RDW_018 - Reading, Discovering and Writing Proofs Version...

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Unformatted text preview: Reading, Discovering and Writing Proofs Version 0.1.8 Steven Furino February 26, 2012 Contents 1 In the beginning 1.1 What Makes a Mathematician a Mathematician? 1.2 How The Course Works . . . . . . . . . . . . . . 1.3 Why do we reason formally? . . . . . . . . . . . . 1.4 Reading and Lecture Schedule . . . . . . . . . . . 1.4.1 Lecture Schedule . . . . . . . . . . . . . . 1.4.2 Reading Schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 8 8 9 11 11 12 2 Our 2.1 2.2 2.3 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 13 13 16 18 3 Truth Tables 3.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Truth Tables as Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Truth Tables to Evaluate Logical Expressions . . . . . . . . . . . . . . . . . 21 21 21 23 4 Introduction to Sets 4.1 Objectives . . . . . . . . . . . 4.2 Describing a Set . . . . . . . 4.3 Set Operations . . . . . . . . 4.3.1 Venn Diagrams . . . . 4.4 Comparing Sets . . . . . . . . 4.4.1 Sets of Solutions . . . 4.4.2 An Example . . . . . 4.5 Showing Two Sets Are Equal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 26 26 30 32 32 32 33 34 5 Discovering Proofs 5.1 Objectives . . . . . . . . 5.2 Discovering a Proof . . . 5.3 Reading A Proof . . . . 5.4 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 36 36 38 39 6 Quantifiers 6.1 Objectives . . . . . . . 6.2 Quantifiers . . . . . . 6.3 The Object Method . 6.4 The Construct Method 6.5 The Select Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 41 41 44 45 46 First Proof Objectives . . . The Language . Implications . . Our First Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Section 0.0 CONTENTS 6.6 6.7 3 Sets and Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A Non-Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Nested Quantifiers 7.1 Objectives . . . . . . . . . . . . . 7.2 Onto or Surjective . . . . . . . . 7.2.1 Definition . . . . . . . . . 7.2.2 Reading . . . . . . . . . . 7.2.3 Discovering . . . . . . . . 7.3 One-to-one or Injective . . . . . . 7.3.1 Definition . . . . . . . . . 7.3.2 Reading . . . . . . . . . . 7.3.3 Discovering . . . . . . . . 7.4 Limits . . . . . . . . . . . . . . . 7.4.1 Definition . . . . . . . . . 7.4.2 Reading A Limit Proof . 7.4.3 Discovering a Limit Proof 47 48 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 50 50 50 52 53 54 54 55 56 57 57 58 60 8 Induction 8.1 Objectives . . . . . . . . . . . . . . . . . . 8.2 Notation . . . . . . . . . . . . . . . . . . . 8.2.1 Summation Notation . . . . . . . . 8.2.2 Product Notation . . . . . . . . . . 8.2.3 Recurrence Relations . . . . . . . . 8.3 Introduction to Induction . . . . . . . . . 8.4 Principle of Mathematical Induction . . . 8.4.1 Why Does Induction Work? . . . . 8.4.2 Two Examples of Simple Induction 8.4.3 A Different Starting Point . . . . . 8.5 Strong Induction . . . . . . . . . . . . . . 8.5.1 Interesting Example . . . . . . . . 8.6 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 62 62 62 64 64 65 66 67 67 69 70 72 73 . . . . 76 76 76 80 82 10 Properties Of GCDs 10.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Some Useful Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 84 84 11 Linear Diophantine Equations 11.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Select Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Linear Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 90 90 90 92 9 The 9.1 9.2 9.3 9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Greatest Common Divisor Objectives . . . . . . . . . . . . . . . . . . . Greatest Common Divisor . . . . . . . . . . Certificate of Correctess . . . . . . . . . . . The Extended Euclidean Algorithm (EEA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Practice, Practice, Practice: Quantifiers and Sets 100 12.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 12.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 4 Chapter 0 13 Congruence 13.1 Objectives . . . . . . . . . . . . . 13.2 Congruences . . . . . . . . . . . . 13.2.1 Definition of Congruences 13.3 Elementary Properties . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 103 103 103 104 14 Modular Arithmetic 14.1 Objectives . . . . . . . . . . . . . . . 14.2 Modular Arithmetic . . . . . . . . . 14.2.1 [0] ∈ Zm . . . . . . . . . . . . 14.2.2 [1] ∈ Zm . . . . . . . . . . . . 14.2.3 Identities and Inverses in Zm 14.2.4 Subtraction in Zm . . . . . . 14.2.5 Division in Zm . . . . . . . . 14.3 Fermat’s Little Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 112 112 114 114 114 116 116 116 . . . . 120 120 120 122 123 15 Linear Congruences 15.1 Objectives . . . . . . . . 15.2 The Problem . . . . . . 15.3 Extending Equivalencies 15.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Chinese Remainder Theorem 125 16.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 16.2 An Old Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 16.3 Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 126 17 Practice, Practice, Practice: Congruences 129 17.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 17.2 Linear and Polynomial Congruences . . . . . . . . . . . . . . . . . . . . . . 129 17.3 Preparing for RSA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 18 The 18.1 18.2 18.3 RSA Scheme Objectives . . . . . . . . . . . . . Why Public Key Cryptography? Implementing RSA . . . . . . . . 18.3.1 Setting up RSA . . . . . . 18.3.2 Sending a Message . . . . 18.3.4 Example . . . . . . . . . . 18.3.3 Receiving a Message . . . 18.4 Does M = R? . . . . . . . . . . . 18.5 How Secure Is RSA? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Negation 19.1 Objectives . . . . . . . . . . . . . . . . 19.2 Negating Statements . . . . . . . . . . 19.3 Negating Statements with Quantifiers 19.3.1 Counterexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 135 135 136 136 136 136 137 139 140 . . . . 141 141 141 143 144 20 Contradiction 146 20.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 20.2 How To Use Contradiction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 Section 0.0 CONTENTS 5 20.2.1 When To Use Contradiction . . . . . . . . . . . . . . . . . . . . . . . 20.2.2 Reading a Proof by Contradiction . . . . . . . . . . . . . . . . . . . 20.2.3 Discovering and Writing a Proof by Contradiction . . . . . . . . . . 21 Contrapositive 21.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The Contrapositive . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.1 When To Use The Contrapositive . . . . . . . . . . . . . . . 21.3 Reading a Proof That Uses the Contrapositive . . . . . . . . . . . 21.3.1 Discovering and Writing a Proof Using The Contrapositive 22 Uniqueness 22.1 Objectives . . . . . . . . 22.2 Introduction . . . . . . . 22.3 Showing X = Y . . . . . 22.4 Finding a Contradiction 22.5 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Introduction to Primes 23.1 Objectives . . . . . . . . . . . . . . . 23.2 Introduction to Primes . . . . . . . . 23.3 Induction . . . . . . . . . . . . . . . 23.4 Fundamental Theorem of Arithmetic 23.5 Finding a Prime Factor . . . . . . . 23.6 Working With Prime Factorizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 147 148 . . . . . 151 151 151 151 152 153 . . . . . 155 155 155 156 157 158 . . . . . . 160 160 160 161 162 164 166 24 Introduction to Fermat’s Last Theorem 168 24.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 24.2 History of Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . . . . 168 24.3 Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 25 Characterization of Pythagorean Triples 174 25.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 25.2 Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 26 Fermat’s Theorem for n = 4 26.1 Objectives . . . . . . . . . 26.2 n = 4 . . . . . . . . . . . 26.3 Reducing the Problem . . 26.4 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 177 177 179 180 27 Problems Related to FLT 181 27.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 27.2 x4 − y 4 = z 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 28 Practice, Practice, Practice: Prime Numbers 184 28.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 28.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 29 Complex Numbers 185 29.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 29.2 Different Equations Require Different Number Systems . . . . . . . . . . . . 185 6 Chapter 0 CONTENTS 29.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Properties Of 30.1 Objectives 30.2 Conjugate 30.3 Modulus . 186 Complex Numbers 189 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 31 Graphical Representations of Complex Numbers 31.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . 31.2 The Complex Plane . . . . . . . . . . . . . . . . . 31.2.1 (x, y ) . . . . . . . . . . . . . . . . . . . . . . 31.2.2 Modulus . . . . . . . . . . . . . . . . . . . . 31.3 Polar Representation . . . . . . . . . . . . . . . . . 31.4 Converting Between Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 192 192 192 193 193 194 32 De Moivre’s Theorem 197 32.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 32.2 De Movre’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 32.3 Complex Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 33 Roots of Complex Numbers 200 33.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 33.2 Complex n-th Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 33.3 More Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 34 An Introduction to Polynomials 204 34.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 34.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 34.3 Operations on Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 35 Factoring Polynomials 208 35.