since (a + b) + c = a +
(b + c) 'd a, b, C E Q. Identity. The element 0 is the identity element. Hence a + 0 = 0 + a = a 'd a E Q. Inverse. The inverse of every element a E Q is  a E Q. Hence the inverse of every
element exists. Since, the algebraic system a group.
(Q, +) satisfies all the properties of a group, hence (Q, +) is
Example 2. Which of the following are groups under addition N, Z, Q, R, C ? Sol. The set of integers Z, the set of rationals Q, the set of reals R, the set of complex
numbers C, are all groups under addition. (Prove yourself as in ExampleI) But N) the set of natural numbers donot form a group under addition. Since) N does not have additive identity. (0 'l N).
Example 3. Let S be the set of n x n with rational entries under the operation of matrix multiplication. Is S a group ? Sol. We know that matrix multiplication is associative. But inverse does not always exist. As we know that if I A I '" 0, then AI exists. Example 4. Prove that G = {l, 2, 3, 4, 5, 6} is a finite abelian group of order 6 under multiplication modulo 7. (P.T.V. B.Tech. May 2010, 2009) Sol. G = {I, 2, 3, 4, 5, 6, x7} Consider the multiplication modulo 7 table as shown below (Table 8.5). Recall that a x 7 b = The remainder when ab is divided by 7 X7
1
2
3
4
5
6
1 2 3 4 5 6
1 2 3 4 5 6
2 4 6 1 3 5
3 6 2 5 1 4
4 1 5 2 6 3
5 3 1 6 4 2
6 5 4 3 2 1
Table 8.5
MONOIDSAND GROUPS
239
From the table, we observe that each element inside the table is also an element of G. It means that G is closed under multiplication modulo 7.
a) b) C E G a x 7 (b x 7 c) = (ax 7 b) x 7 c i.e., associative law hold.
Also for each
From the table, we observe that the first row inside the table is identical with the toprow of the table. Therefore, 1 is the identity (multiplicative) of G. Hence) each element G has an inverse)
i.e.,
Inverse of 2 is 4 and of 4 is 2 Inverse of 3 is 5 and of 5 is 3 Inverse of 6 is 6 Hence, G is a group under the multiplication modulo
7.
Example 5. Consider an algebraic system (Q, *),where Q is the set of rational numbers and * is a binary operation defined by a * b = a + b  ab \;j a, b E Q. Determine whether (Q, *) is a group. Sol. Closure property. Since the element a * b E Q for every a, b E Q, hence, the set Q is closed under the operation *. Associative property. Let us assume a, b, C E Q, then we have (a * b) * c = (a + b  ab) * c = (a + b  ab) + c  (a + b  ab)c = a + b  ab + c  ac  be + abc = a + b + c  ab  ac  be + abc Similarly, a * (b * c) = a + b + c  ab  ac  be + abc. Therefore, (a * b) * c = a * (b * c) * is associative. Identity. Let e is an identity element. Then we have a * e = a \;j a E Q a + e  ae = a or e  ae = 0 or e = 0, if 1  a te 0 e(l  a) = 0 or Similarly, for e * a = a v a E Q, we have e = 0 Therefore, for e = 0, we have a * e = e * a = a Th us, 0 is the identity element.
or
Inverse. Let us assume an element a E Q. Let a 1 is an inverse of a. Then we have [Identity] a * a 1 = O a + a 1  aa1 = 0 a a 1 ( 1  a) =  a or a 1 = a te 1 
a  1' a E Q, if a te 1 aI Therefore, every element has inverse such that a t: 1 . 
group.
Since, the algebraic system
(Q, *) satisfy all the properties of a group.
Hence,
(Q, *) is a
240
DISCRETE STRUCTURES Theorem III. Show that the identity element in a group is unique. Proof. Let us assume that there exists two identity elements i.e., e and e' of G. Since) e E G and e' is an identity. We have e'e = ee' = e Also) e' E G and e is an identity. We have e'e = ee' = e' e = e' Hence) identity in a group is unique.
Theorem IV. Show that inverse of an element a in a group G is unique. Proof. Let us assume that a E G be an element. Also) assume that a1 1 and a2 1 be two inverse elements of a. Then we have) a1 1 a = aa1 1 = e and a2 1 a = aa2 1 = e Now, a1 I = a1 I e = a1 l (aa2 1) = (a1 1 ala2 1 = ea2 1 = a2 1 Thus) the inverse of an element is unique.
ofa.
Theorem V. Show that (a it i = a for all a E G, where G is a group and a i is an inverse Proof. Given that a I is an inverse of a. Then) we have aa 1 = a 1 a = e This implies that a is also an inverse of a I . Therefore (a l) 1 = a. Theorem VI. Show that (ab) i = b i a i for all a, b E G. Proof. We have to prove that ab is inverse of b 1 a 1 . We prove (ab)(b l a I) = (b l a l)(ab) = e Consider (ab)(b l a I) = [(ab) b l ]a l = [a(bb l)] a I = (ae) a I = aa 1 = e Similarly, (b l a l)(ab) = e . . From (1) and (2), we have (ab) (bl aI) = e = (bl aI) ab
... (1) ... (2) Hence proved.
Theorem VII. Prove the left cancellation law in a group G holds i.e., ab = ac
\;j a, b, c E G.
Proof. Consider
Then) we have
b
=c
ab = ac. b = eb = (a 1 a)b = a I (ab) = a I (ac) = (a l a)c = ec = c ab = ac => b = c.
Proof. Consider ba = ca. Then, we have b = be = b(aa l) = (ba)a l = (ca)a l = c(aa ) = ce = c ba = ca => b = c. '
b =c
[ .: ab = ac] Associativity
Theorem VIII. Prove the right cancellation law in a group G holds i.e., ba
\;j a, b, c E G.
=>
=
ca
=>
[ .: ba = cal I Associativity
MONOIDSAND GROUPS
241
Theroem IX. Let G be a group and a, b E G. Then the equation a * x = b has a unique solution given by x = ai * b. Proof. Given a) b E G and G is a group under *) therefore) aI exists in G G is closed aI * b E G. Hence Consider a*x=b = (a * aI) * b I a * aI = e l = a * (a * b) Associativity Left cancellation law x = aI * b Uniqueness. Let the equation a * x = b has two solutions) say) Xl and x2) then we have a * x, = b ... (1) a * x2 = b ... (2) (1) and (2) gives a * x, = a * x2 Xl = X2 Left cancellation law ::::::} 8.15. Zm (THE INTEGERS MODULO m (m � 1 » The integers modulo m, denoted by Zm ' is the set given by Zm = {O) 1, 2, ... m  1 ; +m ' x m } where the operations +m (read as addition modulo and x m (read as multiplication modulo m) are defined as
m)
a +m b = remainder after a + b is divided by m a x m b = remainder after a x b is divided by m. Theroem X. For each m :> 1, [Zm; +m1 is a group known as group of integers modulo m. Proof. By definition, If a, b E Zm) then a +m b is remainder after a + b is divided by m, which is again an element of Zmo Hence Zm is closed under +m' Also the addition modulo m is always associative. 0 is the identity element for +m and every element of Zm has an additive inverse. :. Zm is a group under addition modulo m. 8.16. FINITE AND INFINITE GROUP
*) is called a finite group if G is a finite set. A group (G, *) is called an infinite group if G is an infinite set. For Example 1. The group (I, +) is an infinite group as the set I of integers is an infinite set. 2. The group G = {I, 2, 3, 4, 5, 6, 7, xs} under multiplication modulo 8 is a finite group as
A group (G,
the set G is a finite set.
group.
3. The group G = {I, w, w2} , is a finite group under multiplication 4. The group G = {± 1 , ± i, ±j, ± k} where i2 = P = k2 = 1 and i . j = k,j.i = k etc. is a finite
8.17. (a) ORDER OF GROUP or
The order of the group G is the number of elements in the group G. It is denoted by I G I 1 has only the identity element i.e., ((e)).
o(G). A group of order
8.17. (b) ORDER OF AN ELEMENT Let G be a group under multiplication and a E G. If there exists a least positive integer n such that an = e. Then a is said to be order n and written as o(a) = n
242
DISCRETE STRUCTURES
If G is a group under +, then a is said to be of order n if n is a least positve integer such na = e. Here e is the identity element of G. A group of order 2 has two elements i.e., one identity element and one some other element. Example 6. Let ({e, xj, ) be a group of order 2. The table of operation is shown in (Fig. 8.3).
that
*
*
e
x
e x
e x
x e
Fig. 8.3
The group of order 3 has three elements i.e., one identity element and two other elements.
Example 7. Let ({e, x, yj, ) be a group of order 3. The table of operation is shown in (Fig. 8.4). *
*
e
x
y
e x y
e x y
x y e
y e x
Fig. 8.4
Example 8. Consider an algebraic system ({O, 1). +) where the operation + is defined as shown in (Fig. 8.5).
The system own lnverse.
+
o
1
o 1
o 1
1 o
Fig. 8.5
({O. l), +) is a group. In this 0 is identity element and every element is its
Theorem XI. If G is a finite group of order n and a E G, then there exists a positive integer m such that am = e and m ::; n. Proof. Consider the elements of the group G as a, a2, a3, ... an+1 . These are n + 1 elements. Since I G I = n. Therefore two of its elements, say, aJ\ aq must be equal, i.e., aP = aq, p < q. Take m = q  p am = aqp = uP . aP = aq . (al')l = aq . (aq)l =e Further, since p, q are among n + 1 , l S p < q S n + l � q  p = m S n. .. 8.18. SUBGROUP
(P.T. U. B.Tech. May 2013, May 2007, May 2006)
Let us consider a group (G, *) . Also, let S c G ; then (S, *) is called a subgroup iff it satisfies following conditions :
(i) The operation * is closed operation on S. (ii) The operation * is an associative operation.
MONOIDSAND GROUPS (iii) As e is an identity element belonged to G. It must belong to the set S element of (G, *) must belongs to (S, *) .
243 i.e., The identity
(iv) For every element a E S, a I also belongs to S. Since Z c Q, Z is a subgroup of Q under addition. For example, let (G, +) be a group, where G is a set of all integers and (+) is an addition operation. Then (H, +) is a subgroup of the group G, where H = {2m : m E G}, the set of all even integer.
For example,
indentity element.
let G be a group. Then the two subgroups of G are G and G,
= (e), e is the
Example 9. Let (L +) be a group, where I is the set of all integers and (+) is an addition operation. Determine whether the following subsets of G are subgroups of G. (a) The set (G1, +) of all odd integers. (b) The set (G2 , +) of all positive integers. Sol. (a) The set G, of all odd integers is not a subgroup of G. It does not satisfy the
closure property, since addition of two odd integers is always even. (b) Closure property. The set G2 is closed under the operation +, since addition of two even integers is always even. Associative property. The operation + is associative since (a + b) + c = a + (b + c) for every a, b, C E G2" Identity. The element 0 is the identity element. Hence, 0 E G2. Inverse. The inverse of every element a E G2 is  a tl G2. Hence, the inverse of every element does not exists. Since the system (G2 , +) does not satisfy all the conditions of a subgroup. Hence, (G2 , +) is not a subgroup of (I, +).
Example 10. Consider the group Z of integers under addition. Let H be the subset of Z consisting of all multiples of a positive integer m i.e., H = {...... ,  3m,  2m,  m, 0, m, 2m, 3m, ..... } Show that H is a subgroup of Z. Sol. For r, s E Z, rm, sm E H. Consider rm + sm = (r + s) m E H Le., H is closed under addition. For rm E H,  rm E H and consider rm + ( rm) = (r  r) m = 0 E H i.e., 0 is the identity of H and  rm is the inverse of rm. Hence, H is a subgroup of Z. Theorem XII. A subset H of a group G is a subgroup of G iff (i) The identity element e E H (ii) H is closed under the same operation as in G (iii) H is closed under inverses i.e., if a E H, then a1 E H. Proof. Given G is a group and H is a subset of G. Let H is a subgroup of G, then, by definition, (i), (ii), (iii) are true. Converse. Let (i), (ii), (iii) hold. We show H is a subgroup ofG. We show the associativity of elements of H. Let a, b, C E G and since H c G :. a, b e E H Since elements of G are also elements of H . . Associativity holds for H. Hence the Theorem. Another statement : A subset H of a group G is a subgroup of G iff
\f a, b E H
a * bI E
H.
244
DISCRETE STRUCTURES
Theorem XIII. Let Hi and H2 be subgroup of group G, neither of which contains the other. Show that there exist an element of G belonging neither to Hi nor H2 . Proof. Given H, and H2 are subgroups of G. Also H , q; H2 and H2 q; H" We show that
there exists an element belonging neither to H I nor H2 o Let) if possible) there is an element a belonging to H , and H2 i.e., a E H , n H 2 . Now a E H I and since H I is a subgroup of G . . . .. (1) But a E H2 and since H 2 is a subgroup of G .. . .. (2) (1) and (2) gives H , c H2 , a contradiction. Hence the theorem. Theorem XIV. If H and K are two subgroups of G, then H n K is also a subgroup of G. (P.T.U. B.Tech. Dec. 2012, May 2010, Dec. 2007) Proof. We know that a subset H of a group G is a subgroup of G iff abI E H \;f a, b E H. Let a, b E H n K. We show abI E H n K. Now a E H n K => a E H and a E K Also b E H n K => b E H and b E K Since H is a subgroup of G and a, b E H abI E H (Using theorem X) => Also K is a subgroup of G and a, b E K abI E K => I From (1) and (2), ab E H n K. Hence H n K is a subgroup of G. Cor.
... (1) ... (2)
IfH and K are two subgroups of a group G, then give an example to show that H K may not be a subgroup of G Consider G :::: The group of integers under + HI :::: {...  6,  4,  2, 0, 2, 4, 6 ...} H2 :::: {...  12,  9,  6,  3, 0, 3, 6, 9, 12, ...} are subgroups of G under +. But H , U H2 = {...  4,  3,  2, 0, 2, 3, 4, 6, ...j Since 2 E H , U H2 , 3 E H , U H2 => 2 + 3 = 5 'l H , U H 2 i.e., H , u H2 is not closed under +. Hence H , u H2 is not a subgroup of G under +. u
Theorem XV. If H is a nonempty finite subset of a group G and H is closed under multiplication. Then H is a subgroup of G. Proof. We know that a nonempty subset H of a group G is a subgroups of G iff (i) a E H, b E H => ab E H (ii) a E H => aI E H The condition (i) is true since it is given that H is closed under multiplication. To show (ii), Let a E H, a E H => a2 E H I H is closed under multiplication 3 Again a E H, a2 E H ===> a E H and so on. 2 3 Thus the infinite collection of all the elements a, a , a , ... am, ''' , belongs to H. But H is finite. :. there must be repetetion. Let ar = as r > s > 0 ar . aS = e ars = e E H 1 r s Take y = a   and consider ya = ars1 a = ars = e Similarly, ay = e Hence ya = e = ay ===> y is the inverse of a. Hence the theorem. •
MONOIDSAND GROUPS
245
Theorem XVI. Let H be a subgroup of G. Then (a) H = Ha
l ea = a Conversly, Let a E H. As H is a subgroup and h E H, a E H haE H I H is closed under multiplication. => HaeH ... (1) I Again) if h E H) a E H and since H is a subgroup of G) h a E H (Theorem X) I (ha ) a E H a => h(aI a) E Ha => h e E Ha hE Ha HeHa ... (2) From (1) and (2) Ha = H (b) Let Ha = Hb and we show abI E H Now a = e a E Ha => a E Ha = Hb a E Hb => a = hb, h E H => => abI = (hb)bI = h(bbI) = he = h E H => abI E H Conversly, Let abI E H => abI = h, h E H a = hb => Ha = Hhb = Hb I For h E H, Hh = H => (c) Proceed yourself as in Part (b). h E H. Then, (d) Let H = Hh V h E H I Using part (a) H e HH e H HH = H. 8.19. ABELIAN GROUP Let us consider) an algebraic system (G) *») where * is a binary operation on G. Then the system (G, *)is said to be an abelian group if it satisfies all the properties of the group plus an additional following property : (i) The operation * is commutative i.e.,
a * b = b * a V a, b E G For example, consider an algebraic system (I, +), where 1 is the set of all integers and + is an addition operation. The system (I, +) is an abelian group because it satisfies all the properties of a group. Also the operation + is commutative for every a, b E l.
246
DISCRETE STRUCTURES
Example 11. Consider an algebraic system (G, *), where G is the set ofall nonzero real ab numbers and * is a binary operation defined by a * b = 4' Show that (G, *) is an abelian group. ab Sol. Closure property. The set G is closed under the operation *. Since, a * b = "4 is a real number. Hence) belongs to G. Associative property. The operation
( ) ( )
* is associative. Let a, b, ab (ab)c abc (a * b) * c = "4 * c = 16 = 16 '
Similarly,
a * (b * c) = a *
C
E
G) then we have
C
b a(bc) abc = = . 4 16 16
Identity. To find the identity element, let us assume that e is a positive real number. G, ea  = a or e = 4 4 Similarly, a*e=a ae  = a or e = 4. 4 Thus, the identity an element in G is 4. Inverse. Let us assume that a E G. If a I E Q is an inverse of a) then a * a I = 4
Then for a E

1
 1
16 aa = 4 or a = 4 a Similarly, a * a = 4 gives 16 a I a  = 4 or a = a 4 Th us, the inverse of an element a in G is �. a Commutative. The operation * on G is commutative. ab ba a * b =  =  = b * a. 4 4 Thus, the algebraic system (G, *) is closed, associative, has identity element, verse and commutative. Hence, the system (G, *) is an abelian group. 
1
1
has in
Example 12. Let (Z, *) be an algebraic structure, where Z is the set of integers and the operation * is defined by n * m = maximum (n, my. Determine whether (Z, *) is a monoid or a group or an abelian group. Sol. Closure Property We know that n * m = max. (n,m) E Z \;j n, m E Z Hence * is closed. Associative property. Let us assume a, b, c E Z.
MONOIDSAND GROUPS
247
Then, we have a Similarly,
* (b * c) = a * max. (b, c) = max. (a, max. (b, c» = max. (a, b, c) (a * b) * c = max. (a, b, c)
* is associative. Identity. Let e be the identity element. Then max. (a, e) = a Hence
Hence) the minimum element is the identity element.
Inverse. The inverse of any element does not exist. Since) the inverse does not exist) hence (Z) *) is not a group or abelian group but a monoid as it satisfies the properties of closure) associative and identity. Example 13. Let S = {O, 1, 2, 3, 4, 5, 6, 7} and multiplication modulo 8, that is x 0 y = (xy) Mod 8 (i) Prove that ({O, 1), 0) is not a group. (ii) Write three distinct groups (G, 0) where G c S and G has 2 elements. Sol. (i) (a) Closure property. The set {O, I} is closed under the operation 0, as shown
in table of operation (Fig. 8.6).
o 1
o
1
o o
o 1
Fig. 8.6
(b) Associative property. The operation 0 is associative. Let a, b, c E G, then we have (a 0 b) 0 c = a 0 (b 0 c) e.g., (0 0 1) 0 1 = (0) 0 1 = 0 Similarly, 00 (1 0 1) = 0 0 (1) = O. (c) Identity. The element 1 is the identity element as for every a E {O, I}: We have 1 0 a = a = a 0 1. (d) Inverse. There must exist an inverse of every element a E {O, I}, such that a 0 a 1 = 1 But the inverse of element 0 does not exist. Therefore, since the inverse of every element a E {O, I} does not exist. Hence ({O, I), 0)
is not a group.
(ii) The three distinct groups (G, 0), where G c S and G has 2 elements is as follows (a) ({I, 3), 0) (b) ({I, 5), 0) (c) ({I, 7), 0).
Example 14. Determine whether a semigroup with more than one idempotent element can be a group. (P.T.V. B.Tech. Dec. 2010) have
Sol. Let (A, *) be a semi· group with two idempotent elements a and b (a ", b). Then we ... (1) ... (2)
b * b = b. Now assume that A is a group with identity element e.
a*e=a
and
b*e=b
248
DISCRETE STRUCTURES From
(1) and (2), ::::::}
a*e=a=a#a Also b*e=b=b"b a=e=b which is a contradiction to a i: b. Hence (A, *) can not be group.
=>
e=a e=b
I I
Left Cancellation Law Left Cancellation Law
Example 15. Let (G, 0) be a group. Show that if (G, 0) is an Abelian group then, (P.T.V. B.Tech. Dec. 2010) (a 0 b) 2 = a2 0 b 2 for all a and b in G. Sol. Let us assume G is an Abelian group, then o is associative] (a 0 b)2 = (a 0 b) 0 (a 0 b) = a 0 (b 0 a) 0 b I G is abelian = a o � 0 � o b = � 0 � 0 (b o � = � o b 2 2 2 2 Hence, (a 0 b) = a 0 b \;j a, b E G. Example 16. Let G be a group of 2 x 2 matrices with rational entries and nonzero determinant. Let H be a subset of G consisting of matrices whose upper right entry is zero. Then show that H is a subgroup of G.
[(� �): [(� �):
G=
Sol. Given
H
=
H is a subgroup of G iff
a, b, c, d E Q a nd ad  bc ,, 0 a , c, d E Q
]
]
(i) H is closed under multiplication (ii) For A E H, A' E H Let A, B
E
H where A =
Consider
i.e.,
AB
( (
a1
c,
�)
a1
= c,
d1 2 E
H is closed under multiplication. Further) For A
E
H) we have A=
Also
All
.
I
I� �I 0, ) ( 0)
A
I =
= d) A 2 =  C, A2 =
adj A =
"
(� �)
0
(
1
A ll A 12 A 21 A 22
�'��
Hence H is a subgroup of G.
1
T
A22
d = c
{�, ;]
= ad
a
e
n
=a
H
MONOIDSAND GROUPS
249
Example 17. Let G be a group of real numbers under multiplication. Let H = { I, I}. Then show that H is a subgroup of G under multiplication. Sol. Consider the multiplication table of H under multiplication. 1
1
1 1 1 1 From the table, we observe that each element 1 1
in the table belongs to H.
Hence H is closed under multiplication. Also, the inverse of  1 is  1 and of 1 is 1. Thus each element of H has its inverse. Therefore H is a subgroup of G under multiplication.
Example 18. Consider the group of integers Z under +. Let E = The set of even integers. Then show that E is a subgroup of Z under +. Sol. Given E = {2m : m E Z} i.e., the set of even integers. Clearly E is a subset of Z. Let a, b E E => a = 2m, m E Z b = 2n, n E Z a + b = 2m + 2n = 2(m + n) E E 1 m, n E Z m+nE Z i.e., E is closed under +. Also for each a E E) we have a = 2m) m E Z  a =  2m = 2( m) = 2 t, t =  m E Z aE E Th us each element belonging to E has additive inverse. E is a subgroup of Z under +. Example 19. Let Z be a group of integers under +. Let Z+ is the set of nonnegative integers. Is Z+ a subgroup of Z ? Z+ = {O, 1, 2, 3, ...} Sol. Clearly Z+ is a subset of Z. But Z+ is not a subgroup of Z. Since the elements of Z+ do not have additive inverses. For e.g., 2 E Z\ but  2 tl Z+ . Example 20. Consider Z12 = [0, I, 2, ... I ll, the group under addition modulo 12. Let H = [0, 3, 6, 9]. Show that H is a subgroup of Z1 2 under +1 2' Sol. Given H = [0, 3, 6, 9]. Clearly H is a subset of Z, 2 ' Let a, b E H => a + 1 2 b is also in H. :. H is closed under + 1 2 ' Also we have 3 + ,2 9 = 0, 0 is the identity of Z, 2 6 +,2 6 = 0 9 +,2 3 = 0 each element of H has its inverse.
12. Example 21. Consider the group of integers Z under +. Let 2Z and 3Z are two subgroups of [Z; +]. Is 2Z n 3Z a subgroup of Z ? Sol. We know that if H and K are two subgroups of a group G. Then H n K is also a subgroup of G. Using this result, we can say that 2Z n 3Z is a subgroup of Z. (Theorem XII) H is a subgroup of Z, 2 under addition modulo
DISCRETE STRUCTURES
250 H2
=
Example 22. Consider Z,5' the group under addition modulo 16. Let H, = [0, 6, 4, 8, 12]. Are H , and H 2 subgroups of Z , 5 under +,5 ? Sol. To check whether H, is a subgroup of Z,5' compute the following Table 8.6. Table 8.6
[0,
0 0 5 10
+,5
0 5 10
10] ,
10 10 0 5
5 5 10 0
From the Table 8.6, we observe that each element which is in the interior of the addition table is also in H" :. H , is closed under + ' 5 ' Also we have 6 + , 5 10 = 0, 10 + , 5 6 = 0, 0 is the identity . . each element of H, has its inverse. :. H I is a subgroup of Z 15" To check, whether H2 is a subgroup of Z , 5 compute the following Table 8.7. '
Table 8.7
0 0 4 8 12
+,5
0 4 8 12
8 8 12 1 5
4 4 8 12 1
12 12 1 5 9
From the Table 8.7, we observe that there are some elements in the interior of the addition table, which are not in H2 (e.g., 9 'l H ) . Hence H2 is not closed under + 5' :' H2 is ' not a subgroup of Z , 5 '
Example 23. Let H and K be groups. (a) What do you mean by the direct product of H and K? (b) What is the identify element of H x K ? (c) What is the order of H x K ? (d) Describe and write the multiplication table of the group G = Z2 X Z2' Sol. (a) Let G = H x K be the cartesian product of H and K with the operation * defmed by (h, k) * (h', k') = (hh', kk') Then G is group called the direct product of H and K. (b) Let eR is the identity element of H and eK be the identity element of K, then (eR , e0 is an identity element of H x K. (c) o(G) = o(H x K) = o(H) x o(K) (d) As, Z 2 = {O, I} has two elements . . o(Z) = 2 .. o(G) = o(Z2 x Z) = o(Z) x o(Z) = 2 x 2 = 4 . . G has four elements namely a = (1, 0), b = (0, 1), c = (1, 1), e = (0, 0). The multiplication table of G, = Z2 X Z2 is shown below (Table 8.8): Table 8.8 *
e
a
b
c
e a b c
e a b c
a e c b
b c e a
c b a e
MONOIDSAND GROUPS Here)
251
a2 = e) b2 = e) c2 = e
Since the table is symmetric, therefore
L
I
G is abelian.
TEST YOUR KNOWLEDGE 8.2
If a, b, c are elements of a group G and a * b :::: c * a. Then b :::: c ? Explain your answer.
(P. T. U. B.Tech, Dec. 2006, May 2005)
2. 3.
4.
5. 6.
Discuss the relation between groups and monoids ? Is every monoid a group ? Is every group a monoid ? Which of the following are groups ? (i) M2xS(R) with matrix addition (ii) M2x2(R) with matrix multiplication (iii) The positive real numbers with multiplication (iv) The nonzero real numbers with multiplication (v) The set [ 1, 1] with multiplication. Give an example of (i) a finite abelian group (ii) an infinite nonabelian group. Let V :::: {e, a, b, c}. Let * be defined by x * x:::: e for all x E V. Write a complete table for * so that 0/, *) is a group. Which of the following subsets of the real numbers is a subgroup of [R, +] ? (a) The rational numbers (b) The positive real numbers (c) H {� K is an integer} (d) H {2K : K is an integer} (e) H {x :  100 a*Hcb*H ... (1) => Similarly, if x E b * H) we can easily show that x E a * H. b*Hca*H ... (2) => from (1) and (2), a*H=b*H Case II. Let x E H * a. We show x E H * b. Now x E H * a => There exists an element h, E H such that x = h, * a Also b E H * a => There exists an element h2 E H such that b = h2 * a => h21 * b = h21 * (h2 * a) = (h21 * h;) * a =e*a=a a = h21 * b X = h 1 * a = h 1 * (h21 * b) = (h 1 * h21) * b Since hp h2 E H and H is a subgroup of G. . . h, * h21 E H Hence x = (h, * h21) * b E H * b XE H * b => H*a cH*b => Similarly) if x E H * b) we can easily show that x E H * a H*bcH*a => Hence H *a=H*b Theorem II. Let H be a subgroup of a group G. Then the right cosets Ha form a partition of G. Or Any two right (left) cosets in a group G are either disjoint or equal. Proof. Let Ha and Hb be two right cosets and suppose Ha n Hb '" q,. We show Ha = Hb Let x E Ha n Hb => x E Ha and x E Hb x = h,a and x = h 2 b for some h I ' h2 E H => => h, a = h2 b E Hb h,a E Hb => Ha c Hb => Also h2 b = h,a E Ha h2 b E Ha => Hb c Ha => Hence Ha = Hb Lagrange's Theorem (P.T. U. B.Tech. Dec. 2007, 2006, May 2006, May 2012, May 2013) Theorem III. If G is a finite group and H is a subgroup of G, then o(H) 1 0(G). Proof. Since H is a subgroup of a finite group G) . . H is also finite) say) H = {h I ' h2 , ... hnl . where each hi is distinct. We claim H and any coset Ha have the same number of elements.
MONOIDSAND GROUPS
255
Consider Ha = {h , a, h2a, ... hna}. We claim all hia's are distinct. For if, hp = hja h, = hJ 1 Right cancellation law =>
a contradiction, since h/s are distinct. Thus, H and Ha have same number of elements.
k.
Now G is finite :. The number of distinct right cosets of H in G is also finite, say, G = Ha, u Ha2 u ... u Hak =
Let
o(G)
k
i�' Ha
i
= Number of elements in Ha, + number of elements in Ha2 + ... + number of elements in HaK = n + n + ... k Times = nk n l o(G) o(H) l o(G)
Hence the Theorem.
Converse of Lagrange's theorem is however, not true i.e., If o(H) l o(G).
may not have a subgroup of order o(H).
Consider the alternating group A4. Here o(A4)
Then G
= !..!. =
12 and 6 1 12. But there is no 2 subgroup of A4 whose order is 6. (see the topic "Symmetric Group" in this chapter)
8.21 . INDEX OF A SUBGROUP Let G be a group and H be a subgroup of G. Then the number of right (left) cosets of H in G is called the index of H in G. The index of H in G is denoted by [G : H] .
:�� .
Theorem IV. If G is a finite group and H is a subgroup of G. Then fG: Hi = Proof. Proceeding in the same way as in the proof of Lagrange's theorem, we have o(G) = nk, where k is the number of distinct right cosets of H in G k=
oCG)
=
oCG)
[G: H]
n
=
oCG)
oCH)
oCH) '
Theorem V. Let G be a finite group of order n show that an = e for any element a E G.
(P.T.V. B.Tech. Dec. 2010)
G has order m, then am = e Let H be a subgroup of G, of order m. By Lagrange's theorem,
Proof. Let a E
o(H) l o(G)
min ::::::} n = mr, for some r an = amr = (am) r = er = e Hence the Theorem.
8.22. NORMAL SUBGROUP
(P.T. U. B. Tech. Dec. 2007)
A subgroup H of a group G is called normal subgroup of G if for every g H.
=> ghg' E
E
G, h
E
H,
DISCRETE STRUCTURES
256 or
A subgroup H of a group G is called a normal subgroup of G iff for g E G, we have gHgI = H \;j g E G Example
O. Let H =
[(�
4. Let G be the group of two by two invertible real matrices
�} a ", 0] . Then H is a normal subgroup of G.
(� �).
ad  bc cj'
(P.T.V. B.Tech. Dec. 2010)
Sol. We first show that H is a subgroup of G. Let h I ' h2 E H such that
(� �} (� �J )( ) =( ) (o ) I I (� �r (� �) , ' o ��� o "', (; ; hl =
i.e.,
al 0 0 a1
0 a
Now
1
h2 =
a ", o, a, ,,, o
aal 0 0 aal
I aa , '" 0
E H
H is closed under matrix multiplication. Further) For A E H)we have
a 0 a 0 2 a ) I A I = 0 a =a All = a, A 2 = 0, A2 = 0, A22 = a A=
1
Also
1
adj A =
Hence
=
( )
:)�[:
( )
]
C H ",e
Thus each element belonging to H has multiplicative inverse. Hence H is a subgroup of G. Further, For
g= a b c d
E G,
h= a 0 0 a
.
E H, ConsIder
(: J ( J ( J ( J ( J
[
ad�bC c
�
ad bc ghgl = a ad  bc ad  bc a c a 2 ba ad bC ab bC = ad  bc = ea da = c a cad  dae  cab + da 2 ad  bc ad  bc ad  bc ad  bc  C) a O b E H = a (ad _ bC) = ad  bc b a 0 a b I a b a 0 = c d 0 a d 0 a c d
(
J
[
[ ::d � �
�
�
: [ :� :
:(
:
� �)
Hence H is a normal subgroup of G under matrix multiplication. Example
5. Let G be a group of two by two invertible real matrices
under matrix multiplication. Let
H=
[(� �J :
1
ab '" 0
l
(� �).
ad  bc cj' 0
Is H a normal subgroup of G ?
MONOIDSAND GROUPS
257
Sol. We first show that H is a subgroup of G. Let A, B
(
Consider
::::::}
)
a 0 ° b , ab '" 0,
A_ AB =
(� 0b) (a,O
0 b,
B=
) ( =
(
)
E
H such that
a, 0 ° b, ' a, b, '" °
�J E
aa, ° b
H is closed under multiplication of matrices
H,
Thus, every element of H has multiplicative inverse. Thus H is a subgroup of G under matrix multiplication. Also, For g =
(� �)
E
G,
(� �) E
h=
b2 db
)
[
�
ad bc c ad  bc
H, Consider
b ad  bc a ad  bc
:[ =
[)
:
d b 0 ad  bc ad bc c a b ad  bc ad  bc a 2 d  b 2c  a 2b + b 2 a ad  bc ad  bc "H cad  dbc  cab + dab ad  bc ad  bc
Hence H is not a normal subgroup of G under matrix multiplication.
:
Example 6. Let G be the group of nonsingular 2 x 2 matrices under matrix multiplica tion. Let H be the subset of G consisting of the lower triangular matrices i.e.; matrices of the
form
(�
�) where ad '" O. Show that H is a subgroup of G, but not a normal subgroup.
Sol. Let A, B
E
H such that A=
Consider
AB =
( ( (
a, c, a, c,
)
a,a2 °  c 1a 2 + d1c2 d1d2 E
H is closed under matrix multiplication. Also for any M
E
H, we have
I M I
=
M=
I� �I
(�
=
�)
ad '" °
H
(given)
DISCRETE STRUCTURES
258 Ml exists. Further
(� �) E
H is the identity of H. Hence, H is a subgroup of G.
But H is not a normal subgroup of G. Since) for example) Take
( )( 1 0 )( 1 2 ) ' 1 1 1 3 ( 1 2) ( 1 0 ) ( 3 2 ) =
ghg1 = 1 2 1 3
Consider


1 3
1 1
( ) ( 23 (7 ) =
2)
1 2 = 1 3 9


4 5
1
1
1
'l H.
Example 7. Let G be the group of nonsingular 2 x 2 matrices under matrix multiplica tion. Let H be a subset of G consisting of matrices with determinant 1. Show that K is a normal subgroup of G. Sol. We know that if !
=
E
H
Now, =>
AB
..
H is a subgroup of G.
E
H
Further, . . AI E H.
Let X E G and A Consider
then
(I) = 1 I E H. i.e., H has an identity. => det (A) = 1 , det (B) = 1 det (AB) = (det A) (det B) = 1.1 = 1 i.e., H is closed under matrix multiplication. Let A E det(A1) = 1/det(A) = III = 1 i.e., H has an inverse. det
Let A, B
(� �).
E
H
=>
det(A)
=
1
H such that det A = 1
det (XA XI)
= det (X) det (A) det (XI)
= det (X) .
1 .
1 =1 det (X)
XAXl E H for all X E G H is a normal subgroup of G. Example
gE
8. Every subgroup of an abelian group is normal.
Sol. Let H be a subgroup of a normal group G. We show H is normal. Let G. Consider
hE
H and
MONOIDSAND GROUPS
259 ghg' = ggl h = eh =hE H ghgl E H.
Hence)
h E H e G => h E G Also h, gl E G and SInce G is abelian hgl = gl h
H is a normal subgroup of G.
Theorem VI. Let H be a subgroup and K be a normal subgroup ofa group G. Show that HK is a subgroup of G. (P.T.V. B.Tech. Dec. 2013) Proof. We show that (i) HK is closed under multiplication (ii) For x E HK, we should have yl E HK (iii) e E HK x = hi kp Y = h2 k2 ) where hp h2 E H; kp k2 E K Let x, y E HK => Consider xy = h , k , h2 k2 = h, h2 (h21 k, h) k2 E HK h2 E H => h21 E H e G => h21 E G Let Since K is a normal subgroup of G) and k l E K) .. h21 k, h2 E K (h 21 k, h) k2 E K => h , h 2 (h21 k, h) k2 E HK => Thus HK is closed under multiplication. Further) For x E
HK) we have XI = (h, k,)' = k,' h,' = h,'(h, k,' h,')
h, E H e G => h , E G. Also k, E K k,' E K Since K is a normal subgroup of G h 1 k11 h 11 E K h,' (h, k,') h,' E HK => Thus yl E HK. Finally, e E H, e E K => e . e E HK => e E HK Thus HK is a subgroup of G. Theorem VII. Let H is a subgroup of a group G. Then H is a normal subgroup of G iff aH = Ha \j a E G. Proof. Let H is a normal subgroup of G. Then for a E G, we have aHa1 = H (aH a1)a = Ha aH(a1 a) = Ha aH e = Ha aH = Ha Theorem VIII. The intersection of any number of normal subgroups of G is a normal subgroup of G. Let
=>
260
DISCRETE STRUCTURES Proof. Let HI ' H2, H3, .... H=
,(1
be a collection of normal subgroups of a group G. Let Hi' We show H is a normal subgroup of G.
Now the intersection of any collection of subgroups is a subgroup. :. of G. We show H is a normal subgroup of G. Let g E G,
hE
H
But each Hi is normal in G
ghgI E
Hi
ghgI E
=>
Vi
H
=>
gh gI E
n
i= 1
=>
hE
n
i=1
H is a subgroup Hi
=>
hE
Hi
Vi
= H,
H is a normal subgroup of G. Cor. Prove that intersection of two normal subgroups is again a normal subgroup.
(P.T. U. B.Tech. May 2007, May 2006)
Union of two normal subgroups may not be a normal subgroups Consider G :::: The group of integers under + H I :::: { ...  6,  4,  2, 0, 2, 4, 6 ...} H2 :::: { ...  12,  9,  6,  3, 0, 3, 6, 9, 12, ...} are subgroups of G under +. But H , U H2 = {...  4,  3,  2, 0, 2, 3, 4, 6, ...} Since 2 E H , U H2 , 3 E H , U H2 2 + 3 = 5 'l H, U H 2 i.e., H , u H2 is not closed under +. Hence H , u H2 is not a subgroup of G under +. Remark.
=>
Theorem IX. Let G is a group and H is a normal subgroup of G. Let G IH denotes the collection of right (left) cosets of H in G. Show that G IH is a group under the coset multiplica tion defined by aHbH = abH V a,b E G Proof. (i) Closure Property. By definition, G/H = {aH : a E G} Let aH , bH E G/H and consider (aH) (bH) = a(Hb)H = a(bH)H I Ha = aH Thus associativity holds in G/H. (iii) Identity. Let aH E G/H for a E G. Consider (aH)H = a(HH) = aH I HH = H =>
Le.,
H is the identity element of G/H.
(iv) Inverse. Let aH E G/H and consider (aI H) (aH) = (aI a)H = eH = H aI H is the inverse of aH
Thus G/H is a group under coset multiplication.
MONOIDSAND GROUPS
261
8.23. QUOTIENT GROUP Let G be a group and H be a normal subgroup ofG. Let G/H denotes the set of right (left) cosets of H in G. Then G/H is a group (Proved in above theorem IX) called quotient group, or factor group under the coset multiplication defined by
(aH) (bH) = abH. 8.24. (a) CYCLIC SUBGROUP Let G be any group and a E G. Then all the integral powers of a, given by 3 3 O ... ) a , a2 , a\ a = e, a, a2 , a , form a subgroup of G called the cyclic group generated by a and is denoted by gp(a) or < a >. The element a is called generator of G. Order of a, a E G. The smallest positive integer m such that am = e, is known as order of a and is denoted by o(a) or (a). If I a I = m, then the cyclic subgroup gp(a) has m elements given by •••
e.g., i.e.,
gp(a) = {e, a, a2 , a3 , ... , amI}
If G = (±1, ±i), then G is a cyclic subgroup generated by i 4 3 O Since i = 1, i1 = i, i2 = 1, i =  i, i = 1 n every element of G is of the form i , n E Z. Also, the element i is a generator of G. Example. The group G = {I, w, w2} is a cyclic group. Since
WI = w w2 = w2 w3 = 1
i.e., Hence each element of G is some power of w. Therefore, G is a cyclic group and generator of G. The group G = {± 1, ± i} is cyclic with 'i' as generator.
w
is a
8.24. (b) CYCLIC GROUP A group whose all elements are integral powers of one or more elements is called cyclic. Remark. e.g., Z 12 :::: [Z 1 2 ; +1 2]
Sol.
The order of a generator of the cyclic subgroup is equal to the order of the group. is a cyclic subgroup.
Z, 2 = {O, 1, 2, ... 11, + , 2}' 5=5 5 +, 2 5 = 10 5 + , 2 5 + ,2 5 = 3 5 + , 2 5 + , 2 5 + ,2 5 = 8 5 + , 2 5 + ,2 5 + , 2 5 + 2 5 = 25 = 1 etc. ' Thus we see that every element of Z, 2 is of the form 5n for some n E Z. Thus 5 is a generator of Z 12 ' Hence [Z, 2' + 12] is a cyclic subgroup with 5 as generator. Since inverse of 5 is 7 (5 +,2 7 = 0), therefore, 7 is also a generator. (theorem X below) Example 9. The group of integers Z is cyclic under addition. Sol. Z = {O, ± 1, ± 2, ± 3, ... } Since 1=1 1+1=2 Consider
262
DISCRETE STRUCTURES
1+1+1=3 1 + 1 + 1 + ... 1 = n etc n times Thus we see that every element of Z is of the form Z = < 1 >. Also Z = <  l >.
n (l). Thus Z is cyclic group. Hence
Theorem X. If a is a generator of a cyclic group G, show that inverse of a is also a generator. Proof. Let G = i.e., G is a cyclic group and a is its generator. Let g E G, then g = ar for some r E Z Take r =  s) E Z) we have g = a' = (aI), for some s E Z Thus every element g E G is of the form (aI),. Hence aI is a generator. Theorem XI. Every cyclic group is abelian. (P.T.V. B. Tech. May 2006, May 2005) Proof. Let G be a cyclic group with a as its generator. i.e., let G = and g, E G. Then gl = ar for some r E Z Let g2 E G, then g2 = a' for some s E Z Consider gl . g2 = ar as = ar+s I r + s = s + r as Z is abelian = as+r S
•
=> G is abelian. Theorem XII. Every subgroup of a cyclic group is cyclic. of G.
Proof. Let G = i.e.,
(P.T.V. B.Tech. Dec. 2009) G is a cyclic group with a as its generator. Let H be a subgroup
Case I. If H = (e), then H = i.e., H is a cyclic group with e as a generator. Case II. If H '" [e], then o(H) :> 2 i.e., there exists e '" a E H. Since H is a subgroup) it must be closed under inverses and so contains positive powers of a. Let m is the smallest power of a such that am E H. We claim b = am is a generator of H. Let X E H. But H c G . . X E G. Since G is a cyclic group G with a as its generator. :. x = an for some n E Z. Dividing n by m) we get a quotient q and remainder r. i.e., n = mq + r) 0 :::; r < m Now an = amq+r = amq . ar = bq . ar n ar = bq a => Here an ) b E H and since H is a subgroup . . bq an E H which means ar E H. But m was the least positive integer of a such that am . E H and r < m . . . We must have r = 0 Hence an = b q for some q E Z X = an = bq i.e.) every element x E H is of the form bq for some q E Z ::::::} . . H is cyclic. •
Theorem XIII. If G is a cyclic group of order n and a is a generator of G. Let (n, k) = d. Then the order of the cyclic group generated by ka is !!:. where d is the greatest common divisor d of n and k. Proof. Proof of this theorem is beyond the scope. Example 10. Find the order of the cyclic subgroup generated by 18 in Z30'
MONOIDSAND GROUPS
263
Sol. We know that 1 is a generator of Z30' Also 18 = 18(1) The greatest common divisor of (n, k) = (30, 18) = 6 = d ..
The order of cyclic subgroup generated by
i.e., k =18, a = 1, n = 30
30
18 = 6 = 5.
(Theorem XIII)
Theorem XIV. Every group ofprime order is cyclic. Proof. Let G be a group of order p, p is prime. It means G must contain at least two elements. Since 2 is the least positive integer which is prime i.e., if a E G, then o(a) :> 2. Let o(a) = m and H be a cyclic subgroup of G generated by a, then o(H) = o(a) = m I The order of a cyclic group is equal to the order of its generator Also By Lagrange's theorem, o(H) l o(G)
=> m I p p = 1 or p = m p # l .. p = m o(H) = o(G) => H = G.
=>
But
Hence G is cyclic since H is cyclic.
Theorem XV. Let G is a cyclic group of order p (p is prime). Show that G has no proper subgroups except G and {e}. Proof. Let G is a cyclic group of order p. Let H be any subgroup of G and o(H) = m. By Lagrange theorem, o(H) I o(G) => m I p p = 1 or p = m => But p # 1 .. p = m i.e., o(H) = m = p => H is a group of prime order and hence cyclic. Also o(G) = m .. G = H i.e. G has no proper subgroups.
Let G be any group and G. Define aO :::: e the denoted by where denotes the set of all powers of a, is defined .....} {...... contains the identity element closed under group operation, contains inverses. is a subgroup of G and is called the by
Remark. Cyclic subgroup generated by a. aE cyclic subgroup generated by a, , < a> 2 3 2 :::: , a , aI , e, a, a , a ,
e, . . cyclic subgroup generated a.
by
;
Example 11. Consider the group G = {I, 2, 3, 4, 5, 6} under multiplication modulo 7. (a) Find the multiplication table of G (b) Find ;51, [51, 61 (c) Find the orders and subgroups generated by 2 and 3 (P.T.V. B.Tech. Dec. 2013) (d) Is G cyclic ? Sol. (a) By definition, a x7 b = The remainder when ab is divided by 7 For e.g., 5 x7 6 = 30 = 2 (when 30 is divided by 7, the remainder is 2) The multiplication table is shown below Table 8.10 Table 8.10 X7
1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 1 3 5
3 3 6 2 5 1 4
4 4 1 5 2 6 3
5 5 3 1 6 4 2
6 6 5 4 3 2 1
264
DISCRETE STRUCTURES
(b) The identity element of G is 1. (As the first row inside the table is identical with the top most row). 21 = 4 (In the table, the intersection of 2 and 4 is 1)
31 = 5 61 = 6 2=2 (c) We have 2 x7 2 = 4 2 x7 2 x7 2 = 8 = 1 0 (2) = 3 Hence = The subgroup generated by 2 = { 1 , 2, 4} Also 3=3 3 x7 3 = 9 = 2, 3 x 7 3 x 7 3 = 27 = 6 3 x 7 3 x 7 3 x 7 3 = 81 = 4 3 x 7 3 x 7 3 x 7 3 x7 3 = 243 = 5 3 x 7 3 x 7 3 x 7 3 x 7 3 x 7 3 = 729 = 1 0(3) = 6. . . The group generated by 3 is given as .. = {1 , 2, 3, 4, 5, 6} = G (d) Since 0(3) = 6 = o(G) => G is cyclic. Recall that a group G is cyclic if there exists an element a E G such that o(a) = o(G). Example 12. Let G = [1, 5, 7, 11J under multiplication modulo 12. (a) Find the multiplication table of G (b) Find the order of each element (c) Is G cyclic ? Sol. (a) We know a x ,2 b = The remainder when the product ab is divided by 12 5 x , 2 7 = 35 = 1 1 etc. Le., The multiplication table is shown below (Table 8.11) Table 8.11 X, 2
1 5 7 11
1 1 5 7 11
5 5 1 11 7
7 7 11 1 5
(b) Order (1) = 1 (since 1 is the identity element) To find order of 5. 5 x,2 5 = 25 = 1 .. 0(5) = 2 To find order of 7. 7 x,2 7 = 49 = 1 .. 0(7) = 2 To find order of 11. 11 x ,2 11 = 121 = 1 .. 0(11) = 2 (c) We know that a group G is cyclic if there exists o(a) = o(G). Since 0(1) = 1, 0(5) = 2, 0 (7) = 2, 0(11) = 2 i.e., There is no element of G whose order = 4 ..
G is not cyclic.
11 11 7 5 1
an element
aE
G such that
Example 13. Consider the group G = {l, 2, 3, 4, 5, 6} under multiplication modulo 7. (a) Find the multiplication table of G (b) Prove that G is a group
MONOIDSAND GROUPS
265
(c) Find ;51, ,,1, ()1 (d) Find the orders and subgroups generated by 2 and 3 (P.T.V. B.Tech. Dec. 2009) (e) Is G cyclic ? Justify your answer Sol. (a) and (b). Proceed yourself as in above example 12. (c) 21 = 4, 31 = 5, 61 = 6 (d) < 2 > = {I, 2, 4}, < 3 > = {I, 2, 3, 4, 5, 6} (e) Since o(B) = 6 = o(G) => G is cyclic. Example 14. Let G be a reduced residue system modulo 15 say, G = {I, 2, 4, 7, 8, 11, 13, 14}. Then G is a group under multiplication modulo 15 (a) Find the multiplication table of G. (b) Find ;51, 71 , 1 1 1 (c) Find the orders and subgroups generated by 2, 7 and 1 1 . (d) Is G cyclic ? Sol. (a) The multiplication of G is shown below (Table 8.12): Table 8.12
x 1 2 4 7 8 11 13 14
1 1 2 4 7 8 11 13 14
2 2 4 8 14 1 7 11 13
4 4 8 1 13 2 14 7 11
7 7 14 13 4 11 2 1 8
8 8 1 2 11 4 13 14 7
11 11 7 14 2 13 1 8 4
13 13 11 7 1 14 8 4 2
From the table, 2 x 1 5 8 = 1 i.e., 8 is the inverse of 2. Hence 21 7 x 1 5 13 = 1 => 13 is the inverse of 7. Hence 71 = 13 Also, Further 11 x 1 5 11 = 1 => 1 1 is the inverse of 1 1 . Hence 111 = 11. (c) (i) We have 2 x 1 52 = 4 2 x 1 52 x 1 52 = 8 2 x 1 52 x 152 x 1 52 = 1 order of 2 = 0(2) = 4 The group generated by 2 is given by
(b)
< 2 > = {2°, 2\ 2 2 , 2 3} = {I, 2, 4, 8} 7 x 157 = 4 (ii) We have 7 X 15 7 X 1 5 7 = 4 X 1 5 7 = 13 7 x 1 5 7 X 1 5 7 X 1 5 7 = 13 X 1 5 7 = 1 order of 7 = 0(7) = 4 The group generated by 7 is given by = {70 , 7\ 72 , 73} = {I, 7, 4, 13} (iii) 1 1 x 15 1 1 = 1 order of 1 1 = 0(11) = 2 The subgroup generated by 1 1 is = { 11 0 , 11'} = {I, l l}
=8
14 14 13 11 8 7 4 2 1
. .
(d) The group G is cyclic if 3 an element whose order equals to the order of G. Here O(G) =
8
266
DISCRETE STRUCTURES But we have proved that
2 x 15 2 X 1 5 2 X 1 5 2 = 1 0(2) = 4 . . 0(4) = 2 4 X 15 4 = 1 7 X ,5 7 X 15 7 X 15 7 = 1 . . 0(7) = 4 8 X 1 5 8 = 64 = 4 8 X 15 8 X 15 8 = 4 X 15 8 = 2 0(8) = 4 8 X 15 8 X 15 8 X 15 8 = 2 X ,5 8 = 1 1 1 x 1 5 11 = 1 0(11) = 2 Also, 13 X , 5 13 = 4 13 x 1 5 13 X 1 5 13 = 4 x 1 5 13 = 7 . . 0(13) = 4 13 x 1 5 13 x 1 5 13 x 1 5 13 = 7 x 1 5 13 = 1 14 X 1 5 14 = 1 0(14) = 2 Hence, there is no element a E G such that o(a) = o(G) = 8 . . G is not cyclic. 8.25. MORPHISMS
'morphism ' is a combination isomorphism, actomorphism, endomorphism etc. The word
of various terms like,
8.25. 1 . Group Homomorphism
homomorphism,
(P. T. U. B. Tech. May 2007, May 2006)
A mapping
G is a function defined by f(x) = gXgl for each x E G Let x, y E G and consider f(xy) = g(xy)gl = g(xg1gy)g1 = (gXgl)(gygl) = f(x)f(y) i.e., f is a homomorphism of G to G. To show f is oneone : Let f(x) = fry) gxg' = gygl (gxg1)g = (gygl)g
268
DISCRETE STRUCTURES => => => => => => =>
gX(glg) = gy (g'g) gxe = gye gx = gy gl (gX) = gl (gy) (glg)X = (g'g)y ex = ey x=y
i.e., f is oneone. To show f is onto : Let z E G and consider f(gl zg) = g(g l zg)gl = (ggl) Z(ggl) = eze = z Thus for each z E G, we have gl zg E G such that f(g' zg) = z i.e., f is onto. Hence G = G.
Example 19. Define 8 : Z6 to Z3 by 8 (n) = n(1) (the sum of n ones in Z3) Show that 8 is a homomorphism. If so, find Ker 8 and Im(8). Sol. By definition, Z6 = {O, 1 , 2, 3, 4, 5} and Z3 = [0, 1 , 2] Consider 8(0) = 0, 8(1) = 1, 8(2) = 1 + 1 = 2 8(3) = 1 + 1 + 1 = 3 = 0, 8(4) = 1 + 1 + 1 + 1 = 4 = 1, 8(5) = 1 + 1 + 1 + 1 + 1 = 5 = 2 Thus for each n, m E Z6) we have 8(n+6 m) = (n + m) (1) = n(l) +3 m(l) = 8(n) + 3 8(m) Le., 8 is a homomorphism To find ker a : Let x E Ker 8 8(x) = 0, the identity of Z3 x(l) = 0 ::::::} The sum of x ones which equals to zero, which means x = 0, 3 Le., Ker 8 = {O, 3} Also Im(8) = (8(n) : n E Z6} = (n(l) : n E Z6} = [0, 1 , 2] = Z 3 Theorem XVI. Let f: G � G ' is a group homomorphism. Then (a) f(e) = e: e E G, e' E G' (b) f(ai) = (f(a)fi \;j a E G. Proof. (a) Given f : G � G' is a homomorphism from G to G'. For x E G, consider f(x) e' = f(x) I e' is identity of G' I f is homomorphism = f(xe) = f(x) f(e) I Left cancellation law e' = f(e) f(e) = e'
MONOIDSAND GROUPS
269
e' = f(e) = f(aa1) = f(a) f(a1) f(a) f(a1) = e' (f(a» l f(a) f(a1) = (f(a» l e' f(a1) = (f(a» l
(b) From Part (a),
=> => =>
I f is homomorphism
Theorem XVII. If f is a homomorphism of G to G with Ker f = K. Show that K is a normal subgroup of G. (P.T.V. B.Tech. Dec. 2007) Proof. By definition, Ker f = (x E
G: f(x) = e', e' E
Gj
=K
We first show that Ker fis a subgroup of G
x, Y E Ker f => f(x) = e', fry) = e' Consider I Homomorphism f(xyl) = f(x) f(yl) = f(x) (f(y» l = e' (e;' = e' Xyl E Ker f => Ker f is a subgroup of G. Let g E G and x E Ker f, consider I f is homomorphism f(gxg1) = f(g) f(xg1) l l = f(g) f(x) f(g ) = f(g) f(x) (f(g» = f(g) e' (f(g» l = f(g) (f(g» l = e' gXgl E Ker f => Ker f is a normal subgroup of G. => Theorem XVIII. Let f be a homomorphism of a group G to a group Gc Let Im(f) be the homomorphism image of G in Gc Then Im(f) is a subgroup of Gc Proof. B y definition, 1m(/) = ([(x) : x E Gj Take e E G => e' = f(e) E 1m(/) i.e. 1m(/) '" , we first show that 1m(/) is a subgroup of G'. Let X, y' E 1m(/) => There exists x, y E G such that f(x) = x, fry) = y' xy'l = f(x) (f(y» l Consider = f(x) f(yl) I f is a homomorphism = f(X(.,.') E Im(f) I x, Y E G and G is a group :. Xyl E G x y'l E Im (f) => 1m(/) is a subgroup of G'. Theorem XIX. State and prove Fundamental Theorem ofGroup Homomorphism (P.T.V. B.Tech. Dec. 2012) Statement. Let f : G ; G' is a group from G to G' homomorphism. Then G/K '" G', where K = Ker f Proof. Givenfis a group homomorphism of G to G'. Also, Ker fis a normal subgroup of G. :. G/Ker f is defined. Define 8 : G/K ; G' by 8(Kx) = f(x), K = Ker f We show 8 is welldefined) homomorphism) oneone and onto. Let
270
DISCRETE STRUCTURES S
is welldefined : Consider Kx = Ky Xy' E K = Ker f Ha = Hb x = y 'd x, Y E Z => . . f is oneone To show f is onto: Let y E Z and consider x E Z such that y = f (x) = 2x y x= ' 2
MONOIDSAND GROUPS If y =
271
3
3 E Z, then x = "2
'l
Z.
f is not onto.
Hence f is not a group isomorphism. Example 21. (a) Let G be a group of nonzero complex numbers under multiplication. Let G/ be the group of nonzero real numbers under multiplication. Consider the mapping f : G � G'defined by f(z) = I z I . Show that G 1 Ker f,= G� (b) Define 8 : Z � Zl O by 8(n) = The remainder when n is divided by 10. Show that 8 is a homomorphism from Z to Zl O ' Also find Ker 8 and prove that Z 1 Zl O Zl O' Sol. (a) Given f : G � G' defined by f(z) = 1 z 1 Let zp z2 E G and consider f(Z, z) = 1 Z, z2 1 = 1 z, 1 1 z2 1 = f(Z,) f(z) => f is a homomorphism of G to G'. By fundamental theorem of group homomorphism, G 1 Ker f '= G'. (b) Given 8 : Z ; Z , D ' defined by 8(n) = The remainder when n is divided by 10 Let m) n E Z and consider 8(m + n) = m + 1 0 n = 8(m) + 10 8(n) Thus 8 is a homomorphism Now, Ker 8 = (n : 8(n) = 0, The identity of Z l O} = {IOn : n E Z} = 10Z By fundamental theorem of group homomorphism. Z 1 Ker 8 '= Z l O Z 1 10Z '= Z l O ' => Example 22. Let S = N x N and * be the operation on S defined by (a, b) * (a', b') = (a + a', b + b') (a) Show that * is associative. Also, show that (S, *) is a semigroup (b) Define f : (S, *) ; (Z, +) by f(a, b) = a  b. Show that f is a homomorphism. (c) Find the congruence relation  in S determined by the homomorphism f. i.e., x  y if f(x) = f(y) (d) Describle SI. Does SI have an identity element? Does SI have an inverse element? Sol. (a) Let x, y, z E S be such that x = (a, b), y = (c, d), z = (e, f) where a, b, c, d, e, fare =
elements of N. Consider.
and
x * (y * z) = (a, b) * «c, d) * (e, f)) = (a, b) * (c + e, d + f) . . . (1) = (a + (c + e), b + (d + f)) (x * y) * z = «a, b) * (c, d» * (e, f) = (a + c, b + d) * (e, f) = «a + c) + e, (b + d) + f) . . . (2) = (a + (c + e), b + (d + f)) since a, b, c, d, e, f are elements of N, therefore associativity under addition holds in N
272
DISCRETE STRUCTURES (1) and (2) x * (y * z) = (x * y) * z \;f x, y, z E S Also) given that S is closed under *. Therefore, (8, *) is a semigroup under *. f (x * y) = f«a, b) * (c, d) Consider = f(a + c, b + d) = a + c  (b + d) = (a  b) + (c  d) = f(a, b) + f(c, d) = f(x) + fry) f is a homomorphism (c) We know that a relation � on a nonempty set S is said to be a congruence relation on From
S if
(i) � is an equivalence relation (ii) If a  a', b  b'. Then ab  a'b' Also, given x  y if f(x) = fry) \;f x, Y E S Here S = N x N and if X E S, then 3 a, b E N s.t. x = (a, b) \;f a, b E N Similarly, y = (c, d) \;f c, d E N Therefore, "x  y iff(x) = f(y)" means (a, b)  (c, d) iff(a, b) = f(c, d) ab=cd a+d=b+c . . . (1) Thus, (a, b)  (c, d) if a + d = b + c \;f a, b, c, d E N Here, the relation  defined by (1) is an equivalence relation (see example 17 (b), page
65 in chapter2 "Relations")
We next show that the relation � is a congruence relation on (a, b)  (a', bi and (c, d)  (c', di. Then, we must have (a, b) * (c, d)  (a', bi * (c', di Now (a, b)  (a', bi => a + b' = b + a' (c, d)  (c', di => c + d' = d + c' and
S. For this, we show if . . . (2) . . . (4) I
. . . (3) Using (1)
Therefore (2) will be true if
or if or
if
or if which is true Hence,
(a + c, b + d)  (a' + c', b' + di a + c + b' + d' = b + d + a' + c' a + b' + c + d' = b + a' + d + c' a + b' + c + d' = a + b' + c + d'
by the definition of * I Using (1) I Using (3) and (4)
� is the required congruence relation on S.
(d) We know that if R is a congruence relation on a nonempty set S, then SIR, the set of all equivalence classes of elements of S, form a semigroup under the operation on the equivalence classes, defined by [a] [b] = lab] To describe S/�. Since � is a congruence relation on
S, S/� is a semigroup under the operation on the equivalence classes as defined above. By using fundamental theorem of semigroup homomorphism, S/ is isomorphic to f(Z) where
273
MONOIDSAND GROUPS f:
(S, *)
; (Z, +) is a semigroup homomorphism defined by f(a, b) = a  b where S = N x N and a, b E N f(Z) = (m E Z: There exists X E S for which f(x) = m} = (m E Z: f(a, b) = m \;f a, b E N} = {m E Z: a  b = m \;f a, b E N} = {m E Z = a = b + m \;f a, b E N} =Z Hence, S/ is isomophic to (Z, +). It implies that S/ and Z have same identity as well as same inverse (additive). Since (Z, +) is an additive group under addition, (Z, +) has additive identity as well as
additive inverse. Therefore, S/� has additive identity as well as additive inverse.
Example 23. Let S = N x N be the set of ordered pairs of positive integers with the operation* such that (a, b) * (c, d) = (ad + be, bd) (a) Find (3, 4) * (1, 5) and (2, 1) * (4, 7). (b) Show that * is associative. Also S is a semigroup. a (c) Define r (S, *) ; (Q, +) by f(a, b) = b ' Show that f is a homomorphism.
(d) Find the congruence relation  in S determined by the homomorphism f, i.e., x  y if f(x) = f(y)
(e) Describe SI. Does SI have on identity elements? Does it have inverses? (a) (3, 4) * (1, 5) = (15 + 4, 20) = (19, 20) (2, 1) * (4, 7) = (14 + 4, 7) = (18, 7) (b) Let (a, b), (c, d), (e, f) E N x N, then «a, b) * (c, d» * (e, f) = (ad + be, b d) * (e, f) = «a d + be) f + bde, bdf) = (adf + bcf + bde, bdf) Also, (a, b) * «c, d) * (e, f) = (a, b) x (cf + de, df) = (adf + b(cf + de), bdf) = (adf + bcf + bde, bdf) From (1) and (2), «a, b) * (c, d» * (e, f) = (a, b) * «c, d) * (e, f) * is associative Since, S is closed under * and * is associative. Hence S is a semigroup. x = (a, b), y = (c, d) (c) Let f (x * y) = f«a, b) * (c, d» = [(ad + be, bd) Then Sol.
ad + bc a c +bd b d = f (a, b) + f (c, d) = [(x) + f ry) =
Th us, f is a homomorphism.
... (1)
... (2)
274 S if
DISCRETE STRUCTURES (d) We know that a relation � on a nonempty set S is said to be a congruence relation on (i) � is an equivalence relation (ii) If a � a' b � b', then ab � a'b' Given x � y if f(x) = fry) \;j x, Y E S Here S = N x N and if X E S then, 3 a, b E N such that x = (a, b) \;j a, b E N Similarly, y = (e, d) \;j e, d E N Therefore, "x � y iff(x) = f(y)" means (a, b) � (e, d) if f(a, b) = f(e, d) a
e b d ad = be Thus, (a, b)  (e, d) if ad = be \;j a, b, e, d E N =
. . . (1)
Here, the relation  defined by (1) is an equivalence relation (See example
65 in chapter·2 "Relations")
17(b), page
We next show that the relation "'" is a congruence relation on S. For this) we show If (a,
or if or if or if or if or if or if
b) � (a', bi and (e, d) � (e', di, then we must have (a, b) * (e, d) � (a', bi * (e', di (a, b) � (a', bi, then ab' = ba' [Using (1)] Again, if (e, d) � (e', di, then cd' = dc' [Using (1)] Also, (2) will be true if (ad + be, bd)  (a'd' + b'e', b'di (ad + be) b'd' = bd(a'd' + b'ei (ad) (b'di + (be) (b'di = (bai (ddi + b(dei b' (be) (b'di + (be) (b'di = (abi (ddi + b(edi b' 2beb'd' = (ad) b'd' + (be) b'd' 2beb'd' = (be) b'd' + (be) b'd' 2bcb'd' = 2bcb'd') which is true Hence) � is the requid congruence relation on S. (e) We know that if R is a congruence relation on a non·empty set
. . . (2) . . . (3) . . . (4) by the definition of * I Using (1) I Using (3) and (4) I Using (1)
ad = be
S, then SIR, the set of all equivalence classes of elements of S) form a semigroup under the operation on the equivalence classes) defined by [a] [b] = lab] To describe S/�. Since � is a congruence relation on S) therefore) Sf,..., is a semigroup under the operation on the equivalence classes defined above. Also, by using fundamental isomorphic to f(Q) where
f:
(S, *)
....;
theorem of semigroup homomorphism,
(Q, +) is a homomorphism defined by f(a, b) = � b
where
S/ is
MONOIDSAND GROUPS
275 S = N x N and a, b E N f(Q) = (q E Q: there exists X E S for which f(x) = q} = (q E Q: f(a, b) = q V a, b E N}
Now
{
= qE Q :
f = q V a, b E N}
= Q\ the set of +ve rational numbers
Hence, S/� is isomorphic to ( Q+ , +). Further, since S/ and (Q + , +) are isomorphic, both as well as same inverse.
S/� and (Q + , +) have same identity
But, Q + , the set of positive rational number has no identity (additive) as well as no inverse (additive) therefore) S/� has no identity and no inverse.
Example 24. Show that G = {I,  1, i,  i} is a group under multiplication. Also G '" Z4 by giving an explicit isomorphism r G ; Z4' Sol. Given G = {I,  1 , i,  i} such that i2 =  1 . Clearly G is a group under multiplication. (prove yourself)
Z4 = {O, Define
f:
G ;
1 , 2,
3}
Z4 by f (l) = 0, [ ( 1) = 2 f (i) = 1 , [ ( i) = 3 G ", Z4'
2 ,fT+ 3
G
Z4
Theorem xx. Any finite cyclic group of order n is isomorphic to Zn' (P.T.V. B.Tech. Dec. 2012)
Proof. Let G = be a finite cyclic group, with a as its generator and let o(G) = n Define f : Z ; G by f(m) = am Let m, r E Z such that f(m) = am, f(r) = a' Consider f(m + r) = am+, = am . a' = f(m) f(r) Thus fis a homomorphism of Z to G. By fundamental theorem of group homomorphism. Z I Ker f", G But if s E Ker f, then by definition, f(s) = e, e E G 2 and G is cyclic, then G has atleast two generators. Which of the following functions are homomorphism ? defined by [(a) I a I (i) [ : R* > (ii) [ : Z 5 > Z2 defined by [(n) 0 if n is even and [(n) 1 if n is odd, (iii) [ : Zs > Z2 defined by [(n) 0 if n is even and [(n) 1 if n is odd, How many homomorphism are there from ? (ii) Z 1 2 into Z4 (iiL) Z9 into Z8· (i) Z 1 2 into Z 12 If G is an abelian group and let f is a function from G into G, defined by [(x) x" Is [ a homomorphism ? an isomorphism ? Consider the group G :::: {a, 1, 2, 3, 4, 5} under addition modulo 6. (a) Find the multiplication table of G. (b) Prove that G is a group, (c) Find 21 , 31 , 51 (d) Find the orders and subgroups generated by 2 and 3. (e) Is G cyclic ? Justify your answer. Let (G, *) be a group and a G, Define [: G > G by [(x) a * x (a) Prove that f is a bijection (b) On the basis of part (a), describe a set of bijection on the set of integers. (P, T, U, B, Tech, May 2005) Consider Z20 :::: {O, 1, ... , 19} under addition modulo 20. Let H be the subgroup generated by 5. (a) Find the elements and order of H. (b) Find the cosets of H in Z20. Consider G :::: {1, 5, 7, 11, 13, 17} under multiplication modulo 18. (a) Construct the multiplication table of G. (b) Find 51, 71, and 171 (c) Find the order and subgroups generated by: (i) 5, (ii) 13, (d) Is G cyclic ? Consider G :::: {1, 5, 7, 11) under multiplication modulo 12. (a) Find the order of each element (b) Is G cyclic? (c) Find all subgroups, Let 8 N x N. Let * be the operation on 8 defined by (a, b) * (a', b') (aa', bb'), (a) Show that * is associative. (Hence S is a semigroup.) (b) Define [: (8, *) > (Q, x) by [(a, b) alb, 8how that [is a homomorphism, (c) Find the congruence relation � in S determined by the homomorphism f, i.e., x � y if f(x) :::: fCy). (d) Describe S/�. Does S/� have an identity element? Does it have inverses? Consider the set N of positive integers and let * denote the operation of least common multiple (l.c,m,) on N, (a) Find 4 * 6, 3 * 5, 9 * 18, and 1 * 6, (b) Is (N, *) a semigroup ? Is it commutative ? (c) Find the identity element of *. (d) Which elements in N, if any, have inverse and what are they ? W
W
=
=
=
=
=
=
E
=
=
=
=
21.
MONOIDSAND GROUPS 22.
L 2. 8. 10. 14. 15.
279
Let Q be the set of rational numbers and let * be the relation on Q defined by a * b :::: a + b  ab (a) Find 3 * 4, 2 * ( 5) and 7 * "21 (b) Is (Q, *) a semigroup ? Is it commutative ? (c) Find the identity element for *. (d) Do any of the elements in Q have an inverse ? why ? Answers
4Z, 1 + 4Z, 2 + 4Z, 3 + 4Z, o(G) :::: 36 :::: 6. Take G :::: Z36 and H :::: i.e., a cyclic group with 6 as its generator :. [G : H] :::: o(H) 6 Also by Lagrange's theorem, o(H) I o(G) is true. The answer is not unique. 1,  1 9. (iii), 6 is a generator 12. (i) (i,) No (iii) Yes (i) 5, (ii) 32 Yes, Yes (a) Table 8,13 Table 8.13
0 1 + 2 3 4 0 0 1 2 3 4 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 4 4 5 2 0 1 5 5 2 3 1 1 1 4, 3 3, 5 1 (c) 2(d) {I, 2, 4} and {I, 3} (e) Yes, Required bijection is { : Z > Z defined by {(x) x + a " x Z (a) H {O, 5, 10, 15} and o(H) 3 (b) H H H + 1 {I, 6, 11, 16} H + 2 {2, 7, 12, 17} H + 3 {3, 8, 13, 18} H + 4 {4, 9, 14, 19} (a) Table 8.14 1 11 5 7 XIS 1 1 11 5 7 17 1 5 5 7 17 7 7 13 5 11 11 1 5 13 11 1 17 13 13 17 17 11 13 7 1 1 1 (b) 5 11, 7 13, 17 17 (c) (i) the subgroup generated by 5 G and order « 5» 6 =
=
=
16. 17.
5 5 0 1 2 3 4
=
=
=
=
=
E
=
= = =
=
18.
=
=
=
=
=
=
13 13 11 1 17 7 5
17 17 13 11 7 5 1
280
19. 20. 21.
22.
DISCRETE STRUCTURES
(ii) = the subgroup generated by 13 = {I, 7, 13} and order « 13» = 3 (d) Yes, (a) 0(1) = 1, 0(5) = 2, 0(7) = 2, 0(11) = 2 (b) No (c) G, {1J, {I, 7}, {I, 5}, {I, ll} (c) The congruence relation is given as: (a, b) (c, d) if ad :::: be (d) yes (a) 4 * 6 = 12, 3 * 5 = 15, 9 * 18 = 18, 1 * 6 = 6 (b) Yes, since the operation of least common multiple is associative. Also a * b :::: b * a i.e., the operation of least common multiple is commutative. Hence, (N, *) is a commutative semi group. (c) 1 (d) The only element which has an inverse is 1 and it is its own inverse 1 (a) 3 * 4 = 5, 2 * (5) = 7, 7 * "2 = 4 (b) Yes, (Q, *) is a commutative semigroup (c) 0 is the identity element (d) Yes, if a 1 Q, then a has an inverse and it is _ �
t:
�
E
Hints
aaI_ .
4. See Proof of Lagrange's theorem. 9. (i), (ii) Let Q :::: is a generated by g i.e., :::: {nq : n Z}. But this set contains only integral multiples of q, not every element in Q. Hence Q cannot be cyclic. 10. See example 10. 12. (ii) Consider [(2 +5 4) = [(6) = [(I) = 1 Also [(2) +, [(4) = 0 + 0 = O. 14. [(xy) = (xy)' = x'y' I Since G is abelian = [(x) fry) 15. (d) (i) 2 +6 2 +6 2 = 0 .. 0(2) = 3 = {I, 2, 4} (ii) 3 +6 3 = 0 .. 0(3) = 2 = {I, 3} (e) 0(2) = 3, 0(3) = 2 4 +6 4 :::: 2, 4 +6 4 +6 4 :::: 2 +6 4 :::: 0 0(4) = 3 5 +6 5 :::: 4, 5 +6 5 +6 5 :::: 4 +6 5 :::: 3 5 +6 5 +6 5 +6 5 :::: 3 +6 5 :::: 2 5 +6 5 +6 5 +6 5 + 6 5 :::: 2 +6 5 :::: 1 .. 0(5) = 5 Also, 1 +6 1 +6 1 +6 1 +6 1 + 6 1 :::: 0 0(1) = 6 = o(G) .. G is cyclic as there is an element 1 G such that 0(1) = o(G). E
E
MONOIDSAND GROUPS 16.
Given G is a group under * and a E G. Also f: G G is defined by f(x) a * x \j E G (a) We show fis a bijection. i.e., fis oneone and onto. f is oneone. Let x, Y E G such that f(x) fry) t
X
=
281
=
I Left cancellation law
f is oneone. ::::::} t is onto. Let a, x E
G and since G is a group under * , G must be closed under * i.e., G. Also given f(x) :::: a * x . Therefore for each x E G, there exists an element a * x E G, such that f(x) a * x. Hence f is onto. It implies f is a bijection. (b) Let Z denotes the set of integers and a E Z. Define f : Z Z such that f(x) + a x E Z. We show f is oneone and onto. Let f(x) fry) for x, Y E Z x + a :::: y + a I Right cancellation law x :::: y. Hence f is oneone Let Y E Z such that f(x) Y a*xE =
t
:::: X
=
=>
'r:/
=>
=
x + a :::: y
Z Also for each Y E Z, we can find x E Z such that f(x) :::: y. Hence f is onto. Since f is oneone and onto, f is a bijection. See examples on "cosets" and example 14, Art. 8.24 (cyclic group) in this chapter See examples 11, Art. 8.24 (cyclic group) in this chapter See examples 12,13 Art. 8.24 (cyclic group) in this chapter (c) We know that a relation on a nonempty set S is said to be a congruence relation on S if (i) is an equivalence relation on S (ii) If a a', b b', then ab a'b' Given, x Y if f(x) fry) \j x, Y E S. Here S N x N and if E S N x N a, b E N such that x (a, b) Similarly, Y (c, d) where c, d E N. Therefore, "x Y if f(x) f(Y)" means (a, b) (c, d) if f(a, b) f(c, d) x :::: y  a E
17. 18. 19. 20.
�
�
�
�
�
�
=
X
=
=
=
�
=>
3
=
�
= a
c
b
d
=
ad = be
Thus, (a, b) (c, d) if ad be ... (1) Here, the relation defined by (1) is an equivalence relation (example 17 (b) page 65 chapter 2 "Relations" ) We next show that is a congruence relation on S. For this, we show if (a, b) (a',b') and (c, d) (c', d,), then we must have . . . (2) (a, b) * (c, d) (a', b') (c', d') �
=
�
�
�
�
�
282
DISCRETE STRUCTURES
Now if (a, b) � (a', b') then ab' :::: ba' ...(3) [Using (1)] Again, if (c, d) � (c', d,), then cd' :::: dc' ...(4) [Using (1)] Also (2) will be true if by the definition of * (ac, bd) � (a'c', b'd') or if acb'd' :::: bda'c' I Using (1) or if acb'd' :::: ba'dc' or if acb'd' :::: ab' cd' I Using (3) and (4) or if acb'd' :::: acb'd' which is true Hence, � is the required congruence relation on S. (d) We know that if R is a congruence relation on a nonempty set S, then SIR, the set of all equivalence classes of elements of S, form a semigroup under the operation on the equivalence classes, defined by [a] [b] = lab] To describe S/'"'. Since � is a congruence relation on S, therefore, S/� is a semigroup under the operation on the equivalence classes as defined above. Also, by using fundamental theorem of semigroup homomorphism, S/� is isomorphic to
f(Q)·
where f= (8, *) > (Q, x) is a homomorphism defined by f(a, b) = � where 8 = N x N and a, b E N But, f(Q) = {q E Q: There exists E 8 for which f(x) = q} = {q E Q: f(a, b) = q \j a, b E N} x
{qE Q:�=q \ja,bE N} :::: Q+, the set of positive rational numbers Hence, S/� is isomorphic to (Q+, x). Further, since S/� and Q+ are isomorphic, therefore both S/� and Q+ have same identities as well as same inverses. But, since (Q+, x) is a set of +ve rational numbers and hence it forms a group under multiplica tion and hence has identity as well as inverse. Therefore, S/� has also identity (multiplicative) as well as inverse (multiplicative). 2 4, 6 Let Y E N, Then * Y means of andy. =
21.
(a)
l.c.m.,
4 * 6 = l.c.m., of 4 and 6 = 12 2 2, 3 (c) 1 * a :::: l.c.m., of 1 and a :::: 1 for any positive integer a. 3 1, 3 Also, a * 1 :::: l.c.m., of a and 1 :::: 1 1, 1 .. 1 is the identity element of N l.c.m., of 4 and 6 = 2 x 2 x 3 = 12 (d) Let a E N and b is the inverse of a such that a * b :::: 1 ::::} l.c.m. of a and b :::: 1, which is possible iff a :::: 1, b :::: 1 . . The only number which has an inverse is 1 and it is its own inverse. 22. (c) Let e is the identity element for *, then for every 1 a E Q, we have x,
x
a * e :::: a ::::} a + e  ae :::: a e  ae :::: O e(l  a) = => e= o.
0 Hence, the identity element is
=>
x
0
:f
283
MONOIDSAND GROUPS (d) Let a, x E
Q are such that a * x 0 (identity) + x  ax :::: 0 a + x (1 a) :::: 0 a x , a" 1 a  x (1  a) a1 This a a_1 is the inverse of a (a t: 1). =:::}
=>
=:::}
a
=
=>
=
=