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 35.2 Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 36 The 36.1 36.2 36.3 36.4 36.5 Shortest Path Objectives . . . The Problem . Abstraction . . Algorithm . . . Extensions . . . Problem ...... ...... ...... ...... ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 212 212 214 215 215 37 Paths, Walks, Cycles and Trees 216 37.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 37.2 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 37.3 Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 38 Trees 222 38.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 38.2 Properties of Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 39 Dijkstra’s Algorithm 226 39.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 Section 0.0 CONTENTS 7 39.2 Dijkstra’s Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39.3 Certificate of Optimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Certificate of Optimality - Path 40.1 Objectives . . . . . . . . . . . . 40.2 Certificate of Optimality . . . . 40.3 Weighted Graphs . . . . . . . . 40.4 Certificate of Optimality - Tree 41 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 231 233 233 233 234 239 240 Chapter 1 In the beginning 1.1 What Makes a Mathematician a Mathematician? Welcome to MATH 135! Let me begin with a question. What makes a mathematician a mathematician? Many people would answer that someone who works with numbers is a mathematician. But bookkeepers for small businesses work with numbers and we don’t normally consider a bookkeeper as a mathematician. Others might think of geometry and answer that someone who works with shapes is a mathematician. But architects work with shapes and we don’t normally consider architects as mathematicians. Still others might answer that people who use formulas are mathematicians. But engineers work with formulas and we don’t normally consider engineers as mathematicians. A more insightful answer would be that people who find patterns and provide descriptions and evidence for those patterns are mathematicians. But scientists search for and document patterns and we don’t normally consider scientists as mathematicians. The answer is proof - a rigorous, formal argument that establishes the truth of a statement. This has been the defining characteristic of mathematics since ancient Greece. MATH 135 is about reading, writing and discovering proofs. If you have never done this before, do not worry. This course will provide you with techniques that will help, and we will practice those techniques in the context of some very interesting algebra. 1.2 How The Course Works He who seeks for methods without having a definite problem in mind seeks for the most part in vain. David Hilbert Let me describe how the course works. Throughout the course, we will work on three problems all of which illustrate the need for proof. The first problem resolves a very important practical commercial problem. The second problem begins work on one of the most notorious problems in all of mathematics. The last problem yields a surprising and beautiful result. Here are the three problems. 8 Section 1.3 Why do we reason formally? 9 How do we secure internet commerce? Have you ever bought a song or movie over iTunes? Have you ever done your banking over the web? How do you make sure that your credit card number and personal information are not intercepted by bad guys? Number theory allows us to enable secure web transactions. And that theory is backed by proof. How many solutions are there to xn + y n = z n where x, y and z must be positive integers and n is an integer greater than or equal to three? This is one of the most famous problems in the history of mathematics and it took over 350 years to solve. It was first conjectured by the French mathematician Pierre de Fermat in 1637 and was only solved in 1995 by Andrew Wiles. Why does eiπ + 1 = 0 ? e is a very unusual number. Of all the real numbers a, there is ax exactly one where ddx = ax . And that number is e. i is a very unusual number with its defining property of i2 = −1. π is a very unusual number even if it is common. It is the unique ratio of the circumference of a circle to its diameter. Why should that ratio be unique? One is the basis of the natural numbers, hence the integers, hence the rationals. Zero is a difficult number and was only accepted into the mathematics of western Europe because of the influence of Hindu and Islamic scholars. Why should all of these numbers be connected in so simple and elegant a form? To work with these problems we will need to learn about congruences, modular arithmetic, primes and complex numbers. And to work with these topics, we must learn about proof techniques. The proof techniques will be introduced as we need them. 1.3 Why do we reason formally? Since many people dislike proofs, and think that humans already know enough mathematics, let me deal with the question: “Why bother with proofs?” There are quite a few reasons. To prevent silliness. In solving quadratic equations with non-real roots, some of you will have encountered the number i which has the special property that i2 = −1. But then, √ √√ √ −1 = i2 = i × i = −1 −1 = −1 × −1 = 1 = 1 Clearly, something is amiss. To understand better. How would most of us answer the question “What’s a real number?” We would probably say that any number written as a decimal expansion is a real number and any two different expansions represent different numbers. But then what about this? Let x = 0.9 = 0.999 . . . . Multiplying by 10 and subtracting gives 10x = 9.9 − x = 0.9 9x = 9 which implies x = 1, not x = 0.9. 10 Chapter 1 In the beginning Or suppose we wanted to evaluate the infinite sum 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ... If we pair up the first two terms we get zero and every successive pair of terms also gives us 0 so the sum is zero. 1 − 1+1 − 1+1 − 1+1 − 1+... On the other hand, if we pair up the second and third term we get 0 and all successive pairs of terms give 0 so the sum is 1. 1 −1 + 1 −1 + 1 −1 + 1 −1 + 1 + . . . Or suppose we wanted to resolve Zeno’s paradox. Zeno was a famous ancient Greek philosopher who posed the following problem. Suppose the Greek hero Achilles was going to race against a tortoise and suppose, in recognition of the slowness of the tortoise, that the tortoise gets a 100m head start. By the time Achilles has run half the distance between he and the tortoise, the tortoise has moved ahead. And now again, by the time Achilles has run half the remaining distance between he and the tortoise, the tortoise has moved ahead. No matter how fast Achilles runs, the tortoise will always be ahead! You might object that your eyes see Achilles pass the tortoise, but what is logically wrong with Zeno’s argument? To make better commercial decisions. Building pipelines is expensive. And lots of pipelines will be built in the next few decades. Pipelines will ship oil, natural gas, water and sewage. Finding the shortest route given physical constraints (mountains, rivers, lakes, cities), environmental constraints (protection of the water table, no access through national or state parks), and supply chain constraints (access to concrete and steel) is very important. How do pipeline builders prove that the route they have chosen for the pipeline is the shortest possible route given the constraints? To discover solutions. Formal reasoning provides a set of tools that allow us to think rationally and carefully about problems in mathematics, computing, engineering, science, economics and any discipline in which we create models. Poor reasoning can be very expensive. Inaccurate application of financial models led to losses of hundreds of billions of dollars during the financial crisis of 2008. To experience joy. Mathematics can be beautiful, just as poetry can be beautiful. But to hear the poetry of mathematics, one must first understand the language. Section 1.4 Reading and Lecture Schedule 1.4 1.4.1 Reading and Lecture Schedule Lecture Schedule Here is a proposed lecture schedule. Lec. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Ch. 1 2 5 6 7 8 8 9 9 10 11 11 12 13 13 14 14 15 16 17 19 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 Topic In The Beginning Our First Proof Discovering Proofs Quantifiers Nested Quantifiers Simple Induction and the Binomial Theorem Strong Induction Greatest Common Divisor Extended Euclidean Algorithm Properties of the GCD Linear Diophantine Equations Linear Diophantine Equations Practice, Practice, Practice Congruence Congruence Modular Arithmetic Modular Arithmetic Linear Congruences Chinese Remainder Theorem Practice, Practice, Practice RSA Introduction to Primes Introduction to Fermat’s Last Theorem Characterization of Pythagorean Triples The Case n = 4 Related Problems Practice, Practice, Practice Introduction to Complex Numbers Properties of Complex Numbers Graphical Representations of Complex Numbers DeMoivre’s Theorem Roots of Complex Numbers Practice, Practice, Practice Introduction to Polynomials Factoring Polynomials Practice, Practice, Practice 11 12 Chapter 1 1.4.2 In the beginning Reading Schedule Since one of the goals of this course is to help you become comfortable reading mathematics, there are ten short chapters for you to read. After you have completed the reading, an online assignment will help you consolidate what you know. Ch. 3 4 6 7 8 18 20 21 22 23 Topic Truth Tables Sets Quantifiers Nested Quantifiers Induction Introduction to Cryptography Negation Contradiction Contrapositive Uniqueness Before Lecture 3. Discovering Proofs 3. Discovering Proofs 4. Quantifiers 5. Nested Quantifiers 8. Simple Induction 21. RSA 22. Introduction to Primes 22. Introduction to Primes 22. Introduction to Primes 22. Introduction to Primes Chapter 2 Our First Proof 2.1 Objectives The technique objectives are: 1. Define statement, hypothesis, conclusion and implication. 2. Learn how to structure the analysis of a proof. 3. Carry out the analysis of a proof. The content objectives are: 1. Define divisibility. 2. State and prove the Transitivity of Divisibility. 2.2 The Language Mathematics is the language of mathematicians, and a proof is a method of communicating a mathematical truth to another person who speaks the “language”. (Solow, How to Read and Do Proofs) Mathematics is an unusual language. It is extraordinarily precise. When a proof is fully and correctly presented, there is no ambiguity and no doubt about its correctness. However, understanding a proof requires understanding the language. This course will help you with the basic grammar of the language of mathematics and is applicable to all proofs. Just as in learning any new language, you will need lots of practice to become fluent. 13 14 Chapter 2 Our First Proof With respect to learning proof techniques, the broad objectives of the course are simple. 1. Explain and categorize proof techniques that can be used in any proof. This course will teach not only how a technique works, but when it is most likely to be used and why it works. 2. Learn how to read a proof. This will require you to identify the techniques of the first objective. 3. Discover your own proofs. Knowledge of technique is essential but inadequate. Or, as we would say in the language of mathematics, technique is “necessary but not sufficient”. Discovering your own proof requires not only technique but also understanding, creativity, intuition and experience. This course will help with the technique and experience. Understanding, creativity, and intuition come with time. Talent helps of course. 4. Write your own proofs. Having discovered a proof, you must distill your discovery into mathematical prose that is targeted at a specific audience. Hopefully, in the previous lecture, I convinced you of why we need to prove things. Now what is it that mathematicians prove? Mathematicians prove statements. Definition 2.2.1 A statement is a sentence which is either true or false. Statement Example 1 Here are some examples of statements. 1. 2 + 2 = 4. (A true statement.) 2. 2 + 2 = 5. (A false statement.) 3. x2 − 1 = 0 has two distinct real roots. (A true statement.) 4. There exists an angle θ such that sin(θ) > 1. (A false statement.) Example 2 Now consider the following sentences. 1. x > 0. 2. ABC is congruent to P QR. These are statements only if we have an appropriate value for x in the first sentence and appropriate instances of ABC and P QR in the second sentence. For example, if x is the number 5, then the sentence “5 > 0” is a statement since the sentence is true. If x is the number −5, then the sentence “−5 > 0” is also a statement since the sentence is false. The key point is that a statement is a sentence which must be true or false. If x is the English word algebra, then the sentence “algebra > 0” is not a statement since the sentence is neither true nor false. Sentences like the two above are called open sentences. Section 2.2 The Language Definition 2.2.2 15 An open sentence is a sentence that Open Sentence • contains one or more variables, where • each variable has values that come from a designated set called the domain of the variable, and • where the sentence is either true or false whenever values from the respective domains of the variables are substituted for the variables. For example, if the domain of x is the set of real numbers, then for any real number chosen and substituted for x, the sentence “x > 0” is a statement. In this course, we will treat all open sentences as statements under the assumption that the values of the variables always come from a suitable domain. Self Check 1 For each of the following, choose one of the following possible answers. (a) This is not a statement. (b) This is a true statement. (c) This is a false statement. (d) This is an open sentence. 1. {1, 3, 5, 7, 9} 2. sin2 θ + cos2 θ 3. For all real values θ, sin2 θ + cos2 θ = 1 4. Therefore, x = π/2. 5. If x is a positive real number, then log10 x > 0. 6. x2 + 1 = 0 has two real roots. 16 Chapter 2 2.3 Definition 2.3.1 Implication Our First Proof Implications The most common type of statement we will prove is an implication. Implications have the form If A is true , then B is true where A and B are themselves statements. An implication is more commonly read as If A, then B or A implies B or A⇒B Definition 2.3.2 Compound Statement An implication is a compound statement, that is, it is made up of more than one statement. In the statement “A implies B ”, A is a statement which may be true or false. B is a statement which may be true or false. “A implies B ” is also a statement which may be true or false. Definition 2.3.3 The statement A is called the hypothesis. The statement B is called the conclusion. Hypothesis, Conclusion REMARK To prove the implication “A implies B ”, you assume that A is true and you use this assumption to show that B is true. A is what you start with. B is where you must end up. To use the implication “A implies B ”, you must first establish that A is true. After you have established that A is true, then you can invoke B . It is crucial that you are able to identify 1. the hypothesis 2. the conclusion 3. whether you are using or proving an implication Here are some examples of implications. Example 3 If x is a positive real number, then log10 x > 0. Hypothesis: x is a positive real number. Conclusion: log10 x > 0. Section 2.4 Implications Example 4 17 Let f (x) = x sin(x). Then f (x) = x for some real number x with 0 ≤ x ≤ 2π . Hypothesis: f (x) = x sin(x). Conclusion: f (x) = x for some real number x with 0 ≤ x ≤ 2π . Example 5 In plane geometry, ∠ABC = ∠XY Z whenever Hypothesis: All figures are in the plane. ABC is similar to ABC ∼ XY Z. XY Z. Conclusion: ∠ABC = ∠XY Z . Self Check 2 Identify the hypothesis and the conclusion in each of the following statements. 1. If a, b and c are real numbers, and b2 − 4ac > 0, then ax2 + bx + c = 0 has two distinct, real roots. (a) a, b and c are real numbers. (b) b2 − 4ac > 0. (c) a, b and c are real numbers and b2 − 4ac > 0. (d) ax2 + bx + c = 0. (e) ax2 + bx + c = 0 has two distinct, real roots. 2. If the line segment AB intersects the line segment P Q at O, then ∠AOQ = ∠P OB . (a) The line segment AB (b) The line segment AB intersects the line segment P Q. (c) The line segment AB intersects the line segment P Q at O. (d) ∠AOQ = ∠P OB . (e) None of the above. 3. y = ax2 − 1 has no real root if a < 0. (a) y = ax2 − 1. (b) y = ax2 − 1 has no real roots. (c) a < 0. (d) None of the above. 4. When x is an integer, the maximum value of y = −x4 + 4x2 + 0.5 is 4. (a) x is an integer. (b) y = −x4 + 4x2 + 0.5 (c) x is a maximum value. (d) The maximum value of y = −x4 + 4x2 + 0.5 is 4. (e) None of the above. 18 Chapter 2 2.4 Our First Proof Our First Proof Let us read our first proof. We begin with a definition. Definition 2.4.1 Divisibility An integer m divides an integer n, and we write m | n, if there exists an integer k so that n = km. Example 6 • 3 | 6 since we can find an integer k , 2 in this case, so that 6 = k × 3. • 5 6 since no integer k exists so that 6 = k × 5. • For all integers a, a | 0 since 0 = 0 × a. This is true for a = 0 as well. • For all non-zero integers a, 0 a since there is no integer k so that k × 0 = a. Some comments about definitions are in order. If mathematics is thought of as a language, then definitions are the vocabulary and our prior mathematical knowledge indicates our experience and versatility with the language. Mathematics and the English language both share the use of definitions as extremely practical abbreviations. Instead of saying “a domesticated carnivorous mammal known scientifically as Canis familiaris” we would say “dog.” Instead of writing down “there exists an integer k so that n = km”, we write “m | n.” However, mathematics differs greatly from English in precision and emotional content. Mathematical definitions do not allow ambiguity or sentiment. Definition 2.4.2 A proposition is a true statement that has been proved by a valid argument. Proposition REMARK You will encounter several variations on the word proposition. A theorem is a particularly significant proposition. A lemma is a subsidiary proposition, or more informally, a “helper” proposition, that is used in the proof of a theorem. A corollary is a proposition that follows almost immediately from a theorem. There are particular statements that may look like propositions but are more foundational. An axiom is a statement that is assumed to be true. No proof is given. From axioms we derive propositions and theorems. Obviously, choosing axioms has to be done very carefully. Consider the following proposition and proof. Proposition 1 (Transitivity of Divisibility (TD)) Let a, b and c be integers. If a | b and b | c, then a | c. Section 2.4 Our First Proof 19 Proof: Since a | b, there exists an integer r so that ra = b. Since b | c, there exists an integer s so that sb = c. Substituting ra for b in the previous equation, we get (sr)a = c. Since sr is an integer, a | c. Though this is a simple proof, other proofs can be difficult to read because of the habits of writing for professional audiences. Many proofs share the following properties which can be frustrating for students. 1. Proofs are economical. That is, a proof includes what is needed to verify the truth of a proposition but nothing more. 2. Proofs do not usually identify the hypothesis and the conclusion. 3. Proofs sometimes omit or combine steps. 4. Proofs do not always explicitly justify steps. 5. Proofs do not reflect the process by which the proof was discovered. The reader of the proof must be conscious of the hypothesis and conclusion, fill in the omitted parts and justify each step. REMARK When you are reading a proof of an implication, do the following. 