_
8.29. DIRECT PRODUCT OF GROUPS Let G , and G2 be any two groups and G = G, X G2 denotes the cartesian product of G , G2 with the binary operation * defined by (g1 ' g) * (g'1 ' g') = (g,g'1 ' g2g') \j g1 ' g, E G 1 ' g2' g'2 E G2 The group G = G, X G2 , is know as direct product or external direct product of G, and G2 . Theorem. Show that the direct product of Gl and G2' denoted by G = G X G2' is also a group under the binary composition defined by (gl' g,) x (g" g',) = (gl g'l' g2 g',) \j gl' g� E G l' g2' g'2 E G2 Proof. Since G1 and G2 are groups) therefore) g1 ' g, E G, and g2' g2 E G2 => g, g, E G, and g2 g'2 E G2 => (g,g'1' g2g') E G, x G2 = G Th us) G is closed under * Associativity. Let g1 ' g'1 ' g'; E G, and g2' g;, g'� E G2 Consider (g1 ' g) * «(g'1 ' g;) * (g'; , g;) = (g1 ' g) x (g, g'; , g'2 g'�) = (g, (g, g';), g2 Similarly, o(b) = 2, o(c) = 2 Here
Thus, all the elements of G are of order 2. It means that G does not have any element whose order equals to 4, the order of G. Hence G cannot be cyclic. (e) We know that a finite cyclic group of order n is isomorphic to the group of integers modulo n. Here, G is not cyclic. Hence G ;;t'
Z4'
Zn'
Example 2. Let H and K be any two groups and G = H x K be the direct product of the groups H and K under the binary operation * defined by (h, k) (h', k') = (hh', kk') \;j h, h' E H, k, k' E K Let H' = H x {e}. Then show that (i) H' '" H (ii) H' � G where G = H x K (iii) GIH' = K where G = H x K *
MONOIDSAND GROUPS
285 :
Sol. (i) Define a mapping 8 H' ; H such that 8(h') = h \;j h' E H, h E H we show 8 is welldefined) homomorphism) oneone and onto. 8 is welldefined, Let h'I ' h; E H' such that h { = (h I ' e), h'2 = (h2 , e) \;j h I ' h2 E H Consider h'l = h'2 (h I ' e) = (h2 , e) => h, = h 2 => 8(h;) = 8(h;) 8 is welldefined Homomorphism. Consider 8(h', h;) = (h , h2 , e) = (h I ' e) (h2 , e) = 8(h',) 8(h ;) 8 is homomorphism Oneone. Consider 8(h;) = 8(h; ) h, = h2 (h I ' e) = (h2 , e) h'l = h; 8 is oneone Onto, Let h E H, e E Ie}, then by the definition of cartesian product of two sets. (h, e) E H x (e) = H' h' E H' where h' = (h, e) => 8(h') = (h, e) Also 8 is onto Hence
H' = H
(ii) To show H' � subset of G. For e E H, (e, e)
G (read as H' is a normal subgroup of G), we first show H' is nonempty
E
= H' => (e, e) E H'. Thus H' '"
h' E G => H' (;; G => Hence, H' is a nonempty subset of G. We next show H' is a subgroup of G Let h;, h; E H' => h; = (h I ' e), h'2 = (h2 , e) Consider h; h'i .' = (h I ' e) (h2 , e)l = (h I ' e) (hi.'. e) = (hI ' h21 , e) \;j h I ' h2 E H E H x [e] l h; h; E H H' is a subgroup of G H
x
(e)
286
DISCRETE STRUCTURES
We next show H' is a normal subgroup of G. Let g E G such that g = (h I ' k,)
=Hx
K
=>
3
h, E
H,
k, E
K
h' = (h, e) Consider gh' gI = (h I ' k,) (h, e), (h I ' k,)' = (h I ' k ,) (h, e) (h ," k,') e E (e) is the identity = (h I ' k,) (hh," k,') = (h , hh ," k, k ,') = (h, hh," e) E H x (e) = H' Since h I ' h E H and H is a group, therefore h,' E H => h,hh,' E H => gh'g' E H' \j h' E H', g E G
Let
h' E
H'
=>
Hence H' is a normal subgroup of G.
(iii) Since H' is a normal subgroup of G, therefore, G/H' is welldefined Define [ : H x K ; K such that [(h, k) = k \j h E H, k E K [ is welldefined: Consider (h I ' k,) = (h2 , k) k , = k2 [(h I ' k,) = [(h2 , k) . . [is welldefined I [(h, k) = k \j h E H f is a group homomorphism. Let gI ' g2 E G = H x K => 3 hI ' h2 E H and kI ' k2 E K :
.
such that
g, = (h I ' k ,) and g2 = (h2 , k)
By the definition of cartesian product of H and K, we have
Therefore)
g,g2 = (h I ' k ,) (h2 , k) = (h, h2' k, k2) [(g,g) = [(h , h2 , k, k) = k , k2 = [(h I ' k,) [(h 2 , k)
Hence f is a group homomorphism. Ker [ =
{(h, k) E H x K: [(h, k) = e, e is the identity of K = {(h, k) E H x K : k = e, e E K = {(h, e) E H x {ell = H' By fundamental theorem of semigroup homomorphism of groups. H x KlKer [ "" K H x KlH' "" K G/H' "" K Hence the theorem
MONOIDSAND GROUPS
287 TEST YOUR KNOWLEDGE 8.4
L
Consider G :::: Z2 X Z3. Describe and find the multiplication table of the group G :::: Z2 X Z3 (a) Is G abelian? (b) Is G cyclic? (c) Is G = Z6?
Answer L Z2 X Z3 = {(O, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)) (a) Yes (b) Yes, G is cyclic group generated by (1, 1) , (d) Yes, G = Z6 Hints
(b) Consider
= (1, 1) = (1 +2 1, 1 +3 1) = (0, 2) = (1 +2 1 +2 1, 1 +3 1 +3 1) = (1, 0) = (1, 0) + (1, 1) = (1 +2 1, 0 +3 1) = (0, 1) = (1, 1) + (1, 1) + (1, 1) + (1, 1) + (1, 1) (0, 1) + (1, 1) = (1, 2) (1, 1) + (1, 1) + (1, 1) + (1, 1) + (1, 1) = (1, 2) + (1, 1) = (1 +2 1, 2 +3 1) = (0, 0) Hence every element of G can be expressed in some powers of (1, 1). Therefore, G is a cyclic group with (1, 1) as generator 0 (1, 1) = 6 = 0 (Z 2 X Z,) (c) Since G :::: Z 2 X Z 3 is cyclic group of order 6. Also, any finite cyclic group of order n is isomorphic to Zn (1, (1, 1) + (1, (1, 1) + (1, 1) + (1, (1, 1) + (1, 1) + (1, 1) + (1,
"
1) 1) 1) 1)
� X Z 3 = Z6
M
G=�
PERMUTATION AND SYMMETRIC GROUP 8.30. BASIC TERMINOLOGY Before, we define our symmetric groups, students are advised to go through the following basic terminology_ f(l)
Example consider a function f : S ; S defined on a nonempty set S 1
= 2, f(2) = 3, f(3) =
= {I, 2,
3} such that
Fig. 8.6 on S.
Clearly, f is both one·one and onto (see Fig.
8.6) and we say, f :
S ; S is a permutation
288
DISCRETE STRUCTURES
8.30. (a) PERMUTATION Let S be any nonempty and a finite set) then a oneone and onto mapping from S to S is called a permutation. Notation: The set of all permutations on a nonempty set S is denoted by A(S). Further, if A set S has n elements, then, the total number of bijections (oneone and onto mapping) from S to S is n Cn = n!. In the above example, the total number of oneone and 3 onto mappings from S to S is C3 = = Hence, we can say that if a set S contains n elements, thenA(S), the set of all permutations on S) will contain n Cn = n! elements. We next introduce a notation to represent these permutations.
3! 6.
3, (3)
The function f : S ; S defined on a nonempty and a finite set S = {I, f = 1 can be expressed in array from as shown below:
f (1) = 2, f (2) =
5,
[1 3 31 J
f= 2
3,
2
6, [15 3 31 46 5 46 J
2,
3}
such that
3, 4, 5, 6}
6,
Consider another example of a permutation g : S ; S, defined on S = {I, 2, such that g(l) = g(2) = g(3) = 1 , g(4) = g(5) = 2, g(6) = then, g can be expressed in array form as shown below:
g=
2
2
8.30. (b) COMPOSITION OF PERMUTATIONS IN ARRAY FORM
[1 4 33 45 51J [15 4 31 4 35J
Consider two permutations f and g) given in array
f= 2
2
'
form as
g=
2
2
then the composition of f and g (denoted by gof or sometimes, by gf), in array given as below:
, 345 34 (� 4j, 3) :,i (l, 4 3 5 �) 1,
gof =
2
1
2
form is
2
2
I __ J
Fig. 8.7
We wish to find the following elements
(gof)(l), (gof)(2), (gof)(3), (gof)(4), (gof)(5).
To find (gof)(l): Notice that compositions of permutations expressed in array brackets, is carried out from right bracket to left bracket, by going from top to bottom (in right bracket) and then again from top to bottom (in left bracket). Hence, to find (gof)(l), we start from the right bracket in the following manner: 8.7, we notice that
4 4,
4 '4'.
2 (right bracket), 2 ; (left bracket), => 1 ; This is equivalent to say that (gof)(l) = g(f(l» = g(2) = which means '1 ' goes to To find (gof)(2): We start from '2' in the right bracket as shown below: 2 (right bracket), ; 2 (left bracket), => 2 ; 2 This is equivalent to say that (gof)(2) = g(f)(2» = g(4) = 2 => 2 ; 2 In Fig
; 4
1 ;
4
MONOIDSAND GROUPS
289
To find (gof)(3): Start from '3' (in the right bracket), 3 ; 3, 3 ; 1 => 3 ; 1. To find (gof)(4): Here 4 ; 5, 5 ; 3 => 4 ; 3 To find (gof)(5): Here 5 ; 1, 1 ; 5 => 5 ; 5
go[= [ � 1 a� [1 1 J : 2 3 4 4 2
Hence)
2 3 4 _ 4 2 3
2 3 4 4 3 5
�J
8.30. (e) PERMUTATION AS A SINGLE ROW Consider [ =
[1
1 6J
2 3 4 5 4 2 3 5 6
We try to express 'f ' in a single row: Here, 1 � 2, 2 t 3, be represented as (1235).
3 t 5, 5 t 1, i.e., we started from '1 ' and ended in '1 '. This fact can
4 ; 6, 6 ; 4 it can be expressed as (46). (1235)(46), is in a single row representation. The product of (1235) and (46)
Also,
Hence [ =
8.30. (d) ORBIT OF A PERMUTATION
[1
Consider a permutation [ given by [= In cyclic notation,
6, 2
[=
1 6J
2 3 4 5 6 5 4 3 2
(1625)(34)
Here, (625) is called a cycle and the set {I, 6, 2, and 5. Also {3, 4} is an orbit of the elements 3 and 4.
5} is called an orbit of the elements 1,
8.30. (e) DISJOINT PERMUTATIONS Two permutations f and g on a nonempty set S are called any X E S, [(x) '" x => g(x) = x and
g(x) '" x
disjoint
permutations, for
=> [(x) = x \;j X E S
e.g., [ = (12), g = (13) are not disjoint since [ (1) = 2 and g(l) = 3 '" 1 But [ = (132) and, g = (45) are disjoint permutations. 8.30. (f) CYCLIC PERMUTATIONS Let S be a finite set containing n elements, then, a permutation f on S is called a cyclic permutation or a cycle if there exists xp x2, x3, ...... , xn such that:
290
DISCRETE STRUCTURES ..
f(x,) = X2 , f(x,) = X3 , f(x3) = x4 , f(xn_l) = xn, f(xJ = x, A cyclic permutation on n elements: xp x2 ) ..... ) xn is denoted by (x1x2x3 . . . xn) and is called a cycle of length n or nrecycle. In particular, a cycle of length '2 ' it called a transposition or 2cycle. Remark 1: Two permutations f andg are said to be disjoint of there is no common element in any •••
cycles of f and g.
ILLUSTRATIVE EXAMPLES
Example
1.
cycles.
[
J
1 2 3 4 5 6 7 Consider f = 2 4 6 1 7 3 5 . Express f as a product of disjoint
Sol. We wish to find the disjoint cycles of f. .�
Here, 1
>
2, 2
>
4, 4 > 1 => (124) is a cycle of f.
.�
3 > 6, 6 > 3 => (36) is a cycle of f. Further, 5 > 7, 7 > 5 => (57) is a cycle of f
Also
f=
[1
2 3
4
5
1 7
2 4 6
[
= (124)(36)(57).
Example 2. Consider f= cycles.
Sol. Here) ..
4
1 2 3 8 7 5
5 6 7 6 2 1
4
: J . Express fas a product of disjoint
.. 1 t 8, 8 t 3, 3 t 5, 5 t 6, 6 t 2, 2 t 7, 7 t 1
(1835527) is a cycle of f.
Also 4 > 4 :.
(4) is also a cycle off
Hence f= (1835627)(4), is expressed as a product of disjoint cycles.
8.31 . MULTIPLICATION OF CYCLES
[1 [1
We now develop a procedure for multiplication of cycles (not necessarily disjoint). Consider f =
g=
4
2
We find fog in cyclic notations.
2 3
4
5 6
6 8
3
5 2
2
4 5
6
8
3
3
4 6
1
7 8 1 7 7 8 5 7
J J
We first express f and g a product of cycles (not necessarily disjoint).
MONOIDSAND GROUPS
291 f = (14387)(26) g = (124875)(36) '
a
fog = (14387)(26)(124875)(36) = (JToY, say
Now
... (1)
Note that each cycle that does not contain on element fixes the element.
To find fog, Consider the elements 1, 2, 3, 4, 5, 6, 7, 8. Start with '1 ' as shown in the below pattern. 1�4�4�8�8 � 1 � 8 ' ' As '1 goes to '8 , we start with '8' as shown in the below pattern. 8 �7�7 � 5� 5 � 8 � 5 As '8' goes to '5 ' , we start with '5' as shown in the below pattern 5 � 5 �5 �1 �1 This completes the cycle (185). Now) we are left with the remaining elements 2, 3, 4, 6, 7. Start with '2 ' as shown in the below pattern. 2 � 2�6 �6 �3
7 �1�1�2�2 This completes the cycle (237) Now, we are left with the remaining elements 4 and 6 only. Start with '4' as shown in the below patter. 4�3�3�3�6 6 �6�2�4�4 This completes the cycle (46) and there are no elements left with us. Hence, the only cycles of f are (185), (237) and (46), which are disjoint also. fog = (14387)(26)(124875)(34) Hence from (1), = (185)(237)(46), which shows that the composition of any two permutations on a finite set can be expressed as a product of disjoint cycles
3, Let f = (13)(27)(456)(8) and g = (1237)(648)(5) What is the cycle form of fog? or Write fog as a product of disjoint cycles.
Example
SoL
Let
°1
°2
a3
°4
't"l
't"2
"3
fog = (13)(27)(456)(18)(1237)(648)(5) , say
. . . (1)
292
DISCRETE STRUCTURES Start with
'1 ' as shown in the below manner.
1 � 3 � 3 °3 ) 3 °4 ) 3 � 7 � 7 � 7 => 1 ; 7 I A cycle that does not contain an element fixes the element ' Start with '7 as shown in the following pattern. 7 � 7 02 ) 2 "s ) 2 04 ) 2 � 3 � 3 � 3 => 7 ; 3 Start with '3 ' as shown in the below manner. 3 � 1 °2 ) 1 � 1 °4 ) 1 � 2 � 2 � 2 => 3 ; 2 Start with '2 ' as shown in the below manner. ) 7 0, ) 7 °4 ) 7 � 1 � 1 � 1 => 2 ; 1 This completes the cycle (1732). Now) we are left with the elements 4, 5, 6, 8. Start with '4' as shown in the following manner. 2�2
°2
4 � 4 °2 ) 4 � 5 °4 ) 5 � 5 � 5 � 5 => 4 ; 5 Start with '5' as shown in the following manner. 5 03 ) 6 04 ) 6 � 6 � 4 � 4 => 5 ; 4 This completes the cycle (45). Now, we are left with 6 and 8. Start with '6' , as shown in the following manner. 5�5
02 )
6 � 6 02 ) 6 0, ) 4 04 ) 4 � 4 � 8 � 8 => 6 ; 8 Start with '8' , as shown in the following manner. °2
) 8 0, ) 8 °4 ) 8 � 8 � 6 � 6 => 8 ; 6 This completes the cycle (68) and we are left with no element with us. Thus, the only cycles of fare (1732), (45) and (68) and these cycles are disjoint also. Hence, from (1), fog = (13)(27)(456)(8)(1237)(648)(5) = (1732)(45)(68) 8�8
which shows that the composition of any two permutations on a finite set can be expressed as a product of disjoint cycles. Remark 2. 'n' (n > transpositions 2cycles) 2cycles is not unique. (a1a2a3 ... ak ) = (a1ak)(a1ak_1)(a1ak_2) ... (a1a2) = (a2ak)(a2ak_1)(a2ak_2) ... (a2a1) = (a,ak)(a3ak_1)(a3ak_2) ... (a3a2)(a3a1)
a product of
Every permutation on a finite set containing elements 1) can be expressed as (or and the decomposition of a permutation into a product of It means to say that
Example 4. Express (12345) as a product of 2cycles in atleast 4 ways. (12345) = (15)(14)(13)(12) Sol. = (25)(24)(23)(21) = (35)(34)(32)(31) = (45)(41)(42)(43)
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8.32. PROPERTIES OF PERMUTATIONS
We now study the following properties of permutations. Property 1. Any two disjoint permutations on a finite set commute with each other. Proof. Let f and g are any two disjoint permutations. We need to show fog = got. For this, let X E S and consider f(x) '" x and since f and g are disjoint, then g(x) = x. (By definition) Let f(x) = y, then, since f(x) '" x => y '" Consider . . . (1) ([og) (x) = f(g(x» = f(x) = y and . . . (2) (gof)(x) = g([(x» = g(y) We claim g(y) = y. For if, g(y) '" y and since f and g are disjoint, then fry) = y = f(x) => y = x (since fis a permutation), a contradiction. Hence g(y) = y and therefore, from (1) and (2), we have ([og)(x) = (gof)(x) fog = gof Property 2. Any permutation on a finite set can be expressed as a product of transpo sitions (not necessarily disjoint). 1 2 3 4 5 6 7 8 9 Consider f= = (1324)(56)(789) 3 4 2 1 6 5 8 9 7 = (14)(12)(13)(56)(79)(78), (See remark 2 above) Here, the transpositions (14), (12) and (13) are not disjoint (as these 2cycles have '1 ' as a common element). Property 3. Any permutation on a finite set can be expressed as a product of disjoint cycles (see examples 1, 2, 3 above) Property 4. The nth power of a ncycle is 1. Consider a 2cycle (ab) containing two elements, then, we must show (ab)2 = 1. x
[
°1
°2
Consider (ab )(ab) Here, b � a � b => b ; b 2 Hence (ab) = I (since each element goes to itself). Consider another example. Let (1234) be a 4cycle containing 4 elements. We show (1234)4 = 1. For this, Consider
°1 °2 (1234)2 = (1234)(1234)
3 � 4 � 1 => 3 ; 1
J
294
DISCRETE STRUCTURES
This completes the cycle (13). 4 � 1 � 2 => 4 ; 2 This completes the cycle (24). (1234)2 = (1234)(1234) = (13)(24) Hence, °1 °2 a3 (1234)3 = (1234)(1234)2 = (1234)(13)(24)
2 � 3 °2 1 1 0, 1 1 => 2 ; 1 This completes the cycle (1432). Hence, (1234)3 = (1432) Finally, (1234)4 = (1234)(1234)3 =
CJ1
. . . (1) I using (1)
. . . (2)
I using (2)
CJ2
(1234)(1432)
2 �3
02 )
2 => 2 ; 2
4 � 1 � 4 => 4 ; 4 (1234)4 = (1234)(1432) = 1. (Since each element goes to itself). Property 5. The product of a permutation on a finite set and its inverse is always equals to 1. Consider f = (a, b , )(a2 b) (a3 b ) (a4 b4) ' then 1 r' = «a, b ,)(a2 b ) (a3 b ) (a4b4» = (a4b4)(a3 b) (a2 b) (a, b,) fri = (a, b,)(a2 b) (a3 b) (a4b 4)(a4b 4)(a3 b) (a2 b ) (a, b,) = I (Prove it)
8.33. EVEN AND ODD PERMUTATIONS
A permutation on a finite set is called an even (odd) permutation if it can be expressed as a product of even (odd) number of transposition (2cycles). Following results can be trivially proved. I. The product of two even permutations is an even permutation (sum of two even numbers is even)
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295
II. The product of two odd permutations is an even permutation (sum of two odd numbers is even) III. The product of an even and an odd permutations is an odd permutation (sum of an even and odd number is an odd number) IV. Inverse of an even (odd) permutation is an even (odd) permutation. V. Identity permutation is always an even permutation. Example 5. Consider the following permutation on S = {I, 2, 3, 4, 5, 6, 7}. Examine
whether it is an odd permutation 2 or even permutation:
[31
J
[�
2
4
3 1
4
5
5
6
6 7
7
2
J
.
2 3 4 5 6 7 4 1 5 6 7 2 We first express f as a product of transpositions (2cycles) Here f = (13)(24567) = (13)(27)(26)(25)(24) (See remark 2, above) . . Since f is the product of 5 (odd) transpositions, hence f is an odd permutation
Sol. Let
f=
Theorem I. The order of a permutation on a finite set, written as a product of disjoint cycles, is the least common multiple of the lengths of the disjoint cycles. Proof. This is beyond the scope of the syllabus. Example 6. Find the order of the following permutations (i) (12)(13)(14)(15)(1 6) (ii) (24)(26)(28)(13)(15)(1 7) Sol. (i) Let f = (12)(13)(14)(15)(16) = (165432) . . its order = 6 (number of distinct elements in f) (ii) Let f = (24)(26)(25)(13)(15)(17) = (2864)(1753)
Clearly, right hand side is a product of two disjoint cycles. . . Required order = Lc.m. of the lengths of (2864) and (1753) = 4
Example 7. Find the order of the following permutation. f = (123)(234)(456)(67) Sol. Given permutation is not a product of disjoint cycles. Hence) we first express it as
a product of disjoint cycles.
°1
0"2
a3
°4
f = (123)(234)(456)(67) , say Consider The elements of the given permutation fare 1) 2) 3, 4, 5, 6 and 7. Start with '1 ' as shown in the following manner.
1 � 2 02 ) 3 °3 ) 3 04 Start with '3' as shown in the following manner.
)
3 => 1 ; 3
3 � 1 °2 1 1 °3 1 1 °4 1 1 => 3 ; 1 This completes the cycle (13) We are left with 2, 4, 6, 6 and 7 Start with '2' as shown in the following manner. 2�3
°2
)4
°3
)5
°4
) 5 => 2 ; 5
. . . (1)
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DISCRETE STRUCTURES
Start with '5' as shown in the following manner. 5 � 5 °2 ) 5 0, ) 6 Start with '7' as shown in the following 7 � 7 02 ) 7 03 ) 7 Start with '6' as shown in the following manner.
°4
) 7 => 5 ; 7
04
)
6 => 7 ; 6
6 � 6 °2 ) 6 °3 ) 4 °4 ) 4 => 7 ; 6 This completes the cycle (25764) and we are left with no elements. Hence, the only cycles of f are (13) and (25764) which are also disjoint. :. From (1) f = (123)(234)(456)(67) = (13)(25764) Hence, order of f= l.c.m of the lengths of (13) and (25764) = l.c.m of 2 and 5 = 10 Example 8. Let
f= g=
[� [�
�:;: �!:�
�J �J
Compute each of the following: (a) f1 (b) gof
[
1 2 3 4 5 6 Sol. (a) f = 2 1 3 5 4 6
and
(c) fog
J
On expressing f as a product of transpositions) we get f = (12)(3)(45)(6) r' = «12)(3)(45)(6» 1 = (6) (45) (3) (12) We now express [1 in the array form. From (1), We notice that 6 ; 6, 4 ; 5 and 5 ; 4, 3 ; 3, 1 ; 2 and 2 ; 1. 1 2 3 4 5 6 f1 = 2 1 3 5 4 6
[
J ( 1 2 3 4 5 6) : (1 2 3 4 5 6 ) 1               I
(b)
gof = _
J,
: J..
6 1 2 4 3 5 : 2 1 3 5 4 6
[11 62
, , • __ 1
J
3 4 5 6 2 3 4 5 (see Multiplication of cycles)
Example 9. Find ai ba where a = (135)(12), b = (1579). Sol.
0"1
0"2
a = (135)(12) = (135)(12) , say
... (1)
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297
Consider the elements 1) 2) 3) 5. Start with '1 ', 1 � 3 °2 ) 3 => 1 ; 3 Start with '3', 3 � 5 Start with '5', 5 � 1
°2
°2
) 5 => 3 ; 5
) 2 => 5 ; 2
Start with '2', 2 � 2 0, ) 1 => 2 ; 1 This completes the cycle (1352) Hence, a = (135)(12) = (1352) => aI = (1352)1 = (2531) To find aI ba:
Gl
G2
G3
aI ba = (2531)(1579)(1352) = (253 1)(1579)(1352) , say,
Consider the elements 1) 2) 3) 5) 7) 9 Start with '1 ', 1 � 2 0, ) 2 This completes the cycle Start with '2', 2 � 5 0, ) 7 Start with '7',
7�7
Start with '9',
9�9
0,
0,
)9 )1
0,
) 1 => 1 ; 1
0,
) 7 => 2 ; 7
0,
0,
... (1)
... (1) ) 9 => 7 ; 9 ) 3 => 9 ; 3
Start with '3', 3 � 1 0, ) 5 0, ) 2 => 3 ; 2 This completes the cycle (2793) We are left with the element '5'. Therefore, (2531)(1579)(1352) = (1)(2793)(5) = (2793)
Remark 3: Let A(S) denotes the set of all permutations on a finite set S, then, either all permutations of A(S) are even or exactly half are even. 8.34. SYMMETRIC GROUP
Let Sn denotes the set of all permutations on a fmite set S = (1, 2, 3, ...,n) and let 0 S ; S is a oneone and onto mapping defined by 1 2 3 ... n 0= 0(1) 0(2) 0(3) . . . 0(n) Then, Sn forms a group under the composition of mappings and called symmetric group of degree n. Theorem II. Show that Sn (n :> 1) is a group under the composition of mappings. Proof. Let f, g E Sn => f, g are both one·one and onto maps on a a finite set S. Since f, g are both oneone and onto, their compostion maps fog, got are also oneone and onto maps => fog, gof are also permutations on S. Hence, fog, gof E Sn Sn is closed under the composition of mappings.
[
:
J
298
DISCRETE STRUCTURES
Also) Sn has identity mapping I) each permutation on Sn has its inverse. Further) associativity under the composition of mapping also holds in Sn " Hence) Sn forms a group under the composition of mappings. Theorem III. Show that Sn (n :> 1) forms a group under the composition of mapping with order o(Sn) = n!. Proof. By definition, elements of Sn are of the form
[1
nn J
2 3 ... ° = o(1) o(2) o(3) . . . 0( )
where CJ is a oneone and onto map on the set S = (1) 2) 3) . . . ) n) Now, there are 'n' choices of o(1). Once, o(1) is determined, there are (n  1) choices for o(2) (since ° is one·one, we must have o(1) "" o(2» . After o(2) is determined, there are (n  2) choices for o(3). Continuing in this manner, we observe that there is (n  (n  1» = 1 choice for 0(n).
Hence, Sn has n(n 1) :. o(Sn) = n!
(n2) ... (n n 1)
= n! elements
Example 10. (a) Find the elements and the multiplication table of S3' (b) Show that S3 is a nonabelian group. Sol. (a) We know that Sn has n! elements, therefore, S3 has 3 ! = 6 elements. S3 = {E, 0 ' 02' 03' x X4 m = x = m x4 x ... (1) ::::::} and y x3 n = y = n x3 y ... (2) The only elements m in Z4 and n in Z3 which satisfies (1) and (2) are m = 1, n = 1 Hence the unity of Z4 x Z3 is (1, 1). o .
9.9. MORPHISM OF RINGS
The word 'morphism' is a combination of various terms like ring homomorphism) ring isomorphism etc. 9.9. 1 . Ring Homomorphism
Let (R, +, ) and [R', +', '] be two rings. Then a mapping f : R ; R' is called a ring homomorphism (i) f (a + b) = f(a) +' f(b) \;j a, b E R (ii) f (a . b) = f(a) ' f(b) \;j a, b E R. Also) a ring homomorphism and oneone is called monomorphism. A ring homomor phism and onto is called epimorphism. Further) if in addition) fis oneone and onto) thenfis called on isomorphism and R and R' are said to be isomorphic and we write R == R/.
322
DISCRETE STRUCTURES
Remarks : To check whether the two rings are isomorphic, we should check the following :
Both rings should have same cardinality. (b) Both rings should be commutative. (c) Both rings should have same unity. (d) If there exists an equation which is solvable in one ring, but not solvable in another ring, then two rings cannot be isomorphic. (a)
Lemma: If 8 is a ring homomorphism from a ring R to a ring R'. Then (i) 8 (0) = 0, 0 E R' (ii) 8 (a) =  8 (a) \;j a E R Proof. (i) Consider 8 is a ring homomorphism 8 (a) = 8 (a + 0) = 8 (a) + 8 (0) 8 (a) + 0 = 8 (a) + 8 (0) I Adding 0 E R on L.H.S. => 0 = 8 (0) I Left cancellation law => 8 (0) = 0 => (ii) Consider 8 (a + ({I» = 8 (a  a) = 8 (0) = 0 I Using part (i) 8 (a) + 8 (a) = 0 => 8 (a) =  8 (a) => Example 8. Consider the rings [Z, + , 'J and [2Z, + , 'J and define f : Z � 2Z by f(n) = 2n \;j n E Z (P.T.U. B. Tech. Dec. 2010) Is f a group homomorphism ? Is f a ring isomorphism ? Sol. Z and 2Z are groups under addition. Consider f : Z � 2Z defined by f(n) = 2n \;j n E Z For m, n E Z, consider f(m + n) = 2(m + n) = 2m + 2n = f(m) + f(n) \;j m, n E Z Hence f : Z � 2Z is a group homomorphism. To check whether f is a ring isomorphism. For m, n E Z, consider f(mn) = 2 mn and f(m) f(n) = 2m . 2n = 4 mn. f(mn) '" f(m) f(n) \;j m, n E Z . . f : Z � 2Z cannot be a ring isomorphism. Alternatively: We know that Z is a ring with unity and 2Z is also a ring but without unity. Hence both rings donot have same unity. Therefore Z and 2Z are not isomorphic, i.e., Z ", 2Z. Example 9. Examine whether [2Z, + , J and [3Z, +, J are isomorphic rings ? ... (1). Sol. Consider the equation x + x = x . x
This equation makes sense in both rings. For x = 2, (1) gives 2 + 2 = 2 . 2 => 4 = 4, which is true. Thus equation (1) has a solution x = 2 E 2Z. For x = 3, (1) gives 3 + 3 '" 3 . 3. This means x = 3 is not a solution of (1). Hence we conclude that equation (1) has a solution in 2Z, but does not have a solution in 3Z. Therefore 2Z and 3Z cannot be isomorphic rings. (See Remark (d) of Art. 9.9.1) Example 10. Show that following rings are not isomorphic. (ii) [3Z, + , 'J and [4Z, + , 'J (i) [Z, + , 'J and [M2X2 (R), + , 'J (iv) [Z2 x Z2' + , 'J and [Z4' + , l (iii) [R, + , 'J and [Q, + , 'J
RINGS
323
Sol. (i) We know that Z is a commutative ring and M2X2 (R) is a noncommutative ring. (since for A, B E M2X2 (R), AB = BA is not true) . . The rings [Z, + , .] and [M2X2 (R) , + ,.] cannot be isomorphic rings. (Remark (a) of Art. 9.9.1) ... ( 1) (ii) Consider the equation x + x + x = x . x This equation maks sense in both rings 3Z and 4Z. For x = 3, (1) gives 3 + 3 + 3 = 3.3 => 9 = 9, which is true. Hence equation (1) has a solution x = 3 in 3Z. For x = 4, (1) gives 4 + 4 + 4 '" 4 . 4 =>
12 ", 16,
Hence equation (1) has does not have a solution in 4Z Therefore, 3Z and 4Z cannot be isomorphic rings. (Remark (d) of Art. 9.9.1) (iii) We know that the set of real numbers R is uncountable and the set of rationals Q is countable (see chapter on 'sets'). Hence R and Q cannot have same cardinality and therefore cannot be isomorphic. (iv) By definition, Z 2 = [0, 1, +2 ' x 2] Z 2 X Z2 = [(0, 0), (0, 1), (1, 0), (1, 1 )] .. For (m, n) E Z2 x Z2' consider (m, n) . (1, 1) = (m, n) = (1, 1) . (m, n) Le., (1, 1) is the unity (multiplicative identity) of Z2 x Z2' Now Z4 = [0 ) 1) 2) 3) +r x4] 1 is the unity of Z4' Thus Z2 x Z2 and Z4 do not have same unity. Therefore they cannot be isomorphic rings. Example 1 1. Let R and R' be two rings. Define 8 . R '> R' by 8 (a) = 0 \;j a E R. Show that 8 is a ring homomorphism. Sol. Let a, b E R. Since R is a ring and a, b E R it implies a + b E R and ab E R. Consider 8 (a + b) = 0 = 0 + 0 = 8 (a) + 8 (b) Also 8 (ab) = 0 = 0.0 = 8 (a) 8 (b) Hence 8 is a ring homomorphism. Example 12. Let R be a commutative ring and suppose px = 0 \;j E R, p is prime. Show that the mapping f . R '> R defined by f (x) = xP, E R is a ring homomorphism. Sol. Let x, y E R and consider f (x + y) = (x + y)1', using Binomial theorem, = Peo xP + Pel xP  1 Y + PC2 xP  2 y2 + ... + Pc yP X
X
= xP + Pc, xP  1 y + Pc, xP  2 y2 + ... + yP For x, Y E R => xP  1 y E R. (Since R is a ring) Using px = 0 \;j E R, we have pxp1 y = 0 X
As we know that p
I Pc.
=>
Pc. = kp for some k.
Therefore) PC2 xP  2 y2 = kpxP2 y2 = 0 for some k, etc.
,
DISCRETE STRUCTURES
324
Hence Further)
f (x + y) = xl' + yP = f (x) + f (y) f (xy) = (xy)P = xl'yP = f (x) f (y)
f is a ring homomorphism. 9.9.2. Kernal f
If f: R ; R' is a ring homomorphism, then the kernel of f, is the set of all those elements whose image is the zero element of R/. Thus Kerf = (r E R : f (r) = OJ. :
Example 13. Let 8 R ; R' be a ring homomorphism from a ring R to the ring R'. Show that Ker 8 is a subgroup of R under addition. Sol. By definition, Ker 8 = (r E R : f (r) = OJ Since 8 : R 7 R' is a ring homomorphism) 8 (0) = 0, 0 E R' Ker 8 # q, => Let x, y E Ker 8 => 8 (x) = 0, 8 (y) = O. Consider 8 (x y) = 8 (x)  8 (y) = 0  0 = 0 x  Y E Ker 8 \;j x, Y E R => => Ker 8 is a subgroup of R. 9.10. SUBRING
(P. T. U. Dec.
2005)
Let [R, + , 'J be a ring and S be a subset of R. Then S is called a subring of R iff S is itself a ring under the operations of R. Example. Let E = the ring of even integers, Z = the ring of integers. Then E is a subring of Z. Also Z c Q, therefore Z is a subring of Q. Theorem II. A nonempty subset of a ring R is a subring of R iff (i) a, b E S => a  b E S \;j a, b E S (ii) a, b E S => ab E S \;j a, b E S. Proof. Let S be a subring of R. We prove (i) and (ii).
As S is a subring of R, S is itself a ring under the operations of R. Hence S is additive group under +. that is) S is closed under addition. i.e.) For a, b E S, a + bE S \;j a, b E S Also for each b E S, there exists  b E S such that  b is the additive inverse of b. Now a E S,  b E S => a + ( b) E S a  b E S, which proves (i) => Further) as S is a subring of R, it must be a ring under the operations of R. Thus, S is closed under multiplication i.e.; For a, b E S => a . b E S \;j a, b E S, which proves (ii) Converse. Let (i) and (ii) hold. We show S is a subring of R under the operations of R. For a, a E S => a  a E S => O E S I Using (i) Le., S has additive identity.
RINGS
325
Again o E S, a E S => O  a E S =>  a E S Leo) S has additive inverse. For a E S, b E S =>  b E S From (i), a  ( b) E S a + b E S \;j a, b E S Leo) S is closed under addition. Since S c R, elements of S are also in R . . Associativity under addition holds in S For a, b E S C R => a, b E R a+b=b+a
I Using (i)
(Proved above)
I R is additive group Hence we can say that S is an additive group. From (ii), a, b E S => a. b E S \;j a, b E S Le., S is closed under multiplication. Finally, a, b, C E S C R ::::::} a, b, C E R a. (b + c) = a . b + a . c Distributive laws hold in R (a + b) . c = a . c + b . c Le., left distributive law and right distributive law holds in S. Hence S is a ring under the operations of R. Example 14. The set of integers Z is subring of Q. Sol. We know that "A nonempty subset S of a ring R is a sub ring ofR iff (i) a, b E S => a  b E S \;j a, b E S (ii) a, b E S => a . b E S \;j a, b E S. Since Z c Q i.e., Z is a subset of Q. a, b E Z => a  b E Z \;j a, b E Z is true. For Also for I Theorem II a, b E Z => a . b E Z \;j a, b E Z. Hence Z is a subring of Q.
Example 15. (a) Show that 3Z is a subring of Z. (b) Find all subrings of Zs ' Sol. (a) We know that "A nonempty subset S of a ring R is a subring ofR iff (i) a, b E S => a  b E S \;j a, b E S (ii) a, b E S => a . b E S \;j a, b E S Let x, y E 3Z ::::::} x = 3m, m E Z, Y = 3n, n E Z x  y = 3m  3n = 3(m  n) E 3Z Consider nce m, n E Z => m  n E Z (Z is a ring) 3(m  n) E 3Z => x  Y E 3Z Also x.y = 3m . 3n = 3(3mn) = 3k E 3Z I since 3, m, n E Z and Z is a ring :. 3mn E Z I k = 3 mn where Hence we can say that 3Z is a subring of Z. (b) Consider the following subsets of ZS ' Z2 = [0, 1, +2' x 2l ; Z3 = [0, 1, 2, +3' X 3l Z4 = [0, 1, 2, 3, +4 ' x 4] ; Z5 = [0, 1, 2, 3, 4, +5' x 5] Z6 = [0, 1, 2, 3, 4, 5, +6) x6] ; Z7 = [0, 1, 2, 3, 4, 5, 6, +7 ) x7]
I:
DISCRETE STRUCTURES
326
For a, b E Z2' we observe that a  b E Z2 V a, b E Z2 Also a . b E Z2 V a, b E Z2 Hence Z2 is a subring of Zs For a, b E Z3' we observe that a  b E Z3 V a, b E Z3 Also a . b E Z3 V a, b E Z3 Hence Z3 is a subring of Zs Similarly) we can prove that Z4' Z5) Z6) Z7 are all subrings of ZS" 9.1 1 . UNITS
Let (R, + , .) be a ring with unity. An element a E R is said to be a unit (or invertible) if for 0 '" a E R, 3 b E R such that a . b = 1 = b . a or An element is a unit if a has multiplicative inverse, aI E R such that aa1 = 1 = a1a Consider the rings (R, + ,.) and (Q, + , .). Every nonzero element in R (set of reals) and 4 Q (set of rationals) has a multiplicative inverse. For example, "3 E R has multiplicative inverse 3 4 3  E R since  .  = 1. 4 3 4
The only elements in Z that have multiplicative inverses are  1 and 1.
Theorem III. An element a in Zn is a unit iff a and n are relatively prime. Proof. By definition, Zn = [0, 1, 2, 3, ... n  1, +n' xn] Let a E Zn be a unit. It means there exists an element b E Zn such that a xn b = 1 i.e., when ab is divided by n, the remainder is 1. i.e., a xn b = 1 ::::::} ab = nq + 1, where q is the quotient. => ab  nq = 1 I For a, b E R, if3 x, y E R such that ax  by = 1, then g.c.d. (a, b ) = 1 => (a, n) = l Example 16. Determine the units (those elements which have multiplicative inverses) for each of the following rings .(b) [Q, + , 'J (a) [Z, + , 'J (c) [C, + , 'J (d) [M2x2(R) , + , 'J (e) [Z2 ' +2' x 2J if) [Z6 ' + 6' x 6J (h) [Z5' + 5' X 5J (g) [Zs ' +s' x sJ (i) [Z x Z , +, 'J U) [Z/ , +, l Sol. (a) The only units of Z are  1 and 1. Since for each a E Z, a .l = a = 1 . a a . ( 1) =  a = ( 1) . a (b) Every nonzero element in Q has multiplicative inverse (unit) (c) Every nonzero complex number in C has multiplicative inverse (unit) (d) Every nonsingular or invertible matrix is a unit of M2 2(R). Z2 = [0 , 1, +2 ' x 2J (e) Consider For each x E Z2) we have x X2 1 = x = 1 x2 x . . The only unit of Z2 is 1 . Z6 = [0 , 1, 2, 3, 4, 6, +6' x6J if) The only elements in Z6 which are relatively prime to 6 are 1 and 6. ( : (1, 6) = 1, (6, 6) = 1 x
RINGS
327
Hence the units of Z6 are 1 and 5. Also 1 x6 1 = 1, 5 x6 5 = 1 I Theorem III Zs = [0, 1, 2, 3, 4, 5, 6, 7, +s, xs] (g) The only units of Zs are those elements which are relative prime to 8. The elements relative prime to 8 are 2) 3) 5) 7. Hence the units of Zs are [1, 3, 5, 7]. Z5 = [0, 1, 2, 3, 4, +5 ' x 5] (h) The only units of Z5 are those elements which are relative prime to 5. The elements relative prime to 5 are 1) 2) 3) 4. Hence the units of Z5 are [1, 2, 3, 4] . (i) The units of Z are 1 and  1. Therefore units of Z x Z are (1, 1) (1,  1), ( 1, 1), ( 1,  1). 3 (j) Z 2 = Z2 X Z2 X Z2 where Z2 = [0, 1, +2' x2] The only unit of Z2 is 1. Therefore the only unit of Z23 is (1, 1, 1). Theorem IV. For all a, b E R, (ii) a.( b) = ( a).b =  a.b (i) a.O = O.a = 0 (iii) ( a) . ( b) = a.b (P.T.V. B.Tech. May 2012) (iv) If R has a unit element 1, then ( 1) . a =  a, ( 1) . ( 1) = 1. Proof. (i) For a E R, consider a . 0 = a . (0 + 0) I 0 is the identity Left distribution law =a.O+a.O a.O + ( a) . 0 = a.O + a.O + ( a) . 0 I Right distributive law (a + ( a» . O = a.O + (a + ( a» . 0 O=a.O+O o = a. 0 (Try yourself) Similarly, 0=0.a I by part (i) (ii) Consider a . (b + ( b» = a.O = 0 a.b + a.( b) = 0 I Left distributive law a. ( b) =  a.b (Try yourself) Similarly, ( a) . b =  a.b (iii) ( a) . ( b) =  (a . ( b» I by part (ii) =  ( (a . b» =a.b (iv) If R has a unit element 1, then l 1 .a=a a + ( 1) . a = 1 . a + ( 1) . a Left distributive law = (1 + ( 1» . a I by part (i) =O.a=O ( 1) . a =  a In particular) if a =  1) then ( 1) . ( 1) =  ( 1) = 1.
328
DISCRETE STRUCTURES
9.12. INTEGRAL DOMAIN
Zero divisor
A nonzero element a E R is called a zero divisor if there exists a nonzero element b E R such that ab = 0 A commutative ring R is called an integral domain if for every o '" a. b E R. ab = 0 => a = 0 or b = 0 Thus) a commutative ring R is called an integral domain if R has no zero divisor. Example (i) Z. the ring of integers is an integral domain. Also, Q, R, C are integral domains. (ii) Z5) Zs) Zl O etc. are not integral domains. Z3) Z5) Z7 are integral domains.
Theorem V. The element a in the ring [Zn' +n' xn1 is a zero divisor iff a is not relative prime to n (i.e.; g.c.d. (a, n) '1" 1). Proof. The proof of above theorem is beyond the scope of this book. Theorem VI. (Zp' +P' \) has no zero divisors iffp is a prime number. Proof. Let Zp has no zero divisors. We show p is prime. 1 < a < p) 1 < b < p For if) p = ab) a Xp b = 0 where a) b are nonzero numbers ::::::} ::::::} Zp has a zero divisor) a contradiction) hence p is prime. Converse. Let p is prime) we show Zp has no zero divisor. Let a xp b = 0 for a) b E Zp => ab = 0 (modp) => plab => pia or plb I p is prime But a, b E Zp . . a, b < p. Hence a = 0 or b = 0 . . Zp has no zero divisor. Example 17. Consider (Zs' +s' xsJ· Find the zero divisors ofZs' Is Zsan integral domain ? Sol. Zs = [0, 1, 2, 3, 4, 5, 6, 7] . Since 4 xs2 = 0 = 2 x8 4 2 and 4 are zero divisors.
Also
4 and 6 are zero divisors.
Zs cannot be an integral domain since 0
with zero divisors.
t: 2) 4 E Zs
===>
2xs 4 = 0 Le.) Zs is a ring
Example 18. Show that Z, the set of integers is an integral domain. Sol. We know that Z is a commutative ring. Also if a, b E Z then, gives a . b = 0 either a = 0 or b = 0 Z is a commutative ring without zero divisors i.e.) Z is an integral domain. Example 19. Consider m =
�) b, c, d E R] as a ring under matrix addition and (� �) and B (� �) are zero divisors.
[(�
matrix multiplication. Show that A =
a,
=
RINGS
329
(� �) # 0, B = (� 0 AB  (0 �) (� �) = (�
Sol. Given
A=
But
A and B are zero divisors.
Example 20, Show that the ring Z29 of integers modulo 29 is an integral domain. Sol. By using theorem VI, Zp has no zero divisors iffp is prime. Here p = 29, which is a
prime. Therefore) Z29 has no zero divisors. Consequently) Z29 is an integral domain.
Example 21. Show that the ring Z1 05 of the integers modulo 105 is not an integral domain. Sol. By using theorem VI, Zp has no zero divisors iff p is prime. Here p = 105, which is
a composite number. Consequently) Zl05 has zero divisors. . . Zl05 cannot be an integral domain.
Theorem VII. The cancellation laws hold in a ring (R, + ,.) iffR has no zero divisors.
Or
A commutative ring R is an integral domain iff for a, b, c E R (0 '" a) ab = ac => b = c. Proof. Let R be an integral domain and consider ab = ac (0 # a) ab  ac = O => Left distribution law a.(b  c) = 0 => either a = 0 or b  c = O Since R is an integral domain) =>
so it has no zero divisors
bc=O l a#O => b = c. => Converse. Let cancellation laws hold in R. We show R is an integral domain. Let a, b E R and 0 # a. Consider a.b=0 a.b=a.O I a.0=0 Cancellation law b=O
Hence R is an integral domain.
Example 22. Consider X = [0, 2, 4, 6, 8, +1!Y xl!). Is X an integral domain ? Justify your answer. Sol. We first check whether X is a commutative ring under addition modulo 10 and
multiplication modulo 10. The addition modulo 10 table is shown in Table I. Table I +10
0 2 4 6 8
0
0 2 4 6 8
2
2 4 6 8 0
4
4 6 8 0 2
6
6 8 0 2 4
8
8 0 2 4 6
330
DISCRETE STRUCTURES
From Table I, we observe that every element inside the table is also in X. It means that
X is closed under addition modulo 10. i.e., a, b E X => a+l O b E X \;ja, b E X Addition modulo 10 is associative i.e.) For a, b, c E X, a+ l O (b+ lOc) = (a+ l O b) + 10 c \;j a, b, c E X
The first row inside the table coincides with the topmost row of the Table 1. It means 0 is the additive identity of X. Also each element of X has an additive inverse. For example Inverse of 2 is 8 (the intersection of 2 and 8 at zero) 1 2+ 1 0 8 = 0 Inverse of 4 is 6 etc. 1 4+ 1 0 6 = 0 Table I is symmetrical w.r.t. +10' It means a+lOb = b+ Oa \;j a, b E X. ' Hence X is an additive group under +10" Now consider the multiplication modulo 10 table as shown in Table II. Table II XlO
0 2 4 6 8
0
0 0 0 0 0
2
0 4 8 2 6
4
0 8 6 4 2
6
0 2 4 6 8
8
0 6 2 8 4
From Table II, we observe that each element inside the table is also in X. It means that X is closed under multiplication modulo 10 i.e., For a, b E X, ax lObE X \;j a, b E X Multiplication modulo 10 is associative. Also for a, b, C E X, I Left distributive law ax l O (b+ , Oc) = aX , o b+ l Oax l Oc (a+, O b) x l O c = ax l O c +, O ax l O c \;j a, b, c E X Right distributive law Hence X is a ring under addition modulo 10 and multiplication modulo 10. Now to check commutativity of X, from Table II, we observe that the table is symmetri cal w.r.t. X 10. It means X is a commutative ring. Finally, X is a ring without zero divisors as it is clear from Tabe II, i.e., there do not exist nonzero elements whose product is zero. Hence (X, +10' x lO) is an integral domain.
Example 23. Consider X = [0, 1, 2, 3, 4, 5, +6' xi Is X an integral domain ? Justify your answer. Sol. Proceeding as in example 4, we can prove that X is a commutative ring. Also 2, 3 E X and 2 x6 3 = 0 i.e., product of two nonzero elements in X is a zero element. Thus) X is a commutative ring with zero divisors. So X cannot be an integral domain. Example 24. (i) Give an example of a finite integral domain (ii) Give an example of an infinite integral domain ? (iii) Give counter example to illustrate the fact that product of two integral domain may not be an integral domain ?
RINGS
331
Sol. (i) We know that Zp is an integral domain iff p is prime. Thus Z2 = [0, 1, +2' x2] , Z 3 = [0) 1) 2 ) +3 ) x 3] are finite integral domain. (ii) Z) the set of integers is an example of an infinite integral domain. (iii) Consider Z2 = [0, 1, +2' x 2l ; Z3 = [0, 1, 2, +3 ' x 3l Clearly) Z2 and Z3 are integral domains as 2 and 3 are primes. Consider the product Z2 x Z3' We know that Z2 and Z3 are commutative rings with unity so their product Z2 x Z3 is also a commutative ring with unity. But (1, 0), (0, 2) E Z2 X Z3 are two nonzero elements and (1, 0) . (0, 2) = (0, 0) . i.e., Z 2 x Z3 has zero divisors and hence cannot be an integral domain. Example 25. Find all zero divisors of Z1 5' Z6' Z20' Z,5 = [0, 1, 2, ...... 14, + 5 ' X, 5l Sol. (i) ' We know that an element m, in [Zn) +n' xn] is a zero divisor iff m is not relative prime to n. Here n = 15. The only elements which are not relative prime to 15 are 3, 5, 6, 9, 10, 12. Hence 3, 5, 6, 9, 10, 12, are zero divisors. 3 x 1 5 5 = 0, 9 x 1 5 10 = 0, 5 X , 5 6 = 0 10 x 1 5 12 = 0 etc. Also Z6 = [0, 1, 2, 3, 4, 5, +6 ' x 6l (ii) The only elements which are not relative prime to 6 are 2, 3, 4 The zero divisors of Z6 are 2, 3, 4 2 x 63 = 0, 3 x 64 = 0 etc. Also Z20 = [0, 1, 2, 3, ...... 19, +20 ' x 20l (iii) The only elements which are not relative prime to 20 are 2, 4, 5, 6, 8, 10, 12, 14, 16, 18. Hence the zero divisors of Z20 are 2, 4, 5, 6, 8, 10, 12, 14, 16, 18. Example 26. Consider the ring Zl O = {D, 1, 2, 3, ....., 9} of integers modulo 1 D. (a) Find the unit of Zl O (b) Find  3,  8, [51 (c) Let f(x) = 2x2 + 4x + 4. Find the roots off(x) over Zl O' By finding roots off(x), Conclude that can a polynomial of degree n have more than n roots ? Sol. (a) The units of ZlO are those integers which are relatively prime to 10. Clearly, the units of ZlO are 1, 3, 7, 9. (b) By  a in a ring, we mean that element such that a + ( a) = 0 = ( a) + a. Therefore,  3 = 7 (Since 3 + 7 = 0 = 7 + 3)  8 = 2 (Since 8 + 2 = 0 = 2 + 8) By aI in a ring, we mean that element such that a . aI = 1 = aI a Therefore, 31 = 7 (Since 3.7 = 1 = 7.3) (c) The roots of f(x) will be those elements from 0 to 9 which will yield O. Put, x = O, f (0) = 4, f (1) = 2 + 4 + 4 = 10 = 0 f (2) = 8 + 8 + 4 = 20 = 0 x = 2, x = 3, f (3) = 4, x = 4, f (4) = 2, [ (5) = 4, [ (6) = 0, [ (7) = 0, [ (8) = 4, [ (9) = 2 Thus, f (1) = 0, f (2) = 0, [ (6) = 0, [ (7) = 0 Hence, the roots of f(x) are 1, 2, 6, 7. . .
332
DISCRETE STRUCTURES
Conclusion. This example shows that a polynomial of degree n can have more than n roots over an arbitrary ring. But this cannot happen if the ring is a field. TEST YOUR KNOWLEDGE 9.1
Consider the following sets. The operations involved are the usual operations defined on the sets. (c) [C, + , J (a) [Z, + , 'J (b) [Q, + , J (e) [Z 2' +2' X2J (d) [M2+2 (R), + , J (j) [Z 6' +6' X6J (i) (h) [Z + [Z + x [Z x Z, + , 'J J J X (g) s ' s ' s 5' 5' 5 3 (J) [Z2 , + , 'J (i) Which of the above sets are rings ? (ii) Which of the above rings are commutative ? Are they rings with unity? Determine the unity of the above rings. 2. Perform the indicated operations on the [Z8; +8' XsJ: (ii) ( 3) Xs 5 (i) 2x s ( 4) (iii) ( 2) Xs ( 4) (iv) ( 3) xs 5 +s ( 3) Xs ( 5) 3. (a) Determine all solutions of the equation x2  5x + 6 :::: 0 in Z 12. Find all elements of Z12 which satisfy this equation. (b) Find all solutions of the equation .x2  5x + 6 :::: 0 in Z. Can there be any more than two solutions to this equations in Z ? 4. Solve the equation x2  4x + 4 :::: ° (a) in Z 1 2 (b) in Z (c) in M2x2 (R) (d) in Z3 ' 5. For any ring [R; + , 'J, simplify (i) (a + b) (c + d) for a, b, c, d E R (ii) If R is commutative, show that (a + b)2 :::: a2 + 2ab + b2 a, b E R (iii) Simplify (a + b) 5 in Z5 ' 6. Consider the ring Z10 :::: [0, 1, 2, 3, ...... 9] of integers modulo 10. (P.T. U. B.Tech. May 2008) (a) Find the units of Z 1 0 1 (b) Find  3,  8 and 3(P.T. U. B.Tech. May 2008) (c) Let [(x) = 2X2 + 4x + 4. Find the roots of [(x) over Z IO o] ...... 29, + o Xs [0, 1, 2, s :::: 7. Consider Zso ' (a) Find  2,  7 and  11 (b) Find 71 , 111 and 261 8. Suppose a2 :::: a for every a E R (such a ring is called a Boolean ring.) Prove that R is commutative given that x + y = O x = y for allx, Y E R. (P.T. U. B.Tech. May 2009) 9. Let G be any additive group. Define a multiplication in G by a.b :::: ° for every a, b E G. Show that this makes G into a ring. 10. Let R be a ring with a unity element. Show that R*, the set of units in R is a group under multiplication. 11. Prove that if x2 :::: 1 in an integral domain D, then x :::: ° or x :::: l. 12. If R is a ring with unity, then this unity is unique. 13. Prove that the ring Z2 x Zs is commutative and has unity. L
'r:/
=>
333
RINGS Answers
L
2. 3.
4. 7.
All are rings except (d), all rings are commutative. The unity for (a), (b), (c), (e), (f), (g), (h) is l. The unity for (d) is I, = (� �). The unity for (i) is (1, 1). The unity for (j) is (1, 1, 1). (iii) 0 (iv) O. (i) 0 (ii) 1 (b) x = 2, 3, No (a) x = 2, 3, x = 2, 3, 6, 11 (c) Any 2 x 2 matrix satisfying (X + 21)' = 0 (a) x = 4, 10 (b) x =  2 (d) x = 1 5. (iii) (a + b)5 = a5 + b5 6. (a) 1, 3, 7, 9 (b)  3 = 7,  8 = 2, 31 = 7 (c) 1, 2, 6, 7. (a)  2 = 28,  7 = 23,  11 = 19 (b) 71 :::: 13, 111 :::: 11, 261 does not exist since 26 is not a unit. (i) (ii)
Hints
By  a in a ring Z8' we mean that element, say, a, such that a + ( a) :::: 0 :::: (a) + a ..  4 = 4,  3 = 5 etc. 3. (b) Z l = [0, 1, 2, 3, ...... 11, + 1" X l ] If x ='6, then x'  5x + 6 = 36  30'+ 6 = 12 = 0 If x = 11, then x'  5x + 6 = 121  55 + 6 = 72 = 0 4. (a) x' + 4x + 4 = 0 (x + 2) (x + 2) x =  2,  2. But  2 = 10 Also if x = 4, then ex' + 4x + 4 = 16 + 16 + 4 = 36 = 0 (d)  2 :::: 1 .. x :::: 1 is the only solution. 5. (i) (a + b) (c + d) = a.(c + d) + b.(c + d) = a.c + a.d + b.c + b.d I Left distributive law (ii) (a + b)' = (a + b) (a + b) = a.(a + b) + b(a + b) :::: a.a + a.b + b.a + b.b :::: a2 + a.b + b.a + b2 :::: a2 + a.b + a.b + b2 a2 + 2a.b + b2 I R is commutative 2 2 :::: a + 2a.b + b (iii) (a + b)5 :::: 5co a5 + 5c1 a4b + 5c2 a3b24 + 5cs a2b3 + 5c4 ab4 + 5c5b5 :::: a5 + 5a4b + lOa3b2 + 5ab + b5 I In Z5 ' 5 = 0, 10 = 0 etc. 2 8. Given a :::: a for all a E R ... (1) Let a, b E R. Since R is a ring, it is closed under addition. a + b E R Using (1), (a + b)' = a + b (a + b) (a + b) = a + b Right distributive law (a + b).a + (a + b). b = a + b a.a + b.a + a.b + b.b :::: a + b a + b.a + a.b + b :::: a + b I a.a :::: a2 :::: a b. a + a . b + b = b Left cancellation law Right cancellation law b. a + a . b = 0 a.b = b. a Ifx + y = O x :::: y 9. Given (G, +) is an additive group. Also a.b :::: 0 E G 'r:j a, b E G . G is closed under multiplication. For a, b, c E G, a.(b.c) = a.O = 0 ... (1) 2.
=>
=>
=>
=>
=> =>
=> =>
=>
:
=>
334
DISCRETE STRUCTURES
Also (a.b) . c o.c 0 ... (2) From (1) and (2), a.(b.c) (a.b) . c \j a, b, c E G I b, c E G b.c 0 :. Associativity holds in G. Also a.(b + c) a.b + a.c 0 + 0 0 a.b + a.c :::: 0 + 0 :::: 0 I a, E G a.c :::: 0 a.(b + c) a.b + a.c Similarly, (a + b) . c a.c + b.c (G, +,.) is a ring. Let R* be the set of units in R. We show R* is a group under multiplication. Let a, b E R* i.e., a and b are units in R there exist aI , b1 E R such that aa1 :::: 1 :::: aI a and bb1 1 b1 b Consider (ab) (b1 aI) a(bb1) aI a . 1 . aI aa1 1 (b1 a1 ) (ab) b1 (a1 a) b b1 . 1 . b b1 b 1 Also (ab) (b1 a1) 1 (b1 aI) (ab) .. Hence ab is also a unit in R. Consequently ab E R* i.e., R* is closed under multiplication. Since R is associative and elements of R* are from R R* is associative under multiplication. Finally, if a is a unit in R, then aI is also a unit in R. Consequently aI E R*. Hence R* is a group under multiplication. =
=
=>
=
=
=
=
C
=
=
=:::}
=
10.
=:::}
=
=
=
=
=
=
=
=
=
=
=
=
(P.T. U. B.Tech. Dec. 2009)
9.13. FIELD
A commutative ring F with unity such that each nonzero element has a multiplicative inversy is called a field. It is denoted by F. Alternatively, F is a field if its nonzero elements form a group under multiplication. ILLUSTRATIVE EXAMPLES
Example L Show that the following sets are fields. (ii) [R; + , J (iii) [C; +, ]. (i) [Q; + , J SoL (i), (ii), (iii). We know that the sets Q, R and C and commutative ring with unity (see prob. 1. Exercise 9.1). Also each nonzero element in Q, R and C has multiplicatve in
verse. Hence they form fields.
( � �)
Example 2_ Consider the set M of all 2 x 2 matrices of the type _ the conjugates of a and b. Is M a field ? Justify your answer.
(� n B = (_ � �) AB = (� 32) (11 11) = ( 11 55) BA = (_ � �) (� �) = (� _ �) # AB
SoL Consider A, B E M where A =
Then Also