1. Explicitly identify the hypothesis and the conclusion. If the hypothesis contains no statements write “No explicit hypothesis”. At the end of the proof, you should be able to identify where each part of the hypothesis has been used. 2. Explicitly identify the core proof technique. When reading a proof, the reader usually works forward from the hypothesis until the conclusion is reached. Specific techniques will be covered later in the course. 3. Record any preliminary material needed, usually definitions or propositions that have already been proved. Judgement is needed here about how much to include. 4. Justify each step with reference to the definitions, previously proved propositions or techniques used. 5. Add missing steps where necessary and justify these steps with reference to the definitions, previously proved propositions or techniques used. Let’s analyze the proof of the Transitivity of Divisibility in detail because it will give us some sense of how to analyze proofs in general. First, observe that “If a | b and b | c, then a | c.” is an open sentence, and that the domains for the variables a, b and c are specified in the first sentence, “Let a, b and c be integers.” Professional mathematicians do all of these things implicitly but for the first part of this course, we will do these things explicitly. We will do a line by line analysis, so to make our work easier, we will write each sentence on a separate line. 20 Chapter 2 Our First Proof Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Since a | b, there exists an integer r so that ra = b. 2. Since b | c, there exists an integer s so that sb = c. 3. Substituting ra for b in the previous equation, we get (sr)a = c. 4. Since sr is an integer, a | c. Let’s analyze the proof. What we do now will seem like overkill but it serves two purposes. It gives practice at justifying every line of a proof, and it gives us a structure that we can use for other proofs. Lastly, recall that the author is proving an implication. The author assumes that the hypothesis is true, and uses the hypothesis to demonstrate that the conclusion is true. Here goes. Analysis of Proof We begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: a, b and c are integers. a | b and b | c. Conclusion: a | c. Core Proof Technique: Work forwards from the hypothesis. Preliminary Material: The definition of divides. An integer m divides an integer n, and we write m | n, if there exists an integer k so that n = km. Sentence 1 Since a | b, there exists an integer r so that ra = b. In this sentence, the author of the proof uses the hypothesis a | b and the definition of divides. Sentence 2 Since b | c, there exists an integer s so that sb = c. In this sentence, the author uses the hypothesis b | c and the definition of divides. Sentence 3 Substituting ra for b in the previous equation, we get (sr)a = c. Here, the author works forward using arithmetic. The actual work is: sb = c and ra = b implies s(ra) = c which implies (sr)a = c. Sentence 4 Since sr is an integer, a | c. Lastly, the author uses the definition of divides. In this case, the m, k and n of the definition apply to the a, sr and c of the proof. It is important to note that sr is an integer, otherwise the definition of divides does not apply. At the end of each proof, you should be able to identify where each part of the hypothesis was used. It is obvious where a | b and b | c were used. The hypothesis “a, b and c are integers” was needed to allow the author to use the definition of divides. This completes the analysis of our first proof. Between the readings, lectures, quizzes, assignments and tests, you will work your way through roughly one hundred proofs. Chapter 3 Truth Tables 3.1 Objectives The technique objectives are: 1. Define not, and, or, implies and if and only if using truth tables. 2. Evaluate logical expressions using truth tables. 3. Use truth tables to establish the equivalence of logical expressions. 3.2 Truth Tables as Definitions Throughout this course we work with statements. Definition 3.2.1 A statement is a sentence which is either true or false. Statement Definition 3.2.2 Compound, Component All of the statements we need to prove will be compound statements, that is, statements composed of several individual statements called component statements. For example, the compound statement If a | b and b | c, then a | c. contains three component statements a | b, b | c, and a|c Suppose we let X be the statement a | b and Y be the statement b | c and Z be the statement a | c. Then our original statement 21 22 Chapter 3 Truth Tables If a | b and b | c, then a | c. becomes X and Y imply Z . If we knew the truth values of X , Y and Z , then we would be able to determine the truth value of the compound statement “X and Y imply Z ”. And that is where truth tables come in. Truth tables contain all possible values of the component statements and determine the truth value of the compound statement. Truth tables can be used to define the truth value of a statement or evaluate the truth value of a statement. For logical operations like not, and, or, implies and if and only if, truth tables are used to define the truth value of the compound statement. Definition 3.2.3 The simplest definition is that of NOT A, written ¬A. NOT A ¬A TF FT In prose, if the statement A is true, then the statement “NOT A” is false. If the statement A is false, then the statement “NOT A” is true. Two very important and common logical connectives are AND and OR. Note that these do not always coincide with our use of the words and and or in the English language! Definition 3.2.4 AND Definition 3.2.5 OR The definition of A AND B , written A ∧ B , is A T T F F B A∧B T T F F T F F F The definition of A OR B , written A ∨ B , is A T T F F B A∨B T T F T T T F F This is an opportune moment to highlight the difference between mathematical language and the English language. If you are visiting a friend and your friend offers you “coffee or Section 3.3 Truth Tables to Evaluate Logical Expressions 23 tea”, you interpret that to mean that you may have coffee or tea but not both. However, the logical A ∨ B results in a true statement when A is true, B is true or both are true. In mathematics, or is inclusive. Definition 3.2.6 The definition of A implies B , written A ⇒ B , often seems strange. Implies A T T F F B A⇒B T T F F T T F T The first two rows in the table make sense. The last two make less sense. How can a false hypothesis result in a true statement? The basic idea is that if one is allowed to assume an hypothesis which is false, any conclusion can be derived. We will shortly see that implies is closely related to if and only if. Definition 3.2.7 The definition of A if and only if B , written A ⇐⇒ B or A iff B , is If and Only If A T T F F 3.3 B A ⇐⇒ B T T F F T F F T Truth Tables to Evaluate Logical Expressions We can construct truth tables for compound statements by evaluating parts of the compound statement separately and then evaluating the larger statement. Consider the following truth table which shows the truth values of ¬(A ∨ B ) for all possible combinations of truth values of the component statements A and B . Example 1 Construct a truth table for ¬(A ∨ B ). A T T F F B A ∨ B ¬(A ∨ B ) T T F F T F T T F F F T In the first row of the table A and B are true, so using the definition of or, the statement A ∨ B is true. Since the negation of a true statement is false, ¬(A ∨ B ) is false, which 24 Chapter 3 Truth Tables appears in the last column of the first row. Take a minute to convince yourself that each of the remaining rows is correct. Here is another example. Example 2 Construct a truth table for A ⇒ (B ∨ C ). A T T T T F F F F Definition 3.3.1 Logically equivalent B T T F F T T F F C B ∨ C A ⇒ (B ∨ C ) T T T F T T T T T F F F T T T F T T T T T F F T Two compound statements are logically equivalent if they have the same truth values for all combinations of their component statements. We write S1 ≡ S2 to mean S1 is logically equivalent to S2 . REMARK Equivalent statements are enormously useful in proofs. Suppose you wish to prove S1 but are having difficulty. If there is a simpler statement S2 and S1 ≡ S2 , then you can prove S2 instead. In proving S2 , you will have proved S1 as well. Example 3 Construct a single truth table for ¬(A ∨ B ) and (¬A) ∧ (¬B ). Are these statements logically equivalent? A B A ∨ B ¬(A ∨ B ) ¬A ¬B (¬A) ∧ (¬B ) TT T F F F F TF T F F T F FT T F T F F FF F T T T T Since the columns representing ¬(A ∨ B ) and (¬A) ∧ (¬B ) are identical, we can conclude that ¬(A ∨ B ) ≡ (¬A) ∧ (¬B ). The preceding example and your assignments demonstrate DeMorgan’s Laws. Proposition 1 (De Morgan’s Law’s (DML)) If A and B are statements, then 1. ¬(A ∨ B ) ≡ (¬A) ∧ (¬B ) 2. ¬(A ∧ B ) ≡ (¬A) ∨ (¬B ) Section 3.3 Truth Tables to Evaluate Logical Expressions 25 REMARK The next example shows the equivalence of A ⇐⇒ B and (A ⇒ B ) ∧ (B ⇒ A). This is particularly important for proofs. Because A ⇐⇒ B is equivalent to (A ⇒ B ) ∧ (B ⇒ A), to prove a statement of the form A ⇐⇒ B , we could prove 1. A ⇒ B and 2. B ⇒ A. Example 4 Show that A ⇐⇒ B is logically equivalent to (A ⇒ B ) ∧ (B ⇒ A). A T T F F Exercise 1 B A ⇐⇒ B T T F F T F F T A ⇒ B B ⇒ A (A ⇒ B ) ∧ (B ⇒ A) T T T F T F T F F T T T Use truth tables to show that for statements A, B and C , the Associativity Laws hold. That is 1. A ∨ (B ∨ C ) ≡ (A ∨ B ) ∨ C 2. A ∧ (B ∧ C ) ≡ (A ∧ B ) ∧ C Exercise 2 Use truth tables to show that for statements A, B and C , the Distributivity Laws hold. That is 1. A ∧ (B ∨ C ) ≡ (A ∧ B ) ∨ (A ∧ C ) 2. A ∨ (B ∧ C ) ≡ (A ∨ B ) ∧ (A ∨ C ) Exercise 3 What logical statement is equivalent to ¬(A ⇒ B )? Provide evidence in the form of a truth table. Chapter 4 Introduction to Sets 4.1 Objectives The technique objectives are: 1. Define and gain experience with set, element, set-builder notation, defining property, subset, superset, equality of sets, empty set, universal set, complement, cardinality, union, intersection and difference. 2. Be able to read and use Venn diagrams. 4.2 Describing a Set Sets are foundational in mathematics and literally appear everywhere. Definition 4.2.1 Set, Element A set is a collection of objects. The objects that make up a set are called its elements (or members). Sets can contain any type of objects. Since this is a math course, we frequently use sets of numbers. But sets could contain letters, the letters of the alphabet for example, or books, such as those in a library collection. It is customary to use uppercase letters (A, B, C . . .) to represent sets and lowercase letters (a, b, c, . . .) to represent elements. If a is an element of the set A, we write a ∈ A. If a is not an element of the set A, we write a ∈ A. Small sets can be explicitly listed. For example, the set of primes less than 10 is {2, 3, 5, 7} When explicitly listing sets, we use curly braces, {}, and separate elements with a comma. Many sets are either too large to be listed (the set of all primes less than 10,000) or are defined by a rule. In these cases, we employ set-builder notation which makes use of a defining property of the set. For example, the set of all real numbers between 1 and 2 inclusive could be written as {x ∈ R | 1 ≤ x ≤ 2} 26 Section 4.2 Describing a Set 27 The part of the description following the bar (|) is the defining property of the set. Some authors use a colon (:) instead of a bar and write {x ∈ R : 1 ≤ x ≤ 2} As when explicitly listing sets, we use curly braces, {}. We will give a more formal description of defining property after we talk about quantifiers. Some letters have become associated with specific sets. natural numbers, 1, 2, 3, . . . integers rational numbers, { p | p, q ∈ Z, q = 0} q irrational numbers real numbers complex numbers {x + yi | x, y ∈ R} N Z Q Q R C Example 1 (Set-Builder Notation) 1. The set of all even integers can be described as {n ∈ Z : 2 | n} There is frequently more than one way of describing a set. Another way of describing the set of even integers is {2k | k ∈ Z} 2. The set of all real solutions to x2 + 4x − 2 = 0 can be described as {x ∈ R | x2 + 4 x − 2 = 0 } and, in general, the set of all solutions to f (x) = 0 can be described as {x ∈ R | f (x) = 0} 3. The set of all positive divisors of 30 can be written as {n ∈ N : n | 30} 4. In calculus, you often use intervals of real numbers. The closed interval [a, b] is defined as the set {x ∈ R | a ≤ x ≤ b} Definition 4.2.2 Subset A set A is called a subset of a set B , and is written A ⊆ B , if every element of A belongs to B . Symbolically, we write A ⊆ B means x ∈ A ⇒ x ∈ B or equivalently A ⊆ B means “For all x ∈ A, x ∈ B We sometimes say that A is contained in B . 28 Chapter 4 Introduction to Sets Example 2 {1, 2, 3} ⊆ {1, 2, 3, 4} Definition 4.2.3 Proper Subset A set A is called a proper subset of a set B , and written A ⊂ B , if every element of A belongs to B and there exists an element in B which does not belong to A. In the previous example, it is also the case that Example 3 {1, 2, 3} ⊂ {1, 2, 3, 4} Definition 4.2.4 Superset A set A is called a superset of a set B , and written A ⊇ B , if every element of B belongs to A. A ⊇ B is equivalent to B ⊆ A. Example 4 {1, 2, 3, 4} ⊇ {1, 2, 3} Definition 4.2.5 Proper Subset As before, a set A is called a proper superset of a set B , and written A ⊃ B , if every element of B belongs to A and there exists an element in A which does not belong to B . Example 5 {1, 2, 3, 4} ⊃ {1, 2, 3} Definition 4.2.6 Set Equality Definition 4.2.7 Empty Set Definition 4.2.8 Universal Set Saying that two sets A and B are equal, and writing A = B , means that A and B have exactly the same elements. Equivalently, and the more usual way of showing A = B , is to show mutual inclusion, that is, show A is contained in B and B is contained in A. Symbolically, we write A = B means A ⊆ B AND B ⊆ A There is a special set, called the empty set and denoted by ∅, which contains no elements. The empty set is a subset of every set. When we discuss sets, we are often concerned with subsets of some implicit or specified set U , called the universal set. In our work on divisibility and greatest common divisors, we will be concerned with integers as the universal set, even when we don’t explicitly say so. Section 4.2 Describing a Set Definition 4.2.9 Set Complement 29 Relative to a universal set U , the complement of a subset A of U , written A, is the set of all elements in U but not in A. Symbolically, we write A = {x | x ∈ U AND x ∈ A} Definition 4.2.10 Lastly, the cardinality of a set A, written |A|, is the number of elements in the set. Cardinality Example 6 For example, if A = {1, 2, 3, 4}, then |A| = 4. Here’s a pair of mind-blowing questions. What is the cardinality of N? How much larger is Q than N? Example 7 Let S = {x ∈ R | x2 = 2} and T = {x ∈ Q | x2 = 2}. 1. Describe the set S by listing its elements. What is the cardinality of S ? 2. Describe the set T by listing its elements. What is the cardinality of T ? 3. List all of the subsets of S . Solution: √ √ 1. S = { 2, − 2}. |S | = 2. 2. T = ∅. |T | = 0. √ √ 3. ∅, { 2}, {− 2}, S Example 8 Let the universal set for this question be U , the set of natural numbers less than twenty. Let T be the set of integers divisible by three and F be the set of integers divisible by five. 1. Describe T by explicitly listing the set and by using set-builder notation in at least two ways. 2. Find a subset of T of cardinality three. 3. Find an element which belongs to both T and F . 4. Find an element which belongs to neither T nor F . 5. Explicitly list the set T . Solution: 1. Explicitly listing the set gives T = {3, 6, 9, 12, 15, 18}. Two set-builder descriptions of the set are T = {n ∈ N : 3 | n, n ≤ 20} and T = {3k | k ∈ N, 3k ≤ 20} 30 Chapter 4 Introduction to Sets 2. {3, 6, 9}. There are several choices possible. 3. 15. Notice that this is an element, not a set. 4. 1. There are several choices possible. 5. {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19} 4.3 Definition 4.3.1 Union Set Operations The union of two set A and B , written A ∪ B , is the set of all elements belonging to either set A or set B . Symbolically we write A ∪ B = {x | x ∈ A OR x ∈ B } = {x | (x ∈ A) ∨ (x ∈ B )} Note that when we say “set A or set B ” we mean the mathematical use of or. That is, the element can belong to A, B or both A and B . Definition 4.3.2 Intersection The intersection of two set A and B , written A ∩ B , is the set of all elements belonging to both set A and set B . Symbolically we write A ∩ B = {x | x ∈ A AND x ∈ B } = {x | (x ∈ A) ∧ (x ∈ B )} Definition 4.3.3 Difference The difference of two set A and B , written A − B (or A \ B ), is the set of all elements belonging to A but not B . Symbolically we write A − B = {x | x ∈ A AND x ∈ B } = {x | (x ∈ A) ∧ (x ∈ B )} If U is the universal set and A ⊂ U then A = U − A. Example 9 Let the universal set for this question be U , the set of natural numbers less than or equal to twelve. Let T be the set of integers divisible by three, F be the set of integers divisible by five and P the set of primes. Determine each of the following. 1. T ∪ F 2. T ∩ F 3. P 4. P ∩ (T ∪ F ) 5. T ∪ F Section 4.3 Set Operations 6. (T ∪ F ) − P Solution: 1. T ∪ F = {3, 5, 6, 9, 10, 12} 2. T ∩ F = ∅ 3. P = {1, 4, 6, 8, 9, 10, 12} 4. P ∩ (T ∪ F ) = {3, 5} 5. T ∪ F = {1, 2, 4, 5, 7, 8, 10, 11} 6. (T ∪ F ) − P = {6, 9, 10, 12} 31 32 Chapter 4 4.3.1 Introduction to Sets Venn Diagrams Venn diagrams can serve as useful illustrations of set relationships. In Figure 4.3.1 below, the universal set is U = {a, b, c, d, e, w}, the set A = {a, b, c, d} and the set B = {d, e}. The element d lies in the intersection of sets A and B . Since d is the only such element, A ∩ B = {d}. The element w does not lie in either set A or B . w A B b d a e c Figure 4.3.1: Venn Diagram Add schematic Venn diagrams for intersection, union, disjoint, subset, superset, complement 4.4 Comparing Sets 4.4.1 Sets of Solutions One common use of sets is to describe values which are solutions to an equation, but care in expression is required here. The following two sentences mean different things. 1. Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. The solutions to the quadratic equation ax2 + bx + c = 0 are x= −b ± √ b2 − 4ac 2a 2. Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. Then √ −b ± b2 − 4ac x= 2a are solutions to the quadratic equation ax2 + bx + c = 0 The first sentence asserts that a complete description of all solutions is given. The second √ sentence only asserts that x = (−b ± b2 − 4ac)/2a are solutions, not that they are the complete solution. In the language of sets, if S is the complete solution to ax2 + bx + c = 0, √ and T = {(−b ± b2 − 4ac)/2a}, Sentence 1 asserts that S = T (which implies S ⊆ T and T ⊆ S ) but Sentence 2 only asserts that T ⊆ S . Section 4.5 Comparing Sets 33 This point can be confusing. Statements about solutions are often implicitly divided into two sets: the set S of all solutions and a set T of proposed solutions. One must be careful to determine whether the statement is equivalent to S = T or T ⊆ S . Phrases like the solution or complete solution or all solutions indicate S = T . Phrases like a solution or are solutions indicate T ⊆ S . Similar confusion arises when showing that sets have more than one representation. For example, a circle centred at the origin O is often defined geometrically as the set of points equidistant from O. Others define a circle algebraically in the Cartesian plane as the set of points satisfying x2 + y 2 = r2 . To show that the two definitions describe the same object, one must show that the two sets of points are equal. 