Hence M is not commutative and therefore cannot be field.
where iT, b are
RINGS
335 Example 3. Consider Zr = [0, 1, 2, 3, ...... 6, +7' xl Show that Zr is a field. Sol. Consider the addition modulo 7 table as shown in Table 1. Table I +,
0 1 2 3 4 5 6
0
0 1 2 3 4 5 6
1
1 2 3 4 5 6 0
2
2 3 4 5 6 0 1
3
3 4 5 6 0 1 2
4
4 5 6 0 1 2 3
5
5 6 0 1 2 3 4
6
6 0 1 2 3 4 5
We first show that Z7 is a ring under addition modulo 7 and multiplication modulo 7. From Table I, we observe that each element inside the table is also in Z7' It means that Z is closed under + 7" Addition modulo is always associative The first row inside the table coincides with the top most row of Table 1. It means 0 is the additive identity. Each element of Z7 has additive inverse. For example) Inverse of 1 is 6. Inverse of 2 is 5 etc. 1 1 +7 6 = 7 = 0 1 2 +7 5 = 7 = 0 Also Table I is symmetrical w.r.t. +7' It means Z7 is additive w.r.t. +7 i.e.) For a, b E Z 7 ) a +7 b = b +7 a V a) b E Z7 ' . . Z7 is an additive group w.r.t +7" Now consider the multiplication modulo 7 table as shown in Table II. x7
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6
2 0 2 4 6 1 3 5
Table II
3 0 3 6 2 5 1 4
4 0 4 1 5 2 6 3
5 0 5 3 1 6 4 2
6 0 6 5 4 3 2 1
From Table II, we observe that each element inside the table is also in Z7' It means Z7 is closed w.r.t. x7" i.e.) for a) b E Z7 ::::::} aX7b E Z7 v a) b E Z7 Finally) For a) b) C E Z7) a x 7 (b+7 c) = aX 7 b +7 a x7 c (a +7 b ) X 7c = a X7c +7 6X 7 C is true for all a, b, C E Z7" Hence Z7 is a ring w.r.t. addition modulo 7 and multiplication modulo 7.
336
DISCRETE STRUCTURES
Also the Table II is symmetrical w.r.t. x7. It means that Z7 is commutative i.e., aX 7 b = bX 7 a \;j a, b E Z7
Further, the second row inside the table coincides with the topmost row of Table II. It means 1 is the multiplicative identity of Z7' Hence) we have shown that Z7 is a commutative ring with unity. To show Z7 is a field) we show each nonzero element of Z7 has multiplicative inverse. The units of Z7 are those elements which are relative primes to 7. (See Topic on 'units') The elements which are prime to 7 are 1) 2) 3) 4) 5) Hence the units of Z7 are 1) 2) 3) 4) 5) We can also check the elements which are units as below :
6.
6.
6 6
1 x 7 1 = 1; 2 X 7 4 = 1; 3 X 7 5 = 1; 4 x 7 2 = 1; 5 x 7 3 = 1; x 7 = 1. Hence) each nonzero element of Z7 has multiplicative inverse. Therefore Z7 is a field.
9.14. GAUSSIAN INTEGERS
Any number of the form a + ib) a) b E Z is called a Gaussian integer. Example 4. Show that the set J[iJ of Gaussian integers form a ring under addition and multiplication. Is it an integral domain ? Is it field ? Sol. Let X = [a + ib, a, b E ZJ be the set of Gaussian integers. Then X is a ring. We check X for integral domain. Let a + ib) c + id E X such that a) b) c) d are nonzero integers. Consider (a + ib) (c + id) = 0 => ac  bd + i(ad + bc) = 0 = 0 + Oi ac  bd = 0, ad + bc = 0, which is possible if either a = 0 = b or c = 0 = d i.e., if either a + ib = 0 or c + id = 0 Hence X is without zero divisor. Therefore, X is an integral domain. Further, if 0 ", a + ib E X be any non·zero element of X where a, b E Z, then the multipli· cative inverse of a + ib is 1 _ 1 a  ib a  ib _ ;; 2 2 .". a + ib a + ib a  ib a + b a Since "2.,,. is not necessary an integer. a + b2 X cannot be a field. Example 5. The set of numbers of the form [a + b..{2, a, b E QJ is a field. 
_ 
X
__
_

X = [a + b,fi; a, b E QJ . We show X is a ring.
Sol. Let
Let x, Y E X =>
X
= a, + b,,fi, al ' b, E Q
Y = a2 + b2 ,fi , a2 , b2 E Q x + y = a, + b,,fi + a2 + b2 ,fi = a, + a2 + (b, + b) ,fi E X I ': al ' a2 E Q => a, + a2 E Q
i.e.) X is closed under addition.
Addition of rationals is associative.
RINGS
337
Further, 0 + 0 ,fi E X is the additive identity since a + b ,fi + 0 + O,fi = a + 0 + (b + 0) ,fi = a + b ,fi
Also
0 + o ,fi + a + b,fi = 0 + a + (0 + b) ,fi = a + b ,fi
Also for a + b ,fi E X,  (a + b ,fi) is the additive inverse of a + b ,fi , since a + b ,fi + ( a  b ,fi ) = a  a + (b  b) ,fi = O + O ,fi Also for a + b ,fi , e + d,fi E X, we have (a + b ,fi ) + (e + d,fi ) = a + e + (b + d) ,fi = e + a + (d + b) ,fi = (e + d ,fi ) + (a + b ,fi )
Hence X is an additive group. Further, (a + b ,fi) (e + d,fi) = ae + 2bd + (ad + be) ,fi E X i.e.) X is closed under multiplication. In Q, the associative laws and distributive laws hold good. Multiplication in Q is com mutative and 1 + o,fi is the multiplicative identity. . . X is a commutative ring with unity. We lastly show that each nonzero element in X has a multiplicative inverse. Let a + b ,fi , a, b '" 0 be any element of X. Let e + d,fi is the multiplicative inverse of a + b ,fi such that (a + b ,fi ) (e + d ,fi ) = 1 + o,fi = (e + d ,fi ) (a + b ,fi ) =>
1 1 a  b,fi x = a + b ,fi a + b ,fi a  b,fi a b ,fi a  b ,fi = 2 = 2 2 2 2 a _ 2b a _ 2b 2 a _ 2b
e + d ,fi =
Hence X is a field.
a b x 2 = 2 ,fi E X 2 2b _ a 2 a _ 2b
·I :
a b ' a 2  2b 2 2b 2 _ a 2 E Q
Example 6. The set of real numbers of the form fa + b,f2, a, b E Z] is an integral domain. Is it a field ? (P.T.V. B. Tech. Dec. 2010) Sol. Proceeding as in example 5, we can show that X = [a + b,fi, a, b E Zl is a commutative ring. We show X is an integral domain. i.e., X has no zero divisor. Let a + b ,fi , e + d,fi E X where a, b, e, d E Z
Consider (a + b ,fi) (e + d,fi) = 0 => =>
ae + 2bd + (ad + be) ,fi = 0 = 0 + o ,fi ae + 2bd = 0, ad + be = 0,
which is possible only if either a = b = 0 or e = d = 0 Le., either a + b ,fi = 0 or e + d,fi = 0 Hence X is an integral domain.
338
DISCRETE STRUCTURES
Finally, we check X for multiplicative inverse. Consider 0 '" 5 + 3,[2 E X and c + d,[2 E X such that (5 + 3,[2) (c + d,[2) = 1 + 0,[2 = (c + d,[2) (5 + 3,[2) 5  3,[2 1 1 x "'::"':'F c + d,[2  5 3,[2 5 + 3,[2 5  3,[2 + =
= X cannot be a field.
5 3 5  3,[2 5 3 . = "7  "7 ,[2 'l X. Smce "7 ' "7 'l Z 25 18 _
Example 7. Let D be the ring of all real 2 x 2 matrices of the form D is isomorphic to the complex number C where D is a field.
Sol. Given D is a ring of real 2 x 2 matrices of the form
( )
a b f b a = a + ib
f
[(� ) (� �)] (�: � b a +
( ) (
Thus f is a homomorphism.
c d a b a =f d c a + ib = c + id
Further, Let f b



�). Show that
�). Define f : D ; C, by
(� ) ( � )
We show fis homomorphism, one·one and onto. Let Consider
(�
(�
)
b c d a ' d c ED
+ d» +c = a + c + i(b + d) = a + ib + c + id =f
)
Equating real and imaginary part, we get a = c, b = d Thus a + ib = c + id. . . f is one·one. Hence f : D ; C, defined by
( )
a b f b a = a + ib is an isomorphism. Theorem VIII. Every field is an integral domain. But the converse is not true. (P.T.V. B.Tech. May 2010, May 2012) Proof. Let F be a field. We show F is an integral domain. As F is a field, F must be
commutative. We show F is without zero divisors. Let a, b E F such that a . b = 0
. .. (1)
RINGS
339
If a t: 0) then as F is a field, each nonzero element of F has multiplicative inverse. i.e., for a E F, there exists aI E F such that a aI = 1 = aI a From (1), a.b = 0 1 a (a.b) = aI . 0 = 0 I a.O = 0 \;j a E R (aI a) . b = 0 lob = 0 b=O Hence if a '" 0, then b=O Similarly, if b '" 0, then a=O
Hence F is without zero divisors. Consequently F is an integral domain. The converse is, however, not true. Example of an integral domain, which is not a field. (P.T.V. B.Tech. Dec. 2013) Z is an integral domain. But for 2 E Z, there is no a E Z such that 2a = 1 = a.2 i.e., 2 has no multiplicative inverse. Therefore, Z cannot be a field.
Theorem IX. Any finite non·zero integral domain is a field. Proof. Let D = [ap a2) an] be a finite nonzero integral domain where each a/s are ••••••
distinct. We show D is a field. For this, we show F is commutative ring with unity and each nonzero element of D has multiplicative inverse. Since D is an integral domain, D must be commutative. Let 0 t: a E D and consider the set [aap aa2, aan] . We claim D = [aap aa2) ....... aan] Now ai E D (1 s i s n) and a E D => aai E D \;j 1 s i s n. aap aa2 ) aan E D ::::::} aan) C D ::::::} (aap aa2) D = [aap aa2) aan] .. We next claim that the elements aap aa2) aan are all distinct. For if) aai = aaj aa  aa = 0 => a(a  a) = 0 => But 0 '" a E D and D is without zero divisors .. ai  aj = 0 ::::::} ai = ap a contradiction as each a/s are distinct. Hence D = (aap aa2) aan) has n distinct elements. a E D ::::::} a = aa,· ) i E (1) 2) ..... n) Now We show aio is the multiplicative identity of x) x E D. ••••••
••••••
••••••
••••••
••••••
r
r
J
J
•••••
c
0
For this) we show that xaio = x = a · x As x E D = (aap aa2) aan) ::::::} x = aaj ) 1 "5:.j "5:. n Consider xai. = (aa) ai = a(a.ai ) '0
••••••
v
J
0
J
0
= a( aio a)
D is commutative
= (a aio ) a
Associativity
J
= aa = x.
J
J
I a a· = a '0
DISCRETE STRUCTURES
340
aio x = ai (aa ) = ( ai al a I D is commutative = (a aio ) a = aa· = x Hence xaio = x = aio x i.e., aio is the multiplicative identity of x. Lastly, 1 E D = [aap aa2 ) aan] l = aa l X = 4m, y = 4n for some: m, n E Z. Consider x  y = 4n  4m = 4 (n  m) E H4 Also for r E Z) x E H4) we have rx = r(4m) = 4rm E H4 xr = (4m) r = 4mr E H4 Hence H4 is an ideal of Z. Further H4 = {... 8, 4, 0, 4, 8, ...} H4 + 1 = {... 7, 3, 1, 6, 9 ...} H4 + 2 = {... 6, 2, 2, 6, 10, ...} H4 + 3 = {... 6, 1, 3, 7, 11, ...} H4 + 4 = {... 4, 0, 4, 8, 12, ...} = H4 H4 + 6 = {... 3, 1, 6, 9, ...} = H4 + 1 H4 + 6 = H4 + 2 H4 + 7 = H4 + 3 H4 + 8 = H4 and so on Z I H4 = {H4 ' H4 + 1, H4 + 2, H4 + 3}. Thus
:
RINGS
347 TEST YOUR KNOWLEDGE 9.3
L 2. 3.
Define proper and improper ideals. Define quotient ring with example. (P.T. U. B.Tech. Dec. 2006) Let I be an ideal of a ring R and RlI :::: {x + I :::: E R} under the addition and multiplication defined by (x + J) + (y + J) x + + J (x + J) (y + J) "y + J \j x, E R. Then RlI is a ring, called the quotient ring. If J be an ideal of a ring R and E J be a unit in J. Then J :::: R. If J and K are ideals in a ring R, then J + K and J K are also ideals in R. X
y
=
=
4. 5.
y
U
II
Hint
2.
Consider H4 {4n ; n E ZJ We show H4 is an ideal of Z, the ring of integers. Let b E H4 4n, n E Z, b :::: 4m, m E Z  b 4n  4m 4(n  m) E H4 n, Also, If r E Z, E H4 ' then (4n) r 4(nr) E H4 r r(4n) 4(rn) E H4 Hence H4 is an ideal of Z H4 { ..... 8,  4, 0, 4, 8, .....J Further, H4 + 1 {.....  7,  3, 1, 5, 9, .....J H4 + 2 {....  6,  2, 2, 6, 10, .....J H4 + 3 {.....  5,  1, 3, 7, 11, .....J H4 + 4 {....  4, 0, 4, 8, .....J H4 Thus Z/H4 :::: {H4' H4 + 1, H4 + 2, H4 + 3} is a quotient ring. =
..
a,
a
=:::}
=
a ::::
=
I mE Z
a
ar = a
=
=
=
=:::}
n, r E Z r, n E Z
n  mE Z =:::} =:::}
nr E Z rn E Z
=
=
=
=
=
=
9.18. PRINCIPAL IDEAL
1. Let a E R, then the set = {ra : r E R} is an ideal, called the principal ideal generated by a. Let R be a commutative ring with an identity element
9.19. PRINCIPAL IDEAL DOMAIN (P.!.D.) A ring R is called a principal ideal domain if
(i) R is an integral domain (ii) Every ideal in R is principal. Example. Show that Z, the ring of integers, is a P.I.D. Sol. We know that Z, the set of integers is an integral domain. Let J is an ideal in Z. We show J is a principal ideal. Case I. If J = {O}, then it is principal ideal and hence the result.
348
DISCRETE STRUCTURES Case II. If J '" {O}. Let 0 ", x E J. then  x = ( 1) x E J for some positive x.
in J.
Hence J contains at least one positive integer. Let a be the smallest positive integer We claim J =
{ra : r E Z}
For x E J) using division algorithm) x = qa
+ r) 0 :::; r :::; a) q E Z Z => r E
But J is an ideal and a E J, q E
J. q a E J and x  q a E J But a is the smallest positive integer in J satisfying x = qa
Z is a P .LD.
J=
Le., {qa : q E Z}
9.20. EUCLIDEAN DOMAIN
0 :::; r :::; a. Hence we must have r = O.
(P.T. U. B.Tech. May 2007, May 2006)
Let R be an integral domain and suppose that for every negative number d(a) satisfying the following:
(i) For a, b E R, a ", 0, b '" 0, d(ab) :> d(a) (ii) For 0 '" a E R, b E R, there exists q, r E d(r) < d(a).
R such that
0 t: a E R) there exists a non
b = aq + r, where either r = 0
or
Then R is called Euclidean domain.
Remark. The condition (1) can be replaced by d(ab) ;> d(b). ILLUSTRATIVE EXAMPLES
Example 1. The ring of integers is a Euclidean domain. (non·negative integer) Sol. For 0 '" a E Z, define d(a) = I a I Let 0 '" a, 0 '" b E Z. consider d(ab) = I ab I = I a I I b I :> I a I = d(a) d(ab) :> d(a) => Also for 0 '" a, b E Z, we can find q, r E Z such that b = qb + r, where r = 0 or d(r) < d(a). Hence Z is a Euclidean domain. Example 2. Every field is a Euclidean domain. Sol. For 0 ", 0 E F, define d(a) = 1 I Since if 0 '" a E F and F is a field, we can find aI E F such that aaI = 1 = d(a) E F Let 0 '" a, 0 '" b E F => ab E F d(ab) = 1 :> d(a) => d(ab) :> d(a) Now Also for 0 t: b E F) we can write a = (abI) b + 0 = qb + r where q = ab\ r = 0 Hence every field is an integral domain.
RINGS
349
9.21 . ASSOCIATE
bE
Let R be a commutative ring with unity. An element a E R is said to be an associate of R if a = bu, where u is a unit in R. If a is an associate of b, we write a "'" b.
Example 3. (a) Find the associates of 4 in ZUY where Zl O denotes the integers modulo 1 D. (b) Find the associates of 5 in Zl O' (c) Find the associates of n in Z. Sol. (a) The units of ZlO are those numbers which are relative prime to 10. The units of Z l O are 1, 3, 7, 9. Multiplying each of the units by 4, we get 1.4 = 4, 3.4 = 12 = 2, 7.4 = 28 = 8, 9.4 = 36 = 6 Thus, 2, 4, 6, 8 are the associates of 4 in Z lO 0 (b) Similarly, the associates of 5 can be obtained by multiplying each of the units by 5, ..
we get
1.5 = 5, 3.5 = 15 = 5, 7.5 = 35 = 5, 9.5 = 45 = 5. Th us, the only associate of 5 in Z l O is 5 itself. (c) The units of Z are  1 and 1. Multiplying each of the units by n, we get  n and n, as the associates of n in Z. Example 4. Consider the ring Z1 2 = {D, 1, 2, 3, ... , I I} of integers modulo 12. (a) Find the units of Z1 2 (b) Find the roots of f(x) = x2 + 4x + 4 over Z1 2 (c) Find the associates of 2. Sol. (a) The units of Z,2 are those elements which are relative prime to 12. Hence, the units are 1, 5, 7, I I . (b) Put x = 0, 1, 2, ..... , 11 in f(x) = x2 + 4x + 4, we get when x = 0, f(O) = 4 x = 1, f(l) = 1 + 4 + 4 = 9 x = 2, f(2) = 4 + 8 + 4 = 16 = 4 x = 3, f(3) = 9 + 12 + 4 = 25 = 1 x = 4, f(4) = 16 + 16 + 4 = 36 = 0 etc. Also, when x = 10, f(10) = 100 + 40 + 4 = 144 = 0 The roots of f(x) are x = 4, 10 (c) The units of Z, 2 are 1, 5, 7, 1 1 The associates of 2 are obtained by multiplying each unit by 2 , we get 1.2 = 2, 5.2 = 10, 7.2 = 14 = 2, 1 1.2 = 22 = 10 The associates of 2 are {2, 7}. TheoremXII. Show that relation ofbeing associates is an equivalence relation in a ring R. Proof. Reflexive. For x E R, we can write x = x . 1 where 1 is a unit in R. Hence x "'" x. i.e., is reflexive. Symmetric. Let a  b => there exist some unit u E R such that b = au. ,...,
DISCRETE STRUCTURES
350 Now
au1 = (bu)u1 = b(uu1 ) = b b = au\ u1 is also a unit in R ba
Hence "'" is symmetric.
Transitive. If a  b => there exists some unit u E R such that a = bu If b  c => there exists some unit u , E R such that b = u , c. We show a  c. Now a = bu = (u,c)u = c(u,u) a "'" c. ::::::} As u 1 are units in R and units form a subgroup. U,
Hence "'" is transitive. Relation of being associates is an equivalence relation.
MULTI PLE CHOICE QUESTIONS (MCQs)
1.
Consider the ring Z l O = {O, 1, 2, ... 9} of integers modulo 10. The
(a) 3 (c) 6
2.
(b) 7
(d) none. Consider the ring Zs = {O, 1 , 2, 3, ... 7} of integers modulo 8 and polynomial in Zs' Then the number of roots of f(t) is
(a) 2 (c) 4
3.
2, 10 2, 4
Consider the ring ZlO
(a) 1, 3, 7, 9 (c) 3. 1
5. 6.
7. S.
(b)
f(t)
=
t2
+
6 t be a
1
(d) none.
Consider the ring Z, 2 = {O, 1 , 2, ... 1 1} of integers modulo 12. The associates of 2 are
(a) (c)
4.
31 is
(b)
2, 2
(d) none. = {O, 1 , 2, ... 9} of integers modulo 10. The units of Zl O are
(b) 3, 7,
9
(d) 1 . Consider the ring Z7 = {O, 1 , 2, ... 6} of integers modulo
7, which of the following is true.
(a) Z7 is an integer domain (c) Z7 has no zero divisors
(b) Z7 is a field
(a) (c)
(b)
(d) all of the above. If F is a field, then the number of proper ideals of F is 2
°
1
(d) none. If I and J are two ideals of a ring R. Which is false ?
(a) (c)
I n J is an ideal of R (b) I u J is an ideal of R I u J may or may not be an ideal of R (d) none. Which of the following is true ? The ring of integers (Z, .) .
(a) (c)
is an integral domain
(b)
is a division ring
(d) none.
is a field
+,
RINGS 9.
351 Which of the following statement is not true ?
(a)
Every field is an integral domain (b) Any finite non·zero integral domain is a field
(c) (d)
10.
Every ideal of a ring R is a subring of R
All are true. Which of the following is a false statement ?
(a)
The ring of integers is an integral domain (b) The ring of integers is a Euclidean domain
(c) The ring of integers is a P.LD.
(d) 1.
2.
3.
None.
Answers and Explanations I (b) By a in a ring R, we mean that element such that a . aI = 1 = a1 a Here 3.7 = 1 = 7.3 . . 31 = 7 2 f (t) = t + 6t (c) f (0) = 0, [ (2) = 4 + 12 = 16 = 0 f (4) = 1 6 + 24 = 40 = 0 f (6) = 36 + 36 = 72 = 0 Hence the roots of f(t) are 0, 2, 4, 6 (d) The units of Z,2 are those numbers which are relative prime to 12. The units of Z,2 are
4.
5. 6. 7.
1, 5, 7, 1 1 Multiplying each of the units by 2 , we get 1 . 2 = 2, 5 . 2 = 10 = 2, 7 . 2 = 14 = 2 1 1 . 2 = 22 = 10 = 2 . . The associates of 2 are 2, 7 (a) The units of Zl O are those elements which are relative prime to 10. Therefore, the units of Zl O are 1, 3, 7, 9 (d) Zp is a field iff p is a prime number. Also every field is an integral domain. (d) A field has no proper ideals. (b) 8. (a) 9. (d) 10. (d).
10
BOOLEAN ALGEBRA
1 0. 1 . PARTIALLY ORDERED RELATION Consider a relation R on a set S satisfying the following properties : 1. R is reflexive i.e., xRx for every X E S. 2. R is antisymmetric i.e., if xRy and yRx, then x = y. 3. R is transitive i.e., if xRy and yRz, then xRz. Then R is called a partially order relation and the set S together with partial order relation R is called apartialiy order set or simply, an ordered set or PO SET and is denoted by (S, s). For example :
1. The set N of natural numbers form a poset under the relation ':s;' because firstly x :::; x, secondly, if x S y and y S x, then we have x = y and lastly if x S y and y S z, it implies x S z for all
x, y, z E N. 2. The set N of natural numbers under divisibility i.e., 'x divides y' forms a poset be cause x/x for every x E N. Also if x/y and y/x, we have x = y. Again if x/y, y/z we have x/z, for every X,y,Z E N. 3. Consider a set S = {I, 2} and power set of S is P(S). The relation of set inclusion C is a partial order relation. Since, for any sets A, B, C in P(S), firstly we have A C A, secondly, if A c B and B e A, then we have A = B. Lastly, if A c B and B e C, then A c C. Hence, (P(S), C) is a poset.
ILLUSTRATIVE EXAMPLES
Example 1. Consider the set Z of integers. Define aRb by b = a', for some positive integer r. Show that R is a partial order relation on Z aRa i.e., R is reflexive Sol. (i) Since a = a1 (ii) Let aRb and bRa. Then there exist positive integers r and s such that b = ar and a = b S Now a = bS = (ar) S = ars • •
Consider the following possibilities
(a) If rs = 1 , then r = 1 , s = 1 and hence a = b. Therefore, R is antisymmetric (b) If a = 1 , then b = l' = 1 = a. Therefore, R is antisymmetric
352
BOOLEAN ALGEBRA
353
(c) If b = 1, then a = 1' = 1
. . R is antisymmetric. (d) If a = 1 , then b = ( R is antisymmetric. 

Hence) in all the cases)
=a
1)'
=

1 (since
b # 1) and hence a = b.
Therefore,
R is antisymmetric.
(iii) Let aRb, bRc, then there exist positive integers r and s such that b = ar) c = bS c = b S = (ary = ar\ for some positive integer rs. Hence aRc R is transitive also. Therefore) R is a partial order relation. 1 0.2. COMPARABLE ELEMENTS
a :::; b
Consider an ordered set A. Two elements a and or b :::; a where :::; is a partial order relation.
b
of set A are called comparable if
1 0.3. NONCOMPARABLE ELEMENTS Consider an ordered set A. Two elements a and neither a :::; b nor b :::; a.
b of set A are called noncomparable if
1 0.4. LINEARLY ORDERED SET OR TOTALLY ORDERED SET Consider an ordered set A. The set A is called linearly ordered set or totally
set, if every pair of elements in A are comparable.
ordered
For example : The set of positive integers 1+ with the usual order S is a linearly ordered set.
Example 2. Let N = {I, 2, 3, 4, .. .} is ordered by divisibility. State whether each of the following subsets of N are linearly (or totally ordered). (a) {24, 2, 6} (b) {3, 15, 5} (c) {I, 2, 3, .. .} (d) {5} (e) {2, 8, 32, 16}. Sol. (a) Consider the set S = {24, 2, 6} Since 2/6, 6/24
S is linearly ordered.
(b) 3 does not divide 5. Therefore, the set {3, (c) Let S = {I, 2, 3, ...}
15, 5} is not linearly ordered.
As 2 does not divide 3 and 4 does not divide
7
This set is not linearly ordered.
(d) Any set consisting of one element is linearly ordered.
(e) As 2 divides 8, ..
8 divides 16, 1 6 divides 32,
This set is linearly ordered set.
Example 3. Consider N = {I, 2, 3, 4 .. .} with the relation '7ess than or equal to". Find all linearly ordered subsets of N. Sol. We know that the set N is a linearly ordered set under 's'. Therefore, every subset
of N is linearly ordered.
DISCRETE STRUCTURES
354
Example 4. Let A = {2, 3, 6, S, 9, is} is ordered by divisibility. (a) Find the noncomparable pairs of elements ofA. (b) Find the linearly ordered subsets ofA with three or more elements. Sol. (a) The noncomparable pairs are {2, 3}, {2, 9}, {3, S}, {6, S}, {6, 9}, {S, 9}, {S, IS}
(b) We know that a set is called linearly ordered if every pair of its elements are compa
rable. The linearly ordered set of A with three or more elements are {2, 6, IS}, {3, 9, IS}. Note that the set {2, 6, S} is not linearly ordered since the elements 6 and S are not comparable.
Example 5. Consider the set Z of integers under the partial order relation defined by aRb (iv) Let a' + b = 1 and we show a.b' = 0 I De Morgan's law Consider 0 = I' = (a' + by = a".b' I Involution law = a.b' a.b' = 0 If a.b' = 0, we show a' + b = 1 De Morgan's law Consider 1 = 0' = (a.b')' = a' + bl! I Involution law = a' + b Hence (iii) and (iv) are equivalent Accordingly, (i), (ii), (iii) and (iv) are equivalent. Example 5. Using Boolean Postulates and theorem, simplify the following expression : a + ab + abc + abed + a + ab + abc + abed . (P.T.V. B.Tech. May 2009) Sol. Given expression is
f(a, b, e, d) = a + ab + abc + abed + a + ab + abc + abed = a(l + b) + abe(l + d) + a(l + b) + a be(l + d) = a . 1 + abc . 1 + a . 1 + a be . 1 = a + abc + a: + a; be = a + a . (be) + a + a . (be) = a + a:
For any Boolean algebra
1 +x=x VXE
B I Idendity law
Absorption law : B
a + a . b = a Va, b E
= 1.
I Complement law
1 0. 1 7. PRINCIPLE OF DUALITY The dual of any expression E is obtained by interchanging the operations + and * and also interchanging the corresponding identity elements 0 and 1, in original expression E.
6. Write the dual of the following Boolean expressions : (ii) (1 + x,) * (Xl + 1) (i) (Xl * x,) + (Xl * x3 ) (iii) (a 1\ (b 1\ c)). Sol. (i) (x, + x) * (x, + xa ) (ii) (0 * x) + (x, * 0) (iii) (a v (b 1\ c» . Example
Note.
The dual of any theorem in a Boolean algebra is also a theorem.
BOOLEAN ALGEBRA
385 TEST YOUR KNOWLEDGE 1 0.4
1.
2. 3.
4.
Show that {M; v. A. } is a Boolean algebra where M = P(A). the power set of A and A = {a, b, c}. Express each Boolean expression E (x, y, z) as a sum of products and then its complete sum of products form (a) E = x(>y' + x'y + y'z) (b) E = z(x' + y) + y' Express E (x, y, z) :::: (x' + y)' + x'y in its complete sum of products form. State the dual of (a) a v (b A a) = a
( ) Write the dual of each Boolean expression (c) a A b A b = a v b
5.
(a) a(a' + b) = ab
6.
7.
8.
9.
10.
(b) (a + 1) (a + 0) = a
(c) (a + b) (b + c) = ac + b
Let [B; + , ., '] be a Boolean algebra and let f(x, y) be a Boolean function of the variables x and y. By using the following table, find the Boolean expression. x
y
f(x, y)
1 1 0 0
1 0 1 0
0 1 1 0
Let [B; +, ., 1 be a Boolean algebra and let f(x, y, z) be a Boolean function of the variables x, y and By using the following table, find the Boolean expression f(x, y, z).
z.
z
x
y
1
1
1
1
1
1
0
0
1
0
1
0
f(x, y, z)
0
1
1
1
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
Suppose B is a Boolean algebra with at least 100 elements. How many elements can B have ? Using Boolean algebra, show that abc + abc + ab c + ab c = ab + be + ca.
(P. T. U. B. Tech. May 2008)
Consider the lattices Dm of divisors of m (where m > 1). (a) Show that Dm is a Boolean algebra if and only if m is squarefree, that is, if m is a product of distinct primes. (P.T. U. B. Tech. Dec. 2013) (b) If Dm is a Boolean algebra, show that the atoms are the distinct prime divisors of m.
386 11. 12. 13.
14. 15. 16. 17.
DISCRETE STRUCTURES
Consider the following lattices: D20 ; D55 ; (c) Dgg ; (d) D13o. Which of them are Boolean algebras, and what are their atoms ? Consider the Boolean algebra Dno. List its elements and draw its diagram. Find all its subalgebras. (c) Find the number of sublattices with four elements. (d) Find the set A of atoms of Dno. (e) Give the isomorphic mapping f : Duo P(A) as defined in representation theorem. Let B be a Boolean algebra. Show that : For any x in B, 0 x :O:;: l. < if and only if < An element x in a Boolean algebra is called a maxterm if the identify 1 is its only successor. Find the maxterms in the Boolean algebra D20. Let B be a Boolean algebra. Show that complements of the atoms of B are the maxterms. Show that any element x in B can be expressed uniquely as a product of maxterms. Let B be a 16element Boolean algebra and let S be an 8element subalgebra ofB. Show that two of the atoms of S must be atoms of B. Let B :::: (B, +, *, " 0, 1) be a Boolean algebra. Define an operation on B (called the symmetric difference) by
(a) (b) (a)
:0:;:
((ab)) a b (b)
t
b' a'. (a)
8
x '" y
18.
(b)
(x * YJ + (x' * y).
= Prove that R :::: (B, D,
is a commutative Boolean ring. Let R :::: (R, EB, . ) be a Boolean ring with identity 1 0. Define *)
:f
x' :::: l EB x, x + y :::: x EB y EB x . y, x * y :::: x . y
Prove that B :::: (R, + , *, " 0, 1) is a Boolean algebra.
Answers 2. 3. 4.
(i) xy' + xy'z
(ii) xy'z + xy'z
(i) x'z + yz + y'
(ii) xyz + xy'z + xy'z' + x'yz + x'y'z + x'y'z
((ab)) b ) ( (a) a( A (b )v a) = a (b) a A A a vb = 0 av vb=aAb (a) aba a'acb =a(a b )b (b) a.O a.=1 = a 4, 6, 8, 16, 32 or 64. = atoms and 13. ((ab)) Thereatomsare eightandelements 11. 1. 10, 11, 110. See Fig. (a). (b) There are five subalgebras : {1. 110}. {1, 110}, {1, 110}, {1. 10. 11. 110}, There are 15 sublattices which include the above three subalgebras. A = {2, 11}. See Fig. (b).
xy'z + xy'z' + x'yz + x'yz' (c)
5. 7. 11. 12.
(c)
+
= +
f(x, y. z)
+
6. f(x, y) 8. 2,
+c
+
(x' + z) . y.
5
D55 ;
(d) D I 3.;
2. 5.
(c)
(d) (e)
5,
2, 5 22. 55, 2, 55.
x . y' + x' . y
5. 22.
D uo .
BOOLEAN ALGEBRA
387
110 10 212 �55 5 >< _______ 1I ____
1 2 5 11 10 22 55 110 l� {2}l {5}l {11l } {2,l5} {2,l11} {5,l11} Al (b)
(a) D 10
F : D, 1O
>
PtA)
Hints
L
We know that a lattice which contains a least element, a greatest element and which is both complemented and distributive is called Boolean Algebra. Here the least element is �, the greatest element is A. The complements of each element of B is given below : B
�
{aj {bj {cj {a, bj {b, cj {c, aj
2.
Complement of B
A
{b, cj {c, aj {a, bj {cj {aj {bj
A � Also {M, v, A, } is a distributive lattice. (a) Given E = x(xy' + x'y + y'z) Express E as a sum of products forms, we get E :::: xxy' + xx'y + xy'z :::: xy' + xy'z is the required sum of products forms. Further E = xy' (z + z') + :q'z I xx' :::: 0 :::: xy'z + xy'z + xy'z :::: xy'z + xy'z,
is the required complete sum of products form. (b) Given E = z(x' + y) + y' :::: x'z + zy + y', is the required sum of products form Further, E = x'z (y + y') + (x + x')zy + (x + x') y' (z + z')
l a + a :::: a
:::: x'zy + x'zy' + xzy + x'zy + y' (xz + xz' + x'z + x'z) = � + � + = + � + ¢ + � + �� + �� :::: x'yz + x'y'z + xyz + xy'z + xy'z' + x'y'z.
3.
E :::: (x' + y)' + x'y :::: xy' + x'y, is the sum of products form I De Morgan's law Further, E = :q'(z + z') + x'y (z + z') :::: xy'z + xy'z' + x'yz + x'yz', is the required complete sumofproducts form of E.
388
DISCRETE STRUCTURES
1 0.18. BOOLEAN EXPRESSION OR BOOLEAN FUNCTION Let {B) +) .) '} is a Boolean algebra. An expression involving the variables xp x2 ) ••• xn and the binary operations '+' , '. ' and '" is called Boolean expression. The Boolean expression or Boolean function is denoted by f(xl , x2' ... , x) .
For example,
f(x. Y. z) = (x + y'z)' + (xyz' + x'y)'
f(x, y, z) = (:ry'z' + y)' + (x'z)', are Boolean expression.
1 0. 1 9. LITERAL SlOn.
A literal is a variable or complemented variable like x, x', y, y' etc in the Boolean expres
1 0.20. FUNDAMENTAL PRODUCT By Fundamental Product, we mean a literal or a product of two or more literals in which no two literals involve the same variable. For example, xy'z, x, y', yz', x'yz are fundamental products. But xyx'z, xyzy are not fundamental prod ucts as the first expression contains x and x' whereas the second expression contains y at two places.
(P.T. U. B.Tech. Dec.
1 0.21 . SUMOFPRODUCTS FORM OR SOP FORM
2013)
Let [(Xl ) X2 ) ... ) X) be a Boolean expression. Then) [is said to be in sumoiproducts form or minterm form if [ is a fundamental product or sum of two or more fundamental products) none of which is contained in another. If P, and P2 are fundamental products. Then, P , is said to be contained in P2 if the literals of P I are also literals of P2 • For example) x'z is contained in x'yz since x' and z are literals in x'yz. However) x'z is not contained in xy'z) since x' is not a literal in xy'z.
1 0.22. COMPLETE SUMOFPRODUCTS FORM A nonzero Boolean expression [(xp x2 ) ... ) x) is said to be in Completesumoiproducts form if f is in sum·of·products form and each product involves all the variables. Remarks L Completesumof Products
Every nonzero Boolean expression f(xl, x2' x,) can be put into form and this form is unique. 2. A Boolean expressionf(xl, x2 ' , x,), in n variables can have a maximum of 2n such products. "
'J
•••
ILLUSTRATIVE EXAMPLES
I
Example 1. Reduce the following Boolean products to either 0 or a fundamental product (i) xyx' z and xyzy (ii) xyz'Xyx and xyz'Xy "z< Sol. (i) I Commutative law xyx'z = xx'yz Also,
=0 xyzy = xzyy = xzy
xx' = 0 (Complement laws)
I Commutative law I Idempotent law
BOOLEAN ALGEBRA xyz'xyx = xxxyyzl = xyzl xyz'xy'z' = xxyy'zlzl = xyy'z' = o. I yy' = 0 Example 2. Let {E, +, ., ] is a Boolean algebra. Show that (ii) (a + b).acb'= 0 (i) a.b + a.b' + a'.b + acb'= 1 (iii) (a + b)' + (a + b ) ' = a' (iv) a.b + ((a + b').b)' = 1. Sol. (i) a.b + a.b' + a'.b + a' . b' = a.(b + bi + a'.(b + bi = a.I + a'.1 = a + a' =1 (ii) (a + b).a'.b' = (a + b).(a + by =0 (iii) (a + by + (a + bi' = a'.b' + a'.bl! = a'.(b' + b'i = a'.(b' + b) = a'.I. = a'. (ii)
389 Commutative law
I
Idempotent law Commutative law
I
(Complement law)
I
Distributive law Complement law I Identity law Complement law I De Morgan's law Complement law De Morgan's law
I
(iv) Please Try Yourself.
Example
Idempotent law
Distributive law
I Involution law Complement law I
Identity law
3. Prove that a Boolean algebra cannot have three elements.
Sol. Let, if possible, {B, +, . , '} has three elements, say,
0, 1 and a (other than 0 and 1) a' t: 0 and a' t: 1 a' = 0 ::::::} a" = 0' = 1 ::::::} a = 1) which is a contradiction. Hence a' ::f: 0 I Involution law Similarly) if a' = 1 ::::::} a" = I' ::::::} a = 0) which is also a contradiction. Hence a' t: 1. Therefore, we must have a' = a Now Complement law a.a' = 0 a.a = 0 I a' = a Idempotent law a=O a contradiction as a t: O. We claim
Hence, there cannot be a Boolean algebra containing three elements.
Example
4. Consider the Boolean expressions
f1 = xz' + y z + xyz' f2 = xz' + xyz' + xyz Reduce these expressions into sumoiproducts form Sol. f1 = XZ' + y'z + xyz' is not in a sumofproducts form since xz' is contained in xyz' f1 = xz' + y'z + xyz'
390
DISCRETE STRUCTURES
= xz' + xyzl + y'z = xz' + y'z I Absorption law which is a sumoiproducts form Also f2 = xz' + x'yzl + xy'z is already in sumofproducts form. Example 5. Consider a Boolean expression fla, b, c) = ((ab)' c)' ((a' + c) (b' + c'))' into a sumoiproducts form (P.T.V. B.Tech. Dec. 20 1 3)
f,(a, b, c) = «ab)' c)' «a' + c) (b' + c')' = «ab)" + c') «a' + c)' + (b' + c')') De Morgan's law I Involution law = (ab + c') «a' + c)' + (b' + c')') De Morgan's law = (ab + c') (aile' + bile") I Involution law = (ab + c') (ac' + be) = abac' + abbe + c'ac' + c'bc = aabc' + abbe + ac'c' + be'c Commutative law = abc' + abc + ac' + 0 By idempotent law aa = a = abc' + abc+ ac' By complement law, e'c = 0 = abc' + ac' + abc = ac' + abc I Absorption law Example 6. Consider a Boolean expression f (x, y, z) = x(y'zY Reduce it to complete sum oiproducts form. Sol. f(x, y, z) = x(y'z)' = x(y" + z') De Morgan's law = x(y + z') I Involution law = xy + xz', which is in sumofproducts form By complement law, a + a' = 1 = xy(z + z') + x(y + y')z' = xyz + xyzl + xyzl + xy'z' I a + a = a (Idempotent law) = xyz + xyz' + xy'z'
Sol.
which is in completesumofproducts form.
Example 7. Consider the Boolean expression (i) flx, y, z) = z(x' + y) + y' (ii) fix, y, z) = (x' + y)' + xy (iv) fix, y, z) = (x + y)' (xy Y (iii) fix, y, z) = x(xy' + xy + y'z) Reduce to sumoiproducts form and hence to complete sumoiproduct forms. Sol. (i) f,(x, y, z) = z(x' + y) + y' I Commutative law = (x' + y) z + y' = x'z + yz + y', which is sumofproducts form = x'z(y + y)' + (x + x') yz + (x + x')y'(z+ z') I By complement law, a + a' = 1 = x'zy + x'zy' + xyz + x'yz + (xy' + x'y') (z + z') = x'yz + x'y'z + xyz + x'yz + xy'z + zy'z' + x'y'z + x'y'z' I Commutative law
= x'yz + x'y'z + xyz + xy'z + xy'z' + x'y'z', which is in complete sumof products form.
BOOLEAN ALGEBRA (ii)
f2(x, y, z) = (x' + y)' + x'y = x"y' + x'y = xy' + x'y,
391 De Morgan's law I Involution law
which is a sumofproducts form.
= xy' (z + z') + x'y(z + z') By complement law, a + a' = 1 = xy'z + xy'z' + x'yz + x'yzl which is in completesumofproducts form. (iii) f,(x, y, z) = x (xy' + x'y + y'z) = xxy' + xxy + xy'z I Complement law and Idempotent law = xy' + O + xy'z = xy' + xy'z, which is in sumof products form. = xy'(z + z') + xy'z = xy'z + xy'z' + xy'z = xy'z + xy'z', a + a = a (Idempotent law) which is in complete sumofproducts form.
(iv) Please try yourself Ans. x'y'z + x'y'z'. Example 8. Let E = xy' + xyz' + x'yz', prove the following : (i) xz' + E = E (ii) x + E '" E (iii) z' + E ", E. Sol. We know that A + E = E, A '" 0, iff the summands in the Complete sumaiproducts form for A are among the summands in the complete sumaiproducts form for E. E = xy' + xyzl + x'Y2, reduce it to Completesumofproducts form, we have Now E = xy'(z + z') + xyz' + x'yz' = xy'z + xy'z' + xyz + X'yzl (i) Express xz' in completesumofproducts form. xz' = x(y + y') z' = xyzl + xy'z' Since the summands of xz' are among those of E, therefore xz + E = E (ii) Express x in the completesumoiproducts form, we have x = x(y + y')(z + z') = x(yz + yz' + y'z + y'z') I y + y' = 1 = z + z' (Complement law) = xyz + xyzl + xy'z + xy'z', which is in completesumoiproducts form Now the summand xyz of x is not a summand of E. Therefore) x + E t: E (iii) Express z' in Complete sumoiproducts form) we have I x + x' = 1 = y + y' (Complement law) z' = (x + x') (y + y') z' = (xy + xy' + x'y + xY)z' = xyz' + xy'z' + :;:'yz' + :;:'y'z') which is in Complete, sumaiproducts form. But the summand :;:'y'z' of z' is not a summand of E. Therefore) z' + E t: E. 1 0.23. MINTERM Let {B) +) . /} is a Boolean algebra. Let xp x2) ... ) xn are n variables. A product of the form . Y2 .... ) Yn) where Yi = Xi or x/) i = 1) 2) ... ) n) is called a minterm in n variables Xl) x2) ... ) xn' n Total number of minterms in a Boolean function of n variables = 2 . For example)
Yl
392
DISCRETE STRUCTURES (i) Minterms in variables x) y are x.y) x.y') x'.y) r.y' (4 terms) (ii) Minterms in variables x) y) z are x.y.z, x.y.z', x.y'.z, x.y.z, x.y'.z', x.y'.z', x.y'.z, x.y'.z'
(8 terms) The following theorem (without proof) gives a method of expressing a Boolean function in terms of the minterms in the corresponding variables.
1 0.24. BOOLEAN EXPANSION THEOREM If {B, + , ./, 0, I} is a Boolean algebra and f(xp x2) is a Boolean function in 2 variables, xp x2• then [(x" x2) = [(1. 1).x,.x2 + [(0, 1).x/.x2 + [(1, 0). x,.x,' + [(0 , O) .x/.x,' Similarly, for 3 variables xp x2) x3) we have [(xl ' x2 ' x,) = [(1, 1, 1).x,.x2.x3 + [(0, 1, 1).x;.x2.x3 + [(1, 0, 1).x,.x,'.x3 + t(l, 1, O)xl x2·x/ + teo, 0, 1) .x/.xz'.x3 + teo, 1, O).x/.x2·x/ + t(l, 0, O).x1·xz'.x/ + teo, 0, O).x/.xz'.x/ Example 9. Let {E, +, ., � 0, I} be a Boolean algebra and let [(x, y) be a Boolean [unction o[ the variables x and y. Find the Boolean expression [rom the truth table given below : '
x
y
f(x, y)
1 1 0 0
1 0 1 0
0 1 0 1
Sol. From the truth Table, [(1, 1) = 0, [(1, 0) = 1, [(0, 1) = 0, [(0, 0) = 1 By Boolean expansion theorem,
[(x, y) = [(1, l).x.y + [(1, O).x.y' + [(0, l).x'.y + [(0, O) .x'.y' = 0 + x.y' + 0 + r.y' = x.y' + r.y'. I Commutative law = y'.x + y'.x' = y'.(x + x') I Distributive law = y'.l = y' I Complement law Example 10. Let {B, +, ., � 0, I} be a Boolean algebra and [(x, y, z) be a Boolean [unction o[ the variables x, y, z. Find the Boolean expression [rom the truth table given below : x
y
z
f(x, y, z)
1 1 1 0 1 0 0 0
1 1 0 1 0 1 0 0
1 0 1 1 0 0 1 0
1 0 1 0 0 0 1 1
BOOLEAN ALGEBRA
393
Sol. From the truth table, [(1 , 1 , 1) = 1 , [(1, 1 , 0) = 0, [(1 , 0, 1) = 1 , [(0, 1 , 1) = 0 [(1 , 0, 0) = 0, [(0, 1 , 0) = 0, [(0, 0, 1) = 1 , [(0, 0, 0) = 1 . By Hoole's expansion theorem, we have [(x, y, z) = [(1 , 1, l).x.y.z + [(1 , 1 , O).x.y.z' + [(1 , 0, l).x.y'.z + [(0, 1 , l).xy.z + [(1 , 0, O).x.y'.z' + [(0, 1 , O).x.y.z' + [(0, 0, l).x'.y'.z + [(0, 0, O).xy'z' = l .x.y.z + O.x.y.z' + l .x.y'.z + O.x'.y.z + O.x.y'.z' + O.x'.y.z' + 1.x'.y'.z + 1 .x'.y'.z' + = x.y.z 0 + x.y'.z + 0 + 0 + 0 + x.y'.z + x.y'.z' f(x, y, z) = x.y.z + x.y'.z + x'.y'.z + x'.y'.z' = (x.z.y + x.z.y') + (x.y'.z + x.y'.z') = x.z.(y + y') + x.y'.(z + z') I Complement law = x.z.l + x.y'.l = X.z + x.y'. 1 0.25. DISJUNCTIVE NORMAL FORM OR SUMOFPRODUCTS (OR SOP) FORM A Boolean expression over ({O) I}) a join of minterms e.g.,
v) A,
') is said to be in disjunctive normal form if it is
(x/ /\ xz ' /\ x3') v (x/ /\ x2 /\ x3 ') V (Xl is a Boolean expression in disjunctive normal form.
/\
X
2
/\
x)
Since there are three minterms x/ /\ xz' /\ x/ x/ /\ x2 /\ X3 and Xl /\ X2 /\ X3• ' Maxterm. A Boolean expression of n variables xp x2' ..... , xn is said to be a maxterm if it is of the form Xl v x2 v ...... V Xn , where xi is used to denote Xi or x/.
1 0.26. CONJUNCTIVE NORMAL FORM OR PRODUCTS OF SUMS (OR POS) FORM
A Boolean expression over ({O) I}) v) /\) ') is said to be in conjunctive normal form if it is a meet of maxterms e.g., (Xl v x2 V X) /\ (Xl V X2 V X) /\ (Xl V X2 V X) /\ (x/ V X2 V X3') /\ (x/ /\ xz' /\ X) is a Boolean expression in conjunctive normal form consisting of five maxterms.
1 0.27. (a) OBTAINING A DISJU NCTIVE NORMAL FORM Consider a function from {O, l}n to {O, I}. A Boolean expression can be obtained in disjunctive normal form corresponding to this function by having a minterm corresponding to each ordered ntuple of O 's and l 's for which the value of function is 1 .
1 0.27. (b) OBTAINING A CONJUNCTIVE NORMAL FORM Consider a function from {O, l}n to {O, I}. A Boolean expression can be obtained in con junctive normal form corresponding to this function by having a maxterm corresponding to each ordered ntuple of O 's and l 's at which the value of function is O.
394
DISCRETE STRUCTURES Example 11. Express the following function in (i) Disjunctive normal form (ii) Conjunctive normal form f
(0, (0, (0, (0,
0, 0, 1, 1,
0) 1) 0) 1)
f
(1, 0, 0) (1, 0, 1)
1 ° 1 °
° 1 ° 1
(1, 1, 0) (1, 1, 1)
Sol. (i) (x/ /\ xz' /\ xa') v (x/ /\ x /\ xa') V ( /\ xz' /\ x) V ( /\ /\ x) (ii) (x/ v xz' v xa') /\ (x/ v V x) /\ ( V xz' v x/) /\ ( V V x/). 2
X2
Xl
Xl
Xl
Xl
X2
X2
Example 12. If f(x, y, z) = (x v y) A (x V y') A (x' V z) be a given Boolean function. Determine its DN form. Sol. First determine all values of f(x, y, z) when x, y, z take values 0, 1 and then from this table, we will find the DN form x
z
y
0 0 0 0
0 0
0 1 0 1 0 1 0
1 1
1
0 0
1
1
1 1
f(x, y, z)
1
0 0 0 0 0 1 0
1
(x v y) A (x V y') A (x' V z) = = = = = = = =
1
(0 v 0) A (0 v 1) A (1 v 0) (0 v 0) A (0 v 1) A (1 v 1) (0 v 1) A (0 V 0) A (1 v 0) (0 v 1) A (0 V 0) A (1 v 1) (1 v 0) A (1 v 1) A (0 V 0) (1 v 0) A (1 v 1) A (0 v 1) (1 v 1) A (1 v 0) A (0 V 0) (1 v 1) A (1 v 0) A (0 v 1)
The disjunctive normal form of the function is
f(x, y, z) = (x A y' A z) V (x A Y A z).
swn :
Example 13. Determine the disjunctive normal form of the following Boolean expresX
A
(y V z).
Sol. First determine all values of f(x, y, z) when x, y, this table we will write disjunctive form. Thus, x
0 0 0 0
y
0 0 1
1
1
0 0
1
1
1
1
1
z
0 1
0 1
0 1
0 1
f(x, y, z)
0 0 0 0 0
z take values 0, 1 and then from X = = = = =
1
=
1
=
1
=
A (y V z)
(0 0) (0 0)
oA V o A v 1) o A (1 v o A (1 v 1) 1A V
(0 0) (0 v 1) 1 A (1 v 0)
1A
1 A (l v 1)
BOOLEAN ALGEBRA
395
Thus, the DNF of the function is
f(x, y, z) = (x /\ Y /\ z) V (x /\ y' /\ z) V (x /\ Y /\ Z').
1 0.28. PRIME IMPLICANTS
We know that a fundamental product is a literal or a product of two or more literals in which no two literals involve the same variable. A fundamental product, say, P, is called a prime implicant of a Boolean expression E if P + E = E, but no other fundamental product contained in P has this property. We now discuss a method known as Karnaugh Map method for finding prime implicants and minimal forms for Boolean expression involving atmost six variables. But in this chapter, we will only discuss the cases of two, three and four variables.
1 0.29. KARNAUGH MAP It is a graphical tool used to simplify a logic equation or to convert a truth table to its corresponding logic circuit. A Karnaugh Map, hence forth, will be abbreviated as Kmap.
1 0.30. ADJACENT FUNDAMENTAL PRODUCTS Let P, and P2 are two fundamental products. Then P, and P2 are said to be adjacent if P, and P2 have the same variables and if they differ in exactly one literal, it must be a comple· mented variable in one product and uncomplemented in another.
I
ILLUSTRATIVE EXAMPLES
I
Example 1. Is P = xz� a prime implicant of E = xy' + xyzl + x'yz'? Sol. We have proved in example 8 above, xz + E = E, but x + E '" E, z' + E '" E .. p = xz' is the prime implicant of E. Example 2. Find the sum of adjacents products PI and P2 where (i) PI = xyz� P2 = xy 'z' (ii) PI = x'yzt, P2 = xyz't (iii) PI = x'yzt, P2 = xyz't (iv) PI = xyz� P2 = xyzt. I Commutative law Sol. (i) PI + P2 = xyzl + XY'Z' = xz'y + xz'y' = xz' (y + y') = xz'.l I Complement law = xz'
less literal :
Note.
(ii)
That sum of two adjacent products (squares) P and P2 is a fundamental product with one 1
P,
+ P2 = xyzt + xyzt = xytz + xytz' = xyt (z + z') = xyt.1 = xyt
I Complement law
(iii) P I and P2 are not adjacent since they have two variables x and z. Here x and Z are complemented in PI and uncomplimented in P2.
396
DISCRETE STRUCTURES
(iv) P, and P2 are not adjacent since they have different variables. Thus, in particular, they will not appear as squares in the same Karnaugh Map. 1 0.31 . KARNAUGH MAP FOR TWO VARIABLES The Karnaugh map corresponding to Boolean expression f(x, y) of two variables is shown in Table given below :
y'
y x
xy
x'
x'y
xy'
Also) any complete sumofproducts Boolean expression f(x) y) of two variables) can be represented in Karnaugh map by placing l 's in the corresponding squares. A prime implicant of f(x, y) will be either a pair of adjacent squares or an isolated square (a square which is not adjacent to any square of f(x, y» . Also, the four possible fundamental products with two literals are :
xy, xy', x'y, x'y'. 1 0.32. KARNAUGH MAP FOR THREE VARIABLES The Karnaugh map corresponding to Boolean expression f(x, shown in the Table given below :
yz'
yz
y'z'
y, z) of three variables is
y'z
x x' Also, the eight possible fundamental products with three literals are :
xyz, xyz', xy'z', xy'z, x'yz, x'yz, :;:'y'z', :;:'y'z. 1 0.33. KARNAUGH MAP FOR FOUR VARIABLES The Karnaugh map corresponding to Boolean expression f(x, y, shown in the Table given below :
zt
zt'
z't'
xy' xy Also, the sixteen possible fundamental with four literals are :
xyzt, xyzt', xyz't', xyz't ; xy'zt, xy'zt', xy'z't', xy'z't ; :;:'y'zt, x'y'zt', x'y'z't', x'y'z't ; x'yzt, x'yzt', x'yz't', x'yz't.
z, t) of four variables is
z't
BOOLEAN ALGEBRA
397
1 0.34. LOOPING The expression for the output Boolean expression can be simplified by properly combin· ing those squares in the K·map which contain l 's. This process for combining these l 's called
looping.
1 0.35. LOOPING GROUPS OF TWO Consider a Kmap for there variables as shown below (Fig. 10.24). This map contains a pair of 1s that are vertically adjacent to each other. The first represents xyz and the second represents xyz . We note that in these two terms) only x appears both in complemented and uncomplimented form while y and z remain unchanged. These two terms can be looped (combined) to give a resultant that eliminates the variable x as it appears both in complemented and uncomplimented form. Thus) the corresponding Boolean expres SlOn IS
xy 3'y xy xy
The first represents ABC and the second represents ABC . These two terms can be looped (combined) to give a resultant that eliminates the variable C as it appears both in complemented and uncomplimented form. Thus) the corresponding Boolean expres SlOn IS f(A, B)
= ABC + ABC = AB (C + C ) = AB . 1 = AB
Consider another example of a Kmap for three variables as shown below (Fig. 10.26). This map contains Is in the top row and bottom row respec tively_ These two squares are also considered to be adjacent. These two 1s can be looped (combined) to give a resultant that eliminates the variables x as it appears both in complemented and uncomplimented form. Thus) the corresponding Boolean expression is
f(x, y, z) = x y z + xyz = (x + xl yz = 1. y z = yz Hence, from above discussion, we conclude that "Looping a pair of adjacent 1s in a Kmap eliminates the vari able that appears in complemented and uncomplimented form". 1 0.36. LOOPING GROUPS OF FOUR
0
0
r
1
0 0
Fig. 10.24
f(x, y) = xyz + xyz = (x + xl yz = 1. yz = yz
Consider another example of a Kmap of three variables as shown below (Fig. 10.25) : This map contains a pair of 1s that are horizontally adjacent.
1 1 1 ''
z
Complement law Xii
AB AB AB
c
c
0
0
0
0
0
0
(1
F'19. 10.25
xy
1)
z o
3'y
o
o
xy
o
o
xy
o
Fig. 10.26
A Kmap may contain a group of four l 's that are adjacent to each other. This group is called a quad. Consider the following Fig. 10.27.
398
DISCRETE STRUCTURES z
zt
it
zt
zl
xy
0
0
0
0
1
xy
0
0
0
0
1
xy
1
1
0
1
xy
0
0
xy
0
1
3'y
0
xy xy
(a) y
y
zt
zt
zl
xy
0
0
0
0
0
xy
0
1
1
xy
0
0
0
xy
0
;;: '0 \.!:.. Y 0
(b) it
it
zt
zl
0
0
0
0
xy
0
0
0
0
xy
0
0
xy
0
0
(Ii)
0 0
xy
zl
it
zt
0
0
0
0
0
�
0
0
0
0
0
0
 j)
(t \.
0
0
(c)
it
� xy y
xy
it
V
(e)
0
Ci
Fig. 10.27
In Fig. 10.27(a), the four Is are vertically adjacent. The four squares that contain a '1 ' are xyz, xyz, xyz, XYz . From these terms, we rate that the only variable that remains un changed is z ( as both x and y appear in complemented and uncomplimented form). Thus, the corresponding Boolean expression is f(x, y, z) = z. We can also prove it mathematically as below : Here
are
In Fig.
f(x, y, z) = xyz + xyz + xyz + xyz = xzy + xyz + xyz + xzy I Commutative law = xz ( y + y) + xz (y + y) I Complement law = x2 . 1 + xz . l = xz + xz = (x + x) . Z = 1.z = z 10.27 (b), the four Is are horizontally adjacent. The four squares that contain 1
xyzt, xyzt, xyzt, xyzt. From these terms, we note that the only variable that remains unchanged is xy (as z and t appear in both complemented and uncomplimented form). Thus, the corresponding Boolean expression is f(x, y, z) = xy. We can also prove it math ematically as below. f(x, y, z) = xyzt + xyzt + xyzt + xyzt = xyz(t + t) + xyz(t + t) = xyz.l + xyz. l = xyz + xyz = xy (z + z) = xy.l = xy In Fig. 10.27 (c). The four Is are in a square and there are considered to be adjacent to each other. The corresponding Boolean expression is f(x, y, z, t) = yt (as x and z appear in both Here
complemented and uncomplimented form).
BOOLEAN ALGEBRA In Fig.
399
10.27 (d). The four
Is are also adjacent. The four squares that contain Is are :
xyzt, xyzt, xyz t, xyz t . Only xl remains unchanged. Therefore, the corresponding Boolean expression is
f(x, y, Z, t) = xt In Fig. 10.27 (e), the top and bottom rows are also considered to be adjacent to each other as are the left most and right most columns. The four squares containing Is are : xyzt, xY z t, xyzt, xY z t . From these terms, we note that the term y t remains un changed. Therefore, the corresponding Boolean expression is
f(x, y, z, t) = y t
Thus, from the above discussion. We conclude that
"Looping a quad of adjacent Is eliminates the two variables that appear in both comple mented and uncomplemented form". 1 0.37. LOOPING GROUPS OF EIGTHS A group of eight Is that are adjacent to one another is called a octet. When an octet is looped in a fourvariable Kmap, "Three of the four variables are eliminated and only one variable remains unchanged". Consider the following Fig. 10.28.
zt
zt
zt
zl
xy
1
0
0
0
xy
1
1
1
xy
1
1
xy
0
0
zt
zt
zt
zl
xy
1
1
0
0
1
xy
1
1
0
0
1
1
xy
1
1
0
0
0
0
xy
1
1
0
0
zl
(a) zt
xy ?
(b)
zt
zt
zl
0
0
Ii'
zt
xy
1
1
1
1)
1
0
0
1
xy
0
0
0
0
xy
1
0
0
1
xy
0
0
0
0
xy
1
0
0
1
xy
1
1
1
J
(Ii)
Fig. 10.28 ..
zt
xy
(c) In Fig.
zt
10.28 (a), only y remains unchanged f(x, y, z, t) = y
1)
400
DISCRETE STRUCTURES In Fig. 10.28 (b), z remains unchanged
.. In Fig. In Fig.
f(x, y, Z, t) = z 10.28 (c), t remains unchanged f(x, y, Z, t) = t f(x, y, Z, t) = y . 10.28 (d),
1 0.38. KARNAUGH MAP METHOD FOR FINDING PRIME IMPLICANTS AND MINIMAL FORM FOR A BOOLEAN EXPRESSION First express the given Boolean expression in complete sum of products form. The following algorithm can be used by which minimized Boolean expression can be obtained. 1. Identify the ones which cannot be combined with any other ones and encircle them. 2. Identify the ones that can be combined in groups of two in only one way and encircle them as groups. 3. Identify the ones that can be combined with three other ones, to make a group of four adjacent ones) in only one way and encircle them as groups. 4. Identify the ones that can be combined with seven other ones, to make a group of eight adjacent ones) in only way and encircle them as groups. 5. After identifying the essential groups of 2, 4 and 8 ones, if there still remains some ones which have not been encircled) then these are to be combined with each other or with other already encircled ones i.e., we should combine the left over ones in largest possible groups and in as few groupings as possible.
Example 3. Use Karnaugh maps to find the prime implicants and minimal form of the following Boolean expression (ii) !lx, y) = xy + xy + xy' (i) flx, y) = xy + xy' (iii) !lx, y) = xy + xy'. y' y Sol. (i) We first express f,(x, y) into complete sum·of·product x form. Here f,(x, y) products form.
= xy + xy', which is already in complete sum·of·
(1
1
o
o
The K·map for the two variables is shown below (Fig. 10.29)
Fig. 10.29 Put 1s in the cells corresponding to the terms xy and xy' and 0 elsewhere. The two squares containing Is are .xy and xy'. From these terms) we note that the variable x remains unchanged (as y appears in complemented and uncomplimented form). Hence) the prime implicant of f,(x, y) is x. Hence f/x) y) = x) is its minimal sum (ii) f,(x, y) = xy + xy + xy', which is already in complete sum·
of·products form.
The K·map for the two variables is shown below (Fig. 10.30). Put 1s in the cells corresponding to the terms xy, xy and as else where. This map contains two pairs of adjacent squares. (designated by the two loops). The vertices pair represents y and the horizontal pair represents x. Hence y and x are the prime implicants.
y
x ,
x
'1'
l(i
Fig. 10.30
y' 0
y
BOOLEAN ALGEBRA
401
Therefore)
fzL�
A r' AND B ,>L�
AND C i::»i._../
AND C pot._../
(a)
Fig. 10.51
(c)
P = y.
(c) E
= x' + z.
(b)
1 0.40. APPLICATIONS OF BOOLEAN ALGEBRA TO SWITCHING CIRCUITS
Switching circuit. It is an arrangement of wires and switches connected together to the terminal of the battery. A Switch is a two state device used for allowing current to pass through it or not to pass through it. If current is allowed to pass through a switch, then it is said to be 'closed' or 'ON'. If the current is not allowed to pass through a switch, them it is said to be 'Open' or 'OFF'. The switches are denoted by the letters x, y, z or a, b, c.
Methods of connecting Two switches. There are basically two methods of connect·
ing two switches :
(i) Connecting switches in parallel (ii) Connecting switches in series.
BOOLEAN ALGEBRA
41 1
If x and y are two switches connected in parallel, then we say, the bulb is 'ON' iff atleast one of the switches x and y are closed. (Fig. 10.52)
Fig. 10.52
x r o,, y
Fig. 10.53
If x and y are two switches connected in series) then we say) the bulb is 'ON' iff x and y are both closed. (Fig. 10.53) Further, if a switch is 'ON' , then its value is denoted by '1 ' and if a switch is 'OFF' . Then its value is denoted by '0' .
1 0.41 . TRUTH TABLE FOR THE SWITCHES CONNECTED IN PARALLEL Let the switches x and y are connected in parallel. Then, the current will flow through the circuit of x and y only when at least one switch is 'ON' . In other words, the value of the circuit of x and y is 1 when at least one switch has value 1 . The circuit having switches x and y in parallel is denoted by x + y. The truth table for the switches connected in parallel is shown Truth table as :
x
Switches connected in parallel
0 1 0 1
y 0 0 1 1
x+y 0 1 1 1
1 0.42. TRUTH TABLE FOR THE SWITCHES CONNECTED IN SERIES Let the switches x andy are connected in series. Then, the current will flow through the circuit of x and y only when both switches are 'ON' . In other words, the value of the circuit of x and y is 1 when both switches have value 1 . The circuit having switches x and y in series is denoted by x.y. Truth table
_I X I
.y
x
• • 
Switches connected in series
0 1 0 1
y 0 0 1 1
x.y 0 0 1 1
If x and y are two switches such that y is closed when x is open and y is open when x is closed, then y is written as x.
412
DISCRETE STRUCTURES ILLUSTRATIVE EXAMPLES
Example 1. Consider the circuit as shown below. Express the circuit as a Boolean func tion of the Boolean algebra of switching circuit (Fig. 10. 54). Y
X i ' ri X z i i X I 1 'o � Y ...... z'..sL. Y' ......
Fig. 10.54 Sol. From the given circuit, the first portion represents x.y + z The second portion represents x+ y.z' The third portion represents x + y'. All these three portion are connected in series. The Boolean function of the given circuit is f(x, y, z) = (xy + z) (x + yz') (x + y').
Example 2. Consider the Boolean function f(x, y, z) = (x.y.z') + x'.(y circuit representing f(x, y, z) as a Boolean algebra of switching circuits. Sol. Given Boolean function is f(x, y, z) = x.y.z' + x.(y + z') From (1), the circuits corresponding to x.y.z' and x.(y + z') are connected in parallel. The term x.y.z' implies that the switches x, y and z' are connected in series. The terms x.(y + z') implies that the switch x' and the circuit corresponding to y + z' are connected in series. The required circuit is shown in Fig. 10.55.
+ z').
Construct a ... (1)
Fig. 10.55
Example 3. Consider the Boolean function (P.T.V. May 2012) f(x, y, z) = (x.y. + z).(x' + y.z).(x' + z) Construct the circuit corresponding to the Boolean function of the Boolean algebra of switching circuits. Sol. Given Boolean function is ... (1) f(x, y, z) = (x.y + z).(x + y.z').(x + z) From (1), we observe that the circuits corresponding to x.y + z ; x' + y.z' and x' + z are connected in series. The term, x.y + z implies that the circuit corresponding to x.y and the switch z are connected in par allel. The term, x + y.z' implies that the switch x' and the circuit corre sponding to y.z' are connected in parallel.
re
......
+
Y
i
i X'1 i X'
z � y """' zl..s_L_. z
Fig. 10.56
BOOLEAN ALGEBRA
413
+
The term) x' z implies that the switches x' and z are connected in parallel. Hence, the circuit of the given Boollean function is shown in Fig. 10.56.
Example 4. Simplify the circuit as shown below. (Fig. Sol. Let f(x, y) denotes the Boolean function of the
given circuit in the Boolean algebra of switching circuits, then
f(x, y) = x.y' + x.y' + x.y
... (1)
We simplify (1). We have
10. 57)
1 Lx
� : = ::+0 .
.
f(x, y) = x.y' + x.y' + x.y = (x + x).y' + x.y = 1.y' + x.y I Complement law = y' + x.y = (y' + X).(y' + y) I Disributive law = (y' + X).1 Complement law = y' + x = x + y' ... (2)
......
y
Fig. 10.57
Commutative law
From (2), we can say that the given circuit is equivalent to the circuit in which the switches x' and y' are connected in parallel. (Fig. 10.58)
Fig. 10.58
Example 5. Simplify the circuits given below Fig. 10.59 by obtaining the Boolean ex pressions. Also construct the switching circuit for the simplified Boolean expression.
X
�.
Y
a
X
z
(b) ..... .....
c· ..../c o+, .....c·...