4.4.2 An Example Given a set S and a set T , there are two very frequent tasks one must perform: one must show S ⊆ T or S = T . In fact, the second task is just two instances of the first task: to show S = T one must show S ⊆ T and T ⊆ S . So, the important message here is that mathematicians must become skilled at demonstrating that S ⊆ T . The plan in all cases is the same: choose a generic element of S and show that it belongs to T . Symbolically S ⊆ T means x ∈ S ⇒ x ∈ T or equivalently S ⊆ T means For all x ∈ S, x ∈ T The element chosen must be completely generic and could, if forced, be instantiated as any element of the set S . Showing that a specific element of S belongs to T is inadequate. Example 10 Consider the statement: Integer multiples of π are roots of f (x) = (x2 − 1) sin x. 1. Explicitly identify two sets used in this statement. 2. Are the two sets equal? 3. Is the statement true? Solution: 1. Let S be the set of all roots of f (x) = (x2 − 1) sin x. (We could write S more symbolically as S = {x ∈ R | f (x) = 0}.) Let T be the set of integer multiples of π . (We could also write T more symbolically as T = {nπ | n ∈ Z}). 2. To show that S = T we must show T ⊆ S and S ⊆ T . Since sin(nπ ) = 0 for all integers n, we know that f (nπ ) = 0. Now, the defining property of S is that a real number x belongs to S if f (x) = 0. Since f (nπ ) = 0, nπ ∈ S . This is equivalent to: if nπ ∈ T then nπ ∈ S , or equivalently, T ⊆ S . Now consider x = 1. The value x = 1 is a solution to (x2 − 1) sin x = 0 and so belongs to S , but it is not an integer multiple of π , so it does not belong to T . That is, S ⊆ T and so the two sets are not equal. 3. The statement is true. The statement only claims that T ⊆ S , not S = T . 34 Chapter 4 4.5 Introduction to Sets Showing Two Sets Are Equal Let’s take a look at two proofs of the same statement about sets. The first uses a chain of if and only if statements, the second uses mutual inclusion. Proposition 1 Let A, B and C be arbitrary sets. A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) Proof: This proof uses a chain of if and only if statements to show that both A ∪ (B ∩ C ) and (A ∪ B ) ∩ (A ∪ C ) have exactly the same elements. Let x ∈ A ∪ (B ∩ C ). Then x ∈ A ∪ (B ∩ C ) ⇐⇒ (x ∈ A) ∨ (x ∈ (B ∩ C )) definition of union ⇐⇒ (x ∈ A) ∨ ((x ∈ B ) ∧ (x ∈ C )) definition of intersection ⇐⇒ ((x ∈ A) ∨ (x ∈ B )) ∧ ((x ∈ A) ∨ (x ∈ C )) Distributive Law of logic ⇐⇒ (x ∈ A ∪ B ) ∧ (x ∈ A ∪ C ) ⇐⇒ x ∈ ((A ∪ B ) ∩ (A ∪ C )) definition of union definition of intersection Proof: This proof uses mutual inclusion. That is, we will show 1. A ∪ (B ∩ C ) ⊆ (A ∪ B ) ∩ (A ∪ C ) 2. A ∪ (B ∩ C ) ⊇ (A ∪ B ) ∩ (A ∪ C ) Equivalently, we must show 1. If x ∈ A ∪ (B ∩ C ), then x ∈ (A ∪ B ) ∩ (A ∪ C ). 2. If x ∈ (A ∪ B ) ∩ (A ∪ C ), then x ∈ A ∪ (B ∩ C ). Let x ∈ A ∪ (B ∩ C ). By the definition of union, x ∈ A or x ∈ (B ∩ C ). If x ∈ A, then by the definition of union, x ∈ A ∪ B and x ∈ A ∪ C , that is x ∈ (A ∪ B ) ∩ (A ∪ C ). If x ∈ B ∩ C , then by the definition of intersection, x ∈ B and x ∈ C . But then by the definition of union, x ∈ A ∪ B and x ∈ A ∪ C . Hence, by the definition of intersection, x ∈ (A ∪ B ) ∩ (A ∪ C ). In both cases, x ∈ (A ∪ B ) ∩ (A ∪ C ) as required. Let x ∈ (A ∪ B ) ∩ (A ∪ C ). By the definition of intersection, x ∈ A ∪ B and x ∈ A ∪ C . If x ∈ A, then by the definition of union, x ∈ A ∪ (B ∩ C ). If x ∈ A, then by the definition of union and the fact that x ∈ A ∪ B , x ∈ B . Similarly, x ∈ C . But then, by the definition of intersection, x ∈ B ∩ C . By the definition of union, x ∈ A ∪ (B ∩ C ). In both cases, x ∈ A ∪ (B ∩ C ). The first of these two proofs also uses mutual inclusion. Do you see how? Section 4.5 Showing Two Sets Are Equal 35 REMARK Which technique is better for proving the equality of two sets: a chain of if and only if statements or mutual inclusion? Though some of the choice is personal style, the choice is primarily determined by the “reversibility” of each step in the proof. A chain of if and only if statements only works if each step in the chain is reversible. That’s pretty unusual. Most of the time when you are proving two sets are equal, you will need to use mutual inclusion. Chapter 5 Discovering Proofs 5.1 Objectives The technique objectives are: 1. Discover a proof using the Direct Proof technique. 2. Write a proof. 3. Read a proof. The content objectives are: 1. Prove the Divisibility of Integer Combinations. 2. Prove the Bounds By Divisibility. 3. State the Division Algorithm. 5.2 Discovering a Proof Discovering a proof of a statement is generally hard. There is no recipe for this, but there are some tips that may be useful, and as we go on through the course, you will learn specific techniques. Consider the following proposition. Proposition 1 (Divisibility of Integer Combinations (DIC)) Let a, b and c be integers. If a | b and a | c, then a | (bx + cy ) for any integers x and y . The very first thing to do is explicitly identify the hypothesis and the conclusion. Hypothesis: a, b, c ∈ Z, a | b and a | c. x, y ∈ Z Conclusion: a | (bx + cy ) 36 Section 5.2 Discovering a Proof 37 Since we are proving a statement, not using a statement, we assume that the hypothesis is true, and then demonstrate that the conclusion is true. This straightforward approach is called Direct Proof. However, in actually discovering a proof we do not need to work only forwards from hypothesis. We can work backwards from the conclusion and meet somewhere in the middle. When writing the proof we must ensure that we begin with the hypothesis and end with the conclusion. Whether working forwards or backwards, I find it best to proceed by asking questions. When working backwards, I ask What mathematical fact would allow me to deduce the conclusion? For example, in the proposition under consideration I would ask What mathematical fact would allow me to deduce that a | (bx + cy )? The answer tells me what to look for or gives me another statement I can work backwards from. In this case the answer would be 1. If there exists an integer k so that bx + cy = ak , then a | (bx + cy ). Note that the answer makes use of the definition of divides. Now I could ask the question How can I find such a k ? The answer is not obvious so let’s turn to working forwards from the hypothesis. In this case my standard two questions are Have I seen something like this before? What mathematical fact can I deduce from what I already know? I have seen a | b in an hypothesis before, in the proof of the Transitivity of Divisibility. I can use the definition of divisibility to assert that 2. There exists an integer r such that b = ra. I also know that a | c so I can again use the definition of divisibility to assert that 3. There exists an integer s such that c = sa. Hmmm, what now? Let’s look again at Sentence 1. 1. If there exists an integer k so that bx + cy = ak , then a | (bx + cy ). 38 Chapter 5 Discovering Proofs There is a bx + cy in Sentence 1 and an algebraic expression for b and c in Sentences 2 and 3. Substituting gives bx + cy = (ra)x + (sa)y and factoring out the a gives bx + cy = (ra)x + (sa)y = a(rx + sy ) If we let k = rx + sy then k is an integer, since adding integers gives integers and multiplying integers gives integers, and so there exists an integer k so that bx + cy = ak . Hence, a | (bx + cy ). We are done. Almost. We have discovered a proof but this is rough work. We must now write a formal proof. Just like any other writing, the amount of detail needed in expressing your thoughts depends upon the audience. A proof of a statement targeted at an audience of professional specialists in algebra will not look the same as a proof targeted at a high school audience. When you approach a proof, you should first make a judgement about the audience. I suggest that you write for your peers. That is, you write your proof so that you could hand it to a classmate and expect that they would understand the proof. Proof: Since a | b, there exists an integer r such that b = ra. Since a | c, there exists an integer s such that c = sa. Let x and y be any integers. Now bx + cy = (ra)x + (sa)y = a(rx + sy ). Since rx + sy is an integer, it follows from the definition of divisibility that a | (bx + cy ). Note that this proof does not reflect the discovery process, and it is a Direct Proof. It begins with the hypothesis and ends with the conclusion. 5.3 Reading A Proof Here is another proposition and proof. Proposition 2 (Bounds By Divisibility (BBD)) Let a and b be integers. If a | b and b = 0 then |a| ≤ |b|. Proof: Since a | b, there exists an integer q so that b = qa. Since b = 0, q = 0. But if q = 0, |q | ≥ 1. Hence, |b| = |qa| = |q ||a| ≥ |a|. Let’s analyze this proof. First, we will rewrite the proof line by line. Proof: (For reference purposes, each sentence of the proof is written on a separate line.) 1. Since a | b, there exists an integer q so that b = qa. 2. Since b = 0, q = 0. 3. But if q = 0, |q | ≥ 1. 4. Hence, |b| = |qa| = |q ||a| ≥ |a|. Section 5.4 The Division Algorithm 39 Now the analysis. Analysis of Proof As usual, we begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: a and b are integers. a | b and b = 0. Conclusion: |a| ≤ |b|. Core Proof Technique: Direct Proof. Preliminary Material: The definition of divides. Now we justify every sentence in the proof. Sentence 1 Since a | b, there exists an integer q so that b = qa. In this sentence, the author of the proof uses the hypothesis a | b and the definition of divides. Sentence 2 Since b = 0, q = 0. If q were zero, then b = qa would imply that b is zero. Since b is not zero, q cannot be zero. Sentence 3 But if q = 0, |q | ≥ 1. Since q is an integer from Sentence 1, and q is not zero from Sentence 2, q ≥ 1 or q ≤ −1. In either case, |q | ≥ 1. Sentence 4 Hence, |b| = |qa| = |q ||a| ≥ |a|. Sentence 1 tells us that b = qa. Taking the absolute value of both sides gives |b| = |qa| and using the properties of absolute values we get |qa| = |q ||a|. From Sentence 3, |q | ≥ 1 so |q ||a| ≥ |a|. Exercise 1 Prove the following statement. Let a, b, c and d be integers. If a | c and b | d, then ab | cd. Exercise 2 Prove the following statement. Let x be an integer. If 2 | (x2 − 1), then 4 | (x2 − 1). 5.4 The Division Algorithm As you have known since grade school, not all integers are divided evenly by other integers. There is usually a remainder. We record this as the Division Algorithm. Proposition 3 (Division Algorithm (DA)) If a and b are integers, and b > 0, then there exist unique integers q and r such that a = qb + r where 0 ≤ r < b. 40 Chapter 5 Discovering Proofs We will not prove this statement now. You will see a proof of the uniqueness part later on and a complete proof is available in the appendix. Add to appendix. Let’s see some examples before a few remarks. Example 1 a=q×b+r 20 = 2 × 7 + 6 21 = 3 × 7 + 0 −20 = −3 × 7 + 1 REMARK • The integer q is called the quotient. • The integer r is called the remainder. • The integer r is always strictly less than b. • The integer r is always positive or zero. • Observe that b | a if and only if the remainder is 0. • Though the proposition is commonly known as the Division Algorithm, it is not really an algorithm since it doesn’t provide a finite sequence of steps that will construct q and r. It turns out that the Division Algorithm is remarkably useful. To see how, we must first define the greatest common divisor which we do soon. Chapter 6 Quantifiers 6.1 Objectives The technique objectives are: 1. Learn the basic structure of quantifiers. 2. Use the Object, Construct and Select Methods. 6.2 Quantifiers Not all mathematical statements are obviously in the form “If A, then B ”. You will encounter statements of the form there is, there are, there exists or for all, for each, for every, for any. The first three are all examples of the existential quantifier there is and the final four are all examples of the universal quantifier for all. The word existence is used to make it clear that we are looking for or looking at a particular mathematical object. The word universal is used to make it clear that we are looking for or looking at a set of objects all of which share some desired behaviour. REMARK All statements which use quantifiers look basically like one of the following two open sentences, though some elements of the sentence may be implicit or appear in a different order. There exists an x in the set S such that P (x) is true. For every x in the set S , P (x) is true. where P (x) is an open sentence that uses the variable x. Some mathematicians prefer a more symbolic approach. The symbol ∃ stands for the English expression “there exists”. The symbol ∀ stands for the English expression “for all”. Symbolically, the two quantified sentences above are written as: ∃x ∈ S, P (x) ∀x ∈ S, P (x) 41 42 Chapter 6 Quantifiers REMARK All statements which use quantifiers share a basic structure. 1. a quantifier which will be either an existential or universal quantifier, 2. a variable which can be any mathematical object, 3. a set which is the domain of the variable, often implicit, and 4. an open sentence which involves the variable, It is crucial that you be able to identify the four parts in the structure of quantified statements. Here are some examples. Let’s begin with something we have already seen. Example 1 1. There exists an integer k so that n = km Quantifier: Variable: Domain: Open sentence: ∃ k Z n = km Our next example could come from any of several branches of mathematics. 2. There exists a real number x such that f (x) = 0. Quantifier: Variable: Domain: Open sentence: ∃ x R f (x) = 0 This is a good point to illustrate the influence of the domain. Suppose in this example we are interested in the specific function f (x) = x2 − 2. Then the statement There exists a real number x such that x2 − 2 = 0. is true since we can find an x, √ 2, so that x2 − 2 = 0. But if we change the domain to integers, the statement There exists an integer x such that x2 − 2 = 0. √ √ is false because neither of the two real roots, 2 or − 2 are integers. So changing the domain can change the truth value of the statement. In practice, the domain is often not explicitly stated and is inferred from context. Section 6.2 Quantifiers 43 3. For every integer n > 5, 2n > n2 . Quantifier: Variable: Domain: Open sentence: ∀ n Z, n > 5 2n > n2 The sentence might appear as “2n > n2 for all integers n > 5”. The order is different but the meaning is the same. 4. There exists an angle θ such that sin(θ) = 1. Quantifier: Variable: Domain: Open sentence: ∃ an angle θ R, inferred from the context sin(θ) = 1 Note that in this example, the domain is implicit. Note also that there can be many objects, many angles θ, which satisfy the statement. 5. For every angle θ, sin2 (θ) + cos2 (θ) = 1. Quantifier: Variable: Domain: Open sentence: ∀ θ R, inferred from context sin2 (θ) + cos2 (θ) = 1 6. If f is continuous on [a, b] and differentiable on (a, b) and f (a) = f (b), then there exists a real number c ∈ (a, b) such that f (c) = 0. This conclusion of this implication uses an existential quantifier. The hypothesis and the conclusion are: Hypothesis: f is continuous on [a, b] and differentiable on (a, b) and f (a) = f (b). Conclusion: There exists a real number c ∈ (a, b) such that f (c) = 0. For the conclusion, the parts of the quantified statement are given below. Quantifier: Variable: Domain: Open sentence: ∃ c (a, b) R f (c) = 0 It takes practice to become fluent in reading and writing statements that use quantifiers. 44 Chapter 6 6.3 Quantifiers The Object Method REMARK We use the Object Method when an existential quantifier occurs in the hypothesis. Suppose that we must prove “A implies B ” and A uses an existential quantifier. That is, A looks like There exists an x in the set S such that P (x) is true. We proceed exactly as the English language interpretation would suggest - we assume that the object x exists. We should: 1. Identify the four parts of the quantified statement. 2. Assume that a mathematical object x exists within the set S so that the statement P (x) is true. 3. Make use of this information to generate another statement. For example, let’s look at the proof of the Transitivity of Divisibility again. Proposition 1 (Transitivity of Divisibility (TD)) Let a, b and c be integers. If a | b and b | c, then a | c. Proof: Since a | b, there exists an integer r so that ra = b. Since b | c, there exists an integer s so that sb = c. Substituting ra for b in the previous equation, we get (sr)a = c. Since sr is an integer, a | c. You might first ask “Where is the existential quantifier?”. It isn’t obvious – yet. But recall the definition of divisibility. An integer m divides an integer n, and we write m | n, if there exists an integer k so that n = km. The sentence “there exists an integer k so that n = km” uses the existential quantifier. It is very common in mathematics that sentences contain implicit quantifiers and you should be alert for them. Returning to divisibility, we have already identified the four parts of the quantified sentence. Quantifier: Variable: Domain: Open sentence: ∃ k Z n = km Section 6.4 The Construct Method 45 How would the Object Method work? Consider the statement a | b. It uses an implicit existential quantifier. Since a | b occurs in the hypothesis, we assume the existence of an integer, say r, so that ra = b. And if you return to examine our proof of Transitivity of Divisibility, this is precisely what appears in the first sentence of the proof. Similarly, the Object Method can be used with b | c to assert that there exists an integer s so that sb = c. Together, the first two sentences allow us to derive the third sentence. 6.4 The Construct Method REMARK We use the Construct Method when an existential quantifier occurs in the conclusion. Suppose that we must prove “A implies B ” and B uses an existential quantifier. That is, B looks like There exists an x in the set S such that P (x) is true. We proceed exactly as the English language interpretation would suggest - we show that the object x exists, that x is in the set S , and that P (x) is true. We should: 1. Identify the four parts of the quantified statement. 2. Construct a mathematical object x. 3. Show that x ∈ S . 4. Show that P (x) is true. For example, let us discover a proof of the following proposition. Proposition 2 If n is of the form 4 + 1 for some positive integer , then 8 | (n2 − 1). As usual, let us begin by explicitly identifying the hypothesis, the conclusion and the core proof technique. Hypothesis: n is of the form 4 + 1 for some integer . Conclusion: 8 | (n2 − 1). Core Proof Technique: Since the definition of divisibility contains an existential quantifier, and 8 | (n2 − 1) occurs in the conclusion, we will use the Construct Method. What, precisely should we construct? Again, thinking of the definition of divisibility and the requirement of the Construct Method, we should construct a k and then show that k is an integer and that 8k = n2 − 1. Where is this k going to come from? Let’s start with the 46 Chapter 6 Quantifiers hypothesis, n is of the form 4 + 1 for some integer . Substituting n = 4 + 1 into n2 − 1 gives n2 − 1 = (4 + 1)2 − 1 = 16 2 + 8 + 1 − 1 = 16 2 + 8 = 8(2 2 + ) It seems that a suitable choice for k would be 2 2 + . Since is an integer and the product of integers is an integer and the sum of integers is an integer, k is an integer. It is also clear from the equation above that 8k = n2 − 1. A proof might look like the following. Proof: Substituting n = 4 + 1 into n2 − 1 gives n2 − 1 = (4 + 1)2 − 1 = 16 Since 2 2 2 + 8 + 1 − 1 = 16 2 + 8 = 8(2 2 +) + is an integer, 8 | (n2 − 1). Note that the proof does not explicitly name the Construct Method. Exercise 1 Where was the Construct Method used in the proof of the Transitivity of Divisibility? 