....b. � .
a • •xx2,+c>
{(x,) =
{(x,) = x,
.
{(x,. x,) =
(x,
"2)
+
x,
{(x" x,)
{(x,) =
x,
+ .::t2
{(x" x,)
:::: (xl A X2 )
= x1 · :\2
{(x" x,)
{(x" x,)

{(x" x,)
:::: Xl V X2
:::: xl X2
{(x" x,) :::: Xl · X2
{(x" x,) = xl · x2
:::: Xl
{(x" x,) :::: Xl · X2
ILLUSTRATIVE EXAMPLES
x
(Fig.
Example 1. Find the Boolean expression {or the output o{ the given logic circuit Also draw the truth table {or the given logic circuit.
10. 79).
1 NOT>C>,
x� ,� Y
Y 1 NOT·.:>o' (i)
_ _
(ii)
BOOLEAN ALGEBRA
421 z 1
(ii ) Fig. 10.79
Sol. (i) Given circuit is shown below. Let the inputs are x and y. At point
1, the output is x'
X 1 NOT
. . The inputs to the OR gate at point 3 are x: and y'. Hence, at point 3, the output of the OR gate is x: + y' (Fig. Also, the truth table for the given logic circuit is shown below :
10.S0).
x 0 1 0 1
x' 1 0 1 0
Output
y' 1 1 0 0
x' + y' 1 1 1 0
�
(ii) Given circuit is shown below. Let the inputs are x and y. At point the output is x + y
1,
x
Y
At point 2, the output is x.y At point 3, the output is (x.y)' ..
OR
AND
At point 4, the inputs are x+ y and (x.y)' (Fig.
3
Fig. 10.80
Truth table
y 0 0 1 1
OR
2 ' yj NOTDo"'
At point 2, the output is y'.
Inputs
1
2
NOT
3
AND
Fig. 10.81
The output at point 4 is (x + y).(x.y)'
10.Sl).
Also, the truth table for the given logic circuit is shown below :
x 0 1 0 1
Inputs y 0 0 1 1
Truth table
x +y 0 1 1 1
x.y 0 0 0 1
(x.y)' 1 1 1 0
(iii) Given circuit is shown below. Let the inputs are x) y) z.
At point
1, the output is x + y
Output (x + y).(x.y)' 0 1 1 0
4
DISCRETE STRUCTURES
422 At point 2, the output is z'
(x + y).z' output is «x + y).z')'
At point 3, the output is
At point 4 , the (Fig. 10.82). Also, the truth table for the given logic circuit is shown below :
x 0 1 0 0 1 1 0 1
Inputs y 0 0 1 0 1 0 1 1
4
zi
Fig. 10.82
Truth table
z 0 0 0 1 0 1 1 1
x +y 0 1 1 0 1 1 1 1
(x + y).z' 0 1 1 0 1 0 0 0
z' 1 1 1 0 1 0 0 0
Output ((x + y).z)' 1 0 0 1 0 1 1 1
Example 2. Express the output Y as a Boolean expression in the inputs A, B, C for the logic circuit shown below. (Fig. 10. 83)
� =a:::;;:r:E�� Fig. 10.83
Sol. At point 1, the output is A' BC At point 2, the output is AB'C' At point 3, the output is AB' At point 4, the output is A'BC AB'C' AB' . . The required Boolean expression is Y = A'BC AB'C' AB'. Example 3. Express the output Yas a Boolean expression in the inputs A and B for circuit shown below : (Fig. 10. 84)
+
logic
+
A _�1 B ..+....+.. i
Fig. 10.84
+
+
the
BOOLEAN ALGEBRA Sol. At point 1,
423 (a small circle is the circuit)
the output is AB'
At point 2, the output is (A + BY At point 3, the output is (A' B') At point 4, the output is AB' + (A + BY + (A'BY The required Boolean expression is Y = AB' + (A + B') + (A' BY = AB' + A'B' + A + B'.
1 0.44. CONVERSION OF BOOLEAN EXPRESSION TO LOGIC CIRCUIT We can convert a given Boolean expression to a logic circuit. The process of conversion is explained in the following examples:
Example 4. Find a logic circuit corresponding to the Boolean expression x.(x' + y). Sol. Given Boolean expression is f(x, y) = x.(x' + y) x x . (x' + y) NOT Given expression contains comple x' + OR Y � ment of the variable x. So we draw NOT y ' gate for x. In the next step, combine x' + Y by an OR gate. Next, combine x and x' + Y Fig. 10.85
� ' � 
by an AND gate. The required logic cir· cuit of the given expression x.(x' + y) is shown in Fig.
10.85. Example 5. Find a logic circuit corresponding to the Boolean expression x.(y'+ z) + y. ... (1) Sol. Given Boolean expression is f(x, y) = x(y' + z) + Y As (1) contains the x,
complement of only one vari able y, so we draw NOT gate for y in the first step.
y +
z __
In the second step) combine y' and z by an OR gate.
.J
__ __ __ __
Fig. 10.86
In the third step, combine x and y' + z by an AND gate. In the fourth step, combine x.(y' + z) and y by an OR gate.
10.86. Example 6. Find a logic circuit corresponding to the input!output table given below :
The required logic circuit of the given expression is shown in Fig.
x
0 1 0 1
Inputs
Output
y 0 0 1 1
0 1 1 1
DISCRETE STRUCTURES
424
In
Sol. By Boole's Expansion Theorem, the Boolean expression [(x, y) is given by [(x, y) = [(1, l).x.y + [(1, O).x.y' + [(0, l).x.y + [(0, O).x.y' = 1.x.y + 1.x.y' + 1.x.y + 0 = xy + xy' + xy = (x + x').y + xy' = loy + x.y' I Complement law = y + x.y' ... (1) As (1) contains the complement of one variable y. So, in the first step, draw NOT gate. the second step, combine x and y' by an AND gate. x
Y"'l1� � ,

AND
y,
OR
Fig. 10.87
x y' + y .
In the third step, combine x.y' and y by an OR gate. The logic circuit for the Boolean expression (1) is shown in Fig. 10.87.
Example 7. Find a logic circuit corresponding to the input/output table given below : x
1 1 1 1 0 0 0 0
Inputs
y 1 1 0 0 1 1 0 0
z
1 0 1 0 1 0 1 0
Output 1 0 0 1 0 0 1 0
Sol. Let [(x, y, z) be the Boolean expression corresponding to the given input/output table. By BoDle's expansion theorem,
[(x, y, z) = [(1, 1, l).x.y.z + [(1, 1, O).x.y.z' + [(1, 0, l)x.y'.z + [(1, 0, O).x.y'.z' + [(0, 1, l).x.y.z + [(0, 1, O).x.y.z' + [(0, 0, l).x.y'.z + [(0, 0, O).x.y'.z' = l.x.y.z + O.x.y.z' + O.x.y'.z + l.x.y'.z' + O.x'.y.z' + O.x'.y.z' + l.x'.y'.z + O.x'.y'.z' = x.y.z + 0 + 0 + x.y'.z' + 0 + 0 + x.y'.z + 0 = x.y.z + x.y'.z' + r.y'.z. ... (1) As (1) contains the complement of x, y, z, so, In the first step, draw NOT gate for x, y, z In the second step, draw AND gate to get x.y.z ; X, y'.z' ; r.y'.z.
BOOLEAN ALGEBRA
425
AND x'.y'.z
x +j
x.y'.z K������J AND � X � . y .Z � == == !_� � AND � �
x.y.z + x.y'.z' +x'.y'.z
' LS
y�j z �j
Fig. 10.88 In the third step, combine x.y.z ; x.y'.z' ; x.y'.z by an OR gate to get the required logic circuit as shown in Fig. 10.88. Example 8. Draw the logic circuit with inputs A, B and C and output Y which corre
sponds to the Boolean expression. Y = AB'C + ABC' + AB'C'. Sol. Given expression is Y = AB'C + ABC' + AB'C'
Since it contains the complements of the variables B and C. Draw NOT gate for B and C. In the next step, Combine AB'C by AND gate Combine ABC' by AND gate Combine AB'C' by AND gate Next combine AB'C ABC' AB'C' by OR gate. Hence, the required logic circuit is shown below. (Fig. 10.89)
+
+
AB ..1AND ., t... �+_{)o__j C j.... ttTttL'" AND 
AND Fig. 10.89 Example 9. Find the disjunctive normal form and also the corresponding combinato rial circuit of the following Boolean functions : W
�
A
B
C
I(A, B, C)
A
B
C
I(A, B, C)
0
0
0
1
0
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
0
0
1
1
0
0
1
1
1
1
0
0
0
1
0
0
1
1
0
1
1
1
0
1
0
1
1
0
1
1
1
0
1
1
1
1
1
1
1
1
0
DISCRETE STRUCTURES
426 Sol.
(i) The disjunctive normal form of the Boolean function is
f= AB C + ABc + A B C + A B C + A B C + A B C and the corresponding combinatorial circuit is shown in Fig. 10.90.
A B
C
\ \ \
\ � \ � \ �
I I
D
\ � \ � \ �
Fig. 10.90.
(ii) The disjunctive normal form of the Boolean function is f = A B C + A B C + AB C + A B C and the corresponding combinatorial circuit is shown in Fig. 10.91.
A B
C
"
\ �
K K v
\ � " v ...
� v
\ � \ �
Fig. 10.91.
I I
,
BOOLEAN ALGEBRA
427
1 0.45. EQUIVALENT LOGICAL CIRCUITS Two logical circuits are equivalent iff their input/output tables are same or whenever the two circuits receive the same input, they produce the same unit.
Example 10. Show that the following circuits (Fig.
10. 92)
are logically equivalent.
x
y
(b)
(a) Fig. 10.92
Sol. The inputs for the first circuit are At point 1 , the output of the NOT gate
is x.
AND
•
rv 4
AN D J===:J
At point 2, the output of the NOT gate
is y'. is x.y. is
x
I'r=== 53 =[
At point 3, the output of the AND gate
y
At point 4, the output of the AND gate
r.y'.
is x.y'.
At point At point
L�5 = AND Fig. 10.93
5, the output of the AND gate
6, the output of the OR gate is x.y + x.y' + x.y'. (Fig. 10.93) = x.y + x.y + x.y'
The Boolean expression for the first circuit is f(x, y)
... (1)
The input/output table for the Boolean expression (1) is given below.
Table I
Inputs x 0 1 0 1
y 0 0 1 1
x' 1 0 1 0
y' 1 1 0 0
x�y 0 0 1 0
x�y' 1 0 0 0
Output x.y' x�y + x�y' + x.y' 0 1 1 1 0 1 0 0
DISCRETE STRUCTURES
428 x1
For the second circuit) the inputs are x and y. At point 1, the output of the NOT gate is r. At point 2, the output of the NOT gate is y'. At point 3, the output of the OR gate is (Fig. 10.94).
x' +
y'
y __j
Boolean expression of given circuit is . .. (2) r + y' . The input/output for the Boolean expression (2) is given below :
Fig. 10.94
Table II
Inputs x
0 1 0 1
x
y
0 0 1 1
'
1 0 1 0
y
Output /
'
x
1 1 0 0
+ y/
1 1 1 0
From table I and table II, we observe that the outputs for the Boolean expression (1) and (2) are same. Hence both the given circuits are logically equivalent
Example 11. Consider a logic circuit as shown below (Fig. function representing the logic circuit. Also
10.95).
Find the Boolean
Fig. 10.95 (a) Draw the equivalent logic circuit corresponding to f. Sol. At point 1 , the output is x2 X3 At point 3, the output is x" x2 At point 4) the output is X1·X2, Xa At point 5) the output is X1· X2 ·Xa· X4 At point 2, the output is
The Boolean function of the given logic circuit is
f(xl ) x2) x) = Xl . X2 . Xa . X4
... (1)
We now draw the equivalent logic cir cuit representing (1), As (1) contains the complement of x2 and x3) so In the first step, draw NOT gate representing
x2
In the second step, draw NOT gate representing
X3
Fig. 10.96
BOOLEAN ALGEBRA
429 x,.
In the third step, draw an AND gate representing
In the fourth step, draw an AND gate representing
x2 •
X3 . X4 •
In the fifth step, draw an AND gate representing ( x,. x2 ). logic circuit is shown below. (in Fig. 10.96)
3 · x4
(X
)
.
Hence, the equivalent
Example 12. (a) Write the circuit diagram or gate diagram of f( + ) (( + ) + ) )=( (b) Simplify the function in part (a) by using basic Boolean algebra laws. (c) Write the circuit (gate) diagram of the result obtained in Part (b). Sol. (a) Given Boolean expression is f(xl ' )=( + )« + )+ ) xl , X2' x
Xl · x2
x2 ' x,
X,.X2
X
.
X, .
x2
X2
x3
x,
x3 ·
x,
In the first step) draw an AND gate representing X1.X2•
In the second step) draw an OR gate representing X1·X2
+
+
x3•
+
X
In the third step) draw an OR gate representing x2 x3• In the fourth step, draw an OR gate representing (x2 x,) In the fifth step, draw an AND gate representing ..
+
X,.X2
(
+
X3
.
X2
)«
+
x,
)+ ) x
The circuit (or gate) diagram for the given Boolean expression is shown in Fig.
10.97.
1
X, __
X3
""= "C ==
.J
_ _
Fig. 10.97 (b) To simplify f, we have f(xl ' )=( + )« =( + )( =( + )( + =( + )( + =( + )( + =( + )( + = + (c) The circuit diagram of f(x" x2 ' x,
X,.X2
X, .
)+ ) +( + » )( + ) )( + ) ) )
X2
+
x3
x3
x,
x3
X,.X2 · X2
x3
X, . X3
X2 . X2
x,
x3
X, . X2
X, . X2
x,
x3
X, . X2
x,
x3
X, . X3
x2
X3
x,
I Distributive law and Idempotent law
X1·X2
Idempotent law
... (1)
x2 ' x,
) given by (1) is given in Fig. 10.98.
Fig. 10.98
DISCRETE STRUCTURES
430
x..'_________
Example 13. Consider the given Fig. 10.99. " , (a) Write the Boolean function which represents the given onoff circuit (b) Simplify f algebraically obtained in Part (a). Draw the onoff circuit diagram of this sim plified representation. (c) Construct the circuit (or gate) diagram of the given onoff circuit diagram. (d) Find the minterm normal from by using Venn diagram of the Boolean function obtained in part (a) or part (b). (e) Write the relative simplicity and advantages of the circuit gate diagrams found in (c) and (d). Sol. (a) From the given switching circuit, we note that the last portion represents the circuit Xl + x2 · Also the switch x2 and the circuit Xl + x2 are connected in series) so it represents X2.( X1 + x2 ). Also the switch Xl and the circuit x2 . ( Xl + x2 ) are connected in paralleL Therefore) the _ _
Fig. 10.99
Boolean function representing the given onoff circuit is
f(xl ) x2) x) = Xl + X2 · (X1 + x2 ) (b) To simplify f algebraically, we have Let f(xl ' x2) = x, + (X2 ,X, + X2 · X2 ) = x, + (X2 ,X, + 0) = x, + X2.X, = x,
Fig. 10.100
... (1) Distributive law
I X2 , X2 = 0 Absorption law: a v (a 1\ b) = a 10.100.
... (2) Also, the onoff circuit diagram of (2) is shown in Fig. (c) To draw the circuit (gate) diagram of the given switching circuit, we have, the Boolean expression) representing the given onoff diagram) as
f(xl ' x2) = x, + X2 , (X, + x2 )
•••
(1) contains the complement of the variable x2 ' so In the first step, draw NOT gate representing x2 In the second step) draw an OR gate representing Xl + x2 In the third step, draw an AND gate representing X2 , (X, + x2 ) In the fourth step, draw an OR gate representing x, + X2 , (X, + X2 ) Therefore, the required logic circuit diagram is shown in Fig. 10.101. Here
>"'''''1x, (x, + x )
Fig. 10.101
(1)
BOOLEAN ALGEBRA
431
(d) To find the minterm normal form of (1), consider the Venn diagram representing (2) as given below : I (1) and (2) represents the same f(xl ' x2) From the shaded portion I, the minterm is
X,
From the shaded portion II, the minterm is
X1 .X2 The required minterm normal form is
f(xl , x2) = X1 · X2
+ X1 ·X2
(e) The circuit (or gate diagram) of the Boolean function obtained in part (d) is given in Fig. 10.102.
Venn
Diagram
Fig. 10.102 Relative advantages. There are three levels of gates in graph of part (c) and two levels of gates in part (e) so the time 'cost' of the figure in part (e) is slightly less. For both circuits, hardware costs are same. Example 14. Consider the switching circuit as shown in Fig. 10. 103. (a) Find the Boolean expression representing the switching circuit. X�X2 (b) Construct a logic or gate circuit corresponding to the Boolean expression in Part (a) (c) Simplify f algebraically (d) Find the minterm normal form by using Venn diagram and express it in a circuit gate diagram (P.T.V. B.Tech. Dec. 2010) (e) Interpret the result. Sol. (a) First portion of the given Fig. 10.103 represents that the switches x, and x2 are connected in paralleL Therefore, the circuit is Xl + x2• Second portion of Fig. 10.103 represents that the switches x, and X3 are connected in parallel therefore, the circuit is Xl + x3• Now, both the circuits Xl + x2 and Xl + X3 are connected in parallel, it represents + (x, x2) + (x, + x,). Also, the portion on R.H.S. of Fig. 10.103 represents that the switches x, and x2 are connected in series.
Fig. 10.103
Therefore, the Boolean expression is
f(xl ' x2 ' x,) = [(x, + x2) + (x, + x,)] . x, , X2 ... (1) (b) Since the Boolean expression (1) contains the complement of the variable x2 ' so in the first step, draw NOT gate. In the second step, draw an OR gate representing Xl + x2 In the third step, draw an OR gate representing Xl + x3•
DISCRETE STRUCTURES
432
(x, + X2) + (x, + x,) In the fifth step, draw an AND gate representing X,. X2 In the sixth step, draw an AND gate representing «x, + x2) + (x, + x,» ,X" X2 The logic or gate circuit of the Boolean expression (1) is shown in Fig. 10.150.
In the fourth step, draw an OR gate representing
X, ',"\ x2 fi
Fig. 10.104 (c) To simplify [algebrically or minimize the Boolean function (circuit), we employ Boolean algebra laws. Here [(x"
x2 ' x,) = [(x, + x2) + (x, + x,)l .x,. x2
+ X2,X1 , X2 + X3 ·X1· X2 = Xl X2 + O'XI + X3 ·X1 ·X2 = X1 , X2 + X3 ·X1 · X2 = X"( X2 + X2·x,) = X" X2 (1 + x,)
I I
= X1 ·X1 · X2
(
:

Idempotent law
By distributive law
X" X, = xl ' X2 , X2
=
0)
By distributive law By distributive law
I
Boundedness law
(d) From part (c), the Venn diagram representing [(x" x2) = X" X2 is shown in Fig. 10.105 From this Venn diagram, we find the minterms. From the shaded portion I, the minterm is
X1·X2·Xa" From the shaded portion II, the minterm is
X1·X2·Xa" The required minterm normal form is
Fig. 10.105
BOOLEAN ALGEBRA
433
Also, the circuit (or gate) diagram is shown in Fig.
X,
10.106
...r ,
_ _
X, X2 X3
...r ,
_ _
Fig. 10.106 (e) Interpretation. We notice that f(x" x2 ' x,) = 1 when x, = 1, x2 = 0 and X3 = 0 or X3 = 1. Thus current will b e conducted through the circuit when switch Xl is on) switch x2 is off, and when switch Xl is either off or on.
Example 15. (a) Write all inputs and outputs from the given Fig.
its Boolean function is f(xl'
X y
x,) = ((Xl + X2 ) ' X3 ) . (
X ,
+ x2).
Fig. 10.107
Fig.
1 0. 1 07 and show
that
(P.T.D. B.Tech. Dec. 2005)
(b) Simplify f algebrically. (c) Find the minterm normal form off by using Venndiagram. (d) Draw and compare the circuit (or gate) diagrams ofparts (b) and (c) (e) Draw the on·off switching diagram of f in part (a) (j) Write the truth table of the Boolean function f in part (a) (g) Interpret the result. Sol. (a) The outputs are shown by the points 1, 2, 3 and 4 as shown in the following
10.108 :
X3 ' Fig. 10.108
1, the output of the OR gate is x, + x2 At Point 2, the output of the AND gate is (x, + x2)
At Point
At Point
•
X 3
3, the output of the NOR gate is (x, + x2 ) · X3
At Point 4, the output of the AND gate is
« x, + x2 ) · x3 ) (x, + x2)
The required Boolean function is given as f(xl '
•
x2 ' x,) = (x,
+ x2 ) . X3
•
(x, + x2)
•••
(1)
DISCRETE STRUCTURES
434 (b) To simplify of algebraically, we have [(x" X2 ' x,) = (x, + X2 ) . X3 .(X, + X2)
= [(Xl + X2 ) + X3 ] . (X, + X2)
I
De Morgan's law
= 0 + X3 . (x, + X2)
I
Complement law
= « X, + X2 ) . (x, + X2» + X3 .(X, + X2) = X3 . (x, + X2)
Commutative law
= (x, + X2) X3
... (2)
•
(c) To find the minterm normal form, we draw the Venn diagram of the function [(xl ' x2 ' x,) = (x, + X2). X3 , which is shown in Fig. 10.109
X,
From the shaded portion I, the minterm is
X1·X2 · Xa From the shaded portion II, the minterm is
X1·X2 · Xa From the shaded portion III, the minterm is
X1· X2 · Xa glven as
Fig. 10.109
The required minterm normal form of tis
f(xp x2) x) = X1·X2·Xa + X1 , X2, Xa + X1· X2· Xa (d) From part (b), [(x" x2 ' x,) = (x, + X2). X3 As it contains the complement of the variables x3) so In the first step, draw an NOT gate representing
X3 .
+ x2" In the third step, draw an AND gate representing (x, + X2).X3 • The circuit or gate diagram of the Boolean function (2) is given in Fig. 10.110. In the second step) draw an OR gate representing Xl
Fig. 10.110 (c») the minterm normal form of f(xl , x2 ' x) is given as f(xp x2) x) = X1· X2 , Xa + X1·X2 · Xa + X1. X2 . Xa
Also for part
... (1)
To draw the circuit diagram, we proceed as follow.
Xl x2
Since (1) contains the complement of the variables Xl ' x2 ' x3 ' so, in the first step, draw an
OR gate representing
"
and
x3 .
In the second step, draw an AND gate representing xl'x2 ,xa In the third step, draw an AND gate representing X"X2 . X3
BOOLEAN ALGEBRA
435
In the fourth step, draw an AND gate representing
X,. X2 . X3 In the fifth step, draw an OR gate representing X1.X2 . Xa + X1 , X2, Xa
+
X1. X2 . Xa
Therefore, the logic circuit of the Boolean expression (1) is shown in Fig. 10. 1 1 1 . X,
X2
t'i::::,
Fig. 10.111 Comparing the two circuits diagrams (Fig. 10. 1 10 and Fig. 1 1 1), we observe that the circuit diagram shown in Fig. 10. 1 10 is less expensive as it involves less number of gates as compared to the circuit diagram shown in Fig. 10. 1 1 1 .
(e) The onoff switching diagram off in Part (a) is shown in Fig.
10.112
Fig. 10.112 (f) The truth table of the Boolean function of f in Part (a) is shown below :
X3
Xl
X2
X3
Xl + X2
0 0 0 0 1 1 1 1
0 0 1 1 0 0 1 1
0 1 0 1 0 1 0 1
0 0 1 1 1 1 1 1
(Xl
+ X2
1 1 1 0 1 0 1 0
) · X3
f 0 0 1 0 1 0 1 0
(g) Interpretation_ Current will flow only when one of the switches
is OFF.
x,
or x2 is ON and
DISCRETE STRUCTURES
436 TEST YOUR KNOWLEDGE 1 0.7
L
Consider the logic circuit as shown in Fig. 10.113. Express the output f as a Boolean expression in the inputs A, B, C.
�::�:3E=��::l:::�A�N�D))�' I �1 L�V:::::j�A��N:J D \r1 AND
rO R
2.
J�
Fig. 10.113 Express the output f as a Boolean expression in the inputs A, B and C for the logic circuit in the following figures (Fig. 10.114).
'� AND
B C
3.
OR
AND (a)
f
Fig. 10.114 Determine a Boolean expression for each of the switching circuit given in Fig. 10.115.
[�Jc
1'1'
1'1'
(a)
4.
(b)
(b)
Fig. 10.115 Express the output Y as a Boolean expression in the inputs A, B, C for the logic circuit in the following Fig. 10.116 (b) AB = . ==== 1 AN D 

C ��=t++

=L���� AND AND
Fig. 10.116
BOOLEAN ALGEBRA 5.
437
Express the output Y as a Boolean expression in the inputs A, B and C for the logic circuit in Fig. 10.117 (a) and (b) : A ..1 B tttcL/
c �+1
(b)
(a)
Fig. 10.117 6. Draw the logic circuit L with inputs A, B, C and output Y which corresponds to each Boolean expression : (a) Y AB'C + AC' + B'C. (b) Y NBC + NBC' + ABC' =
=
(P.T. U. B.Tech. Dec. 2013)
7.
8.
9.
Draw a logic circuit corresponding to the Boolean expression f(A, B, C) :::: A + B. C + B. Recall that NOR gate is equivalent to OR gate followed by NOT gate. Draw a logic circuit corresponding to the Boolean expression Y :::: AE + A + C . Show that the logical circuits given in Fig. 10.118 are equivalent : x Y
AND
�
�
x Y
(a)
(b)
Fig. 10.118 10. Show that the logical circuits given in Fig. 10.119 are equivalent :
�1�»'�
�
x
y
(a)
(b)
Fig. 10.119 11. Show that the logical circuits given in Fig. 10.120 are equivalent : x
y
� (a)
� Fig. 10.120
x
y (b)
B
438 12.
Show that the logical circuits given in Fig. 10.121 are equivalent : x
y
II � AND t
(a)
13.
x
y
DISCRETE STRUCTURES
8(b)
Fig. 10.121 (a) Consider the Booleanfunction/(x1 • x2 , xS' x4) :::: xI + (X2.(XI + x4) + xa . (x2 + x4» . Draw its circuit (or gate) diagram. (b) Simplify f algebraically (c) Draw the switching (onoff) circuit of f and the reduction of f obtained in Part (a) (d) Draw the circuit (gate) diagram of the reduction of f obtained in (b).
f(A. B. c) = A.B.C + A R C + AB 2. (a) f(A. B) = AB + BC 3. (a) f(A. B. C) = A(B + A).C 4. (a) Y = NBC + NC' + BC'; 5. (a) Y = (AB')' + (N + B + C)' + AC 6. (a) See Fig. 10.122 1.
AB .....1 + +CX>1 C .+  ...+'1
Answers
(b) f(A. B) = AB + AC (b) f(A. B. C) = A(C + Il) + B. C (b) A + B + C + NBC + AB'C'. (b) Y = (NBC)' + NBC' + (AB'C)' + AB'C' (b) Fig. 10.123
AB .�>f' ..1f...+1 AND C nrtnri../ AND
BOOLEAN ALGEBRA 13.
Fig. 10.126 x,
439
L...j "" >" + ", X4 X,, :.:
f'
_ _
X2_+..L...j :>0\X3 _+y
X4 'I ::::»'
Fig. 10.126
1 0.46. SOME SPECIAL CASES OF BOOLEAN ALGEBRA Consider the Boolean algebra {B, modulo 2. i.e
EB,
*, '} where
EB
is a binary operation of addition
.•
1 EB 1 = 0, 0 EB 1 = 1 ; 1 EB 0 = 1 , 0 EB 0 = 0 EB = X1X2 + X1X2 ·
Generally, Xl
X2
I
ILLUSTRATIVE EXAMPLES
I
1. Using laws of Boolean algebra. prove that xy + xz + yz = xy + (x EB y) z. Sol. R.H.S. = xy + (x EB y) z = xy + (xy' + x'y) z = xy + xy'z + xyz = xy.1 + xy'z + xyz = xy (1 + z) + :ry'z + xyz 1 1+a= l VaE B = xy + xyz + xy'z + x'yz = xy + xz (y + YJ + xyz I Complement law = xy + xz. 1 + xyz = xy + xz + xyz = xy + xyz + xz = Y (x + x'z) + xz = y(x + Xl (x + z) + xz I Distributive law = y.1.(x + z) + xz I Complement law = yx + yz + xz = xy + xz + yz. Example 2. Obtain the sum·oiproducts canonical form of the Boolean expression Example
(x, EB x2 J EB (x, EB x3J. Sol. Here Consider
I
De Morgan's law
DISCRETE STRUCTURES
440
I
=
Involution law
(x, + X2) (x, + X2 ) (XlX3 . XlX3 ) + ( Xl + X2) + (X, + X2») (X,X3 + X,X3 )
= (X, + X2) (X, + X2 ) (X, + X3) (X, + X3) + (X" X2 + X" X2) (X,X3 + X,X3) = (Xl + X2) (Xl + X2) (Xl + Xa) + X1·X2 · X1· Xa + X1· X2 · X1· Xa
I a.a = a \;j a E
+ X1, X2 , X1 , Xa + X1· X2 · X1 · Xa
B
= (XI,XI + X1· X2 + X2 , X 1 + X2 · X2) (Xl + Xa)
I aa
+ O + X1. X2 . Xa + X1 . X 2 . Xa + 0
=O
VaE
B
= (0 + X1· X2 + X1.X2 + 0) (Xl + x) + X1·X2·Xa + X1· X2 · Xa = (X1·X2 + X1·X2) (Xl + Xa ) + X1· X2 · Xa + X1· X2 · Xa
= Xl ' X2 . Xl + Xl . X2 . Xa + Xl . X2 . Xl + Xl . X2 . Xa + Xl . X2 . Xa + Xl . X2 . :fa = 0 + X1· X2 · Xa + X1,X2 + X1·X2 · Xa + X1, X2 , Xa + X1. X2 . Xa = X1, X2 , Xa + X1·X2 ( 1 + xa) + X1·X2 · Xa + X1· X2 · Xa = X1, X2 , Xa + X1, X2 + X1·X2 · Xa + X1· X2 · X3 ) is the required expression.
Example 3. Minimize the Boolean expression f(x, y) = xy EB :>y' EB xy'.
(P.T.V. B.Tech. May 2006)
Sol. We use Karnaugh map to minimize the given Boolean expression f(x, y) = :>y EB xy' EB xy'
The Kmap for two variables is shown in Fig. 10.127
Enter Is in the cells corresponding to the terms X)' ) xy' and
1
xy' and Os elsewhere.
All the Is can be looped (or combined) as shown in the figure. Corresponding to loop (1), the variable y appears in comple mented and uncomplemented form and x remains unchanged. Therefore) the prime implicant is x
x
X'
Similarly, corresponding to loop (2), the variable appears in complemented and uncomplemented form and y' remains unchanged.
Y 1
y' 1 1
2
Fig. 10.127
The prime implicant is y'. The required minimal Boolean expression is X
f(x, y) = EB y' Example 4. Minimize the Boolean expression f (x, y) = xy EB xy EB xy'. SoL Proceed yourself as in example 3(above).
(P.T.V. B. Tech. May 2007) Ans. f(x, y) = x EB y
BOOLEAN ALGEBRA
441
Example 5. Minimize the Boolean expression f(x, y, z) = xy'z EB xyz' EB xyz' (P.T.V. B. Tech. Dec. 2007) Sol. We use Karnaugh map of three variables to minimize the given Boolean expression f(x, y, z) = xy'z EB xyz' EB xyz' The Kmap for three variables x, y, z is shown in Fig. 10.128 Enter Is in the cells corresponding to the terms x'y'z ; x'yzl ; xyzl and Os elsewhere. All the three Is can be looped (or combined) as yz yz' y'z' y'z shown in the figure. The two Is are vertically adjacent o o 1 x 0 and one Is in isolated square. Corresponding to loop (1), the variable x appears 0 xC 1 in complemented and uncomplemented form and y) z' remains unchanged i.e., y) z' appear in the two squares. Therefore) the prime implicant is yz� Fig. 10.128 Also the prime implicant corresponding to loop (2) is xy'z. Hence, the minimal form is f(x, y, z) = yz' EB xy'z.
Example 6. Minimize the following Boolean expression, f(x, y, z, t) = y'z' EB xy' EB zt' Sol. Given Boolean expression is f(x, y, z, t) = y'z' EB xy' EB zt'
... (1)
We first express (1) in completesumofproducts form Consider y'z' = (x + Xl y'z' (t + t') = (xt + xt' + xt + xt') y'z' = xy'z't + xy'z't' + x'y'z't + x'y'z'(' xy' = xy' (z + Z! (t + t') = xy' (zt + zt' + z't + z't') = x'y'zt + x'y'z(' + x'y'zt + x'y'z'(' zt' = (x + Xl (y + y! zt' = (xy + xy' + xy + xy! zt' = xyzt' + xy'zt' + x'yzt' + x'y'z(' Now, the Kmap for the variables x, y, z, t is given in Fig. 10.129. Enter Is in the cells corresponding to the terms given in (2») x'y' as given in (3») and zt' as given and Os elsewhere. All the Is can be looped (or combined) as shown in the figure.
y'z' as in (4),
Corresponding to loop (1), the variables x and y' appear in complemented and uncomplemented form where as z,t' remains unchanged. Therefore, the prime implicant is zt' Corresponding to loop (2), the variables z and t' appear in complemented and uncomplemented form where as x', y' remains unchanged. Therefore, the prime implicant is x'y'.
xy
xy,
zt 0 0
x'y, ( 1 x'y
0
... (2)
... (3) ... (4)
zt' '1
1
z't' 0
1
z't 0
1
1
1
�
0
0
CD
�2
I t
Fig. 10.129
Corresponding to loop (3), the variables x and t appear in complemented and uncomplement form where as y', z' remains unchanged. Therefore, the prime implicant is y'z'. The required minimize Boolean expression is f(x, y, z, t) = zt' EB x'y' EB y'z'.
DISCRETE STRUCTURES
442
Example 7. Minimize the Boolean function L m (2, 3, 4, 7, 10, I I, 12, 15). Sol. Here, the greatest term is 15. Therefore, total number of variables for Kmap is 24 = 1 6 I N = 2' for n = 4, N = 1 5 Let f(A, B , C, D) b e the required Boolean expression. Take A, B along vertical axes and C, D along horizontal axes as shown in Fig. 10.130.
AB
CD CD 00
CD 01
CD 11
CD 10
AB 00
0
1
3
2
AB 01
4
5
7
6
AB 11
12
13
15
14
AB 10
8
9
11
10
Fig. 10.130 Enter 1s in the cells corresponding to the terms in the given expression (Fig. 10.131)
CD AB AB 00
3
CD 01
CD 00
AB 01
1
AB 11
1
AB 10
0
1
4
5
12
13
8
9
CD 11 1 1 1 1
CD 3 7
15 11
1
Fig. 10.131 Corresponding to loop (1), all the four 1s are vertically adjacent. The variables A and B appear in complemented and uncomplemented form whereas the variables C) D remains un changed. Therefore, the prime implicant is CD. Corresponding to loop (2), all the four 1s in the top and bottom row are considered to be adjacent. The variables A and D appear in complemented and uncomplemented form whereas the variables B and C remains unchanged. The prime implicant is B C Corresponding to loop (3), the variable A appears in complemented and uncomplemented form whereas B) C , D remain unchanged. Therefore, the prime implicant is B C D . The required minimize Boolean expression is
f(A, B,
C, D)


= CD + BC + B C D .
BOOLEAN ALGEBRA
443
Example 8. Minimize the following switching function L m (0, 2, 8, 1 2, 13). Sol. The largest term is 13, therefore, total number of variables = 16 (N = 2', n = 4, N = 13) Let f(A, B, C, D) be the required Boolean expression. The Kmap for the variables A, B,
C, D is shown in Fig. 10.132.
C D CD CD CD CD AB 00 01 11 10 1 AB 00 1 0 1 3 AB 01 5 7 6 4 AB 11 1 12 1 13 2 1 5 14 AB 10 1 8 1 9 11 10 Fig. 10.132
3
Corresponding to loop (1), the variable B appears in complemented and uncomplemented
form whereas the variables A) C ) D remains unchanged. Therefore) the prime implicant is
ACD.
Corresponding to loop (2), the prime implicant is AB C . Corresponding to loop (3), the prime implicant is A B D The required minimizing Boolean expression is
f(A, B,
L
I
= A C D + ABC + A B D .
TEST YOUR KNOWLEDGE 1 0.8
Minimize the following Boolean expressions ( ) fl (x, y, z) xyz Ell xyz Ell x'yz' Ell x'y'z (ii) f2(x, y, z) ::: xyz EB xyz' EB xy'z EB x'yz EB x'y'z. Minimize the following switching functions (i) L (1, 2, 3, 13, 15) (ii) L (1, 5, 6, 7, 11, 12, 13, 15) (P.T. U., B.Tech. Dec. 2007) (iii) L (O, 2, 10, 11, 12, 14) (P.T. U., B.Tech. May 2007) Design a threeinput minimal ANDOR circuit with the following truth table T {A, B, C; L} {OOOO, 1111, 0011, 0010, 0101, 0001, 1001, 11m} (P.T. U., B.Tech. May 2013) ,
2.
C, D)
=
m
m
3.
m
=
=
Answers
1. 2.
(ii) f2(x, y, z) z Ell xy Ell yz Ell x'y'z (ii) f(A, B, C, D) BD + AeD ABC + ACD + ABD (iii) f(A, B, C, D) ABD ABD ABC (i) fl(x, y, z) = xy
=
+
=
=
+
+
DISCRETE STRUCTURES
444 Hints 1.
(i) The Kmap is given in Fig. 10.133. yz' y'z' y'z yz
yz
yz
y·z
y'z'
I I I tr::I 1
2.
)
Fig. 10.133 (iL) The Kmap is given in Fig. 10. 134. (i) The Kmap is given in Fig. 10.135. AB
CD


CD 00
CD 0
CD
11
Fig. 10.134
(ii) The Kmap in Fig. 10.136.

CD 0
1
AB
1 1 2 3 1 1 6 AB 0 1 4 r 5 II 7 1 12 111 3 "T 1 5 14 AB 1 1 1 AB 1 0 8 9 11 10 AB 0 0
AB 00
0
3.
AB
J)
AB CI
AB l l
AB
B
0
1
3
4
5
7
6
8
13 9
1 )12

MULTIPLE CHOICE QUESTIONS (MCQs)
xy = 0 ;
+ zw = 0 ; have the following solution for x, y, z and respectively. (a) 0 1 0 0 (b) 1 1 0 1 (c) 1 0 1 1 (d) 1 0 0 0. =1;
y
Fig. 10.137(b)
x+y +z= l ; w
1 5 (1 14 (1 11 1) 10
....+1 .. Do1
The simultaneous equations on the Boolean variables x, y,
xz +
11

Fig. 10.137 (a)
1.
1
CD 0
1 (1 2
1)
AB 10

CD
C IDO+'" D +1
'0
I
CD 0
A .IDo1
1 1
1

CD 00
Fig. 10.136
1''
1
 
AB 01

Fig. 10.135 See Fig. 10. 137(a) and 10.137(b) CD ciS CD C AB if0
CD
.xy
w
z and
w
BOOLEAN ALGEBRA
445
2.
What values of A, B, C and D satisfy the following simultaneous Boolean equations ?
3.
A + AB = 0, AB = AC, AB + AC + CD = CD (a) A = 1, B = 0, C = 0, D = 1 (b) A = 1, B (c) A = 1, B = 0, C = 1, D = 1 (d) A = 1, B Principal of duality is defined as
= 1, = 0,
C C
= 0, = 0,
(a) S is replaced by :> (b) LUB becomes GLB (c) All properties not altered when S is replaced by :>
D D
=0 = O.
(d) All properties not altered when S is replaced by :> other than 0 and
4.
5.
The Boolean function
x y + y + xy
1 element.
is equivalent to
(a) x + y (c) x + y
(b) x + y (d)
x
+ y.
The logic expression for the output of the circuit shown in the figure below is :
A B c ..1 (a) AB C + CD
6.
D 1
8.
9.
10.
+ CD
(c) ABC + C D
(d) A B + C D .
(a) (A + B) (A + C)
(b)
The Boolean expression A + BC equals
(c) (A + B) (A + C)
7.
(b) ABC
(A + B) (A + C)
(d) None of the above.
After minimization of Boolean expression, Y = AC + AB + ABC + BC , we get
(a) A . B + C (c) AB + BC
(b)
AB + C
(d) None of the above. How many truth tables can be made from one function table
(a) 1 (c) 3
(b) 2 (d) 8.
(a) AND function of several OR functions
The term sum·of·product in Boolean algebra means
(b) (c)
OR function of several AND functions
AND function of several AND functions
(d) OR function of several OR functions. Simplifying Boolean expression Y = A B C D + A B C D , we get
(a) ABC
(c) A + BCD
(b)
ABC
(d) AB + CD.
DISCRETE STRUCTURES
446 Answers and Explanations 1. (c) Take x = 1, Y = 0, z = 1, = 1, then (i) x + y + z = 1 + 0 + 1 = 1 (ii) xy = 1.0 = 0 (iii) xz + = 1.1 + 1 = 1 + 1 = 1  (iv) xy + zw = 1.0 + (1.1) = 0 + ( 1 + 1 ) = 0 + 0 + 0 = 0 2. (a) 3. (c) 4. (d) Given expression is x y + y + xy w
w
5.
= x y + xy + y = x(y + y) + y = x + y. (a)
I
y +y = l
Given figure is shown below :
A B
c 1
0 1
AB The output at the point 2 is A B C The output at the point 3 is CD The output at the point 4 is A B C + CD. (a) Using distributive law, (A + B) (A + C) = A + BC. The output at the point
6. 7.
1
is
(a) Given expression is
A C + A B + AE C + BC = A C (B + E) + A B(C + c) + AE C + (A + A) BC = ACB + ACE + ABC + ABC + AB C + ABC + ABC = ABC + AC (B + B) + ABC + AC(B + B) = A BC + A C + ABC + AC I A CB + A BC = A BC, B + E = 1 = A BC + ABC + (A + A)C I A+ A = l = A BC + ABC + C = A B (C + C) + C = A B + C.
8. (b) 10. (b) Given expression is
9. (b)
11
G RAP H S
1 1 . 1 . INTRODUCTION In many problems dealing with discrete objects and binary relations) a graphical repre sentation of the objects and the binary relations on them is a very convenient form of repre sentation. This leads us naturally to a study of the theory of graphs.
1 1 .2. BASIC TERMINOLOGY The graphs consist of points or nodes called vertices which are connected to each other by way of lines called edges. These lines may be directed or undirected.
(P.T.U., B.Tech. Dec. 2013, Dec. 2006, May 2005)
1 1 .3. DIRECTED GRAPH
A directed graph is defined as an ordered pair (V, E) where V is a set and E is a binary relation on V. A directed graph can be represented geometrically as a set of marked points V with a set of arrows E between pairs of points. Also The elements in V are called vertices. The ordered pairs in E are called edges. For e.g., consider the Fig.
11.1 given below. It is a directed graph. a
b Directed graph Fig. Here, the vertices are
11.1
o Loop
Fig.
11.2
a, b, d and the edges are (a, b), (b, a), (b, d), (d, a).
An edge is said to be incident with the vertices it joins. For example, the edge (a, b) is incident with the vertices a and b. Also, we say that the edge (a, b) is incident from vertex a and incident into vertex b.
447
DISCRETE STRUCTURES
448
The vertex a is called the initial vertex and the vertex b is called the terminal vertex of the edge (a, b). An edge that is incident from and into the same vertex is called a loop or selfloop. (Fig. 1 1 .2). Degree of a selfloop is two as it is twice incident on a vertex. Corresponding to an edge (a, b), the vertex a is said to be adja cent to the vertex b and the vertex b is said to be adjacent from the vertex a. b A vertex is said to be an isolated vertex if there is no edge inci dent with it. For example, consider the following graph (Fig. 1 1 .3) The vertex 'a' has a selfloop. deg a = 4 d '*'�,c The vertex 'b' is a Pendent vertex since only one edge is inci dent on it. The vertex 'e' is an isolated vertex as it has no edge incident on it. Also deg e = O. . .
_ _ _ _
Fig. 11.3
1 1 .4. (a) UNDIRECTED GRAPHS An undirected graph G consists of a set of vertices, V and a set of edges E. The edge set contains the unordered pair of vertices. If (u, v) E E then we say u and v are connected by an edge where u and v are vertices in the set V.
For example, let V = {I, 2, 3, 4} and E = {(1, 2), (1, 4), (3, 4), (2, 3)}. Draw the graph. The graph can be drawn in several ways. Two of which are as follows (Fig. 1 1.4 and Fig. 1 1 .5). These are directed graphs
2}_{
}___{4
3
Fig. 11.4
Fig. 11.5
Consider the graph shown in Fig. undirected graph.
1 1 . 6 Determine 2
3
Fig. 11.6
the edge set and the vertex set of this
GRAPHS
449
= {(I, 2), (1, 4), V = {I, 2, 3, 4}.
The edge set is
E
The vertex set is
(2, 3), (2, 4), (3, 4)}
1 1 .4. (b) MIXED GRAPH A graph G = [V, E] in which some edges are directed and some are undirected is called a mixed graph. The graph shown in Fig. 1 1 . 7 is a mixed graph.
1 1 .4. (e) FINITE GRAPH A graph G
=
c
b
Fig. 11.7
[V, E] is said to be finite if V and E are finite sets.
1 1 .4. (d) LINEAR GRAPH A graph G = (V, E) is said to be a linear graph if its edges joining vertices lie along a line. For example) •• ..... . . ..... . . 0... . is a linear graph.
1 1 .4. (e) DISCRETE OR NULL GRAPH
A graph containing only vertices and no edge is called a discrete or null graph. The set E of edges in a graph G = [V, E] is empty in a discrete graph. Also each vertex in a discrete graph is an isolated vertex.
1 1 .5. SIMPLE GRAPH
(P.T.U., B.Tech. Dec. 2006, May 2005 ; M.G.A. May 2007, Dec. 2005, May 2013)
A simple graph is one for which there is no more one edge directed from any one vertex to any other vertex. All other graphs are called multigraphs. (see Figs. 1 1 .8, 1 1 .9)
A .
D ·4
,
B
A
.
,
D
.
,
., Simple graph Fig. 11.8
·3
c
In Fig. 1 1 .9, the edges e4 and e5 are called multi
Br:== ::: ::::;:,== :: 7C(
. Multigraph Fig. 11.9
edges.
1 1 .6. COMPLEMENT GRAPH The complement of a graph is defined to be a graph which has the same number of vertices as in graph G and has two vertices connected iff they are not connected in the graph G.
DISCRETE STRUCTURES
450
(P.T. U., B.Tech. Dec. 2006, May 2005)
1 1 .7. (a) DEGREE
Let v be a vertex of an undirected graph. The degree of v, denoted by d(v), is the number of edges that connect v to the other vertices in the graph. The degree of a graph cannot be negative.
(P.T. U. B.Tech. Dec. 2005)
1 1 .7. (b) INDEGREE AND OUTDEGREE
If v is a vertex of a directed graph, then the outdegree of v, denoted by outless (v), is the number of edges of the graph that initiate v. The indegree of v, denoted by indeg(v), is the number of edges that terminate at v. For e.g., consider the graph shown in Fig. 1 1 . 10. The degrees of A) B) C) D are 3) 3) 5) and 3 respectively.
Multigraph
Fig. 11.10
1 1 .8. SOURCE AND SINK A vertex with indegree 0 is called a outdegree a is called a sink.
sink.
For example,
source
and a vertex with
consider the graph shown in Fig. 1 1 . 1 1 . Here
For example, consider the graph shown below (Fig.
u4 is a
Fig. 11.11
1 1 . 12)
The graph shown in Fig. 1 1 .12 has 7 edges. Indegree of 'a'
= 3, Indegree of 'b' = 2; Indegree of 'c' = 1 , Indegree of 'd' = 1 Also, outdegree of 'a' = 1 , outdegree of 'b' = 3 outdegree of 'c' = 0, . . c is a sink. outdegree of 'd' = 3.
A vertex is said to be number.
even vertex if its degree is an even
A vertex is said to be an odd number.
For example,
odd vertex if its degree is an
consider the graph, as shown in Fig. 1 1 . 13.
The vertices A and D are even vertices since deg(A) deg(D) = 2
= 2,
The vertices B and C are odd vertices since deg(B) deg(C) = 3 A vertex of degree zero is called isolated vertex.
=
Fig. 11.12 A
D
81
3,
A vertex with degree one is called a pendent vertex. The only edge which is incident with a pendent vertex is called the
pendent edge.
c
d
1 1 .9. EVEN AND ODD VERTEX
Fig. 11.13
GRAPHS
45 1
1 1 . 1 0. ADJACENT VERTICES Two vertices are called adjacent if they are connected by an edge. If there is an edge
(ep e2») then we say that vertex e 1 is adjacent to vertex e2 and vertex e2 is adjacent to vertex e 1 .
Theorem I. Show that the sum of degree of all the vertices in a graph G, is even. Proof. Each edge contribute two degrees in a graph. Also, each edge contributes one degree to each of the vertices on which it is incident. Hence) if there are N edges in G) then we have 2N =
d(v,) + d(v ) + ...... + d(v N)
Thus) 2N is always even.
Another statement. The sum of the degrees of the vertices of a graph G is equal to twice the number of edges in G. Theorem II. Prove that in any graph, there are an even number of vertices of odd degree. (P.T.V., M.e.A. Dec. 2005, B.Tech. Dec. 2012) Proof. Consider a graph having vertices of degree even and odd. Now) make two groups of vertices. One with even degree of vertices vp v 2) ... ) v k and other with odd degree of vertices up u2 ) ... , un ' Suppose,
d(v,) + d(v) + ... + d(v k) V = d(u,) + d(u) + ... + d(un). v=
Now, we know that sum of degree of all the vertices is even (Theorem I). So, V + V is even. Since, V is the sum of K even numbers. Hence, it is even. But U is the sum of numbers. So, to be U an even number, n must be even. Hence proved.
ILLUSTRATIVE EXAMPLES
I
n odd
Example 1. Verify that the sum of the degree of all the vertices is even for the graph shown in Fig. 11. 14. v,
V2
Fig. 11.14
V3
DISCRETE STRUCTURES
452
Sol. The sum of degree of all the vertices is = d(v,) + d(v ) + d(v ) + d(v4) + d(v 5) + d(v,) + d(v7) + d(vs) = 2 + 3 + 3 + 3 + 3 + 4 + 2 + 2 = 22, which is even. Example 2. Verify that there are an even number of vertices of odd degree in the graph shown in Fig. 11. 15. c
b d
a
e h
9
Fig. 11.15 Sol. The number of vertices of degree odd are 8 and each have degree three in the above graph. Hence) we have even number of vertices of odd degree. 1 1 . 1 1 . PATH IN A GRAPH
A
path of length n is a sequence of n is an edge of the graph.
+ 1 vertices of a graph in which each pair of vertices
1. A Simple Path
The path is called simple one if no edge is repeated in the path i.e., all the vertices are distinct except that first vertex equal to last vertex. An Elementary Path. The path is called elementary one if no vertex is repeated in the path i.e., all the vertices are distinct.
2.
3. Circuit or Closed Path
The circuit or closed path is a path which starts and ends at the same vertex i.e., v 0 = v n 0 Simple Circuit Path. The simple circuit is a simple path which is a circuit.
4.
Theorem III (a). Suppose a graph G contains two distinct paths from a vertex u to a vertex v. Show that G has a cycle. Proof. Consider two distinct paths from u to v be PI = (ep e2) e3) ...... ) en) and P2 = (e/) e2') e3') ...... ) en')' Now delete from the paths P , and P2 all the initial edges which are identical i.e., of we have e 1 = e l ') e2 = e2') e3 = e3') ...... , ek = ek' but ek 1 t: e'k l ' We will delete all the first k edges of both the paths P, and P2 . Now, after deleting the k edges both the paths start from the same vertex, (let u ,) and end at v. Now, to construct a cycle, start from vertex u, and follow the left over path ofP 1 until we first meet any vertex of the left over path of P2 ' If this vertex is u2' then the remaining cycle is computed by following the left over path of P2 which starts from u2 and ends at v. +
+
GRAPHS
453
Theorem III (b). If a graph has n vertices and vertex v is connected to vertex w, then there exists a path from v to w af length no more than n. Proof. We prove this theorem by method of contradiction. Let us assume that v is connected to w) and the shortest path from v to w has length m) where m is greater than n. We know that, a vertex list for a path of length m will have m + 1 vertices. This path can be represented as v O ) v I ' v2 ... v ) where va = v and v = w. m m Now since there are only n distinct vertices and m vertices are listed in the path after v O ) thus there must be same duplicated vertices in the last m vertices of the vertex list) that
represents a circuit in the path. Thus our assumption is not true and the minimum path length can be reduced) which is a contradiction.
Example 3. Consider the graph shown in Fig. 11.16. Give an example of the following : (i) A simple path from Vi to Ve ' (ii) An elementary path from Vi to Ve ' (iii) A simple path which is not elementary from Vi to Ve ' (iv) A path which is not simple and starting from V2. �+� (v) A simple circuit starting from Vi ' (vi) A circuit which is not simple and starting from V2. Sol. (i) A simple path from V, to VB is
V,
V3
V2
V I ' v2 ) V3 ) V4) V5) V6 '
(ii) An elementary path from V, to VB is (iii) A (iv) A (v) A (vi) A
V I ' V2 ) V3 ) V5) V4) V6 , simple path which is not elementary from V, to VB is V I ' V2 , V3 ) V5) V2 ) V4) V6 , path which is not simple and starting from V2 is V2 ) V3 ) V4) V5) V3 ) V4) V6 , simple circuit starting from V , is V I ' V2 , V4) V6 ) V5) V3 ) VI ' circuit which is not simple and starting from V2 is V2 , V3 ) V I ' V2 ) V5) V4) V2 ·
Fig. 11.16
Example 4. Consider the graph shown in Fig. 11.17. Give an example of the following : (i) An elementary path (ii) A simple path (iii) A path which is not simple (iv) A simple path which is not elementary (v) A simple circuit (vi) A circuit which is not simple.
a
b
d
e Fig. 11.17
c
DISCRETE STRUCTURES
454
Sol. There are many solutions to the above problems, but we will take only one for each. (i) An elementary path is a, b, c, f, e, d. (ii) A simple path is a, f, e, b, d. (iii) A path which is not simple is a, b, e, e b, d. (iv) A simple path which is not elementary is a, b, e, f, c, b, d. (v) A simple circuit is a, b, c, f, a. (vi) A circuit which is not simple is a, b, e, f, c, b, d, a. Example 5. Consider the graph as shown in Fig. 1 1 .18. Determine the following : (i) Pendent vertices (ii) Pendent edges (iii) Odd vertices (iv) Even vertices (v) Incident edges (vi) Adjacent vertices. Sol. (i) The vertex V5 is the pendent vertex. (ii) The edge (V4 ' V5) or e5 is the pendent edge. Fig. 11.18 (iii) The vertices V3 and V5 are odd vertices. (iv) The vertices V V2 and V4 are even vertices. The edge e2 is incident on V, and V3 . (v) The edge e , is incident on V, and V2 . The edge e4 is incident on V3 and V4' The edge e3 is incident on V2 and V3 .
V2
V3
F
The edge e5 is incident on V4 and V5.
(vi) The vertex V, is adjacent to V2 and V3 .
The vertex V3 is adjacent to V, and V4'
The vertex V2 is adjacent to V , and V3 . The vertex V4 is adjacent to V3 and V5.
The vertex V5 is adjacent to V4'
graph.
Example 6. Consider the graph G shown in Fig. 11. 19. Find the complement of this
V2 v,
L_3;;V Fig. 11.19
Sol. The complement of above graph is shown in Fig. 1 1.20. Here, we consider a com· plete graph of vertices and then delete the edges that are in G from the complete graph. The remaining graph is the complemented graph.
4
GRAPHS
455 v,
Fig. 11.20 Complement of Graph G. (P.T. U., B.Tech. May 2013)
1 1 . 1 2. UNDIRECTED COMPLETE GRAPH
An undirected complete graph G = (V, E) of n vertices is a graph in which each vertex is connected to every other vertex i.e., and edge exists between every pair of distinct vertices. It is denoted by Kn' A complete graph with n vertices will have n(n 1)/2 edges. 
The complete graph kn for •
k,
Example
n=
1) 2)
3) 4) 5) 6 are shown below:
6 3
7. Draw undirected complete graphs K4 and K6 .
Sol. The undirected complete graph of K4 is shown in Fig. 1 1.21 and that of K6 is shown in Fig. 1 1.22. v,
V2
v, (, 2 vertices. Because graph G is connected, G and The graph shown in Fig. 1 1.28 is a connected graph.
has no isolated vertices. Suppose G has no vertex of degree 1 . Then the degree of each vertex is at least 2. This implies that the sum of the degrees of vertices of G is at least 2n. Hence, it follows that the number of edges is at least n (': the sum of the degrees of vertices in any graph is twice the number of edges), which is a contradiction. This implies that G contains at least one vertex of degree 1.
(P.T. U., B.Tech. Dec. 2006, Dec. 2013)
1 1 . 1 6. SUBGRAPH
A subgraph of a graph G = (V, E) is a graph G' = (V', E') in which V' c V and E' c E and each edge of G' has the same end vertices in G' as in graph G. Note. A
single vertex is a subgraph.
Example 11. Consider the graph G shown in Fig. 11.29. Show the different subgraphs of this graph.
B
A�+� C
F �+� D
E
Fig. 11.29
DISCRETE STRUCTURES
458
Sol. The following are all subgraphs of the above graph (shown in Figs. 1 1 .30, 1 1 . 3 1 , 1 1 .32, 1 1 .33). There may b e another subgraphs o f this graph.
B
A.� C
A
F �_e D
C Fig. 11.30
Fig. 11.31
B
B
A
C
F
D
A
C
D
E
E
Fig. 11.32
Fig. 11.33
Example 12. Consider the directed graph as shown in Fig. 11.34. Show the four differ ent subgraphs of this graph having at least four vertices.
Fig. 11.34 Sol. The four subgraphs of the directed graph are shown in Figs. 1 1.35, 1 1 .36, 1 1 .37 and 1 1 .38. There may be another sub graphs of this graph.
GRAPHS �
� � __ __ __
__ __
V3
Fig. 11.35
v,
V3
V2
V2
V2
Fig. 11.36
.V�'
�V2
� � __ __
__ __
V3
Fig. 11.37
459
Fig. 11.38
Exrunple 13. Consider the multigraph shown in Fig. 11.39. Show two different subgraphs of this multigraph which are itself multigraphs.
a
Fig. 11.39 shown
Sol. The two different subgraphs of this multigraph which are itself multigraphs are in Figs. 1 1.40 and 1 1.41. There may be another sub graphs of this multigraph.
a
Fig. 11.40
b
a
d
c Fig. 11.41
DISCRETE STRUCTURES
460 1 1 . 1 7. (a) SPANNING SUBGRAPH
A graph G, = (V 1" E ,) is called a spanning subgraph of G = (V. E) if G, contains all the vertices of G and E '" E , . For example : The Fig. 1 1.42 is the spanning subgraph of the graph shown in Fig. 1 1 .29. B A
C
F
D E
Fig. 11.42. Spanning Subgraph. 1 1 . 1 7. (b) COMPLEMENT OF A GRAPH Let G
= (V,
E) be a given graph. A graph G = (V, E) is said to be com
plement of G = (V, E) If V
= V and E does not contain edges of E. i.e., edges in
E are join of those pairs of vertices which are not joined in G. Consider the graph shown in Fig. 1 1 .43.
The complement graph is shown in Fig. 1 1 .44 . Note that a graph and its complement graph have same vertices.
then
If a graph G has
n
vertices and K" is a complete graph with
Consider K4• Then
X G
Consider K6 . Then
G
�D G
G
n
vertices,
D Fig. 11.43
x 11.44 Fig.
GRAPHS
461
1 1 . 1 7. (e) COMPLEMENT OF A SUBGRAPH Let G = (V. E) be a graph and S be a subgraph of G. If edges of S be deleted from the graph G, the graph so obtained is complement of subgraph S. It is denoted by S . S
=GS
Consider the graph subgraph S is
�
and its sub graph
A
Then the complement of
Note that in a complement of a sub graph, the number of vertices donot change.
1 1 . 1 8. (a) CUT SET Consider a connected graph G = (V, E). A cut set for G is a smallest set of edges such that removal of the set, disconnects the graph whereas the removal of any proper subset of this set, left a connected sub graph.
graph.
For example, consider the graph shown in Fig. 11.45. We determine the cut set for this
V2 V3 v,
Fig. 11.45 For this graph, the edge set {(VI V5), (V7' V5)} is a cut set. After the removal of this set, ' we have left with a disconnected subgraph. While after the removal of any of its proper subset, we have left with a connected subgraph.
1 1 . 1 8. (b) CUT POINTS OR CUT VERTICES Consider a graph G = (V, E). A cut point for a graph G, is a vertex v such that Gv has more connected components than G or disconnected. The subgraph Gv is obtained by deleting the vertex v from the graph G and also delet ing all the edges incident on v.
DISCRETE STRUCTURES
462 1 1 . 1 9. EDGE CONNECTIVITY
=
Let G (V. E) be a connected graph. then cardinality of cut set of G is called edge connectivity of graph G. The edge connectivity of a connected graph cannot be more than the smallest degree of a vertex in the graph. It is denoted as Jc(G)
Vertex connectivity Let G be a connected graph. Vertex connectivity of a graph is the least number of verti ces whose removal disconnects the graph. It is written as K(G) and is given by
=n1
K(G) for a complete graph with n vertices
For example, we find edge and vertex connectivity of following graphs (Figs. 1 1.46, 1 1 .47) 2
(i)
N=?I b 3
(ii)
4
Fig. 11.46
Fig. 11.47
In Fig. 1 1.47, removal of vertices 1, 2, 6 disconnects the graph while removal of any two vertices does not. o .
vertex connectivity is 3.
It is a 3regular graph. Only all edges incident on a vertex will disconnect it. o .
edge connectivity is also 3.
In Fig. 1 1 .48, edge and vertex connectivity is 4.
Theorem V. Prove that a simple graph with kcomponents and n vertices can have at th e most 0f (n k) (n k + 1) edges. (P.T.V., M.e.A. Dec. 2006) 
nl '
Proof.

2 Let the number of vertices in each of the kcomponents of a simple graph G be
n2, . . . . . . , nk" Then
k
L ni = n , where ni ;:::: 1 i= l
We know that maximum number of edge in the i component of G
Maximum number of edges in a graph less G
k
=� L.,;
i= l
n! (n!
= ni(ni2  1)
 1)
2
... (1)
GRAPHS
463
Now) we prove that
k
L ni2 S n2  (k  1)(2n  k) i=l k
L n; = n i=l
Since
k
L i=l
=> => =>
==>
k
L ni2  k = n  k i=l
(ni  1) = n  k, squaring both sides
[(n,  1) + (n2  1) + ...... + (nk  1)] 2 = (n  k) 2 (n,  1) 2 + (n2  1) 2 + ...... + (nk  1)2 + 2[n ,  1 ) (n2  1) + ...... + (nk  1)] = (n  k) 2 n , 2 + n22 + ...... + nk2  2(n , + n2 + ...... + nk) + k + Nonnegative terms = (n _ k) 2 k k 2 L ni  2 L ni + k + Nonnegative terms = (n  k)2
i=l
i=l
k
L n;  2n + k + Nonnegative terms = n2 + k2  2nk i=l k
L n; + Nonnegative terms = n2 + k2  2kn + 2n  k i=l = n2  2n(k  1)(2n  k)  n k
L nf S n2  (k  1)(2n  k) i=l From (1), we have Maximum number of edges in the graph G.
=
�
[n2  (k 
1)(2n  k)  n]
1 2 = [n  2nk + k2 + n  k] 2
= .2!.[(n  k)2 + (n  k)] = .2!.(n  k)(n  k + 1) Hence the theorem. Example
points.
14. Give an example of a graph with six vertices that has exactly two cut
Sol. The graph with exactly two cut points is shown in Fig. 1 1 .48.
DISCRETE STRUCTURES
464
b
a c e
Fig. 11.48 The two cut points in this graph are c and d. The other vertices are not cut points since
removal of them does not divide the graph into more than one connected component.
Example 15. Give an example of a graph with six vertices that has no cut points. Sol. The graph with no cut points is shown in Fig. 1 1.49. This graph does not contain
any cut point since removal of any vertex and the edges incident on it does not divide it into more than one connected components.
a
b
e Fig. 11.49 Example 16. Consider the graph shown in Fig. 11.50. Determine the subgraphs (i) G  v 1 (ii) G  v3 (iii) G  v5' V, "'71 V2
3
V
"_ _ _ _ _ _ _
4
"V
Fig. 11.50
Sol. (i) The subgraph GV, is shown in Fig. 1 1.51. (ii) The sub graph Gv3 is as shown in Fig. 1 1.52. (iii) The sub graph GV5 is as shown in Fig. 1 1.53.
GRAPHS
465 v,ec" v2
v, .. v2
v,
v3 ___e v,
V3�� Fig. 1 1.51
Fig. 11.52
Fig. 11.53
Exrunple 17. Consider the graph G shown in Fig. 11.54. Determine all the cut points otG.
b
a
c 9
e �_
d
Fig. 11.54 Sol. (a) The vertex b is cut point for G. Since, Gb has more than one connected compo·
nents as shown in Fig. 1 1 .55.
(b) The vertex e is also a cut point for G. Since Ge has more than one connected compo· nents as shown in Fig. 1 1 .56.
c
a
•
c
a
•
9
9 d
•
e Fig. 11.55
d
.1 Fig. 11.56
1 1 .20. BRIDGE (Cut Edges) Consider a graph G = (V, E). A bridge for a graph G, is an edge e such that connected components than G or disconnected.
Ge has more
DISCRETE STRUCTURES
466
Example 18. Consider the graph shown in Fig. 11.57. Determine the subgraphs (iii) G  e4. (ii) G  e3 (i) G  e J e, :: '__ V, ""'::_
V3��� e3· Fig. 11.57 Sol. (i) The subgraph Ge, is shown in Fig. 1 1 .58. (ii) The Subgraph Ge3 is shown in Fig. 1 1.59. V,
e, V, "":::'e
V3 *"=;:e3fiV,
V3
Fig. 11.58
Fig. 11.59
(iii) The subgraph Ge4 is shown in Fig.
1 1 .60.
e, V, "":::'e
V3 *"=;;e3.. V, Fig. 11.60
GRAPHS
467
Example 19. Consider the graph G shown in Fig. 11.61. Determine all the bridges of G.
Fig. 11.61 Sol. (a) The edge e, is a bridge for G. Since G€, has more than one connected compo·
nents as shown in Fig. 1 1 .62.
(b) The edge e3 is a bridge for G. Since Ge3 has more than one connected components as shown in Fig. 1 1.63. v,
Fig. 11.62 Fig. 11.63 Example 20. Give an example of a graph with six vertices for the following : (a) that has exactly two bridges. (b) that has no bridges. Sol. (a) The graph that has exactly two bridges is shown in Fig. 1 1.64. The two bridges are e , and e2 •
b
c
e,
e Fig. 11.64
a
b
d
e Fig. 11.65
c
(b) The graph with no bridges is shown in Fig. 1 1 .65. This graph does not contain any bridge since removal of any edge does not divide it into more than one connected components.
DISCRETE STRUCTURES
468
Example 21. Draw a graph whose every edge is a bridge. Sol. The graph shown in Fig. 1 1 .66 is a graph whose every edge is a bridge because if any edge is removed from the graph h, we got two components or a disconnected graph. . .... V2 V, ....
V3 Fig. 11.66 1 1 .21 . ISOMORPHIC GRAPHS Two graphs G 1 and G2 are called isomorphic graphs if there is a onetoone correspond ence between their vertices and between their edges i.e., the graphs have identical represen tation except that the vertices may have different labels. Let G , = [VI E , l and G2 = [V2 E2l are two graphs. These graphs are said to be isomor' ' phic if there exists a function f : G, ; G2 such that
(i) f is oneone and onto (ii) f preserves adjancies i.e., If (x, y) E E I ' Then ([(x), f(y» E E2 (iii) f preserves nonadjancies i.e., If (x, y) 'l E I ' Then ([(x), f(y) 'l E2 The function f is called isomorphism between G, and G2 . Theorem VI. lff is an isomorphism ofgraphs G1 and G2, then, for any vertex v in Gl' the degrees of v and f(v) are equal. Proof. Let deg(v) = m : we can find exactly m vertices vp v2) ... , vm adjacent to v. Since fis an isomorphism, f(v ,), f(v) , ... f(v,,) are adjacent to f(v). Also as there is no other vertex adjacent to the vertex v in G F there is no other vertex adjacent to f(v) in G2 .: deg f(v) = m. Hence the Theorem. For example, consider the following graphs shown in Fig. 1 1 .67 and Fig. 1 1 .68. They are
isomorphic graphs.(use above theorem)
V, �� V2
V, i��� V3�� V4 Fig. 11.67
V3 ____Fig. 11.68
GRAPHS
469
Example 22. Show that the graphs shown in Fig. 1 1 . 69 and Fig. 1 1 . 70 are isomorphic. a.__.b
.��'V2
c
V3
e *�d Fig. 11.69
v, Fig. 11.70
Sol. Compare the degrees of vertices of two graphs and find the vertices from both the graphs having same degrees and make the pairs of the vertices in decreasing order of degree. If both the graphs contain vertices having same degree) then they are isomorphic otherwise not. The pairs of vertices in decreasing order of degree are as follows : d(a) H d(v) , d(d) H d(v ) , d(b) H d(v ,), d(e) H d(v4), d(c) H d(v5).
Since) both the graphs contain vertices having same degrees) hence they are isomorphic.
Example 23. Show that the graphs shown in Fig. 11.71 and Fig. 11.72 are not isomorphic. a�" b k __
+

�· d Fig. 11.71
�m

n
Fig. 11.72
Sol. The graphs are not isomorphic because the vertices of the graph shown in Fig. 1 1 .72 is having degree 3 but the graph shown in Fig. 1 1 .71 contains two vertices having degree less than three.
Example 23. (a) List any five properties of a graph which are invariant under graph isomorphism. (P.T.V., B.Tech. May 20 1 3) Sol. The five properties are as follows: 1. Order : The number of vertices. 2. Size : The number of edges 3. Vertex Connectivity: The smallest number of vertices whose removal disconnects
the graph.
4. Edge Connectivity: The smallest number of edges whose removal disconnects the
graph. edges.
5. Vertex Covering Number:
The minimal number of vertices needed to cover all
6. Edge Covering Number: The minimal number of edges needed to cover all vertices.
470
DISCRETE STRUCTURES
1 1 .22. ORDER AND SIZE OF GRAPH Let G be a graph. The number of vertices in a graph G is called order of G. The number of edges in a graph G is called size of G. For example, Consider the graph G shown in Fig. 1 1 .70 Here order of G = 3 size of G = 4 ( : number of edges in G = 4.)
1 1 .23. HOMEOMORPHIC GRAPHS
b '_,,_' C Fig. 11.73
Two graphs G , and G2 are called homeomorphic graphs if G2 can be obtained from G , by a sequence of subdivisions of the edges of G 1 , In other words) we can introduce vertices of degree two in any edge of graph G , . For example, Consider the graph shown in Fig. 1 1 .74 and Fig. 1 1 .75. They are homeomorphic graphs.
Fig. 11.74. G,.
Fig. 11.75. G2.
Example 24. Show that the graphs shown in Figs. 1 1 . 76 and 1 1 . 77 are homeomorphic.
v, ..... ... v2
V3t1_.V, Fig. 11. 76. G,.
Fig. 11. 77.
G2.
Sol.
The two graphs are homeomorphic because G , can be obtained from G2 by introducing vertices of degree 2 on edges (V I Val and (V2 , V4)· '
Example 25. Consider the directed graph G E) as shown in Fig. 1 1 . 78. Determine the vertex set and edge set of graph G. = (V,
Fig. 11.78
GRAPHS
471
Sol. The vertex and edge set of graph G = (V, E) is as follows G = {{1, 2, 3}, {(1, 2), (2, 1), (2, 2), (2, 3), (1, 3)}). Example 26. Let G = {{a, b, c, d} {(a, b), (b, c), (c, c), (d, d), (d, a))}. Draw the graph G. Sol. The graph of G = (V, E) is shown in Fig. 1 1.79.
Fig. 11.79 Example 27. Consider the directed graph shown in Fig. 11.80. Determine the indegree and outdegree of each of vertices of the graph.
}3.r
Fig. 11.80 Sol. The indegree of digraph is indeg (1) = 2, indeg (2) = 1, indeg (3) = 3 The outdegree of digraph is outdeg (1) = 3, outdeg (2) = 2, outdeg (3) = 1 . 1 1 .24. WEAKLY CONNECTED
(P.T. U., B.Tech. Dec. 2005)
A directed graph is called weakly connected if its undirected graph is connected i.e., the graph obtained after neglecting the direction.
1 1 .25. UNILATERALLY CONNECTED DIGRAPH A directed graph is called unilaterally connected if there is a directed path from any
node
u to v or viceversa) for any pair of nodes of the graph.
1 1 .26. STRONGLY CONN ECTED DIGRAPH to
v
A directed graph is called strongly connected if there is a directed path from any node u and viceversa) for any pair of nodes of the graph.
472
DISCRETE STRUCTURES
1 1 .27. DISCONNECTED DIGRAPH A directed graph is called disconnected if its undirected graph is disconnected.
Example 28. Consider the graphs shown in Figs. 11.81, 11.82 and 11.83 which of the graphs are (i) Unilaterally connected digraph (ii) Weakly connected digraph (iii) Strongly connected digraph (iv) Disconnected digraph (also find its connected components). a c b a V, �'V�� 2 d c
9
b V3__.�V4 e d h e Fig. 11.81 Fig. 11.82 Fig. 11.83 Sol. The graph shown in Fig. 1 1 .81 is strongly connected because there is a path from every vertex to v and also there is a path from v to It is also weakly and unilaterally u
u.
connected because a strongly connected digraph is both weakly and unilaterally connected. The graph shown in Fig. 1 1 .82 is weakly connected but not unilaterally connected be cause there is no directed path from vertex a to b or a to c etc. but its undirected graph is connected. The graph shown in Fig. 1 1.83 is a disconnected graph. The components of this graph are (g, a, c, f, h, b) and {e, d}.
1 1 .28. DIRECTED COMPLETE GRAPH A directed complete graph G = (V, E) on n vertices is a graph in which each vertex is connected to every other vertex by an arrow. It is denoted by Kn'
Example 29. Draw directed complete graphs K3 and K5. Sol. Place the number of vertices at appropriate place and then draw an arrow from each vertex to every other vertex as shown in Figs. 1 1 .84 and 1 1 .85. v,
V2 V3 Fig. 11.84. K3.
Fig. 11.85. K5.
GRAPHS
473 TEST YOUR KNOWLEDGE 1 1 . 1
1.
If V = {I, 2, 3, 4, 5} and E {(I, 2), (2, 3), (3, 3), (3, 4), (4, 5)). Find the number of edges and size of graph G (V, E) (b) Find the order and size of the graph G shown in the figure below :
(a)
=
=
d
0 (ii)
(i)
a
2.
3.
4.
5.
a
b
C
b d
What is the difference between directed and undirected graph? (P.T. U., B. Tech. May 2007) Differentiate between paths and circuits. graph G has 16 edges and all vertices of G are of degree 2. Find the number of vertices. (b) A graph G has 21 edges, 3 vertices of degree 4 and other vertices are of degree 3. Find the number of vertices in G. (c) A graph G has 5 vertices, 2 of degree 3 and 3 of degree 2. Find the number of edges. (a) How many nodes (vertices) are required to construct a graph with exactly 6 edges in which each node is of degree 2? (b) Show that there does not exist a graph with 5 vertices with degrees 1, 3, 4, 2, 3 respectively. (c) Can there be a graph with 8 vertices and 29 edges? (d) How many vertices are there is a graph with 10 edges if each vertex has degree 2? (e) Does there exist a graph with two vertices each of degree 4? If so, draw it. (a) Draw a simple graph with 3 vertices (b) Draw a simple graph with 4 vertices (c) Give an example for (ii) nonsimple graph (iii) Multigraph, with suitable diagrams (i) simple graph Show that the maximum number of edges in a graph with n vertices and no multiple edges are (c) (d) (a) A
n(n  1)
2
6. Prove Handshaking theorem which states that the sum of degree of the vertices of a graph is 7.
8.
equal to twice the number of edges. (a) Determine whether it is possible to construct a graph with 12 edges such that 2 of the vertices have degree 3 and the remaining vertices have degree 4. (b) Give an example of each of multigraph, weighted graph, simple graph, nonsimple graph, directed graph with suitable diagrams. (P.T. U., B.Tech. May 2005) (a) Which of the graphs in the given figures are isomorphic ? (i)
_ _ _ _ _ _ _ _ _
(ii) 17
474
DISCRETE STRUCTURES
(iv)
(v)
(vi)
(viii)
(ix)
(x)
(b) Show that the following graphs are not isomorphic. b
b
af;lp c d
e
Fig. I
Fig. II
GRAPHS (c)
475
Determine whether the following graphs are isomorphic or not. a
b
d
c
§
s
t
v
u
§
Fig. I Fig. II (d) Determine whether the following graphs are isomorphic or not U,
U
/ U4
9.
/
2
U3
U,
U
U3
2
U
6
U5
u4
Fig. I Fig. II (a) Draw the graph IT (complement of G) of the graph shown below. Also show that G and IT are isomorphic. (P.T. U., B.Tech. Dec. 2002) e
B f'I D A
10. 11. 12.
c
d
E
a
b
Fig. I Fig. II (b) Draw the complement of the graph shown in the fig. II Draw (a) a graph in which no edge is a cut edge. (b) a graph in which every edge is a cut edge. (c) only one cut vertex. Find k, if a kregular graph with 8 vertices has 12 edges. Also draw kregular graph. Consider the graph G shown below:
Is G simple? (b) What is order and size of incidence matrix for G? (c) Find minimum and maximum degree for G. (a)
476
DISCRETE STRUCTURES
13. 14.
Prove that in a simple graph with n vertices, each vertex has maximum degree (n  1). Prove that maximum degree of edges in a graph G with n vertices and no multiple edges are
15.
Suppose a directed graph has vertices. Show that if there is a path from vertex to v, then there is a path p' of length  1 or less from to v. Construct a graph that has six vertices and five edges but is not a tree
16.
1. 2. 3.
n(n  )
2
m
m
u
u
(P.T. U., B.Teeh. Dec. 2012)
Number of edges 4, Size of graph
(a) (a) 6 (a) 6
(d)
4.
1
Answers
=
= 3 Order = 4, Size = No 5 (b) (i) (b) 1 (c)
10.
(e)
__.b
(a) a.,
_ _ _ _ _
or
a
9.
Not possible and (vi) are isomorphic (iii) and (vii) are isomorphic (viii) and (ix) are isomorphic (d) Isomorphic (c) Not isomorphic (a) (a) (i)
(b)
d e"I\...,,. c
10. (a)
11.
[> n vertices has a Hamiltonian circuit if deg(v ) :> n/2 for each vertex ni. The graph in Fig. 1 1.139 has 5 vertices. Here deg(u,) = 2 :j> 5/2 The graph in Fig. 1 1.139 has no Hamiltonian circuit. Example 17. Consider the graph G shown in the following Fig. 11.140. (a) Find the Euler's path if it exists. (b) Is the graph G Eulerian? (c) Is the graph bipartite? (d) Is the graph hamiltonian? Sol. (a) Consider the vertices 'c' and 'd'. The degree of these ver·
Fig. 11.138
Fig. 11.139
tices is 5(odd) G has Euler's path between c and d. Fig. 11.140 The Euler's path is c, a, b, d, [, b, d, [, e, a, d, e, c, d ' (b) A graph G can have Euler s circuit if each vertices of G are of even degree. Since the vertices 'c' and 'd' are of odd degree, therefore, G cannot have Euler's circuit. (c) The graph G has odd cycle a, c, d, a it cannot be a bipartite graph (d) Yes, the Hamiltonian circuit is a, b, [, e, d, c, a. Example 18. Consider the graph G shown below in Fig. 11.141. (a) Is it a complete graph? (b) Is G connected and regular? (c) Is it a planar graph? If so, find the number of regions. (d) Is G Eulerian? Sol. (a) Since the edge between a and d is not present in the given graph. It cannot be
complete graph. (Every pair of vertices must be joined by an edge.) (b) Given graph is a 4·regular graph. Also it is connected because there is a path from every vertex to other. (c) The given graph is planar graph as it can be re·drawn as shown in Fig. 1 1.142 in which no two edges cross Here V = 6, E = 12 By Euler's theorem, V  E + R = 2 6  12 + R = 2 => => R= 8
GRAPHS
497
a
f+\'lr b e '
(i)
Let each cycle in G is even. Let some vertex) say) x is coloured red) then its adjacent vertex will have different colour) say, blue, and its adjacent vertex will have red colour because every cycle has even length. . . sequence of vertices of even cycles is RBR, RBRBR and so on. Thus only two colours are used to colour the graph. G is 2colourable. Example 4. Determine the chromatic number of the graphs shown in Fig. 1 1 . 1 72.
32
32
(a)
3,
3, (b)
Fig. 11.172
514
DISCRETE STRUCTURES
Sol. The graphs shown in Fig. l1.172(a), has the chromatic number X(G) = 2. The graph shown in Fig. l1.172(b) has the chromatic number X(G) = 2, when n is an even number and X(G) = 3, where n is odd. Theorem II. Ifan undirected graph has a subgraph K3• then its chromatic number is at least three. Proof. Let G be an undirected graph. As G contains a complete graph K3, which is 3· colourable. G cannot be coloured with one or two colours IJf(G) :> 3. Four Colour Theorem. Every planar graph is four colourable. Five Colour Theorem. Every planar graph has chromatic number S 5. . .
Theorem III. The vertices of every planar graph can be properly coloured with five colours. Proof. We will prove this theorem by induction. All the graphs with 1, 2, 3, 4 or 5
vertices can be properly coloured with five colours. Now let us assume that every planar graph with n  1 vertices can be properly coloured with five colours. Next) if we prove that any planar graph G with n vertices will require no more than five colours) we have done. Consider the planar graph G with n vertices. Since G is planar) it must have at least one vertex with degree five or less as shown in theorem V. Assume this vertex to be 'u'. Let G, be a graph of n  1 vertices obtained from G by deleting vertex 'u'. The G, graph requires no more than five colours (Induction hypothesis). Consider that the vertices in G, have been properly coloured and now add to it 'u' and all the edges incident on u. If the degree of u is 1, 2, 3, or 4, a proper colour to u can be easily assigned. Now, we have one case left, in which the degree of u is 5, and all the 5 colours have been used in colouring the vertices adjacent to u, as shown in Fig. 11.173. Colour 1
V,
V o
Colour 2
u Colour 5 V3
V2 Colour 3
Colour 4
Fig. 11.173
Suppose that there is a path in G, between vertices Vo and v3 coloured alternately with colours 1 and 4 as shown in Fig. 11.174.
515
GRAPHS
Colour 4
V o
Colour 1 Colour 2
v4
Colour 5
u
v2
Colour 3
Colour 1
Colour 4
Colour 4
V3
Colour 1
Fig. 11.174
Then a similar path between v, and v" coloured alternately with colours 5 and 3, can not exist ; otherwise, these two paths will intersect and cause G to be nonplanar. Thus, ifthere is no path between v, and v, coloured alternately with colour 5 and 3 of all vertices connected to v, through vertices of alternating colours 5 and 3. This interchange will colour vertex v, with colour 5 and yet keep G, properly coloured. As vertex v, is still with colour 5, the colour 3 is left over with which to colour vertex u which proves the theorem.
Example 5_ Consider the following graphs
Fig. 11.175.
Fig. 11.176
(a) which graph(s) are bipartite graphs ? Colour the vertices of the bipartite graph. (b) If graph is not bipartite,justify your answer.
SoL Consider the graph shown in Fig. 1 1 . 177 Let V, = [R, R, R], V, = [E, E, E]
Let V be the set of vertices of the given graph such that V = V, U V, and V, " V, =
Hence V can be partitioned into two sets V, of red colours and V, of blue colours.
516
DISCRETE STRUCTURES
..
This graph is a bipartite graph . R
B R
R
B Fig. 11.177
R
B
B
R
R
B B Fig. 11.178 R
(b) However, the graph shown in Fig. 1 1. 178 is not a bipartite graph as it is not 2· colourable. If we try to label the vertices using two colours red (R) and Blue (B), we get the graph (Fig 1 1 . 1 78) with adjacent vertices of same colour.
Example 6. Consider the graph (Fig. 11.1 79) G shown below. (a) Find the shortest and longest simple path between A and F (b) What is the diameter of the graph ? (c) Is there Euler's path in G? (d) Is G planar and connected? (e) Find chromatic number of G. Sol. (a) The shortest path between A and F is A, D, E, F. It
is of length 3. The longest path between A and F is A, D, E, B, C, F. It is of length 6. (b) We know that the diameter of a graph is the maximum distance between any two vertices. :. diameter (G) = 3 (c) Since the vertices B and C are of odd degree. :. G has an Euler path between B and C. (d) Since no two edges in the graph intersect and there is a path between each pair of vertices. :. G is connected and planar (e) Consider the subgraph EBC(K,) :. The chromatic number of the graph is 2: 3 But three colours are sufficient to paint the vertices property (see Fig. 11.180). A(Red), B(Blue), C(Red), D(Green), E(Green), F(Blue) :. required chromatic number = 3
B C
E
Fig. 11.179 B (B)
E (G)
Fig. 11.180
GRAPHS
517
Example 7. Write any three applications of colouring ofgraph.
(P.T.V. B.Tech. Dec. 2009)
Sol. (i) Scheduling. Vertex coloring models to a number of scheduling problems. In
the cleanest form, a given set of jobs need to be assigned to time slots, each job requires one such slot. Jobs can be scheduled in any order, but pairs of jobs may be in conflict in the sense that they may not be assigned to the same time slot, because they both rely on a shared resource. The corresponding graph contains a vertex for every job and an edge for every conflicting pair of jobs. The chromatic number of the graph is exactly the minimum makespan, the optimal time to finish all jobs without conflicts. Details of the scheduling problem define the structure of the graph. For example, when assigning aircrafts in band with allocation to radio stations, the resulting conflict graph is a unit disk graph, so the coloring problem is 3·approximable. (ii) Register allocation. A compiler is a computer program that translates one computer language into another. To improve the execution time of the resulting code, one of the techniques of compiler optimization is register allocation, where the most frequently used values of the compiled program are kept in the fast processor register. Ideally, values are assigned to registers so that they can all reside in the registers when they are used. The textbook approach to this problem is to model it as a graph coloring problem. The compiler constructs an interference graph, were vertices are symbolic registers and an edge connects two nodes if they are needed at the same time. If the graph can be colored with k color, then the variables can be stored in k registers. (iii) Other applications. The problem of coloring a graph has found a number of applications, including pattern matching. The recreational puzzle Sudoku can be seen as com pleting a 9·coloring on given specific graph with 81 vertices. 1 1 .53. APPLICATIONS OF GRAPH THEORY 1 1 .53 . 1 . Shortest Path in Weighted Graphs
(P.T. U., B.Tech. May 2007)
Weighted graphs can be used to represent highways connecting the different cities. The weighted edges represent the distance between different cities and the vertices represent the cities. A common problem with this type of graph is to find the shortest path from one city to another city. There are many ways to tackle this problem one of which is as follows : Shortest Paths from Single Source. We will find shortest paths from a single vertex to all other vertices of the graph. The first algorithm was proposed by E. Dijkstra in 1959. Some common terms related with this algorithm are as follows : Path Length. The length of a path is the sum of the weights of the edges on that path. Source. The starting vertex of the graph from which we have to start to find the short· est path. Destination. The terminal or last vertex upto which we have to find the path. 1 1 .54. DIJKSTRA'S ALGORITHM FOR SHORTEST PATH
(P.T. U., M.G.A. Dec. 2005)
This algorithm maintains a set of vertices whose shortest path from source is already known. The graph is represented by its cost adjacency matrix, where cost being the weight of the edge. In the cost adjacency matrix of the graph, all the diagonal values are zero. If there is no path from source vertex Vs to any other vertex Vi' then it is represented by + In this algorithm, we have assumed all weights are positive. 1. Initially there is no vertex in sets. 2. Include the source vertex V, in S. Determine all the paths from V, to all other vertices without going through any other vertex. 00.
518
DISCRETE STRUCTURES
3. Now, include that vertex in S which is nearest to V, and find shortest paths to all the vertices through this vertex and update the values. 4. Repeat the step 3 until n  1 vertices are not included in S if there are n vertices in the graph. After completion of the process, we get the shortest paths to all the vertices from the source vertex. Example 8. Find the shortest path between K and L in the graph shown in Fig. 11. 181 by using Dijkstra's Algorithm. a
7
b
K�+����� L
c
6
d
Fig. 11.181 Sol. Step I. Include the vertex K in S and determine all the direct paths from K to all other vertices without going through any other vertex. Distance to all other vertices a d b
S
K
L K o 4(K) 2(K) 20(K) Step II. Include the vertex in S which is nearest to K and determine shortest paths to all vertices through this vertex and update the values. The nearest vertex is c
c.
Distance to all other vertices a d b
S
K L 2(K) 8(K, ) 18(K, ) 7(K, ) K, 0 3(K, ) Step III. The vertex which is 2nd nearest to K is a, included in S. c
S
K o
c
c
c
c
c
Distance to all other vertices a b d
3(K, ) c
7(K, ) c
c
2(K)
7(K,
c,
Step IV. The vertex which is 3rd nearest to K is b, is included in S. Distance to all other vertices S b K a d c
a)
3(K, ) 7(K, ) 7(K, a) 2(K) Step V. The vertex which is next nearest to K is d, is included in S. o
S
K K, a, b, d c,
o
c
c
c,
Distance to all other vertices b a d
3(K, ) c
7(K, ) c
c
2(K)
7(K,
c,
a)
L 18(K, ) c
L 8(K, b) c,
L 8(K, b). c,
GRAPHS
519
Since) n 1 vertices included in S. Hence we have found the shortest distance from K to all other vertices. Thus, the shortest distance between K and L is 8 and the shortest path is K, b, L. 
c,
Example 9. Find the shortest path between a and z in the graph shown in Fig. b
2 a
1 1 . 1 82.
e
3
2 z
c 4
2 d
3 4
Fig. 11.182 Sol. Step I. Include the vertex a in S and determine all the direct paths from a to all other vertices without going through any other vertices. Distance to all other vertices d e f z a b 2 (a) o l(a) 4(a) Step II. Include the vertex in S which is nearest to a and determine shortest path to all S a
c
the vertices through this vertex. The nearest vertex is S a,
c
a o
c.
Distance to all other vertices d e b 2 (a) l(a) 3(a, ) 6(a, ) c
c
c
f 8(a, ) c
z
Step III. Include the vertex in S which is 2nd nearest to S and determine shortest path to all the vertices through this vertex. The 2nd nearest vertex is b. Distance to all other vertices a d e b b 0 2 (a) 5(a, b) l(a) 3(a, ) Step IV. Next vertex included in S is d. S Distance to all other vertices a d e a, b, d b 0 2 (a) 3(a, ) 5(a, b) l(a) Step V. Next vertex included in S is e. S Distance to all other vertices a d e a, b, d, e b 0 2 (a) 1 (a) 5(a, b) 3(a, ) Step VI. Next vertex included in S is f. S a,
c,
c,
c,
c
c
c
c
c
c
f 8(a, ) c
f 7(a, ) c
f 7(a, ) c
z
=
z
=
z
6(a, b, e)
520
DISCRETE STRUCTURES
S a, c, b, d, e, f
Distance to all over vertices d e f z b c 5(a, b) 7(a, c) 6(a, b, e). 2(a) 1 (a) 3(a, c) o This ends our procedure as n 1 vertices are included in S. Thus) the shortest distance between a and z is 6 and the shortest path is a, b, e, z. Example 10. Using either breadth first search algorithm or Dijkstra's algorithm, find the shortest path from s to t in the following weighted graph. (Fig. 11.183) a

14 23
s oo __ __ � �
__
20��
Fig. 11.183 Sol. Let S be the set of vertices of the given weighted graph. i.e., S = {s, a, b, c, d, t} as shown in the Fig. 11.184.
Fig. 11.184 Step I. Include the vertex s in S and determine all the direct paths from s to all other vertices without going through any other vertex. Distance to all other vertices s a d b c 8 23(8) o 20(s) Step II. Include the vertex in S which is nearest to s and determine shortest paths to S
all vertices through this vertex and update the values. The nearest vertex is C.
Distance to all other vertices s a d b c 39 (s,c) 23(s) s, c o 20(s) 42(s, c) Step III. The vertex which is 2nd nearest to 8 is a. Include this vertex in S S
Distance to all other vertices d s a b c a 23(s) 37(s, a) o 20(s) 42(s, c) Step IV. The vertex which is 3rd nearest to s is b. Include this vertex in S. S
S, C,
S s S,
c, a, b
o
Distance to all other vertices a d b c 23(8) 37(s, a) 20(s) 40(8, a) 63(s, a, b)
GRAPHS
521
Step V. The vertex which is next nearest to s is (d). Include this vertex in S. S
Distance to all other vertices d a b c 23(s) 37(s, a) 40(s, a) 63(s, a, b) 20(s)
s S,
c, a, b, d Since n  1
o
= S vertices are included in S. Hence we have found the shortest distance from s to all other vertices. Thus, the shortest distance between s and t is 63 and the shortest path is s, a, b, t.
Example 11. Show that e :> 3 V  6 for the connected planar graphs shown in Figs. 1 1 .185 and 11. 186.
Fig.
Fig.
11.185
11.186
Sol. (i) The graph shown in Fig. 11.185 contains vertices V = 8 and edges e = 17. Putting the values we have e = 3 x 8  6 = 18 :> 17. Hence proved. (ii) The graph shown in Fig. 11.186 contains vertices V = 5 and edges e = 6. Putting the values, we have 3 x 5  6 = 11 > 6. Hence proved.
L
I
TEST YOUR KNOWLEDGE 1 1 .3
Find the chromatic numbers of the following graphs. (a)
(b) 0
* ��
(d)
(g) Complete bipartite graph KS,4
(P.T.U., B.Tech. May 2013)
522 2.
DISCRETE STRUCTURES
Find the shortest path, by using either Breadth first search or Dijkstra's algorithm, from P to Q in the following weighted graph. Q
3.
Find the chromatic number of the following graphs.
4. Find the shortest path and its length from to by using Dijkstra's algorithm in the following graph. t
s
(a)
A
C
(b)
a
3
6
D
4
F
4
7
b 2
s
3
e
6 3
c
d
4
Answers L 3.
(a) 3 (b) 3 (c) 3 (d) 3 (e) 2 (j)
2 (b)
r
C

d

t.
4 (g) 2
2.
P Aj A2 A, A3 A6 Q.
4. (a)
s

A

D E H 


t.
length = 13
GRAPHS
523
MULTI PLE CHOICE QUESTIONS (MCQs)
1. Which of the following statement is FALSE about undirected graphs? (a) The sum of degrees of all the vertices in a graph is even. (b) There is an even number of vertices of odd degree. (c) The degree of a vertex is the number of edges incident on it. (d) The self loop is counted once, when degree is counted. 2. How many edges do a complete graph contains having n vertices? (a) 2n (b) 2n/2 (c) n(n  1)/2 (d) Any number of edges. 3. Which of the following is FALSE statement about planar graphs? (a) A complete graph of five or more vertices is not a planar graph. (b) A complete bipartite graph having m :> 3 and n :> 3 is a planar graph. (c) Every planar graph has at least one vertex of degree 5 or less than 5. (d) Every planar graph having edges and v vertices has 3v  e :> 6. 4. The chromatic number of a complete graph Kn is (a) n  1 (b) n (c) 2 (d) Any number. Which of the following bipartite graph has Hamiltonian circuit? 5. (a) K" 2 (b) K2, 3 (d) K3, (c) K3, 3 6. Which of the following statement about the directed graphs is not TRUE? (a) The in·degree and the out·degree of the directed graphs are equal. (b) The in·degree and out·degree is not same as the number of edges. (c) The sum of in·degree and out·degree is even. (d) The self loop has one in·degree and one out·degree in the directed graph. 7. Suppose you run Dijkstra's single source shortest path algorithm on the following weighted directed graph with vertex 0 as the source vertex e
5
In what order do the nodes get included into the set of vertices for which the shortest path distances are finalized? (a) 0, 1, 2, 3, 4 (c) 0, 4, 3, 2, 1
(b) 0, 4, 1, 3, 2 (d) 0, 1, 4, 3, 2 8. How many edges are there in a graph with 20 vertices and the sum of the degrees (in· degree and out·degree) is 100? (a) 50 (b) 100 (c) 20 (d) 40
524
DISCRETE STRUCTURES
9. If G is a directed graph with 10 vertices, how many Boolean values will be needed to
represent G using an adjacency matrix?
(a) 50 (c) 200
(b) 100
(d) 1000 10. Which of the following is Not TRUE about the directed graphs? (a) If a digraph is reflexive, then the diagonal elements of the adjacency matrix are 1. (b) If G is a simple digraph whose adjacency matrix is A then the adjacency matrix of GC, is the transpose of A. (c) The diagonal elements of A . AT show the out degree of the vertices. (d) The adjacency matrix of a directed graph is a symmetric matrix. Answers and Explanation
1. (d) The self loop is counted twice, when degree is counted. 2. (c) The number of edges in a complete graph is n(n  1)/2 3. (b) A complete bipartite graph having m :> 3 and n :> 3 is not a planar graph. 4. (b) The chromatic number of a complete graph of n vertices is n. 5. (c) K3• 3 has Hamiltonian circuit. 6. (b) The in·degree and out·degree is same as the number of edges. 7. (b) Apply the algorithm and it will be 0, 4, 1, 3, 2 8. (a) One edge contributes two degree. 9. (b) The adjacency matrix is of N x N dimension, where n is the number of vertices. 10. (d) The adjacency matrix of a directed graph is not a symmetric matrix.
12
TREES
1 2. 1 . INTRODUCTION
In this chapter. we will discuss a special class of graphs. called trees. The concept of trees is frequently used in both mathematics and sciences. To understand the concept of trees) it is essential to know the various common types of trees. Their basic properties and applica tions. 1 2.2. TREE A graph which has no cycle is called an acyclic graph. A tree is an acyclic graph or graph having no cycles. A tree or general tree is defined as a nonempty finite set of elements called vertices or nodes having the property that each node can have minimum degree 1 and maximum degree n. It can be partitioned into n + 1 disjoint subsets such that the first subset contains the root of the tree and the remaining n subsets contains the elements of the n subtree. (Fig. 12.1)
Fig. 12.1. General Tree. 1 2.3. DIRECTED TREES A directed tree is an acyclic directed graph. It has one node with indegree 1, while all other nodes have indegree 1 as shown in Figs. 12.2 and 12.3.
525
526
DISCRETE STRUCTURES
Directed trees
Fig.
12.2
Fig.
12.3
The node which has outdegree 0 is called an external node or a terminal node or a leaf. The nodes which has outdegree greater than or equal to one are called internal nodes or branch nodes. 1 2.4. ORDERED TREES
tree. e.g.,
If in a tree at each level) an ordering is defined, then such a tree is called an ordered the trees shown in Figs. 12.4 and 12.5 represent the same tree but have different orders.
Fig.
12.4
Fig.
12.5
TREES
527
1 2.5. ROOTED TREES
If a directed tree has exactly one node or vertex called root whose incoming degree is 0 and all other vertices have incoming degree one, then the tree is called rooted tree. * A tree with no nodes is a rooted tree (the empty tree). * A single node with no children is a rooted tree.
Example. Suppose 8 people enter a Badminton tournament use a rooted tree model of the tournament to determine how many games must be played to determine a champion if a player is eliminated after one loss. (P.T.V. B.Tech. May 2009) Sol. As there are 8 people in the badminton tournament, there will be four games to be
played in the first round two games to be played in the second round, one game to be played in the final round Hence, total number of games in the tournament is 7. (See Fig. 12.6).
2
3
4
5
6
7
8 Players
Fig. 12.6
1 2.6. PATH LENGTH OF A VERTEX
The path length of a vertex in a rooted tree is defined to be the number of edges in the path from the root to the vertex. For example, we find the path lengths of the nodes b, f, I, q in Fig. 12. 7a.
Fig. 12.7a
tree T.
The path length of node b is one. The path length of node f is two. The path length of node I is three. The path length of node q is four.
Fig. 12.7b
Theorem I. Prove that there is one and only one path between every pair of vertices in a
528
DISCRETE STRUCTURES
Proof. We know that T is a connected graph, in which there must exist at least one path between every pair of vertices. Now assume that there exists two different paths from some node a to some node b of T. The union of these two paths will contain a cycle and therefore T cannot be a tree. Hence, there is only one path between every pair of vertices in a tree. 1 2.7. FOREST
If the root and the corresponding edges connecting the nodes are deleted from a tree, we obtain a set of disjoint trees. This set of disjoint trees is called a forest. (Fig. 12.7b) 1 2.8. BINARY TREE
If the outdegree of every node is less than or equal to 2, in a directed tree then the tree is called a binary tree. A tree consisting of no nodes (empty tree) is also a binary tree. 1 2.9. BASIC TERMINOLOGY
(a) Root. A binary tree has a unique node called the root of the tree. (b) Left Child. The node to the left of the root is called its left child. (c) Right Child. The node to the right of the root is called its right child. (d) Parent. A node having left child or right child or both is called parent of the nodes. (e) Siblings. Two nodes having the same parent are called siblings. if) Leaf. A node with no children is called a leaf. The number of leaves in a binary tree
can vary from one (minimum) to half the number of vertices (maximum) in a tree. (g) Ancestor. If a node is the parent of another node, then it is called ancestor of that node. The root is an ancestor of every other node in the tree. (h) Descendent. A node is called descendent of another node if it is the child of the node or child of some other descendent of that node. All the nodes in the tree are descendents of the root. (i) Left Subtree. The subtree whose root is the left child of some node is called the left subtree of that node. (j) Right Subtree. The subtree whose root is the right child of some node is called the right subtree of that node. (k) Level of a Node. The level of a node is its distance from the root. The level of root is defined as zero. The level of all other nodes is one more than its parent node. The maximum number of nodes at any level N is 2N (l) Depth or Height of a Tree. The depth or height of a tree is defined as the maximum number of nodes in a branch of tree. This is one more than the maximum level of the tree i.e., the depth of root is one. The maximum number of nodes in a binary tree of depth d is 2d  1, where d :> 1. (m) External Nodes. The nodes which has no children are called external nodes or terminal nodes. (n) Internal Nodes. The nodes which has one or more than one children are called internal nodes or nonterminal nodes. lent :
Theorem II. Let G be a graph with more than one vertex. Then the following are equiva
(i) G is a tree. (ii) Each pair of vertices is connected by exactly one simple path. (iii) G is connected, but if any edge is deleted then the resulting graph is not connected.
TREES
529
(iv) G is cycle tree, but if any edge is added to the graph then the resulting graph has exactly one cycle. Proof. To prove this theorem, we prove that (i) => (ii), (ii) => (iii), (iii) => (iv) and finally (iv) => (i). The complete proof is as follows : (i) ::::::} (ii) Let us assume two vertices u and v in G. Since G is a tree, so G is connected and there is at least one path between u and v. More over, there can be only one path between u and v, otherwise G will contain a cycle. (ii) => (iii) Let us delete an edge e = (u, v) from G. It means e is a path from u to v. Suppose the graph result from G e has a path p from u to v. Then P and e are two distinct paths from u to v, which is a contradiction of our assumption. Thus, there does not exist a path between u and v in G e, so G e is disconnected. (iii) => (iv) Let us suppose that G contains a cycle c which contains an edge e = {u, v}. By hypothesis, G is connected but G' = G e is disconnected with u and v belonging to different components of G'. This contradicts the fact that u and v are connected by the path P = C e, 




which lies in Q'. Hence G is cycle free. p Now, Let us take two vertices x and y of G and let H be the graph obtained by adjoining the edge e = {x, y} to G. Since G is connected, there c is a path P from x to y in G ; hence C = Pe forms a cycle in H. Now suppose H contains another cycle C1 0 Since G is cycle free, C1 must con C, tain the edge e, say C, = P,e. Then P and P, are two paths in G from x to y as shown in Fig. 1. p' Thus, G contains a cycle, which contradicts the fact that G is cycle free. Fig. I Hence H contains only one cycle. (iv) => (i) By adding any edge C = (x, y) to G produces a cycle, the vertices x and y must be connected already in G. Thus, G is connected and is cycle is free i.e., G is a tree. Theorem III. Let G be a finite graph with n > 1 vertices. Then the following are equivalent : (i) G is a tree. (ii) G is cycle free and has n 1 edges. (iii) G is connected and has n 1 edges. To prove this. Proof. We use induction on the number of vertices n of G. Let us assume n = 1 i.e., G has only one vertex. Then G has 0 edges and so G is con· nected and cycle free. Thus the theorem holds for n = 1. Now, assume that n > 1 i.e., G has more than one vertex. Assume that (i), (ii) and (iii) are equivalent for all graphs with less than n vertices. 

We have to show that they are equivalent for G. (i) => (ii) Suppose G is a tree. Then G is cycle free, so we have to show only that G has n 1 edges. We know that G has a vertex of degree 1. Deleting this vertex and its edge, we obtain a tree T which has n 1 vertices. Thus the theorem holds for T, so T has n 2 edges. Hence G has n 1 edges. (ii) => (iii) Suppose G is cycle free and has n 1 edges. We have to show only that G is connected. Suppose G is disconnected and has k components) T T2) ... , Tk' which are trees since each is connected and cycle free, i.e., Ti has ni vertices and ni < n. Hence the theorem holds for Ti, so Ti has ni 1 edges. Thus, 





l'
530
DISCRETE STRUCTURES
n = n1 + n2 + ... + n k n 1 = (n , 1) + (n2 1) + ... + (n k 1) = n1 + n2 + ... + nk k = n k
and






Hence k = 1. But it contradicts our assumption that G is disconnected and has k > 1 components. Hence G is connected. (iii) => (i) Suppose G is connected and has n 1 edges. We have to show only that G is cycle free. Suppose G has a cycle containing an edge e. Deleting e, we obtain the graph H = G e) which is also connected. But H has n vertices and n 2 edges and therefore must be disconnected. Thus G is cycle free and hence it is a tree. 


I
ILLUSTRATIVE EXAMPLES
Example 1. For the tree as shown in Fig. 12.8. (i) Which node is the root ? (ii) Which nodes are leaves ? (iii) Name the parent node of each node. A C
Fig. 12.8 Sol. (i) The node A is the root node. (ii) The nodes G, H, I, L, M, N, 0 are leaves. Parent Nodes (iii) B, C
A
D, E
B
F
C
G, H
D
I, J K L, M N, O
E
F J K
TREES
531
Example 2. For the tree as shown in Fig.
12.9.
A c
F
Fig.
12.9
(i) List the children of each node. (ii) List the siblings. (iii) Find the depth of each node. (iv) Find the level of each node. Sol. (i) The children of each node is as follows : Children Node
A
B, C D, E F G, H
B C D E F
I, J K
K
L, M
(ii) The siblings are as follows : Siblings B and C D and E G and H I and J L and M are all siblings.
(iii)
Node
A
B, C D, E, F G) H, I, L, M
(iv)
J, K
Node
A
B, C D, E, F G, H, L, M
I, J, K
Depth or Height 1 2 3 4 5 Level o 1 2 3 4
532
DISCRETE STRUCTURES
Example 3. Show that if in a graph G there exists one and only one path between every pair of vertices, then G is a tree. Sol. The graph G is connected since there is a path between every pair of vertices. A cycle in a graph exists if there is at least one pair of vertices (vp v2) such that there exist two distinct paths from v , to v 2 • But the graph G has one and only one path between every pair of vertices. Thus) G contains no cycle. Hence) G is a tree.
Example 4. Draw two different binary trees with five nodes having only one leaf Sol. The two trees out of many possible trees with five nodes having only one leaf is
shown in Fig. 12.10.
(i)
(ii)
Fig. 12.10 Example 5. (a) Draw two different binary trees with five nodes having maximum number of leaves. (b) Let T be a tree with n vertices. Determine the number ofleafnodes in a tree. (P.T.U. B.Tech. Dec. 2008)
Sol. (a) There are many possible trees, out of which two different binary trees are shown
in Fig. 12.11.
(i)
(ii)
Fig. 12.11 n vertices.
(b) Given T is a binary tree with Therefore, the tree with n vertices has (n  1) edges. Also, if L denotes the number of leaves and I be the number of internal nodes,
then
n=L+1 But
1=
n 1

2
Using (2) in (1), we have
n=L+
graph.
n 1

2
=>
L=n
n 1

2
2n  n 2
+1
n
+1 2
Example 6. (a) How will you differentiate between a general tree and a binary tree ? (b) Define a rooted tree with an example and show how it may be viewed as directed
TREES
533
Sol. (a)
1. 2. 3.
General Tree
There is no such tree having zero nodes or an empty general tree. If some node has a child, then there is no such distinction. The trees shown in Fig. 12.12 are same, when we consider them as general trees. 2
3
2 4
1. 2. 3.
Binary Tree
There may be an empty binary tree. If some nodes has a child, then it is distinguished as a left child or a right child. The trees shown in Fig. 12.12 are distinct, when we consider them as binary trees, because in (i), 4 is right child of 2 while in (ii), 4 is left child of 2.
3
4 (ii)
(i)
Fig. 12.12
(b) Rooted tree: We first define the term 'directed tree' . A directed graph is said to be a
directed tree if it becomes a tree when the directions of the edges are ignored. For example, the Fig. 12.13 is a directed tree.
>
Directed tree �
(operator) < right operand or expression > depending upon precedence of evaluation. The expression tree is a binary tree whose root contains the operator and whose left subtree contains the left expression and right subtree contains the right expression. *
Example 7. Construct the binary expression tree for the expression (a + b) (die). Sol. The binary expression tree for the expression (a + b) * (die) is shown in Fig. 12.15. *
b Fig. 12.15
Example 8. Determine the value of expression tree shown in Fig.
Fig. 12.16
Sol. The value of expression tree is 2.
12. 1 6.
TREES
535
Example 9. Determine the value of expression tree shown in Fig. 12. 1 7.
Fig. 12.17 Sol. The value of expression tree is 7. 1 2. 1 1 . COMPLETE BINARY TREE Complete binary tree is a binary tree if all its levels, except possibly the last, have the maximum number of possible nodes as for left as possible. The depth of complete binary tree having n nodes is log2 n + 1 .
For example : The tree shown in Fig.
12.18 is a complete binary tree.
9
Fig. 12.18 1 2. 1 2. FULL BINARY TREE Full binary tree is a binary tree in which all the leaves are on the same level and every non·leaf node has two children.
Fig. 12.19
536
DISCRETE STRUCTURES
For example : The tree shown in Fig.
12.19 is a full binary tree.
Theorem I. Prove that the maximum number of nodes on level n of a binary tree is 2n, where n ;::: O. Proof. This can be proved by induction. Basis ofInduction. The only node at level n = 0 is the root node. Thus, the maximum number of nodes on level n = 0 is 2° = 1. Induction Hypothesis. Now assume that it is true for levelj, where n ?j ? O. There· fore) the maximum no. of nodes on levelj is
2'.
Induction Step. By induction hypothesis, the maximum number of nodes on levelj  1 is 21  1 . Since) we know that each node in binary tree has maximum degree 2. Therefore, the maximum number of nodes on levelj is twice the maximum number of levelj  1 . Hence, at levelj, the maximum number of nodes is = 2 . 21  1 = 21 . Hence proved. m
Theorem II. Prove that the maximum number of nodes a binary tree of depth d is 2"  1, where d ? 1 . Proof. This can b e proved by induction. Basis of Induction. The only node at depth d = 1 is the root node. Thus, the maximum number of nodes on depth d = 1 is 2 '  1 = 1. Induction Hypothesis. Now assume that it is true for depth K, d > k ? 1. Therefore,
the maximum number of nodes on depth K is 2k  1 .
Induction Step. By induction hypothesis, the maximum number of nodes on depth K  1 is 2 k  1  1. Since, we know that each node in a binary tree has maximum degree 2, therefore, the maximum number of nodes on depth d = K is twice the maximum number of nodes on depth K  1 . So, at depth d = K, the maximum number of nodes is = (2 . 2 K  1)  1 = 2K  1 + 1  1 = 2K  1 . Hence proved. Theorem III. Prove that in a binary tree, if nE is the number ofexternal nodes or leaves and nI is the no. of internal nodes, then nE = nI + 1. Proof. Let n be the total number of nodes in the tree. Then, we may have three types of
nodes in the tree.
So, we have
n E = the number of nodes having zero degree. n I = the number of nodes having two degree. n o = the number of nodes having one degree. n = nE + n] + no
... (1)
Let us assume that the number of edges of the tree is E. So, with these E edges we can connect E + 1 nodes. Hence
n=E+1
. .. (2)
Since) all edges are either from a node of degree one or from a node of degree two) therefore,
E=
no + 2n]
... (3)
537
TREES
Put this value in the eq. (2), we have
n = no + 2n] +
... (4)
1
Subtract eq. (4) from eq. (1), we get Hence proved.
nE = nI + l
1 2.13. TRAVERSING BINARY TREES Traversing means to visit all the nodes of the tree. There are three standard methods to traverse the binary trees. These are as follows :
2.
1. Preorder traversal
Postorder traversal
3. Inorder traversaL
1. Preorder traversal. The preorder traversal of a binary tree is a recursive process. The preorder traversal of a tree is (i) Visit the root of the tree. (ii) Traverse the left subtree in preorder. (iii) Traverse the right subtree in preorder. 2. Postorder traversal. The postorder traversal of a binary tree is a recursive process. The postorder traversal of a tree is
(i) Traverse the left subtree in postorder. (ii) Traverse the right subtree in postorder. (iii) Visit the root of the tree. 3. Inorder traversal. The inorder traversal of a binary tree is a recursive process. The inorder traversal of a tree is
(i) Traverse in inorder the left subtree. (ii) Visit the root of the tree. (iii) Traverse in inorder the right subtree. Example 10. Determine the preorder, postorder, and inorder traversal of the binary tree as shown in Fig. 12.20.
5 10 Fig. 12.20
538
DISCRETE STRUCTURES
Sol. The preorder, postorder and inorder traversal of the tree is as follows : Preorder Postorder
1 3
2 5
3 4
4 2
5 7
6 10
7 9
8 11
9 8
10 6
11 1
Inorder
3
2
5
4
1
7
6
9
10
8
11.
Example 1 1. Give the preorder, inorder and postorder traversals of the tree shown in Fig. 12.21.
M Fig. 12.21
Sol. Preorder
A,
B,
D,
H,
I,
E,
C,
F,
G,
L,
M
H,
D,
I,
B,
E,
A,
F,
C,
G,
L,
M
Postorder
H,
I,
D,
E,
B,
J,
J, K,
J, K,
K,
Inorder
F,
M,
L,
G,
C,
A.
1 2.14. ALGORITHMS
(a) Algorithm to Draw a Unique Binary Tree When Inorder and Preorder Traversal of the Tree is Given 1. We know that the root of the binary tree is the first node in its preorder. Draw the root of the tree. 2. To find the left child of the root node, first use the inorder traversal to find the nodes in the left subtree of the binary tree. (All the nodes that are left to the root node in the inorder traversal are the nodes of the left subtree). After that the left child of the root is obtained by selecting the first node in the preorder traversal of the left subtree. Draw the left child. 3. In the same way, use the inorder traversal to find the nodes in the right subtree of the binary tree. Then the right child is obtained by selecting the first node in the preorder traversal of the right subtree. Draw the right child. 4. Repeat the steps 2 and 3 with each new node until every node is not visited in preorder. Finally) we obtain the unique tree.
Example 12. Draw the unique binary tree when the inorder and preorder traversal is given as follows : Inorder B A D C F E J H K G I Preorder A B C D E F G H J K I.
TREES
539
Sol. We know that the root of the binary tree is the first node in preorder traversal. Now check A, in the inorder traversal, all the nodes that are left of A, are nodes of left subtree and all the nodes that are right of A, are nodes of right subtree. Read the next node in preorder and check its position against the root node, if it is left of root node, then draw it as left child, otherwise draw it as right child. Repeat the above process for each new node until all the nodes of preorder traversal are read and finally we obtain the binary tree as shown in Fig. 12.22.
Fig. 12.22 Example 13. Draw the unique binary tree when the following is given : Inorder Preorder Sol. The
d a
b b
h d
e e
first node In preorder is is shown in Fig. 12.23.
a h
f c
c
f
] g
g ]
a and hence a is the root node. Unique binary tree
h
Fig. 12.23 Example 14. Draw the unique binary tree when inorder andpreorder traversal of tree is given as follows : Preorder 10 7 2 1 9 6 4 3 8 5 Inorder 10 7 2 1 9 6 4 3 8 5
540
DISCRETE STRUCTURES 10
6
Fig. 12.24
Sol. Unique binary tree is shown in Fig. 12.24. (b) Algorithm to Draw a Unique Binary Tree When Inorder and Postorder Traversal of the Tree is Given 1. We know that the root of the binary tree is the last node in its postorder. Draw the root of the tree. 2. To find the right child of the root node, first use the inorder traversal to find the nodes in the right subtree of the binary tree. (All the nodes that are right to the root node in the inorder traversal are the nodes of the right subtree). After that the right child of the root is obtained by selecting the last node in the postorder traversal of the right subtree. Draw the right child. 3. In the same way, use the inorder traversal to find the nodes in the left subtree of the binary tree. Then the left child is obtained by selecting the last node in the postorder traversal of the left subtree. Draw the left child. 4. Repeat the steps 2 and 3 with each new node until every node is not visited in postorder. After visiting the last node) we obtain the unique tree.
Example 15. Draw the unique binary tree for the given Inorder and Postorder traversal Inorder 10 12 2 1 7 11 13 9 3 4 6 8 5 Postorder 12 10 2 7 5 3 1 6 4 13 11 9 8 Sol. We know that the root node is the last node in postorder traversal. Hence one is the root node. Now check the inorder traversal) we know that root is at the centre, hence all the nodes that are left to the root node in inorder traversal are the nodes of left subtree and all that are right to the root node are the nodes of the right subtree. Now, visit the next node from back in postorder traversal and check its position in inorder traversal, if it is on left of root then draw it as left child and if it is on right, then draw it as right child.
TREES
541
Repeat the above process for each new node and we obtain the binary tree as shown in Fig. 12.25.
3
9
13
Fig. 12.25 Example 16. Draw the binary tree when Inorder and Postorder traversal is given : m Inorder n ] o u s v r k p q m Postorder n o u v s r p I k q J. Sol. We know that the last node in Postorder is the root node. hence j is the root. Now applying the algorithm as above, we obtain the tree shown in Fig. 12.26.
Fig. 12.26 Example 17. Draw the unique binary tree when inorder andpreorder traversal oftree is given as follows : + + Inorder a d 3 b 6 c + + Preorder a d 3 b 6 c Sol. Unique binary tree is shown in Fig. 12.27. *
*
*
*
*
*
542
DISCRETE STRUCTURES •
b
Fig. 12.27
Example 18. Draw the binary expression tree, when inorder and postorder traversal of the tree is given as follows : tI + Postorder + n r m k p q Inorder n + k p I +m qlr Sol. Binary expression tree is shown in Fig. 12.28. *
*
*
•
Fig. 12.28 (c) Algorithm to Convert General Tree into the Binary Tree 1. Starting from the root node, the root of the tree is also the root of the binary tree. 2. The first child C , (from left) of the root node in the tree is the left child C, of the root node in binary tree and the sibling of the C , is the right child of C , and so on. 3. Repeat the step 2 for each new node.
Example 19. Convert the following tree as shown in Fig. 12.29 into binary tree.
Q Fig. 12.29
TREES
543
Sol. The root of the tree is the root of the binary tree. Hence A is the root of the binary tree. Now B becomes the left child of A in binary tree, C becomes the right child of B, D becomes right child of C and E becomes right child of D in the binary tree and similarly applying the algorithm we obtain the binary tree as shown in Fig. 12.30. A
Q
Fig. 12.30 Example 20. Convert the general tree as shown in Fig. 12.31 into binary tree.
Fig. 12.31 Sol. The root node 1 in general tree is the root node of the binary tree. Now applying the above algorithm we obtain the binary tree as shown in Fig. 12.32.
544
DISCRETE STRUCTURES
2
7
8
13
Fig. 12.32 Example
21. Convert the forest shown in Fig. 12.33 into binary tree.
Fig. 12.33 Sol. The root of the binary tree is the root node of the first tree (from left) and the root node of the second tree becomes the right son of the root node in binary tree and the root node
TREES
545
Fig. 12.34
of the third tree becomes the right son of the right son in binary tree. Repeat this procedure for each level and we obtain the binary tree as shown in Fig. 12.34.
1 2.15. BINARY SEARCH TREES Binary search trees has the property that the node to the left contains a smaller value than the node pointing to it and the node to the right contains a larger value than the node pointing to it. It is not necessary that a node in a 'Binary Search Tree' point to the nodes whose values immediately precede and follow it.
For example : The tree
shown in Fig. 12.35 is a binary search tree.
Fig. 12.35
1 2.16. INSERTING INTO A BINARY SEARCH TREE Consider a binary search tree T. Suppose we have given an ITEM of information to insert in T. The ITEM is inserted as a leaf in the tree. The following steps explains a procedure to insert an ITEM in the binary search tree T.
546
DISCRETE STRUCTURES
1. Compare the ITEM with the root node. 2. If ITEM > ROOT NODE, proceed to the right child and it becomes root node for the right subtree. 3. If ITEM < ROOT NODE, proceed to the left child. 4. Repeat the above steps until we meet a node which has no left and right subtree. 5. Now if the ITEM is greater than node, then the ITEM is inserted as the right child and if the ITEM is less than node, then the ITEM is inserted as the left child. Deleting in a Binary Search Tree. Consider a binary search tree T. Suppose we want to delete a given ITEM from binary search tree. To delete an ITEM from a binary search tree, we have three cases, depending upon the number of children of the deleted node. 1. Deleted Node has no Children. Deleting a node which has no children is very simple, as just replace the node with null. 2. Deleted Node has Only One Child. Replace the value of deleted node with the only child. 3. Deleted Node has Two Children. In this case, replace the deleted node with the node that is closest in the value to the deleted node. To find the closest value, we move once to the left and then to the right as far as possible. This node is called immediate predecessor. Now replace the value of deleted node with immediate predecessor and then delete the reo placed node by using case 1 or 2.
Example 22. Show the binary search tree after inserting 3, 1, 4, 6, 9, 2, 5, 7 into an initially empty binary search tree. Sol. The insertion of the above nodes in the empty binary search tree is shown in Fig. 12.36.
Insert 3 (i)
I A Insert 1 (ii)
Insert 4 (iii)
~ Insert 6 (iv)
9 Insert 9 (v)
Insert 2 (vi)
Insert 5 (vii) Fig. 12.36
Insert 7 (viii)
TREES
547
Example 23. Show the binary tree shown in Fig. 12.36 (viii) after deleting the root node. Sol. To delete the root node, first replace the root node with the closest element of the
root. For this, first move one step left and then to the right as far as possible to the node. Then delete the replaced node. The tree after deletion is shown in Fig. 12.37.
7
Fig. 12.37 Example 24. A binary search tree contains the values 1, 2, 3, 4, 5, 6, 7, 8, 9. The tree is traversed in preorder and the values are printed out. Determine the sequence of the print out values. Sol. First of all draw the binary search tree shown in Fig. 12.38. Now traverse the tree in preorder and get the output as below 1) 2) 3) 4) 5) 6) 7) 8) 9.
2 3 4 5 6 7 8 9
Fig. 12.38 Example 25. A binary search tree is generated by inserting in order thefollowing integers : 50, 15, 62, 5, 20, 58, 91, 3, 8, 37, 60, 24. Determine the number of nodes in the left subtree and right subtree of the root.
548
DISCRETE STRUCTURES
Sol. First of all draw the binary search tree as shown in Fig. 12.39.
@
I A 15
(i)
15
(ii)
62
(iii)
5 (iv)
(v)
(vi)
50
50
(vii)
(viii)
(ix)
24 (x)
(xi) Fig. 12.39
Thus, the number of nodes in left subtree of the root is 7 and right subtree of the root is 4. Example 26. Consider the binary tree as shown in Fig. 12.40. Draw the binary tree for
each of the following operations, if applied to the binary tree. (i) Delete the node V (ii) Delete the node E (iii) Delete the root node R.
Fig. 12.40
549
TREES
Sol. (i) The binary tree after deleting node V is shown in Fig. 12.41. (ii) The binary tree after deleting node E is shown in Fig. 12.42. (iii) The binary tree after deleting root node R is shown in Fig. 12.43.
T u
Fig. 12.41
Fig. 12.42
Fig. 12.43
1 2.17. SPANNING TREE Consider a connected graph G = (V, E). A spanning tree T is defined as a subgraph of G if T is a tree and T includes all the vertices of G. Example
27. Draw all the spanning trees of the graph G shown in Fig. 12.44. A
E
Fig. 12.44. Graph G. Sol. All the spanning trees of graph G is as shown in Fig. 12.45. A
E
(i)
A
E
(ii) Fig. 12.45
A
E
(iii)
550
DISCRETE STRUCTURES
1 2.18. APPLICATIONS OF TREES 1 2. 1 8 . 1 . Minimum Spanning Tree
Consider a connected weighted graph G = (V. E). A minimal spanning tree T of the graph G is a tree whose total weight is smallest among all the spanning trees of the graph G. The total weight of the spanning tree is the sum of the weights of the edges of the spanning trees. The minimum weight of the spanning tree is unique but the spanning tree may not be unique because more than one spanning tree are possible when more than one edges exist having the same weight.
Theorem IV. Prove that a simple graph is connected iff it has a spanning tree. (P.T.V. B.Tech. Dec. 2008)
Proof. First of all, suppose that a simple graph G has a spanning tree T.
The tree T contains every vertex of G. Further) there is a path in T between any two of its vertices. Since T is subgraph of G) there is a path in G between any two of its vertices. Hence) G is connected. Now) suppose that G is connected. If G is not a tree, then it must contain a simple circuit. Remove an edge from one of these simple circuits, the resulting sub graph has one fewer edge but still contains all the vertices of G and is connected. If this sub graph is not a tree, it has a simple circuit, so again remove an edge that is in simple circuit. Repeat this process untill no simple circuits remain. This is possible because there are only a finite number of edges in the graph. The process terminates when no simple circuits remain. A tree is produced since the graph is still connected as edges are removed. This is a spanning tree since it contains every vertex of G. Hence the theorem.
1 2.19. KRUSKAL'S ALGORITHM TO FIND MINIMUM SPANNING TREE This algorithm finds the minimum spanning tree T of the given connected weighted graph G. 1. Input the given connected weighted graph G with ning tree T) we want to find.
n vertices whose minimum span
2. Order all the edges of the graph G according to increasing weights. 3. Initialise T with all vertices but do not include any edge. 4. Add each of the graph G in T which does not form a cycle until n

1 edges are added.
Example 28. Determine the minimum spanning tree of the weighted graph shown in Fig. 12.46. A B C 6 5 3
4
D
6
2 E Fig. 12.46
4
5
F
551
TREES
Sol. Using KruskaY s algorithm, arrange all the edges of the weighted graph in increasing order and initialise spanning tree T with all the six vertices of G. Now start adding the edges of G in T which do not form a cycle and having minimum weights until five edges are not added as there are six vertices. (Fig. 12.47). Edges
Weights
(B, E) (C, D)
2 3
(A, D) (C, F)
4 4 5 5 6
(B, C) (E, F) (A, B)
Added Added Added Added Added Not added Not added
(A, F)
A
B
5
c
3 4
4 2 D
E
Fig. 12.47
Not added
6 7
(D, E)
Minimum Spanning Tree
Added or Not
Not added.
F
Example 29. Write a shortnote on Prim's and Kruskal's algorithms and execute them (P.T.U. B.Tech. May 2008) by giving a suitable example. Sol. Prim's Algorithm. Let R be a symmetric and connected relation with n vertices.
The Prime's algorithm involves the following steps. Step I. Choose a vertex v, of R. Let V = {v,} and E = { } Step II. Choose a nearest neighbour Vi of V which is adjacent to Vj where Vi' Vj E V and for which the edge (V i ' V) does not form a cycle with members of E. Add V to V and (V ' V) to E. i i Step III. Repeat the step II until we get E = n 1 Then V contains all n vertices of R and E contains the edge of a minimum spanning tree for R. 
Example
30.
We /ind the minimal spanning tree for the graphR shown below (Fig. 12. 48).
B C
3 c
B
A
D
D
4 (i)
F
E
F
( i,)
Fig. 12.48
Sol. This graph R [Fig. 12.48(i)] has 6 vertices, namely, A, B, C, D, E, F. Therefore, any spanning tree of R will have 5 edges. By Prim's algorithm, the edges are ordered by decreasing lengths and are successively deleted (without disconnecting R) until we have five edges remain. This gives the following data.
552
DISCRETE STRUCTURES
Edges Length Deleted edges
AF
BC
9
AC
8
./
7
./
./
BE
7
BF 5
CE
6
X
X
F 4
AE
D
4
./
X
BD
3
X
X
Hence the minimal spanning tree of R will contains the edges {BE, CE, AE, DF, BD}. This spaning tree has length 24 as shown in Fig. 12.48(ii) . Kruskal's algorithm. Let R be a symmetric and connected relation with n vertices and let S = [e" e2 , ... , ek] be the set of weighted edges ofR. The Kruskal "algorithm involves the following steps". Step I. Choose an edge e, in S of least weight. Let E = {e,}. Replace S with S  {e} Step II. Select an edge in S of least weight that will not make a cycle with members of {e) and S with S  {eJ Step III. Repeat step II until we get E = n  1
Example 31. Consider the graph R as shown below (Fig. 12.49). We find the minimal
spanning tree ofR.
A
B
B 6
9
B
C
F
4
9
6
G
()
G F
,
(ii)
Fig. 12.49
Sol. The graph R [Fig. 12.49(i)] has 7 vertices namely A, B, C, D, E, F and G. Therefore, any spanning tree of R will have 6 edges. By Kruskal is algorithm, the edges are odered by increasing length and are successively added (without forming any cycle) until 6 edges are included. This gives the following data.
Edges Length Added edges
CD
4
./
F DG DF BC BE FG DE AB BD AC EG 4 4 5 6 6 6 7 8 8 9 9
C
./
./
X
./
./
X
X
./
X
X
X
Therefore, the minimal spanning tree ofR will contain the edges {CD, CF, DG, BC, BE, AB}. The minimum spanning tree of the graph [Fig. 12.49(i)] is shown in Fig. 12.49(ii).
Example 32. Find a minimum spanning tree of the labelled connected graph shown in
Fig. 12. 50.
A
B 3
C
D
E
Fig. 12.50
F
TREES
553
Sol. Using KRUSKAL'S ALGORITHM, arrange all the edges of the graph in increasing
order and initialize spanning tree with all the vertices of G. Now, add the edges of G in T which do not form a cycle and have minimum weight until n 1 edges are not added) where n is the number of vertices. The spanning tree is shown in Fig. 12.51. 
Edges
Weights
(B, D)
3
Added
(A, E)
4
Added
(D, F)
4
Added
(B, F) (C, E)
A
5
Not added
6
Added
(A, C)
7
Not added
(B, C)
7
Added
8
Not added
(A, F)
Minimum Spanning Tree
Added or Not
B 3
C
4
6
D
4
F
E
Not added (E, B) 9 The minimum weight of spanning tree is
Fig. 12.51
= 24.
Example 33. Find all the spanning trees of graph G and find which is the minimal spanning tree of G shown in Fig. 12.52. a
d
6
2
b
3
3
e
2
c Fig. 12.52
Sol. There are total three spanning trees ofthe graph G which are as shown in Fig. 12.53. a
d e 3 "... 2
2
b
3
c
(i)
d e 3 "....
a
2
b
a
6
3
2
c
(ii)
b
d
6
3
e
2
c
(iii)
Fig. 12.53
To find the minimal spanning tree, use the KRUSKAL' S ALGORITHM. The minimal spanning tree is shown in Fig. 12.54.
554
DISCRETE STRUCTURES
Edges (E, F)
Weights 1
Added
(A, B) (C, D)
2 2
Added Added
(B, C)
3
Added
E)
3
Added
(D,
Minimal Spanning Tree
Added or Not
a
d
b
Example 34. What are the properties of minimum spanning tree. Sol. Properties of Minimum spanning tree
e
2
2
Not added. (B, D) 6 The first one is the minimal spalUling having the mini mum weight = I I .
3
3
c
Fig.
12.54
A minimum spanning tree T of a graph G is a tree whose total weight is the smallest among all the spanning trees of the graph G. It has the following properties. (i) The total weight of the spanning tree is the sum of the weights of the edges of the spanning trees. (ii) The minimum weight of the spanning tree is unique.
TEST YOUR KNOWLEDGE
L 2. 3.
4.
Draw all trees with exactly six vertices. (P.T. U. B. Tech. Dec. 2013) Draw all trees with five or fewer vertices. Find the number of trees with seven vertices. Find a minimum spanning tree of the weighted graph shown below : A
B D
E
F
5.
Find all spanning trees of the graph shown below :
6.
Find all spanning trees of the graph shown in the following figure.
555
TREES
7.
Find the minimal spanning tree of the following graph
A
D
(a) C (c)
(b)
G
Find the minimal spanning tree T for the weighted graph shown below : 2
2
2
2
2 3
3
3
Show that the sum of the degrees of the vertices of a tree with vertices is 3
8.
L
2.
There are six such trees shown below :
3
Answers
(i) (iii) There are eight such trees shown below : (i)
(ii) (vi)
3. 15 4. C
t AD W .A F E
n
(iv)
(iii)
(v)
(vi) L (v)
(iv) (vii)
2n 2.
(viii)
556
5.
DISCRETE STRUCTURES
There are eight such spanning trees shown below : T
"'"
(i)
(ii)
(iii)
V
"'V
�
(v)
7
(vi)
6.
There are twelve such spanning trees shown below.
7.
(a)
A
(b)
V (iv)
V
(vii)
(viii)
B E
G F
MULTI PLE CHOICE QUESTIONS (MCQs)
1.
For the given binary tree) the preorder traversal is
/
D
(a) A B D E C F H G I (c) D E B A C F G H I
2.
/
B
A
"
"
C
/
E F I H
"
G I I
(b) A B C D
EF
GHI
(d) D B E A H F C I G. Let T be a full binary tree with I internal nodes. Then which of the following statements is TRUE? (a) T has 2i + 1 total nodes and i + 1 terminal nodes.
(b) T has 2i total nodes and i + 1 terminal nodes. (c) T has 2i + 2 total nodes and i + 2 terminal nodes.
(d) T has 2i2 total nodes and i2 terminal nodes.
TREES
3.
557
The maximum number of nodes in a binary tree of depth d is
(b) 2d  1 (d) 2d+ l.
(a) 2d  1 (c) 2d + 1
4.
The number of external nodes in a full binary tree with
500 internal nodes are
(a) 501 (c) 500
(b) 1000 (d) Any number. 5. The total number of edges in any tree with n vertices is (a) n(n  1)/2 (b) n/2 (c) n (d) n  l. 6. Suppose T is a binary tree with 20 nodes. What is the minimum possible depth of T? (b) 3 (a) 1 (d) 5. (c) 4 7. If the height of a tree is 15, the highest level of the tree is (a) 15 (b) 14 (c) 3 (d) 5.
8.
In a postorder traversal) the
(a) Left subtree (c) Root 9.
10.
___
is processed first
(b) Right subtree (d) Any of the three.
Which of the following traversal techniques lists the nodes of binary search tree in ascending order?
(a) Preorder (c) Inorder
(b) Postorder (d) Level order.
Which traversal of the EST would print result in original order of input?
(a) Preorder (c) Inorder
1. 2.
(b) Postorder (d) Level order.
Answers and Explanations
(a) Apply the algorithm and check. (a) External nodes in a tree are one more than internal nodes. 3. (b) The maximum number of nodes in a binary tree at depth d is one less than 2"4. (a) The number of external nodes is one more than number of internal nodes. 5. (d) The number of edges needed to connect vertices of a tree is one less than the number of vertices.
6. (d) Draw the tree and see. 7. (b) The level of the tree is one less than its height. S. (a) First the left subtree, then right subtree and then root. 9. (c) Inorder traversal of EST gives nodes in ascending order. (c) Preorder traversal of EST gives nodes in original order.
10.
1
* 3
PROPOSITIONAL CALCULUS
1 3. 1 . BASIC LOGIC OPERATIONS The following are the main logic operations in this chapter : (i) p 1\ q. called as " conjunction ofp and q '" and read as "p and '1". (ii) p v q, called as "disjunction ofp and '1" and read as "p or '1". (iii) p, called as "negation ofp'" and read as "notp'". (iv) T, read as "True" (v) F, read as "False'". 
1 3.2. STATEMENT A statement is any collection of symbols or sounds which is either true or false, but not both. The truth or falsity of a statement is called its truth value. For example, consider the following : (a) Delhi is in France (b) Where are you going ?
(c) 2 + 3 = 5
The expression (a) is a statement and it is a false statement. The expression (b) is not a statement since it is neither true nor false. The expression (c) is a true statement.
(P.T. U., M.G.A. May 2007)
1 3.3. PROPOSITION
A proposition is a statement which is either true or false. It is a declarative sentence.
For example : The following statements are all propositions : (i) Jawahar Lal Nehru is the first prime minister of India. (ii) It rained yesterday. (iii) If x is an integer, then x2 is a + ve integer. For example : The following statements are not propositions : (i) Please report at 11 a.m. sharp (ii) What is your name ? (iii) x2 = 13.
* Not meant for P.T.V. B.Tech., "Discrete Structure" (BTCS402) Course. 558
PROPOSITIONAL CALCULUS
559
1 3.4. PROPOSITIONAL VARIABLES The lower case letters starting from p onwards are used to represent propositions e.g., p : India is in Asia
q : 2 + 2 = 4.
Example. Classify the following statements as propositions or nonpropositions. (i) The population ofIndia goes upto 120 million in year 2012. (ii) x + y = 30 (iii) Come here (iv) The Intel PentiumIII is a 64bit computer. (ii) Not a proposition Sol. (i) Proposition (iv) Proposition. (iii) Not a proposition 1 3.5. TRUTH TABLE The truth value of a proposition depends upon the truth values of its variables. Once the truth values of the variables are known, the truth value of the proposition is also known. The table to show this relationship is known as truth table. For e.g., consider the proposition  (p A  q). The truth table for this proposition is given in Fig. 13.1.
P
q
T T
T F
F
T
F
F
pAq
(p A  q)
F
F
T
T
T
F
F
F
T
F
T T
q
Fig.
13.1
ILLUSTRATIVE EXAMPLES
Example 1. Construct the truth tables of.OO p A 0 � (� � A � v � A � v (r A � (iv) p v q v r. (iii) ( p) v ( q) Sol. The truth table for these propositions are as follows: (Fig. 13.2 to Fig. 13.5) (i)
p
q
q
P A ( q)
T T
T
F
F
F
T
T
F
T
F
F
F
F
T
F
Fig.
13.2
DISCRETE STRUCTURES
560
(ii)
p
q
r
p Aq
q Ar
r Ap
(P A q) v (q A r) v (r A p)
T T T T F F F F
T T F F T F T F
T F F T T T F F
T T F F F F F F
T F F F T F F F
T F F T F F F F
T T F T T F F F
Fig. 13.3 (iii)
p
q
p
q
( p) v ( q)
T T F F
T F T F
F F T T
F T F T
F T T T
Fig. 13.4 (iv)
p
q
r
p vqvr
T T T T F F F F
T T F F T F T F
T F F T T T F F
T T T T T T T F
Fig. 13.5 1 3.6. COMBINATION OF PROPOSITIONS We can combine the propositions to produce new propositions. There are three fundamental and three derived connectors to combine the propositions. These are explained as follows one by one.
(a) Fundamental Connectors 1. Conjunction. It means ANDing of two statements. Assume p and q be two proposi· tions. Conjunction of p and q to be a proposition which is true when both p and otherwise false. It is denoted by p A q. (Fig. 13.6)
q
are true)
PROPOSITIONAL CALCULUS
561
Truth tables are used to determine the truth or falsity of the combined proposition.
p T T F F
q
Fig.
13.6.
T F T F Truth Table ofp A q.
P Aq
T F F F
2. Disjunction. It means ORing of two statements. Assume p and q be two proposi· tions. Disjunction of p and q to be a proposition which is true when either one or both p and q are true and is false when both p and q are false. It is denoted by p v q. (Fig.
p T T F F
3.
q
Fig.
13.7.
T F T F Truth Table ofp v q.
p vq T T T F
13.7)
Negation. It means opposite of original statement. Assume p be a proposition. Nega· tion of p to be a proposition which is true when p is false) and is false when p is true. It is denoted by  p. (Fig.
13.S)
p T F Fig.
13.8.
p F T Truth Table of  p.
Example 2. Consider the following : p : He is rich q : He is Generous. Write the proposition which combines the proposition p and q using conjunction (A), disjunction (v), and negation (). Sol. Conjunction. He is rich and generous i.e., p A q. Disjunction. He is rich or generous i.e., p v q. Negation. He is not rich i.e.,  p He is not generous i.e., q. It is false that he is rich or generous i.e.,  (p v q). He is neither rich nor generous i.e., P /\ q. It is false that he is not rich i.e.,  ( pl. ,...,
,...,
,...,
562
DISCRETE STRUCTURES
Example 3. Let p be "It is hot day" and q be "The temperature is 45"C". Write in simple sentences the meaning offollowing : (iii)  (p /\ q) (i)  p (ii)  (p v q) (v) p v q (iv)  (p) (vi) p /\ q (viii)  ( p v  q). (vii)  p /\  q Sol. (i) It is not a hot day. (ii) It is false that it is hot day or temperature is 45"C. (iii) It is not true that it is hot day and temperature is 45"C. (iv) It is false that it is not a hot day. (v) It is hot day or temperature is 45"C. (vi) It is hot day and temperature is 45"C. (vii) It is neither a hot day nor temperature is 45"C. (viii) It is false that it is not a hot day or temperature is not 45"C. Example 4. Consider the following statements : p : He is coward. q : He is lazy. r : He is rich. Write the following compound statements in the symbolic form. (i) He is either coward or poor. (ii) He is neither coward nor lazy. (iii) It is false that he is coward but not lazy. (iv) He is coward or lazy but not rich. (v) It is false that he is coward or lazy but not rich. (vi) It is not true that he is not rich. (vii) He is rich or else he is both coward and lazy. Sol. (i) p /\  r (iii)  (p /\  q) (ii)  p /\  q (iv) (p v q) /\  r (vi)  ( r) (v)  «P v q) /\  r) (vii) r v (p /\ q). (b) Derived Connectors 1. NAND. It means negation after ANDing of two statements. Assume p and q be two propositions. Nanding of p and q to be a proposition which is false when both p and q are true, otherwise true. It is denoted by p t q. (Fig. 13.9) q p tq p T T F T F T F T T F F T Fig. 13.9 2. NOR or Joint Denial. It means negation after ORing of two statements. Assume p and q be two propositions. NOring of p and q to be a proposition which is true when both p and q are false, otherwise false. It is denoted by p t q. (Fig. 13.5) p .J, q q p T T F T F F F T F F F T Fig. 13.10
PROPOSITIONAL CALCULUS q
563
3. XOR. Assume p and q be two propositions. XORing ofp and q is true if p is true or if is true but not both and vice versa. It is denoted by p EB q. (Fig. 1 3 . 1 1)
ffi q
p
q
T
T
T
F
T
F
T
T
F
F
F
P
F
Fig. 13.11.
Example 5. Generate the truth table for following : (i) A EB B EB C (ii) A t B t C. Sol. The · 1 truth table for above formulas are as shown in Figs. 13.12 and 13.13. A B C A ffi B A ffi B ffi C (i) T T
T T
T F
F
T F
T T
F
T F
T T
F
F
F
T F
T T
F
F
T T
F
F
T
F
T
F
F
F
F
F
C
AtB
AtBtC
F
T T
Fig. 13.12
(ii) Truth table for (ii) is
A
B T
T
T
T
F
F
T
T T
F
T T
F
F
T F
F
T T
T F
T T
F
F F
F
T
T
F
F
F
F
T
T
T
T
F
T T
Fig. 13.13
Example 6. Prove that X EB y", (X 1\  Y) V ( X 1\ Y). Sol. Construct the truth table for both the propositions. (Fig. T T
Y
Xffi Y
Y F
x
F
T
T
F
T
T
F
F
F
x
T
F
13.14)
(X 1\  Y) ( X 1\ Y)
XI\  Y
 XI\ Y
F
T
F
T
F
T
F
T
T
T
T
F
F
F
F
Fig. 13.14
F
F
V
F
DISCRETE STRUCTURES
564
As the truth table for both the proposition are same. X EB Y = (X 1\