6.5 The Select Method REMARK We use the Select Method whenever a universal quantifier occurs. Suppose a statement looks like For every x in the set S , P (x) is true. Observe that this statement is equivalent to If x is in the set S , then P (x) is true. We proceed exactly as the English language interpretation would suggest - we show that whenever an object x in the set S exists, P (x) is true. We should: 1. Identify the four parts of the quantified statement. 2. Select a representative mathematical object x ∈ S . This cannot be a specific object. It has to be a placeholder so that our argument would work for any specific member of S . Note that if the the set S is empty, we proceed no further. The statement is vacuously true. 3. Show that P (x) is true. For example, let us discover a proof of the following proposition. Section 6.7 Sets and Quantifiers Proposition 3 47 For every odd integer n, 4 | (n2 + 4n + 3). Let’s begin by identifying the four parts of the quantified statement. Quantifier: Variable: Domain: Open sentence: ∀ n odd integers 4 | (n2 + 4n + 3) Now we select a representative mathematical object from the set. Let’s call the odd integer that we selected n0 . We could certainly call it n. I am using n0 to emphasize that we have selected a representative element. Now we must show that 4 | (n2 + 4n0 + 3). Since n0 is 0 odd, we can write it as n0 = 2m + 1 for some integer m. Substituting into n2 + 4n0 + 3 0 gives n2 + 4n0 + 3 = 4m2 + 4m + 1 + 8m + 4 + 3 = 4m2 + 12m + 8 = 4(m2 + 3m + 2) 0 which implies 4 | (n2 + 4n0 + 3). 0 A proof might look like the following. Proof: Let n0 be a positive, odd integer. We can write n0 as 2m + 1 for some integer m. Substituting n0 = 2m + 1 into n2 + 4n0 + 3 gives 0 n2 + 4n0 + 3 = 4m2 + 4m + 1 + 8m + 4 + 3 = 4m2 + 12m + 8 = 4(m2 + 3m + 2) 0 Since m2 + 3m + 2 is an integer, 4 | (n2 + 4n0 + 3). 0 The same proof would work if we converted the universal statement into an “If ... then ” form. The equivalent statement would be Proposition 4 If n is a positive, odd integer, then 4 | (n2 + 4n + 3). 6.6 Sets and Quantifiers It is important to emphasize the connection between sets and quantifiers. The basic structures of all quantified statements use sets. There exists an x in the set S such that P (x) is true. For every x in the set S , P (x) is true. To correctly prove or use quantified statements, first correctly identify the set being used. Quantifiers frequently appear in the defining property of a set. For example, the set of even integers {n ∈ Z : 2 | n} uses an implicit existential quantifier in the definition of divides. To show that S ⊆ T , we use the universally quantified statement ∀x ∈ S, x ∈ T Sets and quantifiers are very closely linked. 48 Chapter 6 6.7 Quantifiers A Non-Proof Making mistakes is easy. Let’s take a look at a “proof” which is not a proof. Let’s find out why it fails. Proposition 5 1 If r is a positive real number with r = 1, then there is an integer n such that 2 n < r. Proof: (For reference purposes, each sentence of the proof is written on a separate line.) 1. Let n be any integer with n > 2. It then follows that 1 . log2 (r) 1 < log2 (r). n 1 3. Hence 2 n < 2log2 (r) = r. Analysis of Proof An interpretation of sentences 1 through 3 follows. Sentence 1 Let n be any integer with n > 1/ log2 (r). Since an existential quantifier occurs in the conclusion, the author is using the Construct Method. The four parts of the quantifier are: Quantifier: Variable: Domain: Open sentence: ∃ n Z 1 2n < r In the first sentence of the proof, the author constructs an integer n. Later in the proof, the author intends to show that n satisfies the open sentence of the quantifier. Since r is a real number (not equal to 1), 1/ log2 (r) evaluates to a real number and we can certainly find an integer greater than any given real number. Sentence 2 It then follows that 1 n < log2 (r). Here the author takes the reciprocal of n > 1/ log2 (r). 1 Sentence 3 Hence 2 n < 2log2 (r) = r. 1 Use the left and right sides of n < log2 (r) as exponents of 2 and recall that the x always increases as x increases. function 2 Even the analysis looks good. What went wrong? Let’s look again at Sentence 2. Here we used the statement Statement 6 If a, b ∈ R, neither equal to 0, and a < b, then 1/b < 1/a. Section 6.7 A Non-Proof 49 A proof seems pretty straightforward – divide both sides of a < b by ab. Except that 1 1 the statement is false. Consider the case a = −2 and b = 4. −2 < 4 but 4 −2 . Our proposition really should be Statement 7 If a, b ∈ R, and 0 < a < b, then 1/b < 1/a. Now we can find the problem in our proof. Choose r so that 0 < r < 1, say r = 1/2. That will make log2 (r) negative and hence 1/ log2 (r) negative. Choose n = 1. Now Sentence 1 is satisfied but Sentence 2 fails. Can you think of any way to correct the proposition or the proof? Chapter 7 Nested Quantifiers 7.1 Objectives The technique objectives are: 1. Recognize nested quantifiers. 2. Learn how to parse nested quantifiers. 3. Learn which techniques to apply to a sentence containing nested quantifiers. 7.2 7.2.1 Onto or Surjective Definition We begin with the definition of an onto function. Definition 7.2.1 Onto, Surjective Let S and T be two sets. A function f : S → T is onto (or surjective) if and only if for every y ∈ T there exists an x ∈ S so that f (x) = y . Often S and T are equal to R or are subsets of R. Though you may not understand the definition, the important observation is that the definition contains two quantifiers. Let’s carefully parse the definition beginning with the universal quantifier “For every”. Recall that we must identify the quantifier, variable, domain and open sentence. Quantifier: Variable: Domain: Open sentence: ∀ y T there exists an x ∈ S so that f (x) = y The open sentence itself contains a quantifier! So we must again identify the four parts of this quantifier. 50 Section 7.2 Onto or Surjective Quantifier: Variable: Domain: Open sentence: 51 ∃ x S f (x) = y REMARK Because the existential quantifier is “nested” within the universal quantifier, this definition is an example of nested quantifiers. There are really two basic principles for working with nested quantifiers. 1. Process quantifiers from left to right. (This captures the “nested” structure.) 2. Use Object, Construct and Select methods as you proceed from left to right. Moving from left to right is important. The order of quantifiers matters. For example, consider the following statement about the integers. ∀x ∃y, y > x Translated into prose, this statement can be read as “Given any integer x, there exists a larger integer y .” This is a true statement. Now let’s make a small modification. We will just change the order of the quantifiers. Our new statement is ∃y ∀x, y > x A translation for this statement would be “There exists an integer y which is larger than all integers.” A very different, and false, statement. We should be able to determine the structure of any proof that a function is onto. Let’s keep the definition in mind. Let S and T be two sets. A function f : S → T is onto (or surjective) if and only if for every y ∈ T there exists an x ∈ S so that f (x) = y . The order of quantifiers is For all there exists so we would expect the proof to be structured Select Method Construct Method The Construct Method identifies a mathematical object, shows that the object is within the domain, and that the object satisfies the open sentence. So an onto proof will look like this. 52 Chapter 7 Nested Quantifiers Structure of an “Onto” Proof • Let y ∈ T . (This comes from the Select Method.) • Consider the object x. (This comes from the Construct Method.) • First, we show that x ∈ S . (We show that the constructed object is within the domain.) • Now we show that f (x) = y . (We show that the open sentence is satisfied.) 7.2.2 Reading Let’s work through an example. Notice how closely the proof follows the structure of an onto proof. Proposition 1 Let m = 0 and b be fixed real numbers. The function f : R → R defined by f (x) = mx + b is onto. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let y ∈ R. 2. Consider x = (y − b)/m. 3. Since y ∈ R, x ∈ R. 4. But then f (x) = f ((y − b)/m) = m((y − b)/m) + b = y as needed. Let’s perform an analysis of this proof. Analysis of Proof The definition of onto uses a nested quantifier. Hypothesis: m = 0 and b are fixed real numbers. f (x) = mx + b. Conclusion: f (x) is onto. Core Proof Technique: Nested quantifiers. Preliminary Material: Let us remind ourselves of the definition of the defining property of onto as it applies in this situation. For every y ∈ R there exists x ∈ R so that f (x) = y . Sentence 1 Let y ∈ R. The first quantifier in the definition is a universal quantifier so the author uses the Select Method. That is, the author chooses an element (y ) in the domain (R). The author must now show that the open sentence is satisfied (there exists an x ∈ R so that f (x) = y ). The constructed object in this example is not surprising - we can simply solve for x in y = mx + b. In general, though, it can be difficult to construct a suitable object. Note also that the choice of x depends on y so that it is not surprising that x is a function of y . Section 7.2 Onto or Surjective 53 Sentence 2 Consider x = (y − b)/m. The second quantifier in the definition is a nested existential quantifier so the author uses the Construction Method. That is, the author constructs an element (x). Sentence 3 Since y ∈ R, x ∈ R. Because this step is usually straightforward, it is often omitted. It is included here to emphasize that the constructed object lies in the appropriate domain. Sentence 4 But then f (x) = f ((y − b)/m) = m((y − b)/m) + b = y as needed. Here the author confirms that the open sentence is satisfied. 7.2.3 Discovering Having read a proof, let’s discover one. Proposition 2