Y) V ( X 1\ Y). Hence proved. 
Example 7. Show that (p EB q) v (p J. q) is equivalent to p t q. Sol. Construct the truth table for both the propositions.
p
T T F F
(p .j, q)
(p q)
q
fB
T F T F
F T T F
(p q) v (p .j, q) fB
ptq
F F F F T T F T T T T T Fig. 13.15 Since, the values of (p EB q) v (p J. q) is same as p t q as in Fig. 13.15. Hence, they are
equivalent.
Example 8. Show that ( p t q) EB (p t q) is equivalent to (p v q) 1\ (p J. q). Sol. Construct the truth table for both the propositions
p
T T F F
q
T F T F
ptq F T T T
(p t q) (p t q) fB
p .j, q
pvq
(p v q) 1\ (p .j, q)
F T F F F T F F F T F F F F T F Fig. 13.16 Since, the values of (p t q) EB (p t q) and (p v q) 1\ (p J. q) are same as in Fig. 13.16. Hence,
they are equivalent.
(c) Some Other Connectors 1. Conditional. Statements of the form "Ifp then q" are called conditional statements. It is denoted as p � q and read as "p implies q" or "q is necessary for p" or "p is sufficient for q". Conditional statement is true if bothp and q are true or ifp is false. It is false ifp is true and q is false. The propositionp is called hypothesis and the proposition q is called conclusion. The truth table of conditional statement is (Fig. 13.17)
p
T T F F
q
p > q
T F T F Fig. 13.17. Truth Table ofp > q.
T F T T
For example : The followings are conditional statements : 1. If a = b and b = c, then a = c. 2. If I will get money, then I will purchase computer.
PROPOSITIONAL CALCULUS
565
1 3.7. (a) LAWS OF THE ALGEGBRA OF PROPOSITIONS
I. Idempotent Laws (ii) p /\ p = p (i) p v p = P II. Associative Laws (i) (p v q) v r = p v (q v r) (ii) (p /\ q) /\ r = p /\ (q /\ r) III. Commutative Laws (i) p v q = q v P (ii) p /\ q = q /\ P IV. Distributive Laws (i) p v (q /\ r) = (p v q) /\ (p V r) (ii) p /\ (q V r) = (p /\ q) V (p /\ r) V. Identity Laws (ii) p /\ t = p (i) p v f = p (iii) p v t = t (iv) p /\ f= f where t denotes the true value and f denotes the false value. VI. Complement Laws (ii) p /\  P = f (i) p v  p = t (iv)  t = f,  f = t (iii)   p = P VII. De Morgan's Laws (ii)  (p /\ q) =  p v  q (i)  p(v q) =  P /\  q Example 9. Show by using laws of algebra ofpropositions, the logical equivalence of (p v q) /\ P =  P /\ q. I Commutative law Sol. (p v q) /\  P =  P /\ (p V q) I Distributive law = ( P /\ p) V ( P /\ q) I Complement law = f v ( p /\ q) I Identity law =  P /\ q Example 10. Show that p v (p /\ q) = p by using laws of algebra ofpropositions t I Identity law Sol. p v(p /\ q) = (p /\ t) v (p /\ q) I Distributive law = P /\ (t v q) Identity law = P /\ t I Identity law =p Example 11. Use the laws of algebra of logic propositions, show that  (p v q) v ( P /\ q) =  p. I De Morgan's law Sol.  (p v q) v ( P /\ q) = ( P /\  q) V ( P /\ q) =  P /\ ( q V q) I Distributive law CO  P A t Complement law I Identity law. = p
DISCRETE STRUCTURES
566
1 3.7. (b) VARIATIONS IN CONDITIONAL STATEMENT
Contrapositive. The proposition  q ;  p is called contrapositive of p ; q. Converse. The proposition q ; p is called the converse ofp ; q. Inverse. The proposition  p ;  q is called the inverse of p ; q. Example 12. Show that p ; q and its contrapositive  q ; p are logically equivalent. Sol. Construct truth table for both the propositions. (as in Fig. 13.1S)
p T T
q T F T
F F
p
q
T T
T F T
F F
F
 q > p
p > q
rn rn
F
Fig. 13.18 As, the values in both cases are same, hence both propositions are equivalent.
Example 13. Show that proposition q ; p and  p ;  q is not equivalent to p ; q. Sol. Construct truth table for all the above propositions :
p T T F F
q T F T
p F F
T T
F
q F
T F T
Fig. 13.19
q >p T T F T
p > q T F T T
 p >  q T T F T
As the values of p ; q in table is not equal to q ; p and  p ;  q as in Fig. 13.19. So both of them are not equal to p ; q but they are themselves logically equivalent.
Example 14. Prove that the following propositions are equivalent to p ; q. (iii)  q ;  p. (ii)  p v q (i)  (p 1\  q) Sol. Construct the truth table for all the above propositions :
p T T F F
q T F T F
p
q
F F
F
T T
T F T
p v q
 q > p (p 1\  q)
rn rn Fig. 13.20
F
T F F
 (p 1\
 q)
p > q
rn rn
In the above table (Fig. 13.20) the values ofp ; q is equivalent to (i), (ii) and (iii), hence they are equivalent to p ; q. Hence proved. 2. Biconditional. Statements of the form "if and only if" are called biconditional state ments.
PROPOSITIONAL CALCULUS
567
It is denoted as p H q and read as "p if and only if q". The proposition p H q is true if p and q have the same truth values and is false ifp and q do not have the same truth values. Fig. 13.21. The name of biconditional comes from the fact that p H q is equivalent to (p ; q) /\ (q ; pl· The truth table of p H q is p
T T F F
pHq
q
T F T F Fig. 13.21. Truth Table of p H q.
T F F T
For example : (i) Two lines are parallel if and only if they have same slope. (ii) You will pass the exam if and only if you will work hard. Example 15. Prove that p H q is equivalent to (p ; q) /\ (q ; p). Sol. Construct the truth tables of both propositions :
p
T
T
F
F
T
F
Fig.
q
F T
Fig. 13.22
pHq
IT
p
q
p > q
q > p
F
F
T
T
T
F
F
T F
T
T T T
T F T
Fig. 13.23
(p > q) A (q > p)
rn
Since, the truth tables are same, hence they are logically equivalent. (Fig.
13.23). Hence proved.
13.22 and
1 3.8. PRINCIPLE OF DUALITY Two formulas A, and A2 are said to be duals of each other if either one can be obtained from the other by replacing /\ (AND) by v (OR) and v (OR) by /\ (AND). Also if the formula contains T (True) or F (False), then we replace T by F and F by T to obtain the dual. Note L A v
The two connectives and are called dual of each other. .j, (NOR) are dual of each other. 3. If any formula of proposition is valid, then its dual is also a valid formula.
2. Like AND and OR, t (NAND) and
Example 16. Determine the dual of each of the following : (a) p /\ (q /\ r) (b) p v  q (c) (P /\  q) V ( P /\ q) (d) (p t q) t (p t q) (e) (( p v q) /\ (q /\  s)) V (p v F), here F means false. Sol. To obtain the dual of all the above formulas, replace /\ by v and v by /\, and also
replace T by F and F by T. Also replace t by J. and vice versa. (a) p /\ (q /\ r) = p v (q v r) (b)  p v  q = (c) (p /\  q) V ( P /\ q) = (p v  q) /\ ( P V q)
 P /\  q
(d) (p J. q) J. (p J. q) (e) « p v q) /\ (q /\  s» v (p v F) = « p /\ q) V (q V  s» /\ (p /\ T)
DISCRETE STRUCTURES
568
1 3.9. LOGICAL IMPLICATION A proposition P(P. q ) is said to logically imply a proposition Q (p, q, ....) , written as P(P, q, ...) => Q (p, q, ...) if Q(P, q, ...) is true whenever P(P, q, ...) is true. •
...
Example 17. Show that p H q logically implies p H q. Sol. Consider the truth tables ofp H q and p ; q as in the following table. Fig. 13.24
p ; q is true whenever p H q is true. Hence p H q logically implies p ; q p
pHq
q
T T
T F T
F F
p > q
T
T F T T
F F
T
F
Fig. 13.24 Example 18. Show that p logically implies p v q. Sol. Consider the truth tables for p and p v q as in Fig. 13.25 Now p v q is true when
ever p is true. Hence p logically implies p v q. p
pvq
q
T T
T T
T T T
F
F F
F
F
Fig. 13.25 Example 19. Show that p 1\ q logically implies p H q. Sol. Consider the truth tables of p 1\ q and p H q as in Fig. 13.26. Now p H q is true
whenever p
1\
q is true. Thus p p
1\
T T
q logically implies p H q. q
p l\ q
pHq
F F
F
F F F
T F T
F F
T
T T
Fig. 13.26 Exmaple 20. Show that p H q does not logically imply p ; q. Sol. Construct the truth tables ofp H q andp ; q as in Fig. 13.27. Recall that p H 

q logically implies p ; q ifp ; q is true whenever p H q is true. But p H ; q is false. Hence p H q does not logically imply p ; q. 

q is true when p

p
T T
F F
q
T F T F
q
pHq
F
F
T F T
Fig. 13.27
T T
F
p > q
T F T T
PROPOSITIONAL CALCULUS
569
1 3.10. LOGICALLY EQUIVALENCE OF PROPOSITIONS Two propositions are said to be logically equivalent if they have exactly the same truth values under all circumstances. The table 1 contains the fundamental logical equivalent expreSSlOns :
L
Table 1. 6. Complement Properties
De Morgan's Laws
� (P A q) "" � P V � q  (p V q) "" P A � q 2. Commutative Properties p v q "" q vp ; P A q "" q AP. 3. Associative Properties (p v q) v r ""p v (q v r) (p A q) A r ""p A (q A r) 4. Distributive Properties p A (q V r) "" (p A q) V (p A r) p v (q A r) "" (p v q) A (p V r) 5. Idempotent Laws p v p ""p andp AP ""p
p "" � � P (p > q) "" (� q > � p) 8. Material Implication (p > q) ",, (�p v q) 9. Material Equivalence (p H q) "" [(p > q) A (q > p)] (p H q) "" [(p > q) v (� P A  q)] 10. Exportation [(p A q) > r] "" [p > (q > r)] 7. Transposition
Example 21. Consider the following propositions  p v  q and  (p A q). Are they equivalent ? Sol. Construct the truth table for both (as in Fig. 13.28).
p T T
p
q T F T
F F
F
F
T F T
F
T T
F
�p v � q
q
Fig. 13.28
rn
PAq T
� (p A q)
rn
F F F
Since) the final values of both the propositions are same) hence the two propositions are equivalent.
Example 22. Show that the propositions (p A q) and p v  q are logically equivalent. Sol. Construct the truth tables of  (p A q) and p v  q as in Fig. 13.29. Since the truth tables are the same) i.e.) both propositions are false in the first case and true in the other three cases) the propositions "'" (p /\ q) and ,..., p v ,..., q are logically equivalent and we can write  (p A q) =  p v  q
p T T
F F
q T F T F
P Aq T F F F
(a)

(p
A q)
F
T T T
p T T
F F
Fig. 13.29
q T F T
F
p F
�p v � q
T F T
T T T
F
F
T T
�q
(b)
F
DISCRETE STRUCTURES
570 Example 23. Prove the associative law: (p 1\ q) 1\ r = p 1\ (q 1\ r). Sol. Construct the required truth tables as in Fig 13.30. Since
identical) the propositions are equivalent. p
r
q
T T T T
T T
F F
F
T T
F
F
F F
F
(p 1\ q) 1\ r
q l\ r
F
T
p l\ (q l\ r)
F
T
F
F
F
F
F
F
F
F
F
F
F
F
T
F
F
F
F
F
F
F
F
t
F
j
p 1\ q
T F T F T F T
T T
F
the truth tables are
T
F
Fig. 13.30
F
F
Example 24. Prove that disjunction distributes over conjunction; that is, prove the dis tributive law p v (q 1\ r) = (p v q) 1\ (p V r). Sol. Construct the required truth tables as in Fig. 13.31. Since the truth tables are
identical) the propositions are equivalent. p
q
T T T T
T T
q l\ r
F
F F F
T T F T F T
F F
T T
F F F F
r
F F
T
pvq
F F F
F F
T T T T T
T
F F F
F
p v (q l\ r)
t
p vr
T T T T T T
T T T T T F T
(p v q) 1\ (p T T T T T
V
r)
F F F
i
F
Fig. 13.31
(P.T. U., M.G.A. May 2007)
1 3. 1 1 . TAUTOLOGIES
A proposition P is a tautology if it is true under all circumstances. It means it contains only T in the final column of its truth table.
Example 25. Prove that the statement (p ; q) H ( q ;  p) is a tautology. Sol. Make the truth table of above statement : p
T T
F F
q
p > q
F
F
T T F
T T T
q
p
F
F
T F T
F
Fig. 13.32
T T
 q > p
T F T T
(p > q) H ( q >  p)
As the final column contains all T's, so it is a tautology. (Fig. 13.32)
T T T T
PROPOSITIONAL CALCULUS
571
Example 26. Prove that p v p H P is a tautology. Sol. Construct the truth table of the given statement :
p T
F
p vp T
F Fig. 13.33
p VP H P T T
As the last column contains all T's, so it is a tautology. (Fig. 13.33)
(P.T. U., M.C.A. May 2007)
1 3.12. CONTRADICTION A statement that is always false is called a contradiction.
Example 27. Show that the statement p 1\ P is a contradiction.
p T
F
F F
F
T
Fig. 13.34 Sol. Construct the truth table of above statement. Since, the last column contains all F's, so it is a contradiction. (Fig. 13.34)
1 3.13. CONTINGENCY A statement that can be either true or false depending on the truth values of its vari· abIes) is called a contingency.
p T T
F F
q T
F
T
F
p > q T F
T T
P 1\ q T F F F
(p > q) > (p 1\ q) T T F F
Fig. 13.35 Example 28. Prove that the statement (p ; q) ; (p 1\ q) is a contingency. Sol. Construct the truth table of above statement. As, the value of final column depends on the truth value of the variables, so it is a contingency. (Fig. 13.35)
Example 29. From the following formulae, find out tautology, contingency and contra diction. (i) A '" A 1\ (A v B) (ii) (p 1\  q) V ( P 1\ q) (iii)  (p v q) v ( p v  q).
DISCRETE STRUCTURES
572 Sol. (i) Construct the truth table for A ; A 1\ (A v B). A
A vB
B
T T
T
T T T
F
T
F F
A A (A vB)
F
A ; A A (A v B)
T T
T T T T
F
F
F
Fig. 13.36
Since, the last column of the table contains all T's, hence it is a tautology. (Fig. 13.36)
(ii) Construct the truth table for (p 1\ q) V ( 
p
q
T T
T F T
F F
13.37.
T
F
F
F
F
P A q
F
F
F
T F T
q) as in Fig.
(P A  q) V (p A q)
q
F
P 1\
P A q
p
T T
F

T T
F
T
F
F
Fig. 13.37
Since, the value of the final column depends on the value of the different variables, hence it is a contingency_
(iii) P
Construct the truth table of the proposition q
T T
T
F
F
F
F
P
q
F
F
T
F
T
T T
F
T
P 1\ q
T

 (p 1\ q)
(p v q) v ( p v q) as in Fig. 
P v  q
F
F
T T T
F F F

13.38.
 (p v q) v (p v  q) F
T T T
T T T
Fig. 13.38
Since) the value of final column depends upon the value of different variables) hence it is a contingency_
Example 30. Verify that proposition p v (p 1\ q) is tautology. Sol. Construct the truth table for above proposition. (Fig. 13.39) 
p
T T
F F
q
T F T F
P 1\ q
T
F F F
(p 1\ q)
p V  (p l\ q)
F
T T T
Fig. 13.39
Since) the last column contains all T' s, hence it is a tautology_
T T T T
PROPOSITIONAL CALCULUS
573
Example 31. Determine whether the following is a tautology, contingency and a contra diction : (iii) p 1\ p. (i) p ; (p ; q) (ii) p ; (q ; p) Sol. (i) Construct truth table for p ; (p ; q) as in Fig. 13.40. 
p
q
p > q
p > (p > q)
T T
T
T
T
F
F
F
F
T
F
F
T T
T T
Fig. 13.40
Since) the value of last column depends on the value of different variables) hence it is a contingency_
(ii)
Construct truth table for p
; (q ; p) as in Fig.
13.41.
p
q
q > p
p > (q >p)
T T
T F
T T
F
T
F
F
F
T
T T T T
Fig. 13.41
Since) the last column contains all T's, hence it is a tautology_
(iii) Construct truth table for p 1\

P
as in Fig. 13.42. ' Since, the last column contain all F s, hence it is a contradiction. p
p
T
F
F
F
T
F
Fig. 13.42
1 3.14. FUNCTIONALLY COMPLETE SETS OF CONNECTIVES We have three basic and two conditional connectives i.e., /\, v, ,..., ::::::} and ¢::}. If we have given any formula containing all these connectives, we can write an equivalent formula with certain proper subsets of these connectives. A set of connectives is called functionally complete if every formula can be expresses on terms of an equivalent formula containing the connectives from this set. Example 32. Write an equivalent formula for P 1\ (R P) = ( P v Q) 1\ ( Q v P) = ( (P 1\  Q» 1\ ( (  Q 1\  P»
Hence, {, I\} is functionally complete. Similarly, we can show that (, v) is functionally complete.
Example 35. Show that { , '>} is functionally complete. P => Q '"  P v Q. Sol. We know that So, we have
P v Q '"  P => Q.
Since, { , v} is functionally complete. Hence from above, {, =>} is also functionally complete.
i.e.,
given any formula which involve all the five connectives) we can obtain an equivalence formula using {, '>} by first replacing biconditional and then replacing (1\) ANDing and fi· nally replacing v.
Example 36. Express P P) Sol. (P Q) 1\ (Q => P) = ( P v Q) 1\ ( Q v P) P => Q =  P v Q = ( (  P 1\  Q» 1\ ( (  Q 1\  P» P v Q =  ( P I\  Q). Example 37. Express (P 1\  Q) v ( P 1\  Q) in terms of (, v) only. Sol. (P 1\  Q) v ( P 1\  Q) = ( ( P v   Q» v ( (  P v   Q» .: P 1\ Q =  ( P v  Q) 1 3. 1 5. ARGUMENT An argument is an assertion ; that a group of propositions called premises) yields an other proposition) called the conclusion. Let PF P2 ) P3)''') Pn is the group of propositions that yields the conclusion Q. Then, it is denoted as P l' P2 , Pn" ' " Pn 1  Q.
Conclusion. The conclusion of an argument is the proposition that is asserted on the basis of other proposition of the argument.
Premises. The propositions) which are assumed for accepting the conclusion) are called the premises of that argument.
PROPOSITIONAL CALCULUS
575
(a) Valid Argument An argument is called valid argument if the conclusion is true whenever all the premises are true. The argument is also valid if and only if the ANDing of the group of propositions implies conclusion is a tautology i.e., P (Pp P2 ) P3)''') Pn) � Q is a tautology. Where P (Pp P2 ' P3" '" Pn) is the group of propositions and Q is the conclusion. Some common valid argument forms are given in Table 2.
(b) Falacy Argument An argument is called falacy or an invalid argument if it is not a valid argument.
Elementary Valid Argument Forms Table 2. 2. Modus tollens L Modus ponens p > q p q
p > q q
3.
Hypothetical syllogism
..
p > q q > r p�r
4. Disjunctive syllogism p vq �p q
5. Constructive dilemma
6. Absorption
(p > q) A (r > s)
p > q p >
(p A q)
p vr q vs
7.
Simplification
..
Conjunction
8.
p q
p Aq P
P Aq
9. Addition ..
p p v q.
Example 38. Show that the following rule is valid : p I  p v q or p . . p v q. Sol. We can prove this rule from the truth table (Fig. 1 3.43) p
q
p vq
T T
T F
T T
F
T
T
F
F
F
Fig. 13.43
P is true in line 1 and 2 andp v
q is also true in line 1 and 2. Hence, argument is valid.
DISCRETE STRUCTURES
576 Example
39. Show that the rule modus ponens is valid. p > q p q
Sol. The truth table of this rule is as follows : (Fig. 13.44)
I
p
q
p > q
T
T
T
T
F
F
T
T T
F F
I
F Fig. 13.44 The p is the true in line 1 and 2 andp ; q andp both are true in line 1 andp, p ; q and
q all are true in line Example
1 . Hence) argument is valid.
40. Show that the rule of hypothetical syllogism is valid p > q q > r P t r.
Sol. The truth table of above rule is as in Fig. 13.45. p
q
r
p > q
q > r
p > r
T
T
T
T
T
T
T T T
T
F F
T
F
F F
F
F F
T
F F
T
T
T
T
T
F
T
F
T
F
T
F F
F F
T
T T
T T
T T
F
T T
T
Fig. 13.45 p � q is true in lines 1 , 2, 5, 6, 7, 8. q � r is true in lines 1, 3, 4, 5, 7, 8. Both p 7 q and q � r is true in lines 1 , 5, 7, 8 p � r is also true in lines
argument is valid. Example
41. Show that the rule of Modus Tollens is valid p > q �q � p.
1, 5, 7,
8. Hence,
PROPOSITIONAL CALCULUS
577
Sol. The truth table of above propositions are (Fig. 13.46) p
q
p
q
p > q
T T
T
F
F F
F
T
T
T
F
T
T
F F
I
F
F
T
T T
Fig. 13.46
line 4.
P ; q
I
is true in line 1, 3 and 4.  q is true in line 2 and 4. Both p ; q and  q are true in true in line 4. Hence) the argument is valid.
,..., p is also
Example 42. Show that the rule of disjunctive syllogism is valid p vq p q.
Sol. The truth table of above rule is as follows : (Fig. 13.47) p
q
p
p vq
T T
T
F
F F
T T
T
T
T
F
T
F
I
F F
I
Fig. 13.47. As
q
P v q is true in line 1, 2 and 3.  p is true in line 3. Both p is also true in line 3. Hence) argument is valid.
v q
and  p is true in line 3.
Example 43. Show that the rule of simplification is valid p Aq p.
Sol. The truth table of above argument is as follows : (Fig. 13.48)
I
p
q
P Aq
T
T
T
T
F
F F F
F F
T
F
I
Fig. 13.48 P A q
is true in line 1 and p is also true in line 1. Hence) the argument is valid.
DISCRETE STRUCTURES
578 Example 44. Show that the rule of conjunction is valid. p q P A q.
Sol. The argument is valid if p A q ; P A proposition is as follows : (Fig. 13.49) p
q
T T
F
q is a tautology. The truth table for the above PAq
T
F F
P A q > P A q
T
F F F
T
F Fig. 13.49
[
As the proposition is a tautology_ Hence) the argument is valid.
Example 45. Show that the rule of absorption is valid p > q
p > (p A q).
Sol. We have to show that (p ; q) ; [p above argument is as follows : (Fig. 13.50)
; (p A q)] P >
p
q
P Aq
p > q
T T
T
T
T
F F
F
T
F
F F F
is tautology. The truth table of the
(p A q)
(p > q) > [p > (p A q)]
T
F
F
T T
T T
Fig. 13.50 Since) the argument is a tautology_ Hence) it is a valid argument.
1 3. 1 6. PROOF OF VALIDITY
rn
We can test the validity of any argument by constructing the truth tables. But as the no. of variable statements increases) the truth tables grow unwieldly. So) a more efficient method to test the validity of the argument is to deduce its conclusion from its premises by a sequence of elementary arguments each of which is known to be valid.
Example 46. Prove that the argument p truth tables. p ; � q Sol. (i) (ii) r ; q (iii) � q ; � r (iv) p ;  r (v) r ;  p
;
�
q, r ; q, r f � p is valid without using
(Given) (Given) Contra positive of (ii) Hypothetical syllogism using (i) and (iii) Contrapositive of (iv)
PROPOSITIONAL CALCULUS (vi) (vii)
579 r is true ,..., p is true
(Given)
Modus ponens using
(v) and (vi).
Example 47. Prove that the argument p , q, q , r, r , s,  s, p v t f t is valid without using truth tables. (Given) p , q Sol. (i) (Given) (ii) q , r r , s (Given) (iii) (Given) (iv) s (Given) pvt (v) p , r Hypothetical syllogism using (i) and (ii) (vi) p , s Hypothetical syllogism using (vi) and (iii) (vii) Modus tollens using (vii) and (iv) ( viii) p Disjunctive syllogism using (v) and (viii). (ix) tables.
Example 48. Prove that the argument p, q I (p v r) 1\ q is valid without using truth
Sol. (i) (ii) (iii) (iv) table.
p pvr q (p v r) 1\ q
(Given) Rule of addition using
(Given) Rule of conjunction using
Example 49. Prove that the argument p , q, P Sol. (i) (ii) (iii) (iv)
p , q p l\ r p q
(i)
1\
(ii) and (iii).
r I q is valid without using truth (Given) (Given) Rule of simplification using (i) Modus ponens using (i) and
(iii).
Example 50. Prove that the argument (p , q) 1\ (r , s), (p v r) 1\ (q V r) I q v s is valid (Given) Sol. (i) (p , q) 1\ (r , s) (Given) (ii) (p v r) 1\ (q V r) Simplification using (iii) (iii) (p v r) Constructive dilemma using (i) and (iii). (iv) qvs Example 51. Prove that the argument (p 1\ q) V (r , s), t , r,  (p 1\ q) I t , s is valid without using truth tables. (Given) Sol. (i) (p 1\ q) V (r , s) (ii) (Given) t , r (Given) (iii)  (p l\ q) Disjunctive syllogism using (i) and (iii) (iv) Hypothetical syllogism using (ii) and (iii). (v) Example 52. Prove that the argument (p , q) 1\ (r , s), q , s, (q , s) , (p v r) I q v s is valid using deduction method.
DISCRETE STRUCTURES
580
(p ; q) 1\ (r ; s) q ; s (q ; s) ; (p v r) pvr qvs
Sol. (i) (ii) (iii) (iv) (v)
(Given) (Given) (Given)
(iii) and (ii) Constructive dilemma using (i) and (iv). Modus ponens using
Example 53. Prove that the argument p ; (q v r), (s 1\ t) ; q, (q v r) ; (s 1\ t) I p ; q is valid without using truth table. (Given) p ; (q v r) Sol. (i) (Given) (ii) (s 1\ t) ; q (Given) (iii) (q v r) ; (s 1\ t) Hypothetical syllogism using (i) and (iii) p ; (s 1\ t) (iv) Hypothetical syllogism using (ii) and (iv). (v) p ; q Example 54. Prove that the argument p v (q ; p), p 1\ r I  q is valid without using truth tables. (Given) Sol. (i) p v (q ; p) ""' p A r (Given) (ii) Rule of simplification using (ii) (iii) p Disjunctive syllogism using (i) and (iii) (iv) q ; p Modus tollens using (iv) and (iii). (v) q Example 55. Test the validity of following argument. If I will select in lAS examina tion, then I will not be able to go to London. Since, I am going to London, I will not select in lAS examination. Sol. Let p be "I will select in lAS examination" and q be "I am going to London". Then the above argument can be written in symbolic form as follows : p >  q q Construct the truth table for above argument (Fig. 13.51) p
q
p
q
p >  q
T T
T
F F
F
F
F
T
T
T
T
F
T
F
T
T
T
F F
I
I
Fig. 13.51 P
;  q is true in line 2, 3 and 4. q is true in line 1 and 4 and  p is true in line 3 and 4. Hence) all three are true in line 4. So it is a valid statement.
PROPOSITIONAL CALCULUS
581
Example 56. Consider the following argument and determine whether it is valid. Either I will get good marks or I will not graduate. If I did not graduate I will go to Canada. I get good marks. Thus, I would not go to Canada. Sol. Let p be "I will get good marks" and q be "I will graduate" and r be "I will go to Canada". Thus) the above argument can be written in symbolic form as follows : p vq �q�r p
The truth table of above proposition is (Fig. 13.52)
I
p
q
r
q
r
p vq
 q t r
T
T
T
F
F
T
T
T
T
F
T
T
T
T T
F F
F F
T T
T
T T
F
F F F F
T T
F F
T T
F
T
F
F F
T T
F F
T
F F
T
T T
F
I
T T T T
F
Fig. 13.52
P v ,..., q is true in line 1, 2, 3, 4, 7, 8 and ,..., q � r is true in line 1, 2, 4, 5, 6, 7 andp is true in line 1 , 2, 3 and 4. ,..., r is true in 2, 3, 6 and 8. All the above are true in line 2. Hence, the argument is valid. Example 57. Determine the validity of the following argument without using truth tables. Either I will pass the examination, or, I will not graduate. If I do not graduate, I will go to Canada. I failed : Thus, I will go to Canada. Sol. Let p be "I will pass the examination" and q be "I will graduate" and t be "I will go
to Canada". Thus the above argument, in symbolic form can be written as p vq  q + t p
Thus to prove the validity of the argument, use the standard results as follows :
(i) (ii) (iii) (iv) (v) Hence proved.
pvq  q ; t p q
(Given) (Given) (Given) Disjunctive syllogism using
(i) and (iii) (iv)
Modus ponens using (ii) and
DISCRETE STRUCTURES
582
Example 58. Determine the validity of the following argument using deduction method. If I study. then I will pass examination. If I do not go to picnic. then I will study. But I failed examination. Therefore, I went to picnic. Sol. Let p be "I study" and q be "I will pass examination" and t be "I go to picnic". Then the above argument is written in symbolic form as follows : p > q � t  > p p
Thus to prove the validity of the argument use the rules o f inference.
(i) (ii) (iii) (iv) (v)
p '> q  t '> p p t
(Given) (Given) (Given) Modus tollens using (ii) and (iii) Complement property using (iv)
Hence proved.
Example 59. Prove the validity of the following argument using truth tables as well as deduction method. "If the market is free then there is no inflation. If there is no inflation then there are price controls. Since there are price controls, therefore, the market is free". Sol. Let p be "The market is free" and q be "There is inflation" and r be "There are price
controls". Then the above argument can be written in symbolic form as follows : p > � q �q�r r p
1st Method. By using truth tables Construct the truth table of above argument (Fig. 13.53)
I
p
q
r
q
p > � q
�q�r
T T T
T T
T
F F
F F
F
F F
T T
T
T
F
T
F
T
T
T
T
F F F F
T T
T
F F
T T T T
T T T
F F
F T
F
T T
I
F
Fig. 13.53 P � ,..., q is true in line 3) 4) 5) 6) 7 and 8 ,..., q � r is true in line 1, 2, 4, 5, 6, 7 r is true in 1 , 4, 5, 7. All the above three are true in line 4 and 5. Also p is true in line 4. Hence the
line argument is valid.
PROPOSITIONAL CALCULUS
583
lInd Method. Using deduction method (i) p '>  q (ii) (iii) (iv) (v) (vi)  p (vii) p
(Given) (Given)
(i) and (ii) Transposition using (iii)
Hypothetical syllogism using
(Given) Modus tollens using
(iv) and (v)
Complement of (vi).
1 3. 1 7. QUANTIFIERS The following notation and theorem involving the universal quantifier tential quantifier 3 will be used throughout this section: ('i x E A) p (x) or 'i x, p(x) , for every x E A, p(x) is true.
'i and the exis·
(3 x E A) p (x) or 3 x, p(x) , there exists x E A such that p(x) is true. Here p(x) is a propositional function (or open·sentence or condition) on A, that is, p (a) is true or false for every a in A. Theroem (De Morgan): Let p(x) be a propositional function on A. Then (i) ('i x E A) p (x) = (3 x E A)  p(x) (ii) (3 x E A) p (x) = ('i x E A)  p(x)
The above says, in words, that the following two statements are equivalent: "It is not true that, for every a A, p (a) is true". "There exists an a A such thatp(a) is false". and similarly, the following two statements are equivalent: (i) "It is not true that there exists an a A such thatp(a) is true". (ii) "For all a A, p(a) is false". Remark:
(i) (ii)
E
E
E
E
1 3. 1 8. EXISTENTIAL QUANTIFIER If p(x) is a proposition over the universe U. Then it is denoted as CIx p(x) and read as "There exists at least one value in the universe of variable x such that p(x) is true. The quan· tifier 3 is called the existential quantifier. There are several ways to write a proposition, with an existential quantifier i.e.,
(3x E
A) p(x)
or
CIx E A such that p(x) or (CIx) p(x) or p(x) is true for some x E A.
1 3.19. UNIVERSAL QUANTIFIER If p(x) is a proposition over the universe U. Then it is denoted as 'i x, p(x) and read as "For every x E U, p(x) is true". The quantifier 'i is called the universal quantifier. There are several ways to write a proposition, with a universal quantifier.
'ix E or
A, p(x)
or p(x), 'ix E A 'ix, p(x) or p(x) is true for all x E
A.
Example 60. Let A(x) : x has a white colour, B(x) : x is a polar bear, C(x) : x is found in cold regions, over the universe of animals. Translate the following into simple sentences : (ii) (3x) ( C(x)) (i) 3x (B(x) 1\  A(x)) (iii) (\fx) (B(x) 1\ C(x) '> A(x)).
DISCRETE STRUCTURES
584
Sol. (i) There exists a polar bear whose colour is not white. (ii) There exists an animal that is not found in cold regions. (iii) Every polar bear that is found in cold regions has a white colour. Example 61. Let K(x) : x is a twowheeler, L(x) : x is a scooter. M(x) : x is manufactured by Bajaj. Express the following using quantifiers. (i) Every two wheeler is a scooter. (ii) There is a two wheeler that is not manufactured by Bajaj. (iii) There is a two wheeler manufactured by Bajaj that is not a scooter. (iv) Every two wheeler that is a scooter is manufactured by Bajaj. Sol. (i) (V x) (K(x) ; L(x» (ii) (3 x) (K(x) A M(x» (iii) (3 x) (K(x) A M(x) ;  L(x» (iv) (V x) (K(x) A L(x) ; M(x» Example 62. Determine the truth value of each of the following statements. (Here R is the universal set.) (i) V x, I x I = x (ii) 3 x, x" = x (iv) 3 x, x + 2 = x. (iii) V x, + 1 > x Sol. (i) False. Note that if Xo = 3 then I Xo I '" xo ' (ii) True. For if Xo = 1 then Xo2 = xo ' (iii) True. For every real number is a solution to x + 1 > x. (iv) False. There is no solution to x + 2 = x. Example 63. Negate each of the statements in equation Sol. (i)  V x, I x I = x = 3 x  ( I x I = x) = 3 x, I x I '" x (ii) 3 x, x2 = X = V X (x2 = x) = V x, x" '" x (iii)  V x, x + 1 > x = 3 x  (x + 1 > x) = 3 x, x + 1 m) is that there exists some x such that x2 ;::: m) for every m. The statement is true as there is some greatest x such that x2 ;::: m) for every m. (ii) Negation of 3 m 'i x (x2 < m) is 'i m 3 x (x2 :> m). The meaning of 'i m 3 x (x2 :> m) is that for every m) there exists some x such that (x2 ;::: m). The statement is true as for every m) there exists some greatest x such that x2 � m. X
Example 70. Check the validity offollowing formula under given interpretation. (a) 'i x 3 y P(x, y) under interpretation domain = {1, 2} and P(1, 1) = T, P(1, 2) = F, P(2, 1) = T, P(2, 2) = T where T and F refer to true and false respectively. (b) 'i x P(x) ; Q(a, f(x)) under interpretation domain = {1, 2} and a = 1, f(1) = 2, f(2) = 1, P(1) = F, P(2) = T, Q(1, 1) = T, Q(1, 2) = T, Q(2, 1) = F and Q(2, 2) = T. Here T and F refer to true and false respectively.
PROPOSITIONAL CALCULUS
true.
587
Sol. (a) \;j x 3 y P(x, y) means for every x, there exists some y such that P(x, y) is true. Since P(l, 1) = T and P(2, 1) = T. Hence, the formula is valid under the domain = {I, 2}. (b) \;j x P(x) ; Q(a, f(x» means for every x, whenever P(x) is true implies Q(a, f(x» is also So, under the domain {I, 2}, the different values of \;j x P(x) ; Q(a, f(x» are as follows :
x
1 2 So, when P(x) is true, Q(a, f(x»
I
P(x) F T
Q (a,
{(x))
T T
is also true. Hence, the formula is valid.
ILLUSTRATIVE EXAMPLES
I
Example 1. Translate the following statements in proposition logic : (i) If you study, you will get good marks. If you do not study, you will enjoy. Therefore, either you will get good marks or you will enjoy. (ii) If the catalogue is correct, then if the seeds are planted on April, flowers will bloom in July. (iii) If John is elected class representative, then either Mary is elected treasurer or Alice is elected vicetreasurer. (iv) Either taxes are increased or if expenditures rise, then the debt ceiling is raised. Sol. (i) Let p be "You study" and q be "You will get good marks" and r be "You will enjoy". Then the proposition logic is p > q �p�r q vr
(ii) Let p be "The catalogue is correct" and q be "Seeds are planted in April" and "The flowers bloom in July". Then, proposition logic is
and
r be
p ; (q ; r) (iii) Let p be "John is elected class representative" and q be "Mary is elected treasurer"
r be "Alice is elected vicetreasurer". Then) proposition logic is p ; q v r (iv) Let p be "Taxes are increased" and q be "Expenditures
ceiling is raised". Then proposition logic is
rise" and
r
be "The debt
p v (q ; r).
Example 2. The meaning of proposition p ; q is "If p then q" or "q is a necessary condition for p" or "p only if q" or "p is a sufficient condition for q". Write the following statements in terms of above. (a) p : New Delhi is capital of India q : India is in Asia. q:x=2 (b) p : x" = 4 2 (c) p : x" = y q : x = y. (d) p = he works hard q : he is a Gold Medalist.
DISCRETE STRUCTURES
588
Asia.
Sol. (a) (i) If New Delhi is capital of India, then India is in Asia. (ii) India is in Asia is a necessary condition for New Delhi to be capital of India. (iii) New Delhi is capital of India only if India is in Asia. (iv) The fact that New Delhi is capital of India is a sufficient condition that India is in
(b) (i) Ifx2 = 4, then x = 2 (ii) x = 2 is necessary for x2 = 4 (iii) x" = 4 only if x = 2 (iv) x2 = 4 is sufficient for x = 2. (c) (i) If x" = y2 then x = y (ii) x = y is necessary fo r x2 = y2 (iii) x" = y2 only if x = y (iv) x2 = y2 is sufficient for x = y. (d) (i) If he works hard then he is a Gold Medalist. (ii) Gold medal is necessary for hard work. (iii) He works hard only if he is a Gold Medalist. (iv) Hard work is sufficient condition for Gold Medalist.
Example 3. Construct the truth table for the following statements (ii) ( q ;  p) ; (p ; q). (i) (p ; p) ; (p ;  p) Sol. (i) The truth table for (p ; p) ; (p ; p) is as follows. (Fig. 13.54) p
p
p + p
T
F
F
T
T T
F
F
T
T
Fig. 13.54 (ii) The truth table of statement ( q ;  p) ; (p ; q) is as follows. (Fig.
13.55)
p
q
p
q
 q +  p
p + q
( q  + p)  + (p + q)
T T
T
F F
F
T
F
T
T
F F
T
T T
F
T T
T T
T T T T
F F
T
F
Fig. 13.55 Example 4. Construct the truth table for following statements (i) (p ; (q ; r)) ; ((p ; q) ; (p ; r)) (ii) p H ( p v  q) (iii) (p ; p) v (p ;  p).
PROPOSITIONAL CALCULUS Sol.
589
(i) Truth table for (p ; (q ; r»
;
«P ; q) ; (p ; r»
p
q
r
q > r
p > q
p > r
T T T T F F F F
T T F F T T F F
T F F T T F T F
T F T T T F T T
T T F F T T T T
T F F T T T T T
�
is as follows. (Fig. 1 3.56)
(p > (q > r)) «P > q) >
T F T T T T T T
T F T T T T T T
(p > r))
K > L
T T T T T T T T
Fig. 13.56 (ii) Truth table for p H ( P v  q) is as following. (Fig.
13.57)
p
q
p
q
p v � q
T T F F
T F T F
F F T T
F T F T
F T T T
p H (p v
�
q)
F T F F
Fig. 13.57 (iii) The truth table for (p ; p) v (p ;  p) is as follows. p
p
p > p
T F
F T
T T
F T
(Fig. 13.58)
(p > p) v (p >
�
p)
T T
Fig. 13.58
Example ( p
v
5. Assume the value ofp ; q is false. Determine the value of ( p v  q) ; q.
Sol. Construct the truth table for both the statements and determine the value of  q) ; q against the false values ofp ; q. (Fig. 13.59) ( pv
p
q
p
q
p > q
p v  q
T
T
F
F
T
F
T
T F F
F T F
F T T
T F T
F T T
T T T
F T F
I
Fig. 13.59 So. when p ;
q is false, the value of ( p v  q) ; q is also false.
�
�
q) > q
I
DISCRETE STRUCTURES
590
Example
6. Given the value ofp ; q is true. Determine the value of  p v (p H q).
Sol. Construct truth table for both statements. (Fig. 13.60) p
q
p
T T
T
F
F
F
F
T
T
F
F
T
V
(p H q)
p > q
pHq
I
T
T
T
F
F
F
I
T
F
T
T
T
T
I
P
Fig. 13.60
So, when the value of p also true. Example 7. Prove
p A  q.
; q is true i.e., in line
I I
I
1, 3 and 4. The value of  p
v
(p
H
q) is
that the negation ofconditional statement  (p ; q) is equivalent of
Sol. The truth table of the above propositions are as follows. (Fig. 13.61) p
q
q
p > q
 (p > q)
p Aq
T T
T
F
T
F
F
F
T
F
T
T
F
T
F
F
F
F
F
T
T T
F
F
Fig. 13.61
As the values of both the propositions are same, hence
'=
 (p ; q) p A  q.
Example 8. Prove that the negation of biconditional statement  (p H q) is equivalent to p H  q or  p H q. Sol. The truth table of all the above propositions is as follows. (Fig. 1 3.62) p
q
p
q
pHq
T T
T
F
F
T
F
F
T
F
F
T
F
F
T T
F
F
T
T
 (P H q)
Fig. 13.62
As the values of equivalent.

(p
H
q)
and p
H
q
and
p
p Hq
p H q
rn rn [
H
q
are same, hence they all are
PROPOSITIONAL CALCULUS
591
Example 9. From the following formulae find out tautology. contingency and contradiction. (ii) (H ; (J 1\ J)) ;  (H ; J) (i)  (A ; B) v ( A v (A 1\ B)) (iii) (p H Q) (p 1\ Q) v ( P 1\ Q) . Sol. (i) Construct the truth table for  (A ; B) v ( A v (A 1\ B» as in Fig. 13.63. =
A
T T
B
A
 A v (A I\ B)
A > B
F
F F
F
F
T T
F F
T T T
T T
F
T T T
 (A > B) v ( A v (A 1\ B»
 (A > B)

T T T T
F
T F F
Fig. 13.63 As the last column of the table contains all T's, hence it is a tautology.
(ii) Construct truth table for (H ; I 1\ J) ;  (H ; I) as in (Fig. H
1
T T T T
J
T T
T
F F
F F
(I 1\ J) T F F F
T
H > 1
 (H >
T T
F F
I)
T T
F F
T T T T
F F F
F F
F
H >
F F F
T T T
T F T
F F F F
(I 1\ J) T
T T T T
F F F F
13.64)
(H > 1 1\ J) >

(H >
F
T T T
I)
F F F F
Fig. 13.64 gency.
As the value of last column depends upon the value of the variables, hence it is a contin
(iii) Construct the truth table for (p H q) ; (p 1\ q)
p
T T F F
q
T F T F
p
F F
T T
pHq
P 1\ q
F F
F F F
T T
T
 P 1\ q
(p 1\ q)
F
T T
F
V
V
( P
1\ q) as in (Fig. 13.65)
( p 1\ q)
T T T
F

(p H q) > (p 1\ q)
T T T T
Fig. 13.65 As the last column contains all T's, hence the above formulae is a tautology. Example
10. Prove that following is a tautology A v (B I\ C) oo (A v B) v C.
V
( P 1\ q) 
DISCRETE STRUCTURES
592
Sol. Construct the truth table for A v (B 1\ C) A
B
C
B
C
B I\ C
A v (B I\ C)
T T T T F F F F
T T F F T T F F
T F F T T F T F
F F T T F F T T
F T T F F T F T
F T T T F T T T
T T T T F T T T
; (A v "8) v C
as in (Fig. 13.66)
(A v B) (A v B) v C A v (B 1\ C) > (A v B) v C
T T T T F F T T
T T T T F T T T
T T T T T T T T
Fig. 13.66 As the last column of the table contains all T's, hence it is a tautology. Example 11. Determine whether the following are equivalent, using biconditional statement. (i) p H q '" (p 1\ q) ( P 1\ q) (ii) (p ; q) ; t '" (p 1\ q) ; t. Sol. To prove that the above pairs are equivalent. Prove that (p H q) H ((p 1\ q) ( P 1\ q)) and ((p ; q) ; t) H ((p 1\ q) ; t) are tautologies. (i) Construct truth table for (p H q) H ((p 1\ q) ( P 1\ q)) as in (Fig. 1 3.67) V



V



V
p
q
p
q
T T F F
T F T F
F F T T
F T F T
p H q P 1\ q
T F F T
 p l\  q (p 1\ q)
F F F T Fig. 13.67 H q) H ((p 1\ q) (
T F F F
V
As the proposition (p equivalent. (ii) Construct the truth table for ((p

V


( P 1\  q)
T F F T P 1\
(p H q) H (p 1\ q)
V
( p 1\  q)
T T T T

q))
is a tautology, hence they are
; q) ; t) H ((p 1\ q) ; t) as in Fig. 
13.68.
p l\  q
pA�q�t
((p > q) > t) H ((p 1\  q) > t)
F F T T F F F F Fig. 13.68
T T F T T T T T
T F F T T F T F
p
q
t
q
p > q
(p > q) > t
T T T T F F F F
T T F F T T F F
T F F T T F T F
F F T T T F T F
T T F F T T T T
T F T T T F T F
Since) the proposition is not a tautology) hence they are not equivalent.
PROPOSITIONAL CALCULUS
593
12. Determine the dual of each of the following : (a) (P /\ Q) v ( P v ( P v Q))
Example
W  P /\ 0 Q /\ � v � /\ � v (p /\ � (c) (P 1 Q) t R.
Sol. The dual of above propositions are as follows :
(a) (P /\ Q) v ( Q using Nand t only.
Sol. P => Q '"  P v Q '"  (  P /\  Q) ",  (p /\  Q)
", p t  Q ", p t Q t Q Example Sol. P
P v Q '"  ( P /\  Q)
  p oo p p t Q oo  (p /\ Q)  P = P t p.
14. Express P 1 Q using t only.
1 Q '"  (P v Q) '" (P v Q) t (P v Q) '" [(P t P) t (Q t Q) ] t [(P t P) t (Q t Q) ]
.:
P v Q '" (P t Q) t (P t Q)
15. Express the following formula using only  and /\. (P 1 Q) t R where 1 denotes NOR and t denotes NAND. Sol. (P 1 Q) t R '"  (P v Q) t R P 1 Q ",  (P v Q) Example
'"  « P v Q) /\ R)
Note. t
'"  «   (P /\  Q» /\ R)
P v Q '"  ( P /\  Q)
'"  « (P /\  Q» /\ R)
 P = P
(NAND) and .j, (NOR) are dual of each other.
Example
p t Q ",  (p /\ Q)
16. Show that the connective t (Nand) is functionally complete.
Sol. To show that the connective t is functionally complete. We have to show that the set of connectives (/\ , ,...)., and (v , ,...)., can be expressed in terms of t alone which is shown as follows :  P '" P v  P '"  (P /\ P) '" P t P P v Q '"  (P /\  Q) '"  P t  Q '" (P t P) t (Q t Q) and
p/\ Q '"  (P t Q) '" (P t Q) t (P t Q) Since the above expresses /\, v and ,..., in terms of t alone. Hence t is functionally complete.
17. Prove the validity of following arguments without using truth tables. 2. p, p ; q, q ; r 1  r 1. p v q, p 1  q 3. p, q, (p /\ q) ; r 1  r 4. p, (p /\  q) ;  p 1  p ; q. Sol. 1 . (i) (Given) pvq (ii) (Given) p Disjunctive syllogism (iii) q Example
DISCRETE STRUCTURES
594
(i) (ii) (iii) (iv) (v) 3. (i) (ii) (iii) (iv) (v) 4. (i) (ii) (iii) (iv) (v)
p p ; q q ; r p ; r r p q (p 1\ q) ; r p l\ q r p (p 1\  q) ;  p  (p l\  q)   (p ; q) p ; q
2.
(Given) (Given) (Given)
(ii) and (iii) Modus ponens using (iv) and (i)
Hypothetical syllogism using
(Given) (Given) (Given)
(i) and (ii) Modus ponens using (iii) and (iv)
Rule of conjunction using
(Given) (Given) Modus tollens using (ii) and (i) As
 (p ; q) '" (p 1\  q) (iv).
Complement property using
Example 18. Prove that the argument (p 1\ q) ; r, p ; q 1 p ; «P 1\ q) 1\ r) is valid without using truth table. (p 1\ q) ; r p ; q p ; (p 1\ q) (p 1\ q) ; «P 1\ q) 1\ r) p ; «P 1\ q) 1\ r)
Sol. (i) (ii) (iii) (iv) (v)
(Given) (Given) Rule of absorption using (ii)
(i) Hypothetical syllogism using (iii) and (iv). Rule of absorption using
Example 19. Prove that the following arguments are valid without using truth tables. 1. (p v q) ;  r, r v t, p 1 t 2. (p 1\ q) ; r, (r ; q), (r ; q) ; (q 1\ r) 1 (p 1\ q) ; (q 1\ r) Sol. (ii) (iii) (iv) (v) (vi) 2. (i) (ii) (iii) (iv) (v) (vi)
1.
(i)
(p v q) ;  r rvt p pvq r p l\ q ; r r ; q (r 1\ q) ; (q 1\ r) r ; (r 1\ q) (p 1\ q) ; (r 1\ q) (p 1\ q) ; (q 1\ r)
(Given) (Given) (Given)
(iii) (i) and (iv) Disjunctive syllogism using (ii) and (v) Rule of addition using
Modus ponens using
(Given) (Given) (Given)
(ii) Hypothetical syllogism using (i) and (iv) Hypothetical syllogism using (v) and (iii) Rule of absorption using
PROPOSITIONAL CALCULUS Example
595
20. Prove the validity offollowing argument using deduction system.
(l) A � B (3)  C I\ D :. B. Sol. (i)
(ii) (iii) (iv) (v) (vi) (vii) ( viii) (ix)
OO B �  C (4) A � D A�B B�C  C I\ D A�D A�C C A�C A B
Example 21. Prove (A) (i) P � Q (ii)  Q v R
(iii)  (R 1\  S) (iv) P :. S
Sol. (A)
(ii) (iii) (iv) (v) (vi) (vii) ( viii) (ix) (B) (i) (ii) (iii) (iv) (v) (vi) (C) (i) (ii) (iii) (iv) (v) (vi) (vii)
(i)
Example
(Given) (Given) (Given) (Given) Hypothetical syllogism using (i) and (ii) Simplification of (iii) Transposition of (iv) Modus tollens using (vii) and (vi) Modus ponens using (i) and (viii)
the validity of following by deduction method. (B) (i) P v Q (C) (i) (Q � R) 1\ (S � T) (ii) Q � R (ii) (U � V) 1\ (W � X) (iii) Q v U (iii) R 1\ S :. R v V (iv) P � S (v) P :. S
P�Q QvR  (R I\  S) P Q RvQ R RvS S PvQ Q�R R I\ S P�S P S (Q � R) 1\ (S � T) (V � V) I\ (W � X) QvV (Q � R) (V � V) (Q � R) 1\ (V � V) RvV
(Given) (Given) (Given) (Given) Modus ponens using (i) and (iv) Commutative property using (ii) Rule of Modus tollens using (vi) and (v) De Morgan's Law using (iii) Modus ponens using (viii) and (vii) (Given) (Given) (Given) (Given) (Given) Modus ponens using (iv) and (v) (Given) (Given) (Given) Simplification using (i) Simplification using (ii) Conjunction using (iv) and (v) Constructive dilemma using (vi) and (iii)
22. Prove the validity offollowing argument using truth tables.
"If it rains then it will be cold. If it is cold then I shall stay at home. Since it rains therefore, I shall stay at home".
DISCRETE STRUCTURES
596
Sol. Let p be "It rains" and q be "It will be cold" and r be "I shall stay at home". Then the above argument in symbolic form is p '> q q ,> r p r Construct the truth table of (p
'> q) 1\ (q '> r) 1\ p '> r as in (Fig.
13.69).
p
q
r
p > q
q > r
(p > q) 1\ (q > r)
(p > q) 1\ (q > r) 1\ p
(p > q) 1\ (q > r) I\ p > r
T T T T F F F F
T T F F T T F F
T F F T T F T F
T T F F T T T T
T F T T T F T T
T F F F T F T T
T F F F F F F F
T T T T T T T T
Fig. 13.69 As the last column contain all T's, hence the argument is valid.
Example 23. Translate the following into symbolic form and test the validity of the argument. If 6 is even then 2 does not divide 7. Either 5 is not prime or 2 divides 7. But 5 is prime, therefore, 6 is odd. Sol. Let p be "6 is even" and q be "2 divide 7" and r be "6 is primed". Thus, the above
argument in symbolic form can be written as p > � q �rvq r �p
Construct the truth table to test the validity. (Fig. 13.70) p
q
r
p
q
r
p >  q
�rvq
T T T T
T T F F
T F F T
F F F F
F F T T
F T T F
F F T T
T T F T
F F F F
T T F F
T F T F
T T T T
F F T T
F T F T
T T T T
T T F F
I
I
Fig. 13.70 P t ,..., q is true in line 3) 4) 5) 6) 7 and 8) ,..., r v q is true in line 1 ) 2) 3) 4, 5 and r is true in line 1 , 4, 6 and 7. All the three are true in line 4 and 6 and p is true in line 6. Hence, 
argument is valid.
PROPOSITIONAL CALCULUS
597
Example 24. Determine the negation of the following statements (i) 'i x 'iy 'iz, p(x, y, z) (ii) 'i x 3 y' p(x, y) (iii) 3y 'i x 'iz' p(x, y, z). Sol. (i)  ('ix 'iy'iz, p(x, y, z» 3 x 3y 3z  p(x, y, z) (ii)  ('ix 3y, p(x, y» 3x'iy,  p(x, y) (iii)  (3 y 'ix 'iz' p(x, y, z» 'iy 3x 3z, p(x, y, z). Example 25. Determine the negation of the following statements (i) 3x 'iy (p(x) v q(y)) (ii) 'ix 3y (p(x, y) ; q(x, y)) (iii) 'ix 'iy (P(x) 1\ q(y)). Sol. (i)  (3x 'iy, (P(x) v q(y» 'ix 3y  (P (x) v q(y» 'ix 3y ( p(x) 1\  q(y» (ii)  'ix 3y (P (x, y) ; q(x, y» 3x 'iy  (P (x, y) ; q(x, y» (iii) 'ix 'iy (P (x) 1\ pry»� 3x 3y  (P (x) 1\ pry»� 3x 3y ( p(x) v  pry»�. Example 26. Let U = Q. Use quantifiers to express the following statements (i) J5 is not rational. (ii) Subtraction of any two rational numbers is rational. Sol. (i) (3 x) (x2 = 5) (ii) ('ix) ('iy) (x  y is a rational). =
=
=
=
=
=
=
=
=
TEST YOU R KNOWL EDGE
Statements, Basic Operations
L
2.
3.
Let p be "It is cold" and q be "It is raining". Give a verbal sentence which describes each of the following statements : (a)  p (c) p v q (e)  p l\  q p p (a) (c) (e) p
(b) p l\ q
(d) q v p (j)  ( q)
Let be "He is tall" and q be "He is handsome". Write the following statements in symbolical form using and q. (b) He is tall but not handsome. He is tall and handsome. It is false that he is short or handsome. (d) He is neither tall nor handsome. He is tall or he is short and handsome. (j) It is not true that he is short or not handsome. Let be "Andrey speaks French" and q be" Andrey speaks Danish". Give a simple verbal sentence which describes each of the following (a) p v q (c) p l\  q (e)  ( p)
Truth Value of Compound Statement
(b) p l\ q
(d)  p v  q
(j)  ( P 1\  q)
(P. T. U. M.C.A. Dec. 2002)
Determine the truth value of each of the following statements: (a) 4 + 2 = 5 and 6 + 3 = 9 (b) 3 + 2 = 5 and 6 + 1 = 7 (c) 4 + 5 = 9 and 1 + 2 = 4 (d) 3 + 2 = 5 and 4 + 7 = 1 1 . 5. Determine the truth value of each of the following statements (a) 1 + 1 = 5 or 2 + 2 = 4 (b) 2 + 5 = 9 or 3 + 7 = 8 (c) 1 + 1 = 5 or 3 + 3 = 4 (d) 2 + 5 = 9 or 1 + 7 = 8
4.
DISCRETE STRUCTURES
598
Truth Tables 6.
7.
Find the Truth Table of (a) � p A q (c) � (p v � q) Find the Truth Table for (a) p v � q
(b) �
(d) �
(p v q) (p q) p ¢c>
=
¢c>
�q
p
� ¢c> q. (P.T. U M.C.A. May 2008)
=
.•
Tautologies and Contradictions 8. 9.
Verify that the propositionp v � (p A q) is a tautology. (a) Verify that the proposition (p A q) A � (p v q) is a contradiction. (b) Prove that the sentence "It is raining or it is not raining" is a tautology. (P.T. U (c) Let c denotes contradiction. Then for any statement p, (i) p v c p (ii) P A C c. (P.T. U =
=
.•
M.C.A. Dec. 2006)
.•
M.C.A. Dec. 2006)
Logical Equivalence
10. 11.
Show that the propositions � (p A q) and � p v � q are logically equivalent. Show thatp v q � ( p A � q) by using Truth Tables. 
=
Exclusive Disjunction, Joint Denial
Show that p A q (p .j, p) .j, (q .j, q) by using Truth Tables. (a) Show that � p p .j, P (b) p v q (p .j, q) .j, (p .j, q) Negation and De Morgan's Law 14. Using De Morgan's Law. Show (b) � p A � q (a) � p v � q (p A q) (p v q) (c) � p v � q � P A q (d) � (� p A q) P v � q 15. Simplify each of the following statements. (a) It is not true that his mother is English or his father is French. (b) It is not true that he studies physics but not mathematics. 16. Write the negation of the following statements. (b) He has blond hair or blue eyes. (a) He is tall but handsome. (c) He is neither rich nor happy. (d) He has lost his job or he did not go to work today. (e) Paris is in France and London is in England. (P.T. U M.C.A. Dec. 2006) 12. 13.
=
=
=
=
=

=

=
.•
Conditional Statements and Variations
17. Determine the contrapositive of each of the following statements. (b) Only if Marc studies he will pass the test. (a) If John is a poet, then he is poor 18. Determine the contrapositive of each statement. (b) It is necessary to be strong in order to be a sailor. (a) If he has courage, he will win. (c) Only if he does not tire will he win. (d) It is sufficient for it to be a square in order to be a rectangle.
Biconditional
19. Determine the truth value of the following statements. (b) 2 + 2 = 7 iff (a) 2 + 2 = 4 iff 3 + 6 = 9 (c) 1 + 1 = 2 iff 3 + 2 = 8 (d) 1 + 2 = 5 iff 20. Find the truth table for (p > q) v � (p H � q)
5+ 1=2
3+
1=
4
PROPOSITIONAL CALCULUS
599
Argumen t and State men ts
21. 22.
Show that the following statements are valid statements (a) p H q, q I p
Determine the validity of the argument � p
Logically Implication
23. 24.
(b) p > q, � q I � p
t
q, P � � q
Show that p H q logically implies p q (a) Show that p A q logically implies p H q (b) Show that p q = (p v q) > (p A q) using (a) truth tables (b) algebra of propositions, t
¢c>
(P,T, U,. M,CA May 2008)
Quan tifiers
25.
Determine the truth value of the following statements. (a) \j x, I x I = x (b) x, x' = x 3
\j x, x x \j x, 2x = x (g) x, x'  2x (c) (e)
+ 1>
3
+5
(a) (c) (e)
2.
(a) p A q (c) � (� p v q) (e) p v (� p A q) (a) (b) (c) (d) (e)
4. 5. 6.
=0
+
Answers
(b) It is cold and raining. It is not cold, It is cold or it is raining. (d) It is raining or it is not cold. It is not cold and it is not raining. (/) It is not true that it is not raining or it is false that it is not raining.
L
3.
x, x 2 = x (j) \j x, x  3 < x
(d) 3
(b) p A � q (d) � P A � q (j) � (� p v � q),
Andrey speaks French or Danish. Andrey speaks French and Danish. Andrey speaks French but not Danish. Andrey does not speak French or Andrey does not speak Danish. It is false that Andrey does not speak French. (j) It is false that Andrey speaks neither French nor Danish. The Statement " p and q" is true only when both statements are true. (a) false (b) True (c) False (d) True, The statement "p or q" is false when both statements are false. (a) True (b) False (c) False (d) True, (a)
(c)
p
T T F F
p
T T F F
q
T F T F q
T F T F
�p
�P A q
(b)
�q
pv�q
� (p v � q)
F F T T
F T F T
F F T F
T T F T
F F T F
p
T T F F
q
T F T F
pvq
T T T F
� (p v q)
F F F T
DISCRETE STRUCTURES
600
(d)
7.
11.
15. 16.
17. 18.
19.
20.
22. 25.
(a
)
� (p ¢"> q)
q
p
T T F F
T F T F
F T T F
�q
q
p
T T F F
T F T F
F T F T
T T F F
F T T F
(b)
pv�q
T T F T
T T F F
T T F F False (a) False
q
T F T F (b)
>
p
True
T T F F
q
T F T T
(c)
True
T F T F (d)
False
v
T F T T (e)
�
T F T F
T F F T
P
T T F F
�q
�p
F F T T
�PA�q
F T F T
F F F T
� (� P A � q)
�P A � q
�q
�p
q
p
T F F F T F T F T T F F T T T r Since the truth tables are identical, the propositions are equivalent (a) His mother is not English and his father is not French. (b) He does not study physics or he studies mathematics. (a) He is not tall or not handsome. (b) He has neither bland hair nor blue eyes. (c) He is rich or happy. (d) He has not lost his Job and he did go to work Today. (e) Paris is not in France or London is not in England. (a) If John is not poor, then he is not a poet. (b) If Marc does not study, then he will not pass the test. (a) If he does not win, then he does not have courage. (b) If he is not strong, then he is not a sailor. (c) If he tires, then he will not win. (d) If it is not a rectangle, then it is not a square. p H q is true whenever p and q have the same truth values. (a) True (b) True (c) False (d) False. p
T F T F
� p ¢"> q
F T T F
pvq
q
p
p ¢"> � q
H
F T T F
True (j) True (g) False.
T T T r
�
F T F T
q
T F T F
PROPOSITIONAL CALCULUS
1.
I
601
MULTI PLE CHOICE QUESTIONS (MCQs)
Which of the following is not a proposition?
I
(a) India is in Asia (b) 2 + 2 = 4 (c) The Intel PentiumIII is a 64bit processor
2.
(d) X + Y = 30. Let p: He is coward and q: He is rich. Which of the following is the symbolic form of the statement? "He is neither coward nor rich".
(a)  (p A q) (c)  p  q
(b)  p A  q (d)  (p A  q).
3.
Let p ; q is a proposition, which of the following is false?
4.
The biconditional statement p
5.
(a)  q ;  p is contra positive of p ; q (b) q ; p is converse of p ; q (c)  p ;  q is inverse ofp ; q (d)  (q ; p) is inverse ofp ; q.
(a) (p ; q) A (q ; p) (c) (p v q) A (q ; p)
B
q is equivalent to (b) (p ; q) v (q ; p)
(d) (p v q) v (q ; pl. Which of the following statements is true about the formula? (p ; q) B ( q ;  p)?
(a) It is a contradiction (c) It is tautology
(b) It is contingency
(d) None of these.
Answers and Explanations 1. (d) A proposition is either true or false. 2. (b) The statement is negation of the original one. 3. (d)  p ;  q is inverse of p ; q
4. (a) Biconditional statement is equal to Anding of (p ; q), (q ; pl. 5. (c) Draw the truth table for the statement and then check it.
1
* 4
MATRIX ALGEBRA
1 4. 1 . (a) MATRIX A set of mn numbers (real or complex) arranged in a rectangular array having m rows (horizontal lines) and n columns (vertical lines). the numbers being enclosed by brackets [ l or ( ). is called an m x n matrix (read as "m by n matrix"). An m x n matrix is also called a matrix of order m x n. Each of the mn numbers is called an element of the matrix. For example,
[�  � !] is a 2 x 3 matrix or matrix of order 2 x 3. It has two rows and
three columns. The numbers 2)  1) 5, 3, 0, 4 are its elements.
An m
x n matrix is usually written as
an
a12
al3
. . . . a1n
a21
a22
a23
. . . . a2n
a3 I
a32
a33
. . . . aa n
Here each element has two suffixes. The first suffix indicates the row and the second suffix indicates the column in which the element lies. Thus, a23 is the element lying in the second row and third column, ai is the element lying in the ith row andjth column. j For brevity, a matrix is usually denoted by a single capital letter A or B or C etc. Thus, an m x n matrix A may be written as A = [ai) m x n or A = [ai) , where i = 1 , 2, 3, ..... , m ; j = 1 , 2, 3,
..... , n.
1 4. 1 . (b) KINDS OF MATRICES
[../5o   3 1] IS.
Real Matrix. A matrix is said to be real if all its elements are real numbers. r.; ,,2
e.g.,
7
a reaI rnatn. x.
Square Matrix. A matrix in which the number of rows is equal to the number of columns square matrix, otherwise) it is said to be a rectangular matrix. Thus) a matrix A = [ai) m x n is a square matrix if m = n and a rectangular matrix if m t: n.
is called a
* Not meant for P.T.V. B.Tech., "Discrete Structure" (BTCS402) Course. 602
MATRIX ALGEBRA
603
A square matrix having n rows and n columns is called "a square matrix of order n" or "an nrowed square matrix") a'3 a23
is a 3rowed square matrix.
a31 a32 aa3 The elements of a square matrix are called its diagonal elements and the diagonal along which these elements lie is called the principal diagonal.
alP a22) a33 = [ai) )
In a square matrix A
(i) for elements along the principal diagonal, (ii) for elements above the principal diagonal, (iii) for elements below the principal diagonal, (iv) for nondiagonal elements,
i >j i #j
The sum of the diagonal elements of a square matrix is called its Thus) trace of the n rowed square matrix A
order
trace.
= [ai) all a22 a33 is)
+
+
+ ..... +
n
ann = iL =
1
aij .
Row Matrix. A matrix having only one row and any number of columns i.e., a matrix of 1 x n is called a row matrix. e.g., 5 3 0] is a row matrix. Column Matrix. A matrix having only one column and any number of rows i.e., a
[2
matrix of order
m
x1
is called a column

matrix. e·g·,
l�J
is a column matrix.
Null Matrix. A matrix in which each element is zero is called a null matrix or a zero matrix. A null matrix of order m x n is denoted by Om x
n' � = 03 X 2 ' [� � � �] = 02 X4
l� l
Submatrix. A matrix obtained from a given matrix A by deleting some of its rows or column or both is called a submatrix of A. B
= [� �]
�l
= I� � � 2 II 
is a submatrix of A
6
4
obtained by deleting the first row, second and fourth columns of A.

Diagonal Matrix. A square matrix in which all nondiagonal elements are zero is matrix. 0 for i i: j , x is a diagonal matrix if
called a diagonal Thus, A
= [ai)n n
For example, A
2 =0 o
0 1 o
0 o o
aij =
is a diagonal matrix,
DISCRETE STRUCTURES
604
An nrowed diagonal matrix is briefly written as diag. [dl ' d2 ) ) dnL where dp d2 ) ...... ) dn are the diagonal elements. Thus) the above diagonal matrix A can be written as diag. [2,  1 , 0]. Scalar Matrix. A diagonal matrix in which all the diagonal elements are equal to a scalar) say k) is called a scalar matrix. •••••••
Thus a scalar matrix is a square matrix in which all nondiagonal elements are zero and all diagonal elements are equal to a scalar) say k. Leo)
A=
.
. .
For example,
I�l � o °
{O
when i '" j when i = j
�l r �  � j �tJ
[ai) n x n IS a scalar matnx If aij =
2
0
0
k
re scalar matrices.
5
°
Unit Matrix or Identity Matrix. A scalar matrix in which each diagonal element is unity (i.e., 1) is called a unit matrix or identity matrix. Thus) a unit matrix is a square matrix in which all nondiagonal elements are zero and all diagonal elements are equal to 1. A = [aij ] n x n IS . a unI. t matn. x 1. f aij =
{O
when i '" j 1 when i = j
A unit matrix of order n is denoted by denoted by I.
In ' If the
order is evident, it may be simply
Upper Triangular Matrix. A square matrix in which all the elements below the principal diagonal are zero is called an upper triangular matrix. Thus) A =
[ai)n x n is an upper triangular matrix if aij = 0 for i > j. 4
3
2 For example,
5
is an upper triangular matrix.
3
Lower Triangular Matrix. A square matrix in which all the elements above the principal diagonal are zero is called a lower triangular matrix. Th us, A =
[ai)n x n is a lower triangular matrix if aij = 0 for i < j.
For example,
1
o
5
6
3
2
o o o
is a lower triangular matrix.
MATRIX ALGEBRA
605
Triangular Matrix. A square matrix in which all the elements either below or above the principal diagonal are zero is called a triangular matrix. Thus) a triangular matrix is either upper triangular or lower triangular. Equal Matrices. Two matrices A and B are said to be equal (written as A = B) if and only if they have the same order and their corresponding elements are equaL Thus, if A =
(i) m = p
[ai)m x n
and
n and =q
B=
[bi)p x
q'
then A = B if and only if
(ii)
ail bi} =
for all i andj.
1 4.2. ADDITION OF MATRICES Two matrices are said to be conformable for addition if they have the same order. If A and B are two matrices of the same order) then their sum A + B is a matrix each element of which is obtained by adding the corresponding elements of A and B.
[ai) mx
[ci)mx
[bi)m x
cij aij bi)
= + n ) where n ) then A + B = C = n and B = Similarly, if A and B are two matrices of the same order, then their difference A  B is a matrix whose elements are obtained by subtracting the elements of B from the correspond ing elements of A. In general) if A =
For example) if then
A+B=
and
A
_
B=
[2321 [3 2(12) +
56 0 +5

1 2] = [31 1 111] 12] = [1 11 33] .
+ 4 +7

5  ( 6) 05 4 7
5
5 5
1 4.3. MULTIPLICATION OF A MATRIX BY A SCALAR
[ai) kA [kai) [b,a, ab22 ab; ] [ab: 1) k.
The product of a matrix A = multiplying every element of A by Thus, If In particular,
by a scalar
=
A=
 A = (
k
is denoted by kA and is obtained by
[ka, kb, a b3 ] • b22 a3
, then kA =
A=
_
ka2 ka3 ] kb2 kb3
1 4.4. PROPERTIES OF MATRIX ADDITION
(i) Matrix addition is commutative i.e., A + B = B + A. (ii) Matrix addition is associative i.e., (A + B) + C = A + (B + C). (iii) For any matrix A, there exists a null matrix 0 of the same order as A such that A + O = O + A = A.
(iv) For any matrix A, there exists a matrix A + ( A) = ( A) + A = O.
A of the same order as A such that
DISCRETE STRUCTURES
606
ILLUSTRATIVE EXAMPLES
EXample 1. Find X' Y' Z and W if3 [� �] � [_ � 2 !] + [Z ; W x +3 Y] . . x 4 6 + x + Y] . . 3 3Y Sol. The given equatlOn IS [3zX 3w] _ [ _ 1 ++Z + w 2w +3 Equating the corresponding elements on the two sides) we get 4, � �6 �
3x x +
3y + x + y, 3z  1 + z + w, 2y � 6 + x, 2z =  1 + w, z � 1. y � 4,
3w � 2w + 3 w�3
l 2  :7l l� � Jl l� � �l 0 l � 7l l 0 l l00 00 0l l l  7l l 00 0 �l � 0  7 l 0 0l l �
Example 2. Express
9
15
5
12
 13
as the sum of a lower triangular matrix and an
upper triangular matrix with zero leading diagonal. Sol. Let and
L�
be the lower triangular matrix
U�
be the upper triangular matrix with zero leading diagonal such that
=>
5 12 15  13 2 5 12 9 15  13
4 6
4 6
° c o + � d e f O+ p a + b c+O + � d + e+O a
b
O p q r o+ q O+r f+O
Equating the corresponding elements on the two sides
2 = a, 5 �p,
. .
7� L�
q,
9�
12 15  13
1 4.5. MATRIX MULTIPLICATION
b,
12 = c,
and U �
4
= r,
5
15 � d,  13 = e,
6�
f
4 . (P.T. U., M.C.A. May 2007)
Two matrices A and B are said to be conformable for the product AB (in this very order of A and B) if the number of columns in A (called the pre·factor) is equal to the number
of rows in B (called the post·factor).
m x n and p x q respectively, then (i) AB is defined if number of columns in A � number of rows in B, i.e., if n � p. (ii) BA is defined if number of columns in B � number of rows in A, i.e., if q � m. Thus, if the orders of A and B are
MATRIX ALGEBRA Let A =
607
[ai)m x n and B = [bi) n Xp be two matrices conformable for the product AB, then n
= [ci) m x p where cij = ai l b ) + ai2 b2j + ...... + ainbnj = L aik bkj I k=l (i, j)th element of AB = sum of the products of the elements of ith row of A with the
AB is defined as the matrix C
i.e., corresponding elements of jth column of B.
The rule for multiplication of two conformable matrices is called
method.
l
l
l l
rowbycolumn
bll b12 a ll a'2 a 13 a2 1 a22 a23 and B = b2 1 b22 b3 1 b32 a3 I a32 a33 Orders of A and B are 3 x 3 and 3 x 2 respectively. AB is defined and is of order 3 x 2. Consider
where
A=
l l
Cll C 12 AB = C = C21 C22 C3 l C32 cll = Sum of products of elements of 1st row of A and 1st column of B
= [a21 a22 a23l
l ll I��: lb32
= a2,b ' 2 + a22 b22 + a23b32
C31 = Sum of products of elements of 3rd row of A and 1st column of B b ll = [a31 a32 a33l b2 1 = a31 b ll + a32b21 + a33b31 b3 1 c32 = Sum of products of elements of 3rd row of A and 2nd column of B
DISCRETE STRUCTURES
608
Note.
Another useful notation to remember matrix multiplication. R :::: 1st Row where R21 :::: 2nd Row Ra � 3rd Row where C1 :::: 1st column, C2 :::: 2nd column B :::: I��� l ��� l :::: [C1 C2], b3 1 b3 2
AB
J
I�� 1 [C C21 I��g� ��g:1 . lRa j , lRaC , RaC2 j
Example 3. If A =

and show that AB '" BA. 1 2 Sol. AB = 2 3 1 3 =
BA
�
l � ; 1l l� ! �l ,  l m l � � l l(l) + + + + + + + + + + J�II � � l ll � � �l ++ ++ ++ ll(l) ++ ++ �
=
and B
=
0 1 2
form the products AB and BA
+ ll l + ++ l l 7l + +
 2(0) 3(1) 1(0)  2(1) 3(2) 1(2)  2(2) 3(0) 2(1) 3(0)  1(1) 2(0) 3(1)  1(2) 2(2) 3(2)  1(0)  3(1) 1(0) 2(1)  3(0) 1(1) 2(2)  3(2) 1(2) 2(0) 1 2 2 0 3 0(2) 2( 3) 1( 2) 0(3) 2(1) 1(3) 0( 1) 2(2) 0(1) 1(2) 2( 3) 0( 2) 1(3) 2(1) 0(3) 1( 1) 2(2) 1(1) 2(2) 0( 3) 1( 2) 2(3) 0(1) 1(3) 2( 1) 0(2)
+ +
+ +
�
4 4 2 1 1 10 1 5 4
�
5 0 4 5 3 5 4 1
Orders of AB and BA are the same but their corresponding elements are not equal. Hence
AB tc BA.
Example 4. fllustrate with the help of examples that multiplication of matrices is not commutative in general, i.e., AB ::f: BA. Point out the various possibilities. Sol. Case I. AB is defined, but BA is not defined. Let A be a matrix of order 2 x 3, and B be a matrix of order 3 x 4. Then AB is defined and is a matrix of order 2 x 4, whereas BA is not defined. Case
II. AB and BA are both defined but their orders are different.
x 3 and B be a matrix of order 3 x 2. The product AB is defined and is a matrix of order 2 x 2. The product BA is defined and is a matrix of order 3 x 3. Let A be a matrix of order 2
Since orders of AB and BA are different, AB tc BA.
Case III. AB and BA are both defined and are matrices of the same order, yet AB tc BA (since corresponding elements are not equal).
MATRIX ALGEBRA
609
Let
A=
[� !]
and B =
[! �]
Here AB and BA are both defined and are of order 2
[� BA = [ !
AB =
Clearly, AB '" BA.
!] [! �m
�] = [�� �] !] = [! �]
x 2.
Hence multiplication of matrices is not commutative in general) i.e.) in generalAB Note.
If AB :::: BA, we say 'A and B commute' .
t: BA.
Example 5. Show by means of an example that the matrix AB = 0 does not necessarily mean that either A = 0 or B = 0, where 0 stands for the null matrix. Sol. Consider the matrices A = Here But
A ", 0, B AB =
'" O.
[� �] and B = [_ � �]
[ � �l � �] = [� = � � : �] = [� �] = O.
6. Prove by an example that AB = AC does not imply that B = C. Or Give an example of matrices A, B, C such that B '" C, but AB = AC.
Example
Sol. Let Then
Clearly,
[� �] , B = [� �] , C = [� �] [1 4] [1 5] [1(1) ++ 4(2) 1(5) ++ 4(1)] 0 AB = o 0 2 1 = 0(1) 0(2) 0(5) 0(1) = [9 4] [ 1 1] = [ 1(1) + 4(2) 1(1) + 4(2)] = 9 AC = [� 0(1) + 0(2) 0(1) + 0(2) [0 0 2 2 A=
B ", C but AB = AC.
90] 0]9
1 4.6. PROPERTIES OF MATRIX MULTIPLICATION
(i) Matrix multiplication is not commutative in general i.e., AB '" BA. (ii) Matrix multiplication is associative i.e., (AB)C = A(BC). (iii) Matrix multiplication is distributive with respect to matrix addition. Le.,
A(B + C) = AB + AC.
(iv) If A and I are square matrices of the same order, then AI = IA = A. (v) If A is a square matrix of order n, then A x A = A2, A x A x A = A3, A x A x A ..... m times = Am are all square matrices of order n. Also, we define A = I (vi) For any positive integer n, r n = I. 0
If A is a matrix such that A2 :::: I, then A is called an involutory matrix. If A is a matrix such that A2 :::: A, then A is called Idempotent matrix.
Note. (i) (iL)
(P.T. U., M.C.A. May 2007)
DISCRETE STRUCTURES
610
Example
7. Evaluate
Sol. A2 = A x A
=
A2  3A + 9I =
=
=
is the unit matrix of order
3 and A l3 I� 3 =
1
� �l ll � � �l l l(l) l l�l l 7 I I 7 l l I I �l l l 7 1 1 1 1 �l l l  �l · l
J 3 �1 l 
=
. .
A2  3A + 91, ifI
2



2 3 1 2  2(2) + 3( 3) 1( 2)  2(3) + 3(1) 1(3)  2( 1) + 3(2) 2(1) + 3(2)  1( 3) 2( 2) + 3(3)  1(1) 2(3) + 3( 1)  1(2)  3(1) + 1(2) + 2( 3)  3( 2) + 1(3) + 2(1)  3(3) + 1( 1) + 2(2)  12  5 11 4 11  6 2 3  12  5 11 0 11 4 1  3 2 3  1 + 9 0 1 11  6 3 1 2 0 0 3 6 9 9 0  12  5 11 11 4 1  6 9  3 + 0 9 11  6  9 3 6 0 0 6 1  12  3 + 9  5  ( 6) + 0 11 9 + 0 49+9 11  6 + 0 1  ( 3) + 0 = 5 4 11  3 + 0 66+9 2 8  7  ( 9) + 0
3
l
1
2
.
Note. Mathematical Induction
Mathematical induction is a very useful device for proving results for all positive integers. If the result to be proved involves n, where n is a positive integer, then the proof by mathematical induction consists of the following two steps : Step L Verify the result for n :::: l. Step 2. Assume the result to be true for n :::: k and then prove that it is true for n :::: k + l. Example
A [3
]
[1 +n2n
A [
_
] ][ ]
Sol. We prove the result by mathematical induction. When n
=
]
8. By Mathematical Induction, prove that if  4 then n = 1 + 2n  4n where n �s' any posawe . . . = �nteger. 1 1 ' n 1 2n
1,
 4n 1  2n 1 + 2. 1  4. 1 3  4 A'  [ 1 1  2. 1  1  1 An =
=A
=> The result is true when n = 1 Let us assume that the result is true for any positive integer k. let Now
Ak =
[1 +k2k
l Ak+ = Ak . A =
[
]
 4k 1 2k 1 + 2k  4k [3  4 k 1  2k 1  1
]
]
... (1) [Using (1)]
MATRIX ALGEBRA
_ [3(1 + 2k)  4k  4(1 + 2k) + 4k] 3k + 1 2k  4k  1 + 2k _ [1 + k2(k+ 1+ 1) 1 4(k2(k+ 1)1)] 
=>
The result is true for
n �
k+ 1
+
�
[31++2kk  41 2k4k]
Hence) by mathematical induction) the result is true for all positive integers
I
61 1
n.
TEST YOU R KNOWL EDGE 1 4. 1
Given (xx +yy + 11 zz +tt + 33) (32 :) , find x, y, z, t, 6 1 2 1 �1j , find 3A  4B. 2. IfA J = �  �t nd B � 1 5 3 II 2 2j 3. (a) If Am n and B p are two matrices, state the conditions for the existence of (iii) BA (i) A + B (ii) AB (iv) AB + BA. (b) Matrix A has x rows andx+ 5 columns. MatrixB hasy rows and l1 ycolumns. BothAB and BA exist. Find x and y. 2 3 4 4. If A � � � �1j and B � 1 2 3 , form the products AB and BA and show that AB " BA. 1 1 2 o 0 2 1 2 5. If A J� � �1 and B � 1 0 , obtain the product AB and explain why BA is not defined. l2 3 4 j 2 1 �
L
Xq
x
j
6.
7.
8.
9. 10. 11.
j
IfA J= � � = �tnd B J � � = �1j , verify that AB � BA � I3' l3 0 l  6 9  4j compute AB  BA, whereA � [! �l and B � [� �l l
If A J� � � �1 and B � I� � �1 , find AB or BA, whichever exists, l 1 0 1j l3 1 0 5 j IfA� [_ � �l B [; �l and C � [; �l , verify that (AB)C �A(BC) andA(B + C) � AB + AC,
IfA � [� �l
,
�
[� �l show that (2I  A)(10l  A) � 9L (a) IfA + B � [�  �l and A  B � [� �l , calculate the product AB, l
(b)
and I �
If AB :::: A, BA :::: B where A and B are square matrices of same order, then show that A2 :::: A and B' � B,
612 12.
13.
DISCRETE STRUCTURES
� �] , show that A2  5A + 7I :::: where I is a unit matrix of second order. 0  tan �1 c� sin a] . IfA = show that I + A = (I A) [sm cos J tun 2
If A :::: [
l
0,
_
Cf.
o
14.
Show that A is an involutory matrix where A =
15.
Given A =
16.
18.
19.
20.
21.
22.
23.
1 1 1 1
1
(i) [21  01
17.
l! �3  �] B JI�I 24 '
Evaluate :
CI.
_
'
li
CI.
 52
CI.
1 1 1 1
]
2 � , show that AB = AC.
IfA = [� �] , B = [� �] where i' =  l, verify that (A + B)' = A' + B'.
IfA = [�  �] , B = [i �] , verify that (i) (A + B)' "A' + 2AB + B' If A =
l�
] II ]
(ii) (A + B)(A  B) " A'  B'.
3 3 , B = 0 ! , C = [21 02 ;  i] , verify that A(BC) = (AB)C. 1 2
1 1
o
1
_
IfA = [� !] , B = [! �] , C = [� �] , verify that (i) A(B + C) = AB + AC (ii) (B + C)A = BA + CA. 3 1 2 1 If A = 0 ; , B = � 0 , verify the result (A + B)' = A' + AB + BA + B'. 3 1
l� ] l  �] 2 If A = l� �] , compute A'  4A  51. 2 2 3] Show that A = l� 4 satisfies the matrix equation A3  A2  18A  301 :::: Prove that A3  4A'  3A + 11 1 = where A = l� � !] . 1
(a)
(b)
1 1
1
0,
O.
l� j
MATRIX ALGEBRA 24.
25.
26.
27.
28.
29.
30.
613
o 1 1 3 , 5x + 6, find [(A), where A = 1 0 [Hint. [(x) = x'  5x + 6 [(A) = A'  5A + 61 = A'  5A + 6Is] " COS. sin a] , show that An :::: [ cos. na sin na] , where n IS. a pOSItive . . . If A :::: [SIn mteger. cos a  sm na cos na 1 O 0 0 O If E = O l and F = [ 1 0 0 , calculate EF and FE and show that E'F + FE' = E. o 0 0 1 0 ] If A = [� �] , prove that An = [k; nkknn 1 for all positive integers n. 1 O 1 n l.n(n  1)1 IfA = 1 1 , then An = 0 1 2 � j for all positive integers n, o 1 0 0 IfA = (� �J, 1 2 :::: (� �J and n is a positive integer, then prove that (aI 2 + bA) n :::: anI2 + nan1 bA. Factorise the matrix A :::: I� �  �l into the form LV, where L is lower triangular with each l3 7 4] diagonal element one and U is upper triangular matrix.
f[ x
I ( ) = ex' 
CI.
[� J l� J CI.
[
J
Answers
2. 3.
(a) (i) m = P. n = q (iv) m = n = p = q
j
(ii) n = p (b) = Y = 8
x 3, 1 12 11 !1 , BA J= i 7 8 4. AB J� ; l 2 2 4] l2  1 5 5. AB :::: I� = �l, no. ofcOls. in B :f no. ofroWS in A l7  8 ] 43 4] 8. BA J1� � 1� �� 1 7. [ 3  43 l 4 3 3 9] 16. (i) [: �] (ii) [7(x' + y' + z') + 6xy + 8yz] i  11 103 22. Os 3 24' 5 4 4 2 1 0 1 19/5  32/5 , 30. 7/5 0 327/19 3/5 41/ 19
II x
�m
lj
j
(iii) q = m
1 1 . (a)
[20  26] _
DISCRETE STRUCTURES
614
(P.T. U., M.C.A. May 2008, May 2007)
1 4.7. TRANSPOSE OF A MATRIX
Given a matrix A) then the matrix obtained from A by changing its rows into columns and columns into rows is called the transpose of A and is denoted by A' or A'. 'o, ,�,'e '" 0
[i � ; ;] .
Clearly, (i) if the order of A is
'>en A 0
[i r
m x n then order of A' is n x m.
(ii) (i, j)th element of A' = (j, i)th element of A. In symbols) if A = [ai)m x n) then A' = [ bi) n x m) where
bij = aji"
1 4.S. PROPERTIES OF TRANSPOSE OF A MATRIX If A' and B' denote the transposes of A and B respectively, then
(a) (A'Y = A i.e., the transpose of the transpose of a matrix is the matrix itself. Let A = [ai)m n be any matrix) Order of A' is n x m. . . Order of (A')' is m x n x
..
The matrices (Ai' and A are of the same order.
Also, (i, j)th element of (A')' = (j, i)th element of A' = (i, j)th element of A. Since (A')' and A are of the same order and their corresponding elements are equal (A')' = A.
(b) (A + BY = A' + B' i.e., the transpose of the sum of two matrices is equal to the sum of their transposes. Let A = [ai)m n and B = [ bi) m n b e two matrices. Order of A + B is m x n. . . Order of (A + BY is n x m. A' and B' are both n x m matrices. Order of A' + B' is n x m. x
x
Thus (A + BY and A' + B' are of the same order. Also (i, j)th element of
(A + BY = (j, i)th element of A + B
= (j, i)th element of A + (j, i)th element of B
= (i, j)th element of A' + (i, j)th element of B'
= (i, j)th element of A' + B' Since (A + BY and A' + B' are of the same order and their corresponding elements are equal. (A + BY = A' + B'.
(c) (ABY = B'A' i.e., the transpose of the product of two matrices is equal to the product of their transposes taken in the reverse order. Let A = [ai)m n and B = [ bi) n be two matrices conformable for the product AB. Order of AB is m x p . . Order of (ABY is p x m. Order of B' is p x n and order of A' is n x m. . . Order of B/A' is p x m. x
X
p
Thus, (ABY and B'A' are of the same order.
MATRIX ALGEBRA Also
615
(i, j)th element of (AB)' = U, i)th element of AB = Sum of the products of the elements of ph row of A with the corresponding elements of ith column of B = Sum of the products of elements of ith column of B with the corresponding elements of the ph row of A = Sum of the products of elements of ith row of B' with the corresponding elements of the ph column of A' = (i, j)th element of B'A'.
Since (AB)' and B'A' are of the same order and their corresponding elements are equaL .. (AB)' = B'A'.
1 4.9. SYMMETRIC MATRIX A square matrix A
= [ai)
is said to be symmetric if A' =
A i.e., if the transpose of the
matrix is equal to the matrix itself.
Le.,
Thus, for a symmetric matrix A = [aij L aij = (i, j)th element of A = U, i)th element of A
1 For example,
2
aji
3 h b f
2 3
g�l
are symmetric matrices.
1 4.10. SKEWSYMMETRIC MATRIX (OR AntiSymmetric Matrix) A square matrix A = [ai) is said to be skewsymmetric if A' =  A i.e., if the transpose of the matrix is equal to the negative of the matrix. Thus, for a skewsymmetric matrix A
Le.,
= [aij L aij =  aji
(i, j)th element of A = negative of U, i)th element of A Putting j = i, aii =  aii ::::::} 2aii = 0 or aii = 0 for all i. Thus, all diagonal elements of a skewsymmetric matrix are zero. For example,
l1 �3  �1
 �l,ll � g 0
�
f

Jl
are skewsymmetric matrices.
0
1 4 . 1 1 . EVERY SQUARE MATRIX CAN UNIQUELY BE EXPRESSED AS THE SUM OF A SYMMETRIC MATRIX AND A SKEWSYMMETRIC MATRIX
Proof_ Let A be any square matrix. B = t (A + Ai and C = t (A  Ai Consider B + C = t (A + Ai + t (A  Ai = A Also B' = t (A + AT = t [A' + (AT] = t (A' + A) = t (A + Ai = B
616
DISCRETE STRUCTURES ::::::} ::::::}
B is a symmetric matrix. C' = t(A  Ai' = t [A'  (Ai'] = t(A'  A) =  t (A  Ai =  C C is a skewsymmetric matrix.
Hence every square matrix A can be expressed as A = B + C, where B = t (A + A') is a symmetric matrix and C = t(A  A') is a skewsymmetric matrix. To prove that there is only one way in which A can be expressed as the sum of a symmetric matrix and a skewsymmetric matrix) suppose
A=P+Q is another such representation in which matrix.
... (1) P is a symmetric matrix) and Q is a skewsymmetric
P' = P and Q' =  Q A' = (P + Q)' = P' + Q' = P  Q Then A + A' = 2P and A  A' = 2Q From (1) and (2), P = t(A + Ai = B and Q = t (A  Ai = C so that P + Q = B + C
... (2)
Hence the result.
1 4.12. ORTHOGONAL MATRIX
A square matrix A is called an orthogonal matrix if AA' = I = A'A. 1 4.13. FOR ANY TWO ORTHOGONAL MATRICES A AND B, SHOW THAT AB IS AN ORTHOGONAL MATRIX
Proof. A and B are orthogonal matrices AA' = I = A'A and BB' = I = B'B =>
Now
::::::}
... (1)
(AB)(AB)' = (AB)(B'Ai = A(BBi A' = AIA' = (AI) A' = AA' = I
(1)] [Using (1)] [Using
AB is an orthogonal matrix.
I
ILLUSTRATIVE EXAMPLES
Example 1. IfA and B are symmetric matrices, prove that AB BA is a skewsymmetric matrix. Sol. A and B are symmetric matrices ... (1) A' = A and B' = B =>
(AB  BA)' = (AB)'  (BA)' = B'A'  A'B' = BA  AB =  (AB  BA) (AB  BA) is a skewsymmetric matrix.
Now
[.:
(A  B)' = A'  B'] [.: (AB)' = B'A'] [Using (1)]
MATRIX ALGEBRA
617
Example 2. Express the following matrix as the sum of a symmetric and a skewsymmetric
matrix,
l� :l l 251 7 :l 7 3 0
Sol. Given matrix is
A=
l 711 2 55l 1 5 l 2 5 + 5l �l A' =
3
0
+
3 0
4
=
4 +0
9 6
9 6
4 l
1 t2 B = t(A + A,) = 325 1 + 1 7 2 415 0 l50 05 A  A' = l 27 51 0  4 55l 4  4
l  �l 2 2 �l 1 2 !l 2 2 �l 3
.@. 2 3
=
33
5
0
. .
C = t(A  A,) =
"2
0
"2
9
5
0
"2
.@. 3 2 3
A=B+C= where
5
+
"2
II
0
2
B is a symmetric and C is a skewsymmetric matrix. Example
Sol. Here
3. Show that A =
A' =
AA' =
=
l l l
l
cos 8 0
 sin 8
1 0
0
eas 8
0
sin 8
l
o
0
0
cos 8
1 0
 sin 8
o
l ll
sin 8
eos 8
cos 8
sin 8
1 0
sin 2 8
+ eos2 8
0
+ sin 2 8
0
l
sin 8 O is orthogonal. Find the value of 1 A I . cos 8
 sin 8
0
eas 8
2 cos 8
o
� o
0
0
�
 sin 8
0
cos 8
l
l
l J� � � J lo 0 l
=I
DISCRETE STRUCTURES
618
Since AA' = I) A is an orthogonal matrix. cos 8
o sin 8 0 1  sin 8 0 cos 8
IAI=
0
1
Expanding by second column)
IAI=
1
C?S 8 sin 8 = cos2 8 sin2 8 = + cos 8
 SIn 8
1.
1 4. 1 4. ADJOINT OF A SQUARE MATRIX The adjoint of a square matrix is the transpose of the matrix obtained by replacing each element of A by its co· factor in A
l
I I.
l
Adjoint of A is briefly written as adj A. Thus, if A =
a l b, CI a2 b2 c2 , then adj A = Transpose of
�
�
�
l
A'
AI A2 A,
B,
C,
where the capital letters denote the cofactors of corresponding small letters in
IAI=
l
a 1 b I c1 a2 b2 C2 aa ba ca
1 4.15. AN IMPORTANT RELATION BETWEEN A SQUARE MATRIX A AND adj A
[
.
Theorem. If A is an nrowed square matrix, then A(adj A) = (adj A) A = I A a ll a'2 . . . . . . aln
Proof. Let
A=
��.1 ��� :::::: ���
[
anI an2 . . . . . . ann
then
adj A = Transpose of
[
A ll A 21 A 12 A 22 ... ... A m A 2n
A ll A 21 ...
A nI
. . . . . . A nl l . . . . . . A n2 ..... . . . . . . . A nn
I In '
. . . . . . A ln l n :::::: �. . . . . . . A nn (where Ai is cofactor of aiJ in
�
)
(i, j)th element in A (adj A) = ai lAJ I + ai2AJ2 + ai3AJ 3 + ...... + ainAJn = In the product A (adj A), each diagonal element is element is O.
I
A
{6
AI
I
I A I)
if , = J if i " j
and each nondiagonal
MATRIX ALGEBRA
r: ° iAl
A(adj A) = Similarly, Hence
AI = B, = C, =
IA
.
IA I
II �l , I� � ;l Jl�� �� �� l II 2 3 I � � 1  1, 1 � ; 1 3, 1 � � 1  I � � 1  7, 1 i ; 1 3, 1� �1 5  I i � I = 3, 1 � i 1 1 � � 1 5, + + 2( 1) + 1( 7) + 3(5) 6 I l l l �: �: l J=�l 5 3; � 1l IIl�2(I 2�1) +3�l1(7)ll 5� +33(5); �2(3)1l + 1(3) + 3( 3) 2( 1) + 1(5) + 3( l) 1( 7) + 2(5) 3(3) + 1(3) + 2(3) 3( 1) + 1(5) + 2( 1) 3(1) 1( 1) ++ 2(7) + 3(5) 1(3) + 2(3) + 3( 3) 1( 1) + 2(5) + 3( 1)l I�l � � l J�l �0 �l I I I o°6 o Jl � 3; �1l l�l 1 2� 3�l l7(2)51(2) ++ 3(3)3(3) + 5(1)1(1) 7(1)  1(1) ++ 3(1) 1(2) 7(3)  1(3) + 3(2)  1(3)l 3(1) + 5(2) + 3(2) + 5(3) 5(2) 3(3)  1(1) 5(1) 3(1)  1(2) 5(3) 3(2)  1(3) 6l0 °6 °0l 0I 01 �l I I I o 0 6 I0 0
Example
Sol.
.
619
4. If A
J� i 2
3
verify that A(adj A) = (adj A)A = I A I
A=
a3 b3 c3
A2 = 
=
=
B2 =
=
=
B3 = C3 =
C2 =
=
= a,A, b,B I C, C, = A I B, C I adj A = Transpose of A 2 B2 C 2 A 3 B3 C3
=
=
=
A(adj A) =
=
=
=
= A
=
(adj A)A
=
=
Hence A(adj
=
6
A) = (adj A)A = I A
A3 =
= A
I I.
AI B, C , C 2 C3
=1 =
=1
I.
DISCRETE STRUCTURES
620
1 4.16. SINGULAR MATRICES AND NONSINGULAR MATRICES
A square matrix A is said to be singular if I A I = 0 and non·singular if I A I
# O.
1 4. 1 7. INVERSE (or Reciprocal) OF A SQUARE MATRIX
(P. T. U., M.G.A. May 2007, Dec. 2005)
that
Let A be an nrowed square matrix. If there exists an nrowed square matrix B such
AB = BA = I then the matrix A is said to be invertible and B is called the inverse (or reciprocal) of A.
Only square matrices can be invertible. From the definition, it is clear that if B is the inverse of A, then A is the inverse of B. 3. Inverse of A is denoted by AI, thus B :::: AI and AA1 :::: A1A :::: 1.
Note L 2.
1 4.18. THE INVERSE OF A SQUARE MATRIX, IF IT EXISTS, IS UNIQUE
Let A be an invertible square matrix. If possible, let B and C be two inverses of A. Then AB = BA = I (By definition of inverse) AC = CA = I Now B = BI = B(AC) = (BA)C [.: Matrix multiplication is associative] = IC = C Hence the inverse of A is unique. 1 4.19. THE NECESSARY AND SUFFICIENT CONDITION FOR A SQUARE MATRIX A TO POSSESS INVERSE IS THAT I A I # 0 (i.e., A is NonSingular)
Proof. (a) The condition is necessary :
Given that A has inverse, to show that I A I Let B be the inverse of A, then AB = BA = I => I A I I B I = I B I I A I =l
#0 =>
(b) The condition is sufficient :
Given that I A I
I AB I = I BA I = I I I => I A I # O .
to show that A has inverse. I A I #0 dA Consider B= a j !AT _ ( I A I I) AB = A adj A _l_ (A(ad]' A» = _ IA I IAI IAI [.: A(adj A) = I A I I] =I dA _ ( I A I I) BA = a j A = __ «ad]' A)A) = _ IAI IAI IA I [.: (adj A)A = I A I I] =I Thus AB = BA = I . . and AI = B = adj A The mverse of A eXIsts !AT # 0,
( ) ( ) 1
1
1
'
MATRIX ALGEBRA
AI
=
Note. From the above theorem, we conclude that A has inverse if and only if
adj A
IAI
621
I A I t: O
and then
.
= l� ;l A = l� 331 441l l C'l + = 032 331 441 = 32 011 441 = 1 1� � 1 = 3 2= 1 1 '" 0 = = 1 31 i l = 1, =  1 31 i l = 1, = 1 33 44 1 _ 0 B, = I � i 1 =2, B2= 1 � i 1 = 3, B3 =  I � ! 1 = 4 I � 13 1 = 2, = 1 � 13 1 = 3, = I � 33 1 = 3 2l B2B,B3 l lB, B2 B3 l l221 133 3�l = = I�I l � 133 3�l = 1) = = l� 331 m� 331 ;l = l3(2(0(333)))3(3(1(222))) +++ 4(4(1(000))) 2(3(0( 3)3)3) 3(3(1( 3)3)3) +++ 4(4(1( 1)1)1) 3(2(0(444))) 3(3(1(444))) +++ 4(4(1(111)))l J � = i �l l2 2 3 3
3
Example 5. IfA
1
'
=
Sol. (i)
A
I
A
a, a2 a3
b, b2 b3
(ii) show that A3 = A1
c2 c3
Operating C2 � C2 C3
I
Expanding by third row
o
I
(i) find Al
I
:.
A' exists.
A,
A2
A3
C, =
C2
C3
..
A, adj A Transpose of A A3 A'
(ii)
A2 A x A
adj A = =
C C 2' C3
=
A , A2 A 3 C,
C2
=
C3
C':
I
A
I
l 30 4 1 2 2 4(3)l l 3(3)0(3)4 1(0)(0) 4(2)2) 3(0(4)4)4( 1( 1)1) 40(2)(2) 3(0(44))4(0) 1(0) 3(3) 0(3) 2(3) 2(0) 3(2) 2(4) 2( 1) 3(2) 2(4)  2(0) 0I 001 0°1l I
DISCRETE STRUCTURES
622
A4 = A2 A2 = X
= =
=>
Le.,
+
o
+ +
+
+
+ +
=1
A4 = I A.A3 = A3. A = I 3 A is the inverse of A A3 = AI
Now =>
+ + O(
1 4.20. IF A IS INVERTIBLE, THEN SO IS A' AND (A')'
=
A
A is invertible => AI exists and AA1 = AI A = I By definition of inverse of a matrix) AI is invertible and inverse of AI is A. (A1)1 = A. 1 4.21 . IF A AND B BE TWO NONSINGULAR SQUARE MATRICES OF THE SAME ORDER, THEN (AB)' = B' A'
Proof. A and B are nonsingular square matrices
I A I # O, I B I # O I 1 A and B both exist. Also I AB I = I A I I B I # O => AB is nonsingular and (AB)l exists. Now (AB)(B1A1) = A(BB1) AI = AIA1 = (AI)A1 = AA1 = I (B'A')(AB) = B'(A' A)B = BIIB = Bl (IB) = BIB = I Again From (1) and we get (AB)(B'A') = (B'A')(AB) = I Inverse of AB = BIAl i.e., (AB)l = BIAI => =>
... (1) ...
(2)
(2),
Le.,
Note.
The result can be extended for more than two matrices.
Putting ..
l (AIA2A3 ..... An_1A)1 :::: Anl An_II ..... ASIA21 A 1A I :::: A2 :::: As :::: ...... :::: A. I :::: An :::: A (A.A.A..... n (An)I
=
(kI)n.
we have
times)l :::: AI . AI ..... n times
1 4.22. IF A IS A NONSINGULAR SQUARE MATRIX, THEN SO IS A' AND (A')'
A is a nonsingular square matrix => I A I # 0 and AI exists .. I A' I = I A I #O => A' is nonsingular and (A;' exists.
=
(A'),
MATRIX ALGEBRA
623
AAI = AI A = I Now Taking transpose = (AAI), = (AIA) , = l'
[.:
(A')'A' = A'(A')' = I => Inverse of A' = (AI) , i.e., (A,)l = (AI)',
(AB)' = B'A']
1 4.23. IF A AND B ARE TWO NONSINGULAR SQUARE MATRICES OF THE SAME ORDER, THEN
adj (AB) = (adj B) (adj A) Proof. A) B are nonsingular => I A I '" 0, I B I '" 0 I AB I = I A I I B I ", O ::::::} AB is nonsingular => (AB)l exists. Now (adj AB) AB = I AB I I ... (1) Also, (adj B) (adj A) AB = (adj B) (adj A . A) B = (adj B) ( I A I I) B = I A I (adj B) IE = I A I (adj B) B = I A I I B I I = I AB I I
...(2)
(2),
From (1) and (adj AB) AB = (adj B) (adj A) AB Postmultiplying by (AB)I , we have (adj AB)(AB)(AB)l = (adj B) (adj A) (AB)(AB)l (adj AB) I = (adj B) (adj A) I adj (AB) = (adj B) (adj A).
II j l 02 02 24l l61 40 612l � 0 m� 0 2 0 0 4 + 21 l� 40 6I!l 3 I 02 2�l + 2 �I 1 00l 0 I� 0 I 1 l 3 + 2 4  6 + 2 12 6+126 ++ 00l l00 00 l l66+0 46+2 0 0 � 0 A2  3A + 2I = O
Example 6. If A = Sol.
. .
2
0
0
2
2
4 , 2
0
verify A2  3A + 21 = 0 and hence find A1
A2 = AA =
A2  3A
=
o
=
o
o
=
=>
o
To find AI , multiplying (1) by AI , we get A
=
 31 + 2AI = 0
... (1)
624
DISCRETE STRUCTURES
+l 0
2A1 =  A 31
=>
=
Al =
l l I H I
01 ! 2 2 3 l 0 0 2 0� 0 112oo 2112 . +
TEST YOU R KNOWL EDGE 1 4.2
L 2.
If A is a square matrix, prove that (ii) A  A' is skewsymmetric. (i) A A' is symmetric Prove that (i) if A, B are symmetric, then so is A B (ii) if A, B are skew symmetric, then so is A B (iii) if A is a square matrix, then and A'A are both symmetric (iv) if A is symmetric, then B' AB is symmetric. (v) if A and B are symmetric matrix of order n, show that AB BA is also symmetric. Express each of the following matrices as the sum of a symmetric and a skew symmetric matrix : aab (ii) e b b . (i) � � = �1
+
+ +
AA'
3.
j
�j
6 0 7
4.
 �tnd B J� � , verify that (AB)' B'A'. 6 oj lo 1 (b) If A [! n B [; n find (A B)T If A [� ;] verify that (A")' (Ai'(a) IfA J �
� l4
=
6.
6.
7.
8.
=
,
_
10.
=
+
=
=
�4 �4 �lj is its own adjoint. Given the matrix A :::: I� � �l . Compute adj A and verify that j Show that the matrix

II

6 12
3
A(adj A) = (adj A)A = I A I 1.
l3 3 j
Find the inverse of each of the following matrices : 2 5
� � �j l If diag. D
=
If A
=
t
1 2 1 2 1 d,d2d3 " 0, (ii)
(i)
9.
j lc a e
+
where � � ; 1 prove that [d l ' d2, dsl,
2 2
_
 lj
,
A' =
then show that A'.
II 334 343j .
(iii) 1
Dl
=
diag.
1
[d,' , d21 , d31].
MATRIX ALGEBRA 11.
12. 13,
14,
625
JI i � �1j ,BJ�I � _�1j I I  sin ] [ Show that [cos sin cos IfA
4
1
2
8 8
4
(i)
(ii)
8,
[
3
"2 4
3
J
4
"2
2
3 3 3 7
+
"3
"3
"3
1
1
"2
1
"3
2
0
verify that (AB)l = B1A1  tan 812
1
tan 8/2
Prove that (ABC)l = C lB1Al
Prove that the matrix
,
1
=
8 8
"3
3,
3
1
"2 0
1 3
2 2
"3
 �J
iI
I[
] [
1 tan 812
tan 812 l
(i)
1
1
3
4
4
1
3
11
2
(ii)
'
is orthogonaL Answers
0 a t (a  c) t (b  c) t (a + c) t (b + c) 0 b t(a + c) t(b  a) t(a + b) + t(c  a) c o t (c  b) t (a  b) t (b + c) t (a + b)
2
r
3
1
7
4
4
4
1
1
5
4
4
4
4
4
1
1
5
6
1
13
2
4
4
4
4
4
I
4, (b)
(iii)
l �
[� �] 1
j
3 3 1 1 0 1
O.
1 4.24. SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS BY MATRIX INVERSION METHOD OR MATRIX METHOD
(a) Nonhomogeneous system of linear equations
Let AX = B be a given system of n linear equations in nunknowns. Then (i) if I A I '" 0, we say that the system is consistent and has a unique solution given by
l X = A
B.
(ii) if I A I = 0 and (adj A)B = 0, we say that the system is consistent and has infinitely many solutions. (iii) if I A I = 0 and (adj A)B '" 0, we say that the system is inconsistent and has no solution. (b) Homogeneous system of linear equations
Let AX = 0 be a given homogeneous system of n linear equations in nunknowns. Then (i) if I A I '" 0, we say that the system is consistent and has a unique solution or trivial solution or = 0) y = 0) z = o. (ii) if I A I = 0, we say that the system is consistent and has infinite solution or non trivial solution. x
Remark. A
homogeneous system is always consistent.
626
DISCRETE STRUCTURES
1 4.25. IF A IS NONSINGULAR MATRIX, THEN THE SYSTEM OF EQUATIONS AX = B HAS A UNIQUE SOLUTION GIVEN BY X = A1 B
A AX = B AI A(A''A(A)X)X == A' BB IXX ==AIAIBB AX = B X, X2 AX, = B AX = B, AXAX,2== AXB 2 A(AI' A(A) X,X,) == (AAI' A(A)XX2) IX,Xl == XI�2
Proof. Let
Now
is non·singular.
exists.
. .
A'
Thus the system of equations Uniqueness :
Let => => => =>
and
Example
1.
I
Premultiplying by
has a solution given by
be two solutions of
I X = AI B.
then
i.e., the solution is unique.
ILLUSTRATIVE EXAMPLES
Solve the following equations x +y +z = l + x 2y + 3z = 6 x + 3y + 4z = 6, by matrix inversion method.
I
(8
9)
The given system has a unique solution given by
To find
AX = B)
A = l� 231 H X = m , B =lil I A I = 111 231 341 =   (4  3) + (3 2) =  1  1 + 1 =  1 X = AI B . .(1) AI: Al = 1 � ! 1 =  =  1 ; A'2 =  I � ! 1 =  (4  3) =  1 A'3 = 1 � 23 1 = 3 _ 2= 1'' A21 =  I ; � I = (4  3) = 1 A2 = 1 � 41 1 = 4 _ 1 = 3'' �3 = 1 � ; 1 =  (3  1) =  2
Sol. In matrix notations) given system of equations can be written as
. .
AI
8
9
where
,, 0
MATRIX ALGEBRA
I � ;I = 3 2 = 1 ; A33 = 1 � � 1 = 2  1 = 1 A3I =
A32 =
l
 I � ; I =  (3  1) =  2
627
l  l 3I ll l � = J=� ;  �l J�I  ; �l ll I
All A '2 A 13 T 2 adj A = A 2I A 22 A 23 =  1 1 2 1 A 3I A 32 A 33 
A AI = a l I Hence, from (1),
2
2 1
1
x = 1, y =  5, Z = 5 is the required solution. => Example 2. Solve the following system of simultaneous linear equations by matrix method. x + y + 2z = 4 2x  y + 3z = 9 3x  y  z = 2. Sol. In matrix notations, given system of equations can be written as AX =
where
A=
B,
l� Jl X = m, B = lil 1 1 1
1
1
I A I = 2 1
3
2
3
1 1
= (1 + 3)  ( 2  9) + 2 ( 2 + 3)
X B
= 4 + 11 + 2 = 17 # 0 The given system has a unique solution given by = AI To find AI : A11 = 11  1 = 1 + 3 = 4 ''
1
31
... (1)
1
1 =  ( 1  3) = 4 1
1 11 23 1 = 3 + 2 = 5 '' A32   I 21 � I =  (3  4) = 1 A33 = 1 � _ � 1 =  1  2 =  3
A31 =
_
_
628
DISCRETE STRUCTURES
adj A =
From (1),
l
A ll A '2 A 13 A 21 A 22 A 23 A 3 1 A 32 A 33
rl =
4 1 111 74
Jl
4 1 5 17 17 171 1 7 1 11 AI = ��� = 17 7 4 3 171 174 317 17 17 17 16 9 + 10 17 4 1 5 17 + 2 1717 1711 717 171 44 63 171 174 317 4 + 3617 6 3174 17 17 17 17 17 x 1) y  1) z 2 ) is the required solution.
H
X =A' B l�l =
lil
�l
=
=
=
Hl
= = = ::::::} Example 3. Solve, by matrix inversion method. x +y +z = 6 x + 2y + 3z = 10 (P.T.V., M.e.A. May x + 2y + 4z = 1. Sol. In matrix notations, given system of equations can be written as AX =
where
A=
l� !l X =l�l B =ll�l 1 2 2 1 1 1 123 124
B,
6)  (4  3) + (2  2) 2  1 1 ,, 0 The given system has a unique solution given by AI All 1 4 1 To find AI : ' A'2  I � ! I � A'3 1 � � 1 A21  I �I � A22 1 � 4 1 ' A23 1 1 1  (2  1) 1 � � 1  (3  1)  2 A32 A31 1 1 ' 1 3 1 �
Here
IAI
. .
=
=
3
=
=
=
1
1
=
(8 
X B
=
=
=
=
=  (4  3) =  1
=2 2 = 0 ;
=
=  (4  2) =  2
= 4_ 1 = 3 '
=_
1
=
=
=3_2= 1 '
=_
1
=
=
= 8_ 6 = 2 '
2007)
... (1)
MATRIX ALGEBRA
629
l
A ll adj A = A 21 A 31
A' =
lXl
=

IAI
B
+
Y = z x
II 2 2 l 1 3 2 l 0 1 1 T
 l �1  �l 2l 23 2lll106l l126 23002ll l227l 1 0  1 1 1 0 10 1 J� l0
adj A
X = A'
From (1),
A 13 A 23 A 33
A '2 A 22 A 32
=
=
+
=
+
7) Y = 22)
z
=
9
9) is the required solution.
TEST YOU R KNOWL EDGE 1 4.3
L
Apply Matrix method or matrix inversion method to solve the equations :
((ii)) x3x yy 2z= 0=, 2x3, 2x5y3y3zz ==13,, xx 2y2y ==24 ((iivi)) xx y2y z::3z:: 1=, 6,2x2x3y4y2z::=:: 2,7,3x3x 3y2y 4z::9z =:: 114 ((vvi)) 57xx 6ylOy 4z5z==15,42,7x13x4y6y3z2z= 1=9,31,2x llx 6z14y= 468z = 63 ((vviii)) x2x 3yy 6z::4z =:: 2,2, x3x y3y = 4z::2z :: 9,7, x5z4y8 = 2z::5x :: 73y (i(xx)) x2yzyzx =x6.y6::,:: x3xyz, 3yz= 2.2zx2, x 4xy2y :::: 139zxyz,= 15.6yz2. tzx xy :::: 17xyz. [Hint. w] y = 9,= 2x12, x5ylOy72 = 5z,= 12, x z 1=0z = 12. M.C.A. May (xSolvei ) xlOxthey yfollowing system of equations by matrix method 2Xl x2x2 xs::3xs :::::: 19 3 0 :: :: , by matrix method. Xl x2 X Solve by matrix method xyz=l=yzx=zxy +
+
+
+
Z
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
X
+
+
+
+
Y+Z
+Z
+
+Z
+
+
+
+
3.
+
+
Divide the equations by xyz and put ...!.. :::: u, ...!.. ::::
+
2.
+
+
Z
Z
+Y+
+
+
(xi)
+
Z
+
+
+
+
+
+
+
+
+
+
+
+
+Z
+
Z
2x
+Y
+Y+
o.
v,
� :::: Z
(P.T. U.,
2008)
630 4.
DISCRETE STRUCTURES
(a)
Solve :
x + y z :::: 21 2xy+z x y 2z:: :: 3 by matrix method. Solve for x, y, z, the following equations x+y+z:: :: 3 2xy+z:: :: 22 x2y + 3 z 5. these Does the system x 2y + 3z :::: 2x + y 4z:::: xy + z:::: have nontrivial solutions ? If so, find solutions. Determine the value of). for which the equations 3x2y + z:::: 14y + 15z:::: + 2y 3z:::: have nontrivial solutions.. Find the solutions. Find the values of for which the equations l)x + (3\and+ lfind)y + the2\zratios(\oflx,)y,x +z when(4\ 2)hasy + (the\ +smallest 3)z 2xofthese+ (3\ values. + l)y +What3(\ lhappens )z (are\consistent, when has the greatest of these values ? Prove that the equations x:::: ay + z, y + z:::: + y, are consistent (have nonzero solution), then + 1 Test the consistency of the following equations : xTest+for2y consistency + 3z + 14 the5xfollowing + 9y + llzsystem+ 56 and3findx + 5ythe+solution, + 16 it2xexists+ 7y +: 5z + 30 3x + 3y + 2z 1, x + 2y 4, lOy + 3z 2, 2x 3y z 5, xx 3,1, 2, zz 23 xx 1y , z 1 z 1 xx 3,1, 4,2, zz 16 x 3, 4, z 1 x 2, 1, z (ix) x 1.2, 2,2, z 3,2 xx:::: 11,, y::::3,1,z z::5 :: 1 (xi ) x 1, 1, z 1. (x::x):: x1, y 1::,:: 2,z:::: 3 z 1, 1, z 1 5. Yes\ x5; x::; x1,:: 2y::k,2,:: 2zk,5z::k1, z:: where isxarbitrary. 0Inconsistent. ,3; when x:::: y:::: z; whenConsistent3, equations become identical. ; x:::: 2, y :::: 1, z:::: 4. + =
(b)
=
0,
0,
0
0, AX
6.
7.
0, X
0
A
=
0,
=
0,
=
0
A
A
if
8.
a3
:::: Z
:::: o.
ax,
X
9.
=
0,
=
0,
Z
=
0,
=
0,
if
10.
=
=
=
=
Answers
1.
(i)
=
(iv)
=
(vii)
=
=
0,
Y =
=
=
=
Y =
Y =
=
"2 1
'
=
1
'3
(a)
=
Y =
Y =
=
(iii)
=
Y =
(v)
=
Y =
=
(vi)
=
Y =
(viii)
=
Y =
(xi)
=
Y =
=
7. A ::::
=
(b)
=
k,
6.
=
=
=
1
"2
=
=
0,
=
=
Y=
Y =
=
=
3.
2.
4.
(ii)
Y =
=
=
Y =
=
k
k,
4k,
A :::: 0,
9.
1 4.26. RANK OF A MATRIX
A ::::
10.
(PT. u" M.GA May 2007)
Let A be any m x n matrix. It has square sub·matrices of different orders. The determinants of these square sub· matrices are called minors of A. If all minors of order (r 1) are zero but there is at least one nonzero minor of order r, then r is called the rank of A. Symbolically, rank of A = r is written as p(A) = r or r(A) = r.
MATRIX ALGEBRA
631
From the definition of the rank of a matrix A, it follows that :
(i) If A is a null matrix, then p(A) = O. r·: every minor of A has zero value.] (ii) If A is not a null matrix, then p(A) :> 1. (iii) I f A is a nonsingular n x n matrix, then p(A) = n
[.:
IAI
'"
0 is the largest minor of A.]
n x n unit matrix, then I In I = 1 '" 0 => p(In) = n. (iv) If A is an m x n matrix, then p (A) S minimum of m and n. (v) If all minors of order r are equal to zero, then p (A) < r. If I n is the
1 4.27. TO DETERMINE THE RANK OF A MATRIX A We adopt the following different methods : Start with the highest order minor (or minors) of A. Let their order be r. If any one of them is nonzero, then p(A) = r. If all of them are zero, start with minors of next lower order (r  1) and so on till you
get a nonzero minor. The order of that minor is the rank of A.
This method usually involves a lot of computational work since we have to evaluate several determinants. We generally find the rank of a matrix by reducing it to echelon form.
1 4.28. ECHELON FORM OF A MATRIX
A matrix A is said to be in echelon form if it satisfies the following properties: (i) All entries below the leading element are zero. (ii) The number of zeros before the first nonzero
number of such zeros in the previous row. For example A =
where as
B=
l� l�
0 3 0 0 3 3
�l �l
element in a row is more than the
is in echelon form
is nonechelon form
1 4.29. RANK OF A MATRIX BY USING ECHELON FORM The rank of a matrix is equal to number of nonzero rows in its echelon form.
II 2l 0
For example
r(A) = 3, where A = 0 3 0 . 0 0 4
632
DISCRETE STRUCTURES
I L LUSTRATIVE EXAMPLES
Example
(.) [ ,
1
2
1. Find the rank of the matrices
2
3
4
5
0
7
Sol. (i) Here A = r (A) S 2
]
[�
� � �]
_
is a 2 x 4 matrix.
(smaller of 2 and 4)
Operating R2 ; R2 + 2Rl we get ' A
[�
r(A) = 2
(ii) Here A =
l� 4 l 2
3
6
5
!
3
11
�l
which is in echelon form
1
(No. of nonzero rows in echelon form)
2 is a 3 x 3 matrix.
l� �l  l� �l 2 2 2
A
1
2 2
0
r(A) S 3
operating R3 ; R3  R2
which is in echelon form.
.. r (A) = 2 (No. of nonzero rows in echelon form) . Example 2_ Find the rank of the matrices (i)
l; 4
D [; ���
3
6
3
4 0
2
12
5
3
1
1
1
2
3
Sol. (i) Here
3
0
1
1 1
l
']
4
2
(ii)
.
.
7
A=
l
2
5
4
3
2 5
4 0
12
 ll
" [ 2
4
6
8
3
2
" 3 2 1
7
3
5
,
.
(P.T.V., M.C.A. May 2008)
 1 is a 3 x 4 matrix. 1
r (A) S 3 (smaller of 3 and 4)
l l
Interchanging C , and C4' we get 3
A 1 2 1
5
4 o
12
2

45
l
, operating R2 ; R2  Rl ' R3 ; R3  R,
MATRIX ALGEBRA

l l l l
..
(ii) Here ..
Operating
..
3
4
�l �l
0 1 4 0 2 8 6
r(A) = 2 A0
3
4
0 1 4 0 0 0
633
operating
R3 ; R3 + 2R2
which is in echelon form.
(No. of nonzero rows in echelon form) .
[!
2 3 01 4 3 �2 IS. a 4 x 4 matrix. . 2 1 8 7
r(A) R,  " 0 3 0 3 2 3 4  0 0  38 w,", ill in �,,'"" 'O
p(A) '" p(A : B)
The system has no solution. Case
II. If
A ", 3,
" may have any value, then p(A) = p(A : B) =
3 = number of unknowns
The system has unique solution. Case
III. If
A
=
3,
,, = 10, then p(A) = p(A : B) =
2
< number of unknowns
The system has an infinite number of solutions.
6. Solve the equations by using Guass Jordan method Xl + 3x2 + 2x3 = 0) 2X1  x2 + 3x3 = 0) 3xl  5x2 + 4x3 = 0) Xl + 1 7x2 + 4x3 = o.
Example
MATRIX ALGEBRA
641
Sol. In matrix notation) the given system of equations can be written as
R2  2RI' R3  3RI' R4  Rl' 3  2�1 R3 ; R3  2R2, R4 ; R4 2R2 1414 22 3 21 � � , o 0 2 3 Xl 3x2 2X3  7X2  X3 X3 7k Xl 1 1k, k, 7k, k
Operating
we get
7
_
_

p(A) =
=> Leo)
AX = 0
'
operating
+
which is in Echelon form
= number of unknowns
R 3

��2 R2 J�' �� lo 0
c,
... (4)
644
DISCRETE STRUCTURES
Now if c3" t: 0) from (4») the given system of linear equations is equivalent to a lx b 1y c1z = d1 b 2'y c2'y = d2'
+ + +
Using back substitution
y_
X
and
c3"z = d3" Z = d3''1c3''
1
b'2C3" a
{d2' C3"  C2' d3''l
1
b'c 2 3"
I
I
II
I
{d , b 2 c3  b , d2 c3
+ b,c2 d3  b2 c,d3 } I
II
"
b2' or cs" becomes zero. In such cases by interchanging the rows we can get the nonzero pivots. 2. Partial pivoting. From the first column of (2) [called the pivot column if :f 0, i 1, 2, 3] select the component with the largest absolute value. This is called the pivot. Then at the second stage i.e., from the second column of (3) select once again the component with the largest absolute value as the pivot. Continue this process. This procedure is called partial pivoting. [Refer example (1)] 3. Complete pivoting. If we are not interested in the elimination of in a particular order, then we can choose at each stage the numerically largest coefficient of the entire coefficient matrix. This requires an interchange of equations and also an interchange of the position of the variables.
Notes :
L This method fails if any one of the pivots
II
ap
ai
x, y, z

::::
Example 1. Solve the system of equations 3x + y  z = 3, 2x  8y + z = 5, x  2y + 9z = 8 using Gausselimination method. Sol. The given system is equivalent to
A
X=B [A : B] =
:. The augmented matrix is
l�
1 1 ; 8 2 9 i
3
< 5
8
1
Now we will make A as upper triangular choosing 3 as pivot 2 3 1R R ; R  , 3 3 3
R 2 ; R 2   R,
Now choosing
26
3
R3 ; R3 
3 0 0
1 1 26 3 7 3
3
5
7
3 28 3
7
as the pivot from the 2nd column
7 26
3
R2 
0 0
1 26 3 0
1
5
3 7
3 693
231
78
26
MATRIX ALGEBRA
645
From this we get
3x + y  z = 3 ; 5
26
y+z=7 3 3 Now by back substitution z =
and
693

78
231 Z =  . 26
1 5
26
5
26
y =  7   z =  7   =   :. y = 1 3 3 3 3 1
1
x = "3 [3  y + z] = "3 [3  1 + 1] = 1
and
x = 1, y = 1, z = 1 .
.. Example
2. Solve by Gausselimination method : 2x + 3y + z = 2, x + 2y  z = 6, 3x + z = 2,
Sol. The given equations in matrix notation are
AX = B
or
The augmented matrix is
[A : B] =
l
1 2 2 3 1 2 1 6 3 0 1 2
1
In First column, the component with largest absolute value is R3 ) we get
[A : B] =
l
3. So interchange R, and
3 0 1 2 2 3
3
0
o 2 o 3
1 i 2 16
_� !
3 i 3 1 i 2
3
3
In second column, the component with largest absolute value is and R3 ) we get
[A : B] 
3
0
0
3
0
2
1 1
2 2
3 4
3 16
3
9
operating R3 ; R3 
� R2 3
3.
So interchanging R2
646
DISCRETE STRUCTURES
3
0
1 1
0 3
From third row,
14

9
0
0
44
z �
=>
9
2 2
3 14
3 44
9
9
22 z = 7
z 2 From second row, 3y + "3 = "3
3y =
From first row,
�  % � �  �(

22 7
)
2 22 14 + 22 36 12 =  + = = = 3 21 21 21 7 3x + z = 2
14 + 22 22 3x = 2  z = 2 +  = 7 7 Hence, the required solution is x= Example 3. Solve 2x + 3y + z
=>
36 7
=>
4
y=
7
12 x= 7
12 22 4 Y= ,z= 7' 7' 7
=
9, x + 2y + 3z = 6,
3x + y + 2z = 8, by Gausselimination method. Sol. The given equations in matrix notations are
l� � m�l � lil
or AX = B
The augmented matrix is [A : B] =
R,
li
l�
In first column, the component with largest absolute value is 3. So interchanging and we get
R3,
l
3 [A : B] = 1 2
3
o o
1 2 18 2 3 6 , operating 3 1 19 1
2 7
8 10
3 7
3 1
3 11
3
3
3
5
R3 ; R3  �R, R2 ; R2  �Rl' 3 3
MATRIX ALGEBRA
647
7 In second column) the component with largest absolute value is "'3' So interchaning R2 and R3 ) we get 3 0 0
3 0

0 From third row)
1 7
2 1
8 11
3 5
3 7
3 10
3
3
3
1 7
2 1
8 11
3
3 54
3 15
21
21
0
operating R3 ; R3 
f
R2 ,
15 15 5 54 Z� ::::::} z = �21 21 54 18



7 1 11 7 1 1 z 11 5 From second row , y   z =  ::::::} y =  +  =  + 3 3 3 3 3 3 3 54 7
From first row,
203
203
3
29
3Y � 54 => Y � 54 x 7 � 18 2x + 3y + z � 9 => 2x � 9  3y  z
or
2x � 9 
5 162  87  5 87  � :_::_ 18 18 18



70

18
::::::} x =
35

18
Hence) the required solution is 35 5 29 x � 18 , y � 18 ' z � ' 18 TEST YOU R KNOWL EDGE 1 4.6 Solve the following system of linear equations by (i) Gausselimination method (ii) Gauss Jordan method. L 3x + 4y  z :::: 8 ;  2x + y + Z :::: 3 ; x + 2y  z :::: 2 2. lOx + Y + z 12 ; x + lOy + z 12 ; x + y + 10z 12 3. 3x + Y + 2z 3 ; 2x  3y  z  3 ; x  2y + z  4 4. 2x  6y + 8z 24 ; 5x + 4y  3z 2 ; 3x + Y + 2z 16 5. x + 2y  z 3 ; 3x  Y + 2z 1, 2x  2y + 3z 2 6. 2x  3y + z  1, x + 4y + 5 25, 3x  4y + z 2 7. 2x + Y + 4z 12, 8x  3y + 2z 20 ; 4x + lly  z 33 8. 2x + Y + z 10, 3x + 2y + 3z 18, x + 4y + 9z 16 9. x + 2y + z 8 ; 2x + 3y + 4z 20, 4x + 3y + 2z 16 10. 5x1 + 2X2 + xs :::: 12 ;  Xl + 4X2 + 2xs :::: 2, 2Xl  3x2 + lOxs ::::  45 �
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
648
DISCRETE STRUCTURES
11. 12.
Xl  X2  X3  X4 :::: 2 ; 2Xl + 4X2  3xs :::: 6, 3x2  4xs  2X4 ::::  1 ;  2Xl + 4xs + 3x4 ::::  3 Xl + 2X2 + 3xs + 4x4 :::: 20 ; 3x1  2X2 + 8xs + 4X4 :::: 26, 2Xl + x2  4xs + 7X4 :::: 10 ; 4Xl + 2X2  8xs  4X4 =2 5x1 + x2 + Xs + x4 :::: 4 ; Xl + 7X2 + Xs + x4 :::: 12, Xl + x2 + 6xs + x4 ::::  5 ; Xl + x2 + X3 + 4X4 ::::  6 2Xl + x2 + 5X3 + x4 :::: 5 ; Xl + x2  3X3 + 4X4 ::::  1, 3x1 + 6x2  2xs + x4 :::: 8 ; 2Xl + 2X2 + 2xs  3x4 :::: 2 x + y + 2z  w :::: 5 ; x + 3y + 2z + w :::: 17, x + Y + 3z + 2w :::: 20 ; x + 3y + 4z + 2w :::: 27.
13. 14. 15.
Answers 1. 5. 9.
13.
1, 2, 3  1, 4, 4 1, 2, 3
2. 1, 1, 1 6. 8.7, 5.7,  1.3 10. 2, 3,  4
1, 2,  1,  2
14. 2,  , 0, 
1
4
5
5
3. 1, 2,  1 7. 3, 2, 1 11. 2,  1 ,  2, 3 15. 1, 2, 3, 4.
4. 1, 3, 5 8. 7,  9, 5 12. 4, 3, 2, 1
ARITHMETIC PROGRESSION
1 5A
1 5.1 . SEQUENCE
Consider the following arrangement 3, 7, 11, 16, . Each term of the above arrangement is four more than the proceeding term. We call it
uA
sequence Le.,
succession of numbers formed according to a certain rule, arranged in a certain order ) is called a sequence." The nth term of a sequence is denoted by Tn' tn' an' a(n) etc.
A
A
1 5.2. FINITE AND INFINITE SEQUENCE
sequence containing finite number of terms is called a finite sequence. sequence having infinite number of terms is called an infinite sequence. For e.g., the sequence I, 3, 5, 7, 9 is a finite sequence and the sequence 2, 4, 6, 8, 10, . is an infinite sequence. Consider some more examples given below : ) . . Example 1. Wnte the first three terms oithe sequence < an>' where an =
Sol. For n =
Ix 2 x 3 6
= I,
a2
=
2x3x5 6
= 6,
a3
Example 2. Find Tw ior the sequence < Tn> where T, = 1, Sol. GlVen T, =
1, T2
= 6,
T3
=
Tn
T2 T,
=
Tn 1 , n :> Tn 2

6
= 1 = 6,
T4
=
T3 T2
6
= 6" = 1
T5
3x4x7 6
T2
= 6,
Tn

T4
1
1
T3
6
6
6
T g =  = 1 = 6, T lO = T7 649
Tg Ts
6
= 6" = 1 .
6
= 14. Tn
1 =T ' n
 ==
1
6 Ts
=
3. Put n = 3, 4, 6, . . . . . , so that
1
Similarly,
6
1, 2, 3, we have a1 =
.
n(n + 1) (2n + 1)
2
n :>
3.
.
650
DISCRETE STRUCTURES
Example 3. Find the second term of the sequence where Tn = {_ 1)n1 2n1, n E N Sol. Given Put n = 2 in (1),
(P.T.U., M.B.A. May 2003)
Tn (_ 1) n1 2 n 1 T, = ( 1) ' 2 ' =  2. =
(1)
1 5.3. ARITHMETIC PROGRESSION (A.P.)
is said to be an arithmetic progression (written as AP.) if there exists a number, say, d such that an +1  an = d 'if n ;::: 1 i.e., If the difference of each term of a sequence except the first one, from its proceeding term is always same, then the sequence is called Arithmetic Progression. For e.g., the sequence 4, 8, 12, 16, ... , is an AP. since 8  4 = 12  8 = 16  12 = ... . The constant number d is called the common difference of the concerned A.P. Also the first term of an AP. is denoted by a.
' then Tn
Tn = a + (n  1) d, n E N Step I. For n = 1, L.H.S. = T, = First term = a R.H.S. = a + (1  l)d = a The result holds for n = 1 Step II. Assume the result holds for n = m i.e., Tm = a + (m  1) d ( 1) Step III. Take n = m + 1. By definition, Tm + 1  Tm = d => Tm + 1 = T m + d = a + (m  l)d + d I Using (1) = a + md = a + (m + 1  l)d i.e, the result holds for n = m + 1, which completes the induction. Therefore by the principle of mathematical induction, the given result holds for each n i.e., Tn = a + (n  l)d, n E N.
I
4
=
I L L U STRATIVE EXAMPLES
Example 1. Find the 20th term and nth term of the sequence 4, 9, 14, 19 ... Sol. Given sequence 4, 9, 14, 19, ... , is an AP. Here a = 4, d = common difference = 9 
5. Using T n = a + (n  l)d, we have T20 = 4 + (20 1) (5) = 4 + 95 = 99 Also Tn = 4 + (n  1) (5) = 5n  5 + 4 = 5n  1.
Example 2. Show that the sequence log a, log b' log b2' ... is an A.P. a2
Sol. Consider
a3
I
lOg m  lOg n = IOg
:
ARITHMETIC PROGRESSION
651
a3
3
2
a
a a b x  = log T3  T = log  log  = log 2 ' b b2 b a2 b Thus ..
a
T,  T, = T3  T, = ... = log b
The given sequence is an AP.
Example 3. Which term ofA.P. 3, 15, 27, 39, ... will be 132 more than its 54th term ?
(P.T.U., M.B.A. May 2006)
Sol. Here a = 3, d = 12, Let T, be the required term such that T, + 132 = T5,
=> a + (r  l)d + 132 = a + (54  l)d => (r  l)d + 132 = 53d
12(r  1) + 132 = 53 x 12 = 636 12(r  1) = 636  132 = 504 504 rl= = 42 12 => r = 43 i.e., the required term = T' 3 ' Example 4. Iflog10 2, log10 (2'  1), log10 (2x + 3) form an A.P., /ind the value ofx. Sol. Given loglO 2, loglO (2X _ 1), loglO (2X + 3) are in AP. => logl O (2X_ 1)  logl O 2 = log l O (2X + 3)  logl O (2X  1)
(1)

=>
2" + 3
2X  1
log lO  = log lO 2"  1 2
2"  1 2" + 3 = => 2 2"  1
tl t+3 = t = 2x => (t  I)' = 2t + 6 2 t  1' => t' + 1  2t = 2t + 6 => t'  4t  5 = 0 => t'  5t + t  5 = 0 => t(t  5) + (t  5) = 0 => (t + 1) (t  5) = 0 => t =  I, t = 5 When t =  1, we have 2X =  1, not possible When t = 5, we have 2X = 5 => x = log, 5. Example 5. Show that the sequence {n2 + 4} is not an A.P. Sol. Let Tn = n' + 4 ' n E N . T, = (I)' + 4 = 5, T, = (2)' + 4 = 8, T3 = (3)' + 4 = 13, T, = (4)' + 4 = 20,
=>


.. The sequence is 5, 8, 13, 20, . T,  T, = 8  5 = 3 Here T3  T, = 13  8 = 5 not equal.
The given sequence is not an AP., because the differences T,  T" T3  T" ...... are
652
DISCRETE STRUCTURES
Example 6. For what value of n, the nth term of the series ,' "3 + 10 + 1 7 + ...... " and "63 + 65 + 67 + ...... " are equal. Sol. 1st series. We have 3 + 10 + 17 + . Here a, = 3 and d, = 10  3 = 17  10 = 7 .. Tn = a, + (n  l)d, = 3 + (n  1) 7 = 7n  4. lInd series. We have 63 + 65 + 67 + . Here a, = 63 and d, = 65  63 = 67  65 = 2 Tn = a, + (n  1) d, = 63 + (n  1) 2 = 2n + 61. .. Let nth term af given series be equal. .. 7n  4 = 2n + 6 1 ar 5 n = 65 i.e., n = 13.
Example 7. Determine the Ist term and the 40th term of the A.P. whose 7th term is 34 and 15th term is 74. Sol. Let a be the first term and d, the camman difference af the A.P. T7 = a + 6d = 34 :. We have (1) and (2) T, 5 = a + 14d = 74 Sd = 40 => d = 5. (2)  (1) => Putting d = 5 in (1), we get a + 6(5) = 34 ar a = 34  30 = 4. First term = a = 4 and 40th term = T,o = a + 39d = 4 + 39 (5) = 199.

Example 8. The 2nd, 31st and last terms ofan A.P. are 7 � , 2 and 6 2 respectively. 4
2
Find the first term and the number of terms. Sol. Let a be the first term and d, the camman difference af the A.P. :. We have Naw
(2)  (1) => Putting d =

3 31 13 1 1 T3 1 =  and last term ' I =  6  = . T2 = 7  = 2 2 2 4 4 ' T, = a + d and T3 1 = a + 30d. 31 a + d= 4 1 a + 30d = 2 29 1 29 1 31 2  31 1 29d =    = =   ar d =   x = 2 4 4 29 4 4 4

and
2
 .!.4 in (1), we get a +
Let number af terms be n.
(.!.) ( ) 4
=
31 4
31 4
1 4
ar a =  +  = 8.
1 32  n + 1 ' 33  n 1 = Tn = a + (n  l)d = S + (n  1)   = =  . 4 4 4
  
33  n 13 2 4 33  n =  26 "e.,
n = 59.

(1) (2)
ARITHMETIC PROGRESSION
653
Example 9. If (p + l)th term of an A.P. is twice the {q + l)th term, prove that the {3p + l)th term is twice the (p + q + l)th term. Sol. Let a be the first term and d, the common difference of the AP. We have Tp +1 = 2Tq+1 '
=>
Now
a + (p + 1  l)d = 2[a + (q + 1  l)d] ... (1) a + pd = 2a + 2qd => a = (p  2q)d Top+l = a + (3p + l  l)d = (p  2q)d + 3pd = 2(2p  q)d (Using (1 »
Tp+q+1 = a + (p + q + 1  l)d = (p  2q)d + (p + q)d = (2p  q)d .. T3P+1 = 2Tp+q+1' Example 10. The fourth term of an A.P. is equal to three times the first term and the seventh term exceeds twice the third term by one. Find the first term and the common difference. Sol. Let a be the first term and d, the common difference of the AP. Tn = a + (n  l)d , n E N .. By the given conditions, T, = 3 T, and T7 = 2T3 + 1 (1) (2) 3d or (1) => a + 3d = 3a a=(3) and
(2) => or
a + 6d = 2(a + 2d) + 1 3d 2d =  + 1
2
..
(3) =>
or or
2
2d = a + 1 d  = 1 or d = 2
2
[by (3)]
3 a =  (2) = 3.
2 Example 11. If m times the mth term of an A.P. is equal to n times its nth term,
then show that {m + n)th term of the A.P. is zero. Sol. Let a be the first term and d, the common difference of the A P. Tm = a + (m  l)d and Tn = a + (n  l)d .. mTm = nTn By the given condition, => m{a + (m  l)d) = n {a + (n  l)d) => ma  na + m(m  l)d  n(n  l)d = 0 => (m  n)a + (m2  m  n 2 + n)d = O => (m  n)a + (m  n)(m + n  1)d = O (m  n){a + (m + n  l)d) = 0 => a + (m + n  l)d = 0 ( .,' m r' n) Tm+n = O. .. Example 12. Ifpth, qth, rth terms of an A.P. be a, b, c respectively, then prove that p(b  c) + q (c  a) + r{a  b) = o. Sol. Let A be the first term and d, the common difference of the AP. Tp = a, Tq = b and T, = c We have A + (p  l)d = a, A + (q  l)d = b and A + (r  l)d = c. .. L.HB. = pCb c) + q(c  a) + rea  b) = pleA + (q  l)d)  (A + (r  l)d] + q[(A + (r  l)d)  (A + (p  l)d)] + r[(A + (p  l)d)  (A + (q  l)d)] = p [d(q  1  r + 1)] + q [d (r  1  P + 1)] + r [d(p  1  q + 1)]
654
DISCRETE STRUCTURES
� pd (q  r) + qd (r  p) + rd (p  q) � d(O) 0 �
�
RH.S.
Theorem II. If a, b, c, are in AP., then (i) a + k, b + k, c + k are also in AP.
(iii) ka, kb, ke, k 'l" 0 are also in AP.
�
d [(pq  prj + (qr  qp) + (rp  rq)]
(ii) a  k, b  k, e  k are also in AP.
(iv)
Proof. Given a, b, c are in AP.
� '�'I' k 'l" 0 are also in AP.
.. b a=eb (i) a + k, b + k = e + k will be in AP. if b + k  (a + k) = e + k  (b + k) i.e., if b  a = e  b,
which is true
(ii) a  k, b  k, e  k will be in AP. if b  k  (a  k) = e  k  (b  k) i.e., if b  a = e b,
which is true
(iii) ka, kb , ke are in AP. if kb  ka = ke  kb i.e., if k(b  a) = k(e  b) i.e., if b  a = e  b,
which is true
ba eb a b c b a e b � (iv) k ' k ' k will be in AP if k  k � k  k i.e, if i.e., if b  a = c  b, which k k 
is true

Theorem III. If each term of an AP. is
(ii) decreased by the same number (i) increased by the same number (iii) multiplied by the same number (iv) divided by the same number, then the resulting sequence is also an AP. Proof. Let the given AP. is a, a + d, a + 2d (i) Let each term is increased by "k' , then the new sequence will be a + k, a + k + d, a + k + 2d, . This is again an AP. with common difference d and first term a + k. (ii) Let each term of (1) is decreased by k. The new sequence will be a  k, a  k + d, a  k + 2d, . It is again an AP. with common difference d and first term a  k. (iii) Let each term of (1) is multiplied by k '" 0 the new sequence will be ka, k(a + d), k (a + 2d), . It is again an AP. with common difference kd and first term ka. (iv) Let each term is divided by k, k ", 0 the new sequence will be a a + d a + 2d k' k ' k
or
It is again an AP with common difference
a a d a k'k k 'k
(1)
2d . k '
  +  + 
d , k '" 0 and first term !!:. , k '" O. k k
1 5.4. IMPORTANT OBSERVATIONS
(i) If sum of three numbers in AP. is given, the numbers are a  d, a, a + d. (ii) If sum of four numbers in AP. is given, the numbers are a  3d, a  d, a + d, a + 3d. (iii) If sum offive numbers in AP. is given, the numbers are a  2d, a  d, a, a + d, a + 2d. Note.
If sum of numbers in A,P, is not given, then the numbers are
a, a +
d, a + 2d, '
ARITHMETIC PROGRESSION
655
Example 13. If a, b, c are in A, P, show that ,' 1
1
1
(ii) a2 (b + e), b2 (e + a), e2 (e + b) are in AP,
' A. P. (,')  ,  ,  are �n be ea
ab
, A, P, " ' a(b + e) b(e + a) e(a + b) ' A, P, (w ' ) b + e  a, e + a  b, a + b  e are m (m) , , are m be ea ab 1
1
1
are in A,P r' r (v) vb r + va r ' "Ie h + "Ie h va + vb
(v i) (b + e)2  a 2 , (e + a)2  b2, (a + b)2  e2 are in AP,
Sol. (i) � , � , � are in AP, be ea ab
if
if
abe abe abe , , are in AP, be ea ab
(Multiplying each term by abc)
a, b, c are in AP " which is given to be true,
� , � , � are in AP, be ea ab
Alternative method
� , � , � are in AP, if
if "
(ii)
be ea ab b  a = c  b,
1
1
1
1
1
ea
be
  
1
1
ab
ea
,
if
= 
ba abc

eb abc )
= 
if a, b, c are in AP " which is given to be true,
 ,  ,  are in A.P. be ea ab
a 2 (b + e), b2(e + a), e2(a + b) are in AP,
if
a 2b + a 2C, b2c + b2a, c2a + c2b are in A.P.
if
a2b + a2c + abc , b2c + b2a + abc , c2a + c2b + abc are in A.P. (Adding abe to each term) arab + ac + be), b(be + b a + ac), e (e a + eb + ab) are in AP, a, b, e are in AP, which is given to be true, (Dividing each term by ab + be + ea)
if if "
(iii) if if
a2(b + e), b2(e + a), e2(a + b) are in AP, a (b + e ) b(e + a) e(a + b) ' A. P . are ill , be ea ab
",", """,, , ;
ab + ae be + ba ea + eb ' A. P . , are ill be ea ab be + ba ab + ae ea + eb ' A. P . : + 1 , + 1, + 1 are ln be ea ab
: ,
(Adding ' 1' to each term)
if if
�+�+� �+�+m m+�+� :: are in AP, be ea ab
� , � , � are in AP, be ea ab
(Dividing each term by ab + be + ea)
656
DISCRETE STRUCTURES
if
abc abc abc are in A.P. , , be ca ab
if
a, b, c are in AP., which is given to be true.
(Multiplying each term by abc)
a(b + c) b(c + a) c(a + b) . bc ' ea ' ab are In A.P. if
b + C  a, c + a  b, a + b  c are in AP. (iv) (b + c  a)  (a + b + c), (c + a  b)  (a + b + c), (a + b  c)  (a + b + c) are in AP. (Subtracting a + b + C from each term)
 2a,  2b,  2c are in AP.
if if
a, b, c are in AP. which is given to be true. b + C  a, c + a  b, a + b  c are in AP. (v)
1
1
1
if
$ + ../c  ../c  .,fa (../c + .,fa)($ + ../c)
if
1
, $ + ../c ' ../c + .,fa .,fa + $ are in AP. 1
1
1
../c + .,fa  .,fa  $ (.,fa + $)(../c + .,fa)
$  .,fa ../c  $ if $ + ../c .,fa + $ �
b  a = c  b if a, b, c are in AP., which is given to be true.
if
1 1 1 . are m A. P . h h i ' i r ' i b ", + yc yc + ya va + "b
(vi) if if if P. if if
(b + C)2 a 2 , (c + a)2  b 2 , (a + b)2 c 2 are in AP. _
_
(b + c + a) (b + c  a), (c + a + b) (c + a  b), (a + b + c) (a + b  c) are in AP. b + c  a, c + a  b, a + b  c are in AP. (Dividing each term by a + b + c) (b + c  a)  (a + b + c), (c + a  b)  (a + b + c), (a + b  c)  (a + b + c) are in A  2a,  2b,  2c are in AP. a, b, c are in A.P., which is given to be true.
Example 14. The sum of the first three consecutive terms of an A.P. is 9 and the sum of their squares is 35. Find Tn' Sol. Let the first three consecutive terms of the AP. be a  d, a, a + d. Sum = (a  d) + a + (a + d) = 9
3a = 9 Le., . . The numbers are 3  d, 3 , 3 + d.
a = 3.
(Given)
ARITHMETIC PROGRESSION
657
= (3  d)2 + (3)2 + (3 + d)2 = 35 (9 + d 2  6d) + 9 + (9 + d 2 + 6d) = 35 Le., d 2 = 4 or 2d 2 = 35  27 = 8 The numbers are 3  2 = I , 3, 3 + 2 = 6.
Also, sum of squares
Case I. d = 2. . .
(Given)
d = ± 2.
The AP. is I, 3, 6, . and Tn = a + (n  I)d = 1 + (n  1)2 = 2n  1 . Case II. d =  2. . . The numbers are 3  ( 2) = 6, 3, 3 + ( 2) = 1. The AP. is 6, 3, I , . and Tn = a + (n  l)d = 5 + (n  I) (  2) =  2n + 7. ..
is 120.
Example 15. Find four numbers in A.P. whose sum is 20 and the sum of whose squares Sol. Let the numbers be a  3d, a  d, a + d, a + 3d. ..
= (a  3d) + (a  d) + (a + d) + (a + 3d) = 20 4a = 20 i.e., a = 6
Sum The numbers are
(Given)
6  3d, 6  d, 6 + d, 6 + 3d.
Also, sum of squares
= (5  3d)2 + (5  d)2 + (5 + d) 2 + (5 + 3d)2 = 120
(Given)
(25 + 9d2  30d) + (25 + d2  10d) + (25 + d 2 + 10d) + (25 + 9d 2 + 30d) = 120 20d2 = 20 i.e., d2 = I or d = ± 1.
Case I. d = 1. The numbers are 6  3(1), 6  (1), 6 + (I), 6 + 3(1) or 2, 4, 6, 8. Case II. d =  1. The numbers are 6  3( 1), 6  (  1), 6 + ( I), 6 + 3( 1) or 8, 6, 4, 2. Example 16. If Xi8 = y2i = Z28, prove that 3, 3 logy x, 3 logz y and 7 log z are in A.P. X'8 = y21 = Z28 = k. Sol. Let x
Taking common logarithms, we get
log X'8 = log y 21 = log Z 28 = log k 18 log x = 21 l0g y = 2 8 10g z = log k .
) [ ) [ _ 3 [ lOg y ) _ 3 [ (lOg k)/ 21 ) _ 3 x 28 _ (log k) / 28 21 log [ IOg Z ) = 7 [ (lOg k) / 28 ) = 7 x 18 = � . 7 Z = x log x (log k)/ 18 28 2
IOg X = 3 (lOg k) / 18 = 3 X 2l = 7.. 3 10gy x = 3 log (log k)/ 21 y 18 2 3 10gz y and
7 10g Now the numbers
AP.
4
z
3, 27 '
9
4, 2 are in AP. with
c.d.
1 = 2'
Given numbers are in
DISCRETE STRUCTURES
658
TEST YOUR KNOWLEDGE 15.1
1. 2. 3. 4. 5. 6.
7. 8. 9. 10. 1 1. 12. 13. 14. 15. 16. 17.
Show that 4, 10, 16, 22, """ is an Find its 7th and 9th terms. 1 2 Show that 6,5 ,4 is an Find its 10th and kth terms. 3 ,4, 3 Find the 20th, 25th and nth terms of the given by 21, 16, 11, 6, . Show that log a, log ab, log ah2, is an Find its 7th and nth terms. . the value of k. If 32 ' k, 85 k are In. fmd Each term of an is doubled. Is the resulting sequence also an ? If it is, write its first term, common difference and the nth term. Determine the number of terms in the sequence 17) 14"21 ) 12, """) 38. Which term of the 1, 6, 11, 16, ...... is 301 Which term of the series 20 + 16 + 12 + " is  96 ? Is 310 a term of the sequence 3, 8, 13, "",, ? If the 9th term of an be zero) show that its 29th term is twice its 19th term. In a certain the 24th term is twice the 10th term, Show that its 72nd term is twice the 34th term, Determine the 25th term of the whose 9th term is  6 and common difference is 5/4, The 3rd term of an is 1 and the 6th term is  11, Determine its 15th and rth terms, Determine the 2nd and rth terms of an whose 6th term is 12 and the 8th term is 22, Determine x so that 2x + 1, x2 + + 1 and 3x2  3x + 3 are consecutive terms of an If5 times the 5th term of an is equal to the 10 times the 10th term, find the 15th term of the Which term of the 8 6i, 7 4i, 6  2i, """ is (i) purely real purely imaginary, If a, b, c are in show that : 2 2 2 (i) 5a, (ii)  ,  ,  are in 5e are in bc ca ba b + e 4a ' e + a  4b ' a + b 4e are In. , (iii) b + + a, a + b are in 3 3 3 b + c , 111 c+a a + b (i) If a + b + 0 and , a b c are in show that a , b , c are also in (ii) If the numbers a2 , b 2 , c2 are in show that b +1c ' c +1 a , a+1 b are also in b ' a ' e are in (iii) If the numbers a2 , b2 , c2 are in show that b+ c c + a a+ b If log5 2, log5 (2X  3), log5 ( 127 + 2X1) are in then find the value of x, The sides of a right angled triangle are in show that these are in the ratio 3 : 4 : 5, A,P,
" "
A,P,
'
AP.
A,P,
" " "
A,P
'J
A,P,
A,P,
?
AP.
"
"
A,P,
A,P,)
A,P,
A,P,
A,P,
X
A,P,
A,P,
AP.
18. 19.
A,P"
5b,
AP.
AP.
c, c
20.
(h)
A,P,
(,'v)
A,P,
C :f
A P,


A,P"
A,P"
2 1. 22.

A,P"

A,P,

A,P,
A,P,
A,P"
A,P"
ARITHMETIC PROGRESSION 23. 24. 25.
26. 27.
29. 30. 31. 32.
33. 34. 35.
659
The pth term of an AP, is q and the qth term is show that its rth term is p + q  r. If pth, qth, rth terms of an A.P. be a, b, c respectively, then show that a(q  r) + b(r  p) + c(p  q)�O. Find the 8th and nth terms of the series n1 + 2nn+ 1 + 4nn+ 1 + Can the 8th term be deduced from the nth term. If not, why ? In any A,P, show that Tm+n + Tm_n 2Tm , How many numbers of two digits are divisible by 7 ? P,



=
If a, b, c are inA'P' J show that bc_a2 ,ca_ b2 ,ab _ c2 are in A,P, If the roots of the equation (b c)x2 + (c  a)x + (a b) 0 are equal, then show that a, b, c are in AP. Find three numbers in A,P, whose sum is 3 and product is 8. (i) The sum of three numbers in A,P, is 9 and the sum of their squares is 59.Find the numbers. (ii) The sum of three numbers in A,P, is 15 and the sum of the squares of the two extremes is 58, Find the numbers, Split 69 into 3 parts in A,P, such that the product of two smaller parts is 483, Four numbers are in A,P" their sum is 16 and sum of their squares is 84, Find the numbers, Divide 32 into four parts which are in A,P, such thatthe product of the extremes is to the product of means as 7 15, =
:
Answers
1. 5. 9. 15. 21. 28. 32. 33.
20 2k 2. 0, 3. 74, 99, 26  5n 40, 52 6. yes 7. 23 16/33 10. No 13. 14 30 17. 0 8, 5r  18 16. 1, 2 14n + 1 2n2  2n + 1 25 3 n , n N 31.  4, 2 or 2,   4 4/5 (ii) 3, 5, 7, or 7, 5, 3 (,)  1, 3, 7, or 7, 3,  1 21, 23, 25 3
,
•
 I,
0
4. log
ab6 , log abn 1
61  47, 13 4r 18. 4, 9 27. 13 8.
14.
I,
Hints
6. 10. 17. 18.
Let the A,P, be a, a + d, a + 2d, ' .. New sequence is 2a, 2a + 2d, 2a + 4d, ' This is also an A,P, with first term 2a, common difference 2d and nth term 2a + (n  1)2d, T 310 implies n 62,4, which is not in N, 5Tn 5 � 10TlO 5(a + 4d) � 10(a + 9d) 5a + 70d � 0 a �  14d .. T �a+ 14d � 14d+ 14d �0. Tn � 8156i + (n  1)( 1 + 2,) � 9  n + i(2n  8) (,) If Tn � 0 2n 8 � O. =
�
�
=
�
�
=
660
22. 25. 27. 28. 29. 30. 33.
DISCRETE STRUCTURES
17 + 2x 2 (2" 3) (a + d)' � a2 + (a  d:j2 � a � 4d :. Sides are 4d  d, 4d, 4d + d. Ts cannot be deduced from Tn because 'c.d,' is in terms of Required numbers are 14, 21, 28, """) 98. a4 2 lmp a, + 2 . . If c.d, = dI then =a7 3 les a1 + =3 Le., a1 = 3d be  a2, ea  b2, ab  e2 are in AP. if (ea  b2)  (be  a2) � (ab  e2)  (ea b2). Disc. �O � (ea)2 4(be)(a b) � O � a2 + 4b2 + e2  4ab 4be + 2ac � O � (a  2b + e)' �O. Let the parts be  + where d > O . ..  d) + + + d ) � implies (23  d:j23 � 483.
'0
I ro
n,
3d
I'
a
d,
,
&i
(a
d
a, a
a
a
_
(a
69
1 5.5. SUM OF FIRST n TERMS OF AN A.P.
denotes the nth term of an AP. then the sum of first n terms is denoted by S n ' where S n � T, + T, + T3 + ' " + Tn '
Theorem IV. If 'a' and 'd' be the first term and common difference ofan A.P., then show
that the sum offirst n terms is given by Sn �
� [2a + {n  l)d}, n
E
N
(P.T.U., M.B.A. Dec. 2001) Proof. We apply principle of mathematical induction to prove n
S n � '2 [2 a + (n  I ) d]
(I)
Step I. For n � I in (I), 1
L.H.S. � S, � T, � a, R.H.S. � '2 (2a + 0) � a => L.H.S. � R.H.S. The result holds for n � 1.
m
. Step II. Assume the result holds for n � m "e., Sm = 2 (2a + (m  I)d) (2) Step III. Consider
8m + 1 = T1 + T2 + T3 + ... Tm + Tm + 1 = 8m + Tm +
m
1
= 2 (2a + (m  I)d) + a + (m + I  I) d
= ma +
m2
( m2 1 )
(m  I)d + a + md = (m + I)a + md  + 1
= (m + I)a +
md 2
(m  I + 2) = (m + I)a +
md 2
(m + I)
I Using (2)
ARITHMETIC PROGRESSION
661
m+ l m+l  (2a + md) =  (2a + (m + 1  1)d) =2 2 i.e., the result holds for n = m + 1, which completes the induction. Therefore by the principle of mathematical induction, the given result is true for each n i.e., n 8 n = "2 (2a + (n  l)d).
Particular Cases (a) We know
n n n 8 n =  (2a + (n  1)d) =  (a + a + (n  l)d) =  (a + l) ' 2 2 2
The above form is used when the last term is known to us. (b) We know 8 n = T, + T, + T3 + ". Tn
= (T, + T, + T3 + " . + T n_1 ) + T n = 8 n _ 1 + T n Tn = Sn  Sn _ 1 J n > L For n = I, T l = 8 1 >
===>
I
I L L U STRATIVE EXAMPLES
Example 1. Evaluate (i) 2 + 4 + 6 + " . 50 terms "'
1
2
(ii) 45 + 47 + 49 + ". + 99
1
('")  +  +  + ". 25 terms. 9
9
I
3
Sol. (i) Here a = 2, d = 2, n = 50. Using 8n = 50
� [2a + (n  l)d], we have
8fi) = 2 (2 x 2 + 49 x 2) = 25 (4 + 98) = 25 x 102 = 2550. (ii) Given series is an A.P. with a = 45, d = 2, 1 = 99 1 = 99 . . To find n, => a + (n  l)d = 99 45 +(n  1) x 2 = 99 => => 2(n  1) = 99  45 = 54 => n  1 = 27 => n = 28. n Using 8 n = "2 (a + I), we have 28 88= (45 + 99) = 14 (144) = 2016. ' 2 (iii) Here
2 1 1 1 2 32 1 T,  T, = 9  9 = 9' T3  T, = 3  9 = 9 = 9 l
I
. . Given series is an A.P. with a = 9' d = 9 ' n = 25. we have
n
. Using 8 n = "2 (2a + (n  1)d),
:
662
DISCRETE STRUCTURES
Example 2. Find theA.P. in which the sum ofany number of terms is always three times the square of the number of these terms. Sol. Let the AP. be a, a + d, a + 2d, ...... . By the given condition, 5n = 3n'. Using Tn = Sn  Sn_ l J we have
Tn = 3n 2  3(n _ 1) 2 = 3n2 _ 3(n 2  2 n 1) = 6n  3 T, = 6(1)  3 = 3, T2 = 6(2)  3 = 9, T3 = 6(3)  3 = 15, . . . The AP. is 3, 9, 15, . Example Find the sum ofall integers which are divisible by 7 and lying between 50
+
and 500.
3.
Sol. Integers lying between 50 and 500 and divisible by 7 are 56, 63, 70, . . . . . . , 497.
:. Required sum = 56 + 63 + 70 + . . . . . . + 497 This is an AP. with a = 56 and d = 7 Let n be the number of terms . . . 497 = Tn => 56 + (n  1)7 = 497 Solving, we get n = 64 ..
Required sum = 5 4 =
6
64 2
+
(56 497) = 32(553) = 17696.
Example 4. How many terms of the sequence  12,  9,  6,  3, ...... must be taken to make the sum 54 ? Sol. The given sequence is  12,  9,  6, 3, . Here T2  T, = 9 ( 12) = 3 

T3  T2 =  6  ( 9) = 3 T4  T3 =  3  ( 6) = 3 The given sequence is an AP. with a =  12 and d = 3. Let 54 be the sum of the first n terms.
n  [2a (n  l)d] = 54 2
+
=> n[2( 12) + (n  1)3] = 108
n(3n 27) = 108 => n'  9n  36 = 0 => n =  3, 12 n = 12, because n =  3 is impossible. Example 5. Determine the sum of the first 35 terms ofanA.P' if its second term is 2 and seventh term is 22. Sol. Let the AP. be a, a + d, a + 2d, ...... T, = a + d = 2 (1) T7 = a + 6d = 22 and (2) 5d = 20 i.e., d = 4 (2)  (1) => Putting d = 4 in (1), we get a + 4 = 2 Le., a = 2  4 =  2. ..
35
+
35
53 5 =  [2( 2) (35  1)4] =  x 132 = 2310. 2 2 
ARITHMETIC PROGRESSION
663
Example 6. If the first term of an A.P. is 2 and the sum offirst five terms is equal to onefourth of the sum of the next five terms, then : (,) show that T2 =  1 1 2 (i,) find the sum offirst 30 terms. Sol. Let the AP. be a, a + d, a + 2d, ......
0
1
We have T, + T2 + T3 + T4 + Ts = "4 [T6 + T7 + Ts + Tg + TlO ]
5 5  [a + (a + 4d)] =  [(a + 5d) + (a + 9d)] 2 8 4a + 8d = a + 7d => d =  3a
[
Sum =
(,)
T,o
a = 2 and d =  6. = a + (20  IX/ = 2 + 19 ( 6) =2  114 =  1 12.
(i,)
530
=
� (first term + last term)] 1 4
=> 2a + 4d =  [2a + 14d] => d =
3; [2(2) + (30  1)( 6) 15( 170) ]= =  2550.
 3(2)  6 =
(.: a = 2 is given)
[Sn = � [2a + (n  l)d]]
Example 7. If the sum of the first n terms ofa sequence is of the form An2 + Bn, where A
and B are constants, independent of n, show that the sequence is an AP. Is the converse true ? Justify your answer. Sol. We have 5n = An' + Bn . Replacing n by n  I, we get 5n_1 = A(n _ 1) 2 + B(n  1) = An 2  2An + A + Bn  B ..
Replacing n by n ..
Tn
Tn = 5n  5n_1 = (An' + Bn)  (An'  2An + A + Bn B) I, we get Tn_1 = 2A(n  I)  A + B
 Tn_1 = (2An  A + B)  (2An  2A  A + B) = 2A, a constant independent of n.
The sequence is an AP. Conversely, let a, a + d, a + 2d, ...... be an AP. 5n
5n =
= 2An  A +B.
( )
d n n d n [2a + (n  1)d] =  [nd + (2a  d)] =  n2 + a  =2 2 2 2 d d and B = a  . = An2 + Bn, where A = 2 2
An2 + Bn, where A andB are constants, independent ofn.
Converse is also true.
Example 8. If the sum ofm terms ofanA.P. be n and the sum ofn terms be m, show that the sum of m + n terms is  (m + n). Sol. Let a be the first term and d, the common difference of the AP. m .. 5m = n =>  [2a + (m  l)d] = n => 2am + m(m  1)d = 2n (I)
and
2 n 5n = m =>  [2a + (n  l)d] = m => 2an + n(n  l)d = 2 m 2
(2)
664
DISCRETE STRUCTURES
(I)  (2) => 2a(m  n) + d(m 2  m  n2 + n) = 2n  2m => 2a(m  n) + d«m'  n')  (m  n» =  2(m  n) Dividing by m  n, we get 2a + d(m + n  I) = 2 m + n [2a + (m + n I)d] = 5 2 m + n (2) =  (m + n). = 2
(3)

m+n


(Using (3»

Example 9. If the sum offirst n, 2n, 3n terms ofan A.P. are 51' 52' 53 respectively show that 83 = 3(82  81 ). Sol. Let a be the first term and d, the common difference of the A.P. ..
n
5, = "2 [2a + (n  I)d], 82
R.H.S. = 3(82  8, ) = 3
=
2n
2""" [2a + (2n  I)d]
and 83
[ 2; [2a + (2n  I)d]  � [2a + (n  I)d]]
3n
= 2"""
[2a + (3n  I)d]
3n [2 [2a + (2n  I)d] [2a + (n  l)d]] 2 3n 3n = [4a + 4nd  2d  2a  nd + d] = [2a + (3n  l)d] = 83 = L.H.S. 2 2
=




Example 10. If the sum of m terms of an A.P. is to the sum of n terms of the same A.P. is as m 2 : n2 ; prove that the ratio of its mth and nth terms is 2m  1 : 2n  1. Sol. Let a and d be respectively the first term and common difference of the A.P. 8 m2 By the given condition, S n2
"'  n
m  [2a + (m  l)d] 2 n "2 [2a + (n  l)d]
2a + (m 1)d m 2a + (n 1)d n 

2an + (m  l)nd = 2am + (n  I)md => 2a(n  m) = (nm  m  mn + n)d ']a = d ... (1) T a + (m  l)d a + (m  1).2a = [by (I)] Required ratio T a + (n  l)d a + (n  1).2a 1 + 2m  2 2m  1 = 1 + 2n  2 2n  1 . => =>
m n
(.:
a ", 0. For if a = 0, then d would also be zero.)
TEST YOUR KNOWLEDGE 15.2 1.
(0 Find the sum of first 50 natural numbers Find the sum of first 35 even natural numbers. Find the sum of first 65 odd natural numbers.
(ii) (iii)
ARITHMETIC PROGRESSION 2. 3.
665
Find the sum of indicated number of terms of each of the following A,P, C') 5, 2, 1, ...... n terms. Cii) 0. 9 , 0.91, 0. 9 2, ...... 100 terms. Find the following sums : 2 1 2 C') 2 + 5 + 8 + ..... + 44 C") 6+ 5 + 5 + ...... + . "3 "3 "3 If Sn denotes the sum of n terms of an A,P, whose common difference is d, show that d Sn  2Sn_1 + Sn_2J n 2. (0 Find the sum of 10 terms of the sequence (x + y)2 , x2 + y2 , (x y)2 , ..... x Bx  2y 5x  By (ii) Find the sum of 27 terms of the sequence  , x+y x+y , x+y , (0 Find the sum of all odd numbers between 100 and 200. (ii) Find the sum of first 100 even natural numbers divisible by 5. (iii) Find the sum of all natural numbers between 200 and 400 and divisible by 7. (iv) Find the sum of all positive integers less than 500 and divisible by both 3 and 7. (0 Find the rth term of the A,P., the sum of whose first n terms is 2n + 3n2 , (ii) Find the sum of first n terms of the A, P , whose kth term is given by 5k + 1, (iii) Find the sum of first n terms of an A, P , whose nth term is 5 6n, n N. (0 How many terms of the A,P, 4, 7, """ are needed to give sum 715 ? (ii) How many terms of the A, P , 6,  11/2, 5, """ are needed to give sum 25 ? (0 How answer,many terms of the A,P, 18, 16, 14, """ are needed to give sum 78 ? Explain the double (ii) If the first term of an A, P , is 22, the common difference is 4 and the sum to n terms is 64, find n. Explain the double answer, Solve the equation : 1 + 6 + 11 + """ + x 148, If Sl ' S2' Ss are the sums of n terms of three A,P,s, the first term of each being unity and the respective common differences being 2, 3, show that Sl + Ss 2S2 , (0 Find the sum of first n terms of an A,P, whose 7th term is 30 and 13th term is 54, (ii) If the 5th and the 12th terms of an A, P , are 30 and 65 respectively, what is the sum of the first 20 terms ? If Sn denotes the sum of n terms of an A,P, show that Find the common difference of an A,P, whose first term is 100 and the sum of whose first six terms is five times the sum of the next six terms, Show that the sum of first n even natural numbers is equal to (1 + �) times the sum of first n odd natural numbers, If the 12th term of an A,P, is  13 and the sum of the first four terms is 24, what is the sum ofthe first 10 terms ? The sum f filrst n terms f an A, P . zero, show that the sum f next terms . am (m + , a being the first term, The sum of n terms of two A,P, are in the ratio 5n + 4 : 9n + 6, Find the ratio of their 18th terms, ..
4.
=
5.
>
_
Y
6.
7.
8. 9.
10. 11.
13. 14.
15.
16.
17. 18.
. . . . . •
E
I,
=
I,
12.
.
=
S3 0 = 3 (S2 0  S lO ) '
0
0
, IS
0
m
IS 
n)
nl
666
DISCRETE STRUCTURES
19. The first, second and the last terms of a finite are a, b, respectively. Show that the sum of + c 2a) all terms of the is (a + c)(b 2(b  a) . be a, b, respectively, show that 20. If the sums of the first p, q, r terms of an c  0 �� a �  � + b �  � +(p q p 21. If the sum of first p terms of an is equal to the sum of first qterms, show that the sum of first p + q terms is zero, 22. If the pth term of an is ...!.q . and qth term is ..!.., show that the sum of pq terms is �2 (pq + 1). 23. If Sn denotes the sum of n terms of an and If Sl 6, S7 105, show that _ nn + 33 . AP.
c
AP,

c
A,P, r
A,P,
A,P,
.
A.P.
(,)
Answers
(h)
1275
1260 2. ( ) n ( 13  3n) / 2 139.5 27 [ 27Xx + y14Y ] ( ) 10 (x2 + y2  7xy) (iii) 8729 (iv) 5796 (iii) n(2 3n) 8. (i) 22 (ii) 4. 8 10. 36 14.  10 16. 0 L
,
6.
P
� Sn3
=

4225 3. (i) 345 6. (i) 7500 7. (i) 6r  1 (ii) 5. 20 12. (i) 2n(n + 2) 18. 179 : 321.
(iii)
(h)
(h)
,
=

.. 5 2 3 (ii) 50500 (ii) n(5n + 7)/2 9. ( ) 6. 13 (ii) 1150 (��) 6 
,
Hints 4.
Sn  2Sn_1
(,)
+
Sn_2 = (Sn  Sn_l)  (Sn_l  Sn_2 ) = Tn  Tn_1 = d .
T, � 6. 11. 16 The is 5(l) + 5(2) + 5(3) + ...... n 10. We have x � 1 + (n  1)5 and 148 � (1 + x). 2 6 [� (200 + 5d)] � 5 [� (200 + lld)] 6S � 5S 14. S � 5(S S ) 2 2 4 [2a + 3d] � 24 . 16. Solve a + 11 d �  13 and 2 17. If the be a. a + d. a + 2d then 2 [2a + (n  1)d] � 0 implies d � � n1. Required sum 0 18. Let and be the first terms and common differences of the two ,s respectively, n2 [2a, + (n  1) d,) 5n+ 4 5n+ 4 9n + 6 9n + 6 n  1 � 18 _ 1 . n = 35, Now put 2 7.
(ii)
6
S,  S,_l'
1,
AP.
12
6
=c}
1,
12
6
1,
Le..
•
.
=c}

AP.
aI' a2
•
=
dl, d2
�
. . . . . .
Sm+n  Sn = Sm+n 
Le.,
=
Sm+n "
A,P
ARITHMETIC PROGRESSION
667 =::::}
19.
20. 21.
=:::}
ba
:. Sum � "2n (a + c) . . p D . . a a �  (2A + (p  1)D] lmplIes  � A + (p  1) 2 etc. 2 p = implies � [2 a + (p  1)d] � % [2a + (q  1)d] Sp
Sq
�
23.
ca =n_l b a
SI
2a(p  q) + (p  q)(p + q  l)d � 0 i.e 2a + (p + q  l)d � 0 .•
�
6
� a�
1 5.6. ARITH METIC
6,
S7 �
105 � '2 7
[2(6) + (7  1)d] � 105
�
etc.
d � 3.
MEANS
If three or more than three numbers are in AP., then all the numbers lying between the first and the last numbers are called the arithmetic means (A.M.s) between them. Equivalently, if a, AI ' A2 , . . . . . . , An ' b are in A.P., then AI ' A2 , . . . . . . , An are called the n arithmetic means between a and b .
For example,
10 is the single AM. between 5 and 15. (,) 5, 10, 15 are in AP. (i,) 5, 10, 15, 20, 25, 30 are in AP. . 10, 15, 20, 25 are the four AM.s between 5 and 30. :
1 5.7.
SINGLE A. M . BETWEEN ANY TWO GIVEN N U M BERS
Let a, b be any two numbers. Let A be the single AM. between a and b . By definition, a, A, b are in AP. (each � common difference) Aa� bA 2A
� a + b or A � a + b . 2
Remark The single between any two numbers is simply referred as the numbers. Thus, the between given two numbers is equal to half their sum, 7 + 29 36 For example, the between 7 and 29 is �  = 18. 2 2 A, M ,
AM ,
AM ,
between the
A, M ,
1 5.8. n A. M.s BETWEEN ANY TWO GIVEN NU MBERS Let a, b be any two numbers. Let A" A2 , . . . . . . , An be the n AM.s between a and b . . . By definition, a, A" A2 , ...... , An ' b are in AP. Let d be the common difference of this
AP.
Now
b � Tn +2 a + (n + l)d. �
A1 � a + d � a + 
ba n+ 1
ba
d �  n+l
668
DISCRETE STRUCTURES
[) [ ) � = a + nd = a + n ��: . ba � = a + 2d = a + 2 n + 1
. . The n A.M.s between a and b are a +
[ )
b + 2 a a n+l n+l
ba .
• ......•
[ )
b a a + n n + l .
Theorem V. The sum of n A.M. s between any two numbers is equal to n times the A.M
between them.
Proof. Let A" A2' . . . . . . ' An be the n AM.s between numbers a and b. ..
Sum ofnAM.s between a and b
=A, + A2 + . . . . . . + An = Ca + A, + A2 + . . . . . . + � + b)  Ca + b) n+2 = Ca + b) Ca + b) C': a, A" A2 , ...... , An ' b is an AP. of n + 2 terms) 2 n + 1  1 n a +b . the AM. between a and b. = Ca + b) =  = n t,mes 2 2

[
] ( )
Sum of n A.M.s between a and b = n (A.M. between a and b).
The five A,M,s between a and b are given by a + b, a + 2d, a + 3d, a + 4d, and M.B.A. 2002, May 2002) a + 5d, respectively. CP. Note.
T. U.
Dec.
ILLUSTRATIVE EXAM P L ES
Example 1. Insert three A.M s between 11 and 14. Sol. Let A" A2 , A3 be the three AM.s between 11 and 14. 11, A" A2 , A3 , 14 are in AP. Let d be the common difference of this AP. Now
14 = Ts = 11 + 4d. :. d = 3/4
47 3 25 3 47 A A, + A, = 1 1 + d = 1 1 + 4" = 4' 2 = d = 7 + 4" = 2 and
50 + 3 53 A3 = A2 + d = 4 4" = 4 '
Example 2. Prove that if the number of terms of an A.P. is odd, then the middle term is the A.M. between the first and the last terms. Sol. Let the AP. be a, a + d, a + 2d, ...... C2n + I) terms. CNote that 2n + I is odd) . . Middle term = Tn + , = a + (n + l  l)d = a + nd. AM. between first and last term
a + Ca + C2n + 1  1)d) 2a + 2nd = a + nd = Tn+ , 2 2 The result holds.
ARITHMETIC PROGRESSION
669
3.
Example The sum of two numbers is 13/6. An even number ofA.Ms are being inserted between them and their sum exceeds their number by 1. Find the number of A.M.s inserted. Sol. Let the numbers be a and b. . . a + b = 13/6. Let A" A" ...... , A n be the 2n (even number) AM.s between a and b . '
.. A, + A2 + ....... + A2n= 2n (AM. between a andb) a+b = n(a + b) = n x E = E n =2n 2 6 6 Also A, + A2 + ...... + A2n = 2n + 1 13 n . n =6. 6 n = 2n + 1 or "6 = 1 "e.,
( )
(Given)
No. of AM.s inserted = 2n = 2(6) = 12 .
an
Example 4. Find n such that n a
1
a and b.
+ bn + bn
1
may be the A.M. between distinct numbers a+b
Sol. The AM. between a and b is a + b . Let
2
2
=> a n lea _ b) = bn l(a  b)
(.,'
a # b => a  b # O)
a n1 = 1 => n  l = O Le., n = l. b n1 Example 5. There are n A.M.s between 7 and 85 such that {n  3)th mean ,' nth mean is as 11 " 24. Find n. Sol. Let A" A" ...... , An be the n AM.s between 7 and 85. . . 7, A" A" ...... , An' 85 is an AP. of n + 2 terms. Let d be the common difference of this 
..
AP.
85 = Tn+2 = 7 + (n + l)d
Now We are given
(n  3)th A. M. nth A.M.
1(n3)+ 1  11 Tn + l
=> =>

n+1
 24
24
7n + 7 + 78n  234 11 7n + 7 + 78n  24
11
d =�
l 105n = 5525
7 + (n  2  1)d 11 => => 7 + (n + 1  1)d 24 => =>
78 7 + (n  3) n + 1 11 78 24 7 + n n+ 1
85n  227 11 => 2040n 5448 = 935n + 77 85n + 7 24 n = 5.
670
DISCRETE STRUCTURES
Example 6. If the A.M between pth and qth terms of an A.P. be equal to the A.M between rth and sth terms of the A.P., show that p + q = r + s. Sol. Let the AP. be a, a + d, a + 2d, .
Tp = a + (p  lXl, Tq = a + (q  lXl, T, = a + Cr  lXl, T, = a + (s  lXl. We are given that : AM. between Tp and Tq = AM. between T, and T,
=> => =>
[a + (p  l)d] + [a + (q  l)d] = [a + (r  l)d] + [a + (s  l)d] => p + q2= r+ s2 (p  1 + q  l) d = (r  1 + S  l) d p + q = r + s.
..
The result holds.
TEST YOUR KNOWLEDGE 15.3
1. Findthe A,M. between 5 and 9. 2. Findthe AM. between (x y)2 and (x + y)2 . 3. Find the sum of 500 A,M,s between 2 and 3. 4. If a, b, c are in AP show that (a cJ' = 4(b2 ac). 5. If x, y, z are in AP. show that (x + 2y z) (2y + z  x) (z + x y) = 4xyz. 6. Find the ratio of the sum of m A, M , s between any two numbers to the sum ofnA, M , s between the same numbers. 7. Insert 3 A, M , s between 3 and 19. 8. Insert 5 A, M , s between 8 and 26. 9. Insert 6 A,M,s between 3 and 24. 10. Insert p A,M,s between 1 andp2 , 11. A,Insert 8 A, M , s between 2 and 29. Also verify that the sum of these 8 A,M,s is equal to 8 times the M. between 2 and 29. _
.•
12. 13. 14.

Find n such that ana+1n ++ bbnn+1 may be the A,M. between a and b. n2 A,: 11.M,Find s aren.inserted between 5 and 86 such that the ratio of the first and the last mean is as Thereofaren. nA,M,s between 3 and 17, The ratio of the last mean to the first mean is 3 : 1. Find the value Answers
7 2. x2 + y2 7. 7, 11. 15 8. 11, 14. 17. 20, 23 10. p, p l. 3p 2 . ....... p2 p + l 12. 13. 8
3. 1250 6. m :n 9. 6, 9, 12, 15, 18, 21 11. 5, 8, 14, 17, 20, 23, 26 14. 6.
L
°
2


4. Use b = 2 in the RH.8, a+
C
II,
Hint
GEOMETRIC PROGRESSION
1 58
1 5.9. GEOM ETRICAL PROGRESSION
A sequence of non·zero numbers is said to form a geometric progression (G.P.) if the ratio of each term (except the first one) to its proceeding term is always same. 1 1 1 For example, I, 2' 4' 8' . . . ) is a G.P. or A sequence
< Tn>' with Tn t: 0 is said to be a T Tn
geometric progression if there exists a number r, r fc 0, such that � 'r' is called the common ratio of the concerned G.P.
=
r \;j n :> 1. The number
,p,
The first term of a G is generally denoted by 'a', 2. Incase ofa G,P" a:fO, r:fO.
Remark. 1,
Theorem VI. If 'a' and 'r ' be the first term and common ratio of the G.P. ' then show arn  1 n E N. Proof. We apply principle of mathematical induction to prove Tn arn  1 ... (1) '' Step I. For n 1 in (1), L.H.S. T, a ; R.H.S. ar a
that Tn
=
J
=
=
=
=
=
=
The result holds for n 1. Step II. Assume the result holds for n m i.e., Tm arm 1 Step III. Take n m + 1 and consider ..
=
=
(2)
=
=
Tm + l Tm

=
r ::::::} Tm+1
i.e., the result holds for n
=
r Tm r' arm1 arm arm + 1  1 =
=
I By definition
=
m + 1. It completes the induction. Therefore by the principle of mathematical induction, the given result holds for each n =
Tn = arn 1 , n E N.
Le.,
1 5. 1 0. I M PORTANT OBSERVATIONS
(,) If product of three numbers in a G.P. is given, then the numbers are
a

r
, a, ar.
(i,) If product offour numbers in a G.P. is given, the numbers are '!!:'" !!:., ar, ar3 . r3 ' r
(ii,) If product offive numbers in a G.P. is given, the numbers are '!!:'" !!:. , a, ar, ar'. r2 ' r Remark.
Ifproduct of numbers in G,P, is not given, the numbers to be taken as a, ar, ar2, ' 671
672
DISCRETE STRUCTURES
ILLUSTRATIVE EXAM P L ES
Example 1. Find the 12th term and general term of the sequence 3, 6, 12, 24,... Sol. Given sequence 3 , 6, 12, 24, ... is a G.P. with a = 3, r = 2
Using Tn = arn  1 , we have T 2 = ar " = 3 x 2 " = 6144 ' Tn = arn1 = 3 x 2 n 1 . Also
Example 2. Which term of the series 1
1
Sol. Given series is 4 '2 + 1 +.. T2 T,
Here ..

� � + 1 + ... , is 256 ? 
= 112 = _ 2, 3 = 1_ = _ 2 and so on 114 T2 112 T
_ 
1
Given series is a G.P. with a = 4 ' r = 2. 
3.
n  1 = 10 => n = 11 i.e., Tll = 256. Example For what value ofn, the nth term of the series 5 + 10 + 20 + ... and 1280 + 640 + 320 + ... are equal ? Sol. In first series, a = 5, r = 2 . . Tn = arn 1 = 5 x 2n 1
In second series,
a = 1280, r =
1 2

..
Tn = arn1 =
1
1280 x 2 n 1
If nth term offirst series and second series are equal, then we must have
2n1 + n1 = 256 2n 2 = 8
=> 2n = 10 => n = 5
i.e., 5th term of the two series are equal. Example 4. The 6th, 12th and 18th terms ofa G.P. are a, b and c respectively. Show that b 2 = ac.
Sol. LetA be the first term andR, the common ratio of the G.P. ..
.. ..
The G.P. is A, AR, AR 2 , .
a = T6 = AR 5 , b = T' 2 = AR " and c = T, S = AR ' 7 b2 = (AR ") 2 = A'R22 and ac = (AR5) (AR 17) = A'R 22 b2 = ac.
GEOMETRIC PROGRESSION
673
Example 5. The first term ofa G.P. is 1. The sum ofthird and fifth terms is 90. Find the common ratio of the G.P. Sol. Let r be the common ratio of the G.P. Here a = 1. We are given T3 + T5 = 90. => ar' + ar' = 90 => r' + r' = 90 (.; a = 1)
=> r' + r2  90 = 0 => r2 =

l±�:+ 360
:
 1 19
=
= 9, _ 10
r' = 9 implies r =  3, 3 r2 =  10 is impossible. Common ratio =  3 or 3. Example 6. If the 4th, 10th and 16th terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P. Generalise the result. Sol. LetA be the first term andR, the common ratio of the G.P. .. The G.P. is A, AR, AR', . 3 x = T, = AR , Y = T O = AR9 and z = T = AR 15 . l '6 z AR'5 Y = AR9 = R93 = R 6 and '= = R ' 59 = R 6 
Now
X
AR3
y
z
Y
AR9
. G. P.  =  Le., . x, y, z are ill
Generalisation. Let
X
Y
= T = ARPl ) P
X
A q l AR P
y
G.P.
or if p, q, r are in A.P.
= T = AR ql and z = Tr = AR,l ' q
R p qp and = ':7 1 = Rql +1 = R
x
x, y, z are in
Y
if ;L x
= � Le., if R qp = R'q or if q p = r p Y
. . In general, if p, q, r are in A.P., then T T T, of any G.P. are also in G.P. p'
q'
Example 7. If the 5th term of a G.P. is 16 and the 10th term is find its 15th term. Sol. Let a be the first term and r, the common ratio of the G.P. Tn = arn1 ) n E N. We have
i, find the G.P. Also
1
T5 = 16 and Tl O = '2
1 2
ar 9
__
ar' = 16 and ar9 = 
ar '
1
2 = __
16
1 2
r=ar' = 16 =>
a(�r
= 16
=>
a = 256
674
DISCRETE STRUCTURES
The G.P. is Also,
T15
= ar 14 = 256
4 (�)' 2
or 256, 128, 64, .
=
1
64
Example 8. Ifp, q, r are in A.P. and x, y, z are in G.P', prove that xq'. y'P . zPq = 1. Sol. p, q, r are in A.P. Let d be the common difference ..
q = p + d and z = p + 2d Let R be the common ratio y = xR and z = xR'. Putting the values of q, r, y, z in L.H.S. we get L.H.S. = xq'y 'Pzpq= xip + d)  ip + 2d) (xR) ip + 2d) P (xR'Y'  ip + d)
Also, x, y, z are in G.P.
= ;rd (XR)' d (XR') d = xd . x' dR' d . xdR' d o = xd + 'd  d . R ' d  ' d = xOR = 1 x 1 = 1 = R.H. S. Example 9. Find all the sequences which are simultaneously arithmetic and geometric progressions.
Sol. Let T" Let
3 , . . . . . . be a sequence which is A.P. as well as G.P. Tn = a + (n  l)d ¥ n E N The sequence is a, a + d, a + 2d, . T" T
n E N.
This is also a G.P.
IX/ = a + (n ":::
a'+ nd " ..
a + (n + l)d a + nd
=> (a + nd)' = [(a + nd) + d] [(a + nd)  d]
(a + nd)' = (a + nd)'  d' => d' = 0 i.e., d = 0
The sequence is a, a + 0, a + 2 (0) , ...... Le., a, a, a, .
. . Only a constant sequence can be both A.P. and G.P. Example 10. If a, b, c, d in G.P', show that the following are also in G.P. Sol. a, b, c, d are in G.P.
Let r be the common ratio of this G.P. (,) a'  b', b'  c', e'  d' are in G.P. if if
if
(i,) a2 + b 2 + c 2, ab + be + cd, b2 + c 2 + d2.
b = ar, c = ar2 , d = ar3 , (ar)2 _ (ar 2 )2 (ar2 )2 _ (ar ' )2 if a 2 _ (ar)2 = (ar)2 _ (ar2 )2
a 2r 2 ( 1  r 2 ) a 2r 4 ( 1  r2 ) if r2 = r 2 , which is true. = 2 2 a r (l _ r 2 ) a 2 (l _ r 2 ) (Since a, b, c, d are in G.P. therefore, a ", O. We also assume that r '" 1.)
.. a2  b2 , b2  c2 , c2  d2 are in G,P. (i,) a' + b' + c' , ab + bc + cd, b' + e' + cf2 are in G.P.
ab + bc + cd b2 + c2 + d2 a 2 + b 2 + c2  ab + bc + cd
GEOMETRIC PROGRESSION
675
a 2 + (ar 2 ) + (ar2 )2 a 2r (l + r 2 + r4 ) a 2r 2 ( I + r 2 + r 4 ) if r = which is true. = 2 a (1 + r2 + r 4 ) a2 ( 1 + r2 + r4 )
if
T,
if
..
a' + b' + c' , ab + bc + cd, b' + c' + d' are in G.P.
Example 11. Find four numbers forming a G.P. where the third term is greater than the first by 9 and the second term is greater than the fourth by 18. Sol. Let the faur numbers in G.P. be a, ar, ar' , and ar3 . and ar = ar3 + 18 . . By the given canditians : ar' = a + 9 ... (1) a(r'  1) = 9 (1) =>  ar(r'  l) = 18 => ar ( l  r') = 18 (2) => r = 2 "e., r = 2 Dividing (4) by (3), we get Putting r =  2 in (3), we get a (( 2)'  1) = 9 i.e., 3a = 9 ar a = 3. 3 i.e., 3,  6, 12,  24. . . The numbers are 3, 3(2), 3(2)', 3(2)
... (2) ... (3) ... (4)
Example 12. The sum ofthree numbers in G.P. is 13112 and theirproduct is  1. Find the
numbers.
a r
Sol. Let the numbers be , a and ar. ..
a r
= . a . ar =  l
Praduct
The numbers are Also, sum
(Given)
3 3 if =  1 => a = ( 1) ar a =  l. 1  ,  1 and  l(r) =  r.
r
(
)
( �) + ( l) + ( r)
= 
=
��
(Given)
.  .! + 1 + r = � =>  12  12r  12r' = 13r r 12 12r' + 25r + 12 = 0 => r =  3/4,  413
r =  3/4
and
1 ,  1,  ( 3/4)  314 4/3,  1, 3/4
The numbers are ar

r =  413
The numbers are ar
1
( 4tJ) ,
_ _ _
 1,  ( 413)
3/4,  1, 413.
Example 13. The product of three numbers in G.P. is 1000. If we add 6 to its second number and 7 to its third number, the resulting three numbers form an A.P. Find the numbers (P.T.U., M.B.A March 1999) in G.P.
a
Sol. Let the numbers in G.P. be r , a and ar.
a r
Praduct =  . a . ar = 1000
if
= 1000 "e.,
a3 = (10)3
(Given) ar a = 10
676
DISCRETE STRUCTURES
10 The numbers are  , 10 and lOr.
r
10 New numbers are  , 10 + 6, lOr + 7. These are in A.P.
r
10
10
=> 16  r = lOr  9 => 2r'  5r + 2 = 0 => r = 112, 2.
16   = (lOr + 7)  16 r 16r 1O = lOr'  9r
1 10 Case I. r = 2 ' The numbers in G.P. are  , 10, 10 (112) 112
Le., 20, 10, 5 .
Case II. r = 2. The numbers in G.P. are � , 10, 10(2) Le., 5, 10, 20. 2 Example 14. The sum of the first three terms of a G.P. is 7 and sum of their squares is
21. Determinate first five terms of the G.P. Sol. Let the G.P. be a, ar, ar' , . By the given conditions, a + ar + ar' = 7 (1) => a(1 + r + r') = 7 (3) and
(4)
a 2 (1 + r2 + r ' ) implies [a (1 + r + r2 )] 2
(4)
and (a)' + (ar)' + (ar')' = 21 (2) => a' (1 + r' + r') = 21
(1) (3)
... (2) ...
21
(7)2 ' (1 + r 2 + r) (1 + r 2  r) 
=
1 + r2  r
3 7
3 7
1
7 + 7r'  7r = 3 + 3r + 3r' => 2r'  5r + 2 = 0 => r = 2, 2 '
4) 4 . 4 � => a ( + � + �) 4, 4 (H4(H, 4, 4, 2,
Case I. r = 2. In this case, (3) => a(1 + 2 + . . The G.P. is 1, 1(2), 1(2)2, ...... i.e., 1, 2, The first five terms are 1, 2, , 8, 16. Case ll. r = The G P is
.
.
. In this case,
The fIrst fIve terms are
1
(3)
...... i.e.,
=7
=>
=7
a=1
=> a = 4
2, 1, .
1 1
1, 2' 4'
Example 15. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in this order, we obtain an A.P. Find the numbers. Sol. Let the numbers in G.P. be a, ar, ar'. By the given conditions : a + ar + ar' = 56 (1) and a  I, ar  7, ar'  2 1 are in A.P. (2) a(l + r + r') = 56 (1) => (3) (ar  7)  (a  1) = (ar'  21)  (ar  7) (2) => ar  a  6 = ar'  ar  14 = a  2ar + ar' = 8
GEOMETRIC PROGRESSION
677
a(l  2r + r') = 8 a (l + r + r 2 ) 56 = Dividing (4) by (3), we get 2 8 a (I  2r + r ) 1 + r + r' = 7  14r + 7r' 1
(4)
l + r + r2 =7 1  2r + r 2 => 6r'  15r + 6 = 0 => 2r'  5r + 2 = 0 r = 112, 2. "e.,
7a
Case I. r = 2'
= 56 Le., a = 32
32, 16, 8. Case ll. r = 2. (1) => a (I + 2 + 4) = 56 => 7a = 56 "e., a = 8 . . The numbers are 8, 8(2), 8(2)' i.e., 8, 16, 32. Example 16. There are four numbers such that the first three of these form an A.P. and the last three form a G.P. The sum of the first and the third numbers is 2 and that of the second and fourth is 26, what are these numbers ? Sol. Let the numbers be a, b, c, d. By the given conditions : a+c=2 (1) (2) b + d = 26 and d = 26  b. c =2a (1) => (2) => . . The numbers are a, b, 2  a, 26  b. Also, a, b, 2  a are in A.P. .. b  a = (2  a)  b Le., 2b = 2 or b = 1. . . The numbers become a, 1, 2  a, 26  1 = 25. 25 2a Also, last three numbers are in G.P. =  . 2a 1 a2  4a  2 1 = 0 => a =  3, 7 => (2 _ a) ' = 25 => Case I. a = 3. In this case, the numbers are  3, 1, 2  ( 3), 25 i.e.,  3, 1, 5, 25. Case II. a = 7. In this case, the numbers are 7, 1, 2  7, 25 i.e., 7, 1,  5, 25. ..
The numbers are 32, 32(112), 32(112)' i.e.,
4


1.
I
Findthe common ratio ofthe following G,P, . (,) 0.01, 0. 0 001, 0.000001, . (ii) a, a2, . amn , am , am+n ) , (���) 't/2 , J2 2.J2 Show that each of the following is a G Also find nth term in each case : 5/2,5/4, 5/8, . (,) 128, 64, 32, . (iii) ../3 3 3../3 (iv) ak, ak3 , ak', . Each term of a G is doubled. Is the resulting sequence also a G ,P, ? If a, b, e are in G.P. and k (* 0) E R, then show that : (,) a+ k, b +k, e+ k may not bein G.P. (i,) a  k, b  k, e  k may not be in G.P. (iii) ka, kb, ke are in G. P . (iv) a/k, b/k, elk are in G. P . If each term ofa G,P, is : (0 multiplied by the same nOllZero number (ii) divided by the same nOllZero number, then show that the resulting sequence is also a G,P, ...
2.
r;::
!
,
3. 4.
5.
TEST YOUR KNOWLEDGE 15.4
,
'
!
,
, ......
...... ,p,
,p,
. (w)
I,
(h)

678 6. 7. 8. 9. 10. 11. 12.
DISCRETE STRUCTURES
Find the sequence whose nth term is 3n1 , What type of sequence is it? What is the 5th term ? (ii) Find the sequence whose nth term is 3 n  1. Is it a G,P, ? What is its 6th term ? Determine the number of terms in the sequence 5/2, 5, 10, "" ") 640. (,) Which term of the G.P. 2, 2./2 , 4, ...... is 128? (ii) Which term of the series 5 + 10 + 20 + " " " is 2560 ? Is 512 a term of the sequence 11256, 1164, 1116, "",, ? (0 The 5th, 8th and 11th terms of a G,P, are q and r respectively. Show that q2 = pr. (ii) In any G,P, prove that Tn_r Tn+r = (Tn?' (,) The 3rd term of a G.P. is 24 and 6th term is 192. Find the lOth term. (ii) The 5th term of a G.P. is 16 and lOth term is 1/2. Find the G.P. Also find the 15th term. (0 The fourth term of a G,P, is square of its second term and the first term is 3. Determine the seventh term of the G ,P, (ii) Find the G for which the sum of the first two terms is 4 and the fifth term is 4 times the third term, Ifa, b, c are in A,P" then show that for any nonzero real number k, the numbers ka, kb, ke are in G,P, 1 � 1 are inA, P " show thatx,y, zare in G, P , If x+y 2y y + z (,) Find the value of x if 217, x, 7/2 are in G.P. 13 + k are in G,P, ? (ii) For what value(s) of k, the numbers 1 + k, � + k, 6 18 If a, b,careinG,P" showthat __1_ , __1_ , __1_ areinA,P , If aX = bY= and x, y, z are in G,P" show that 10g a = loge b, If the mth, nth andpth terms of a G,P, are in G,Pb" show that m, n andp are in A,P, If the pth, qth and rth terms of a G,P, are x, y andz respectively, show that : (,) X'' y'P zpq � 1 (q r) log x + (r  p) log y + q) log z � O. Find the value of x for which x + 9, x 6 and 4 are the first three terms of a G and calculate the fourth term of the progression, Ifthe pth, qth, rth and sth terms of an A,P, are in G,P" show that p q, qr, r  are also in G,P, If a, b, c are in G show that : (,) a2, b2, are in G.P. ( " a1 b1 c1 areln. G ,P , (iii) b2c2 ,C2a2 ,a2b2 areinG, P " (iv) a2 + b2 , ab + bc, b2 + c2 are in G, P , If a, b, c, d are in G,P" show that : (0 a + b, b + c, c + d are in G,P, a2 + b2 , b2 + c2 + d2 are in G,P, (iii) (ae)'+(b  e) 2 +(bd)'�(a  d:j2 (iv) (a2 + b2 + c2) (b2 + e2 + d") � (ab + be + ed:j2 If�(aT:)a ben/2 the first term of a G,P" the nth term andP, the product of first nterms, showthatP If the (m +m!n)th term of a G,P, is p and (m  n)th term is q, show that mth and nth terms are ,J"Pq andp(q/p) 2n respectively, The qth and rth terms of anA,P , aswell asofa G,P, are a, b, crespectively, Provethatabe be l r 1
(1)
(2)
682
DISCRETE STRUCTURES
ILLUSTRATIVE EXAM P L ES
Example 1. Find the sum of the following series upto 50 terms. 1 1 1 1 (P T U, M.BA May 2003) 1  '3 + '32  '32 + 37 '"
1
. a =1, r =  3 ' n = 50. Sol. Given series is a G.P. wIth Using S n = a
[_l1__r_rn)
,we have S50 =
Example 2. Evaluate
1
(�r
1+
2 2 2 + + + 625 125 25 . . .
3
1 1 3 50 4 3
=
� 4
( 1 1 ). _
_
350
50 .
2 2 . a= 125 = 5, 1 = 50. Sol. Given series is a G.P. wIth 2625 ' r = 625 2 50 x 5  625 156248 39062 Ir  a = Required sum =  ' r > 1 = r 1 51 625 x 4 625 Example
Sn = 189.
3.
Determine the number of terms in the G.P. (Tj if T1
=
3, Tn = 96 and
Sol. Let the G.P. be a, ar, ar' , . ..... We are given : Tl = 3, Tn = 96, Sn = 189.
a(l  rn) = 189 1r l 3  96r 189 = a arn . r = __ 1r 1r Crass·multiplying, we get 189  189r = 3  96r. Solving, we get r = 2 . .. arn1 = 96 implies 3(2) nl = 96 or 2 n1 = 32 = 2 5 => n  1 = 5 "e., n = 6. ./3 Example 4. How many terms of the G.P. ../3, 3, 3. , . . . . . . sum up to 39 + 13../3 ? Sol. The given G.P. is ../3, 3, 3../3, . . . . . . . 3 Here a = ../3 , r = ../3 = ·../3 , a = 3, arn1 = 96,
Let 3 9 + 13../3 be the sum of n terms of the G.P.
../3 (1 (../3)n ) rn 1  ,,3
=
39 +
rc; 13,,3
_
Sn = 39 + 13../3
GEOMETRIC PROGRESSION
683
.,/3 (1 (.,/3)n )
= 39  39.,/3 + 13.,/3  39 => .,/3 (1 (.,/3 )n) =  26.,/3 3 n = 3 => n = 6. 1  3nl2 =  26 => 3nl2 = 27 = 3 => 
2
Example 5. The second and fifth terms ofa G.P. are 112and 1116 respectively. Find the
sum of the G.P. upto 8 terms. Sol. Let a be the first term and r, the common ratio of the G.P. T5 = ar51 = ar4 . T = ar21 = ar and 2
1 ar =  2
Dividing (2) by (1), we get putting r = 
�
(1)
_ ( �) �
ar 4
1/16 . ;;;:   1/2 Le.,
in (I) We get a '
1 16
ar4 = 
and r
(2)
1 2
1
=  '8 or r =  ,
Le., a = 1.
=
1 185 2 =256 = 255 X _ . 3 256 3 128 2 __
[8n
=
a( l  r n ) r 1r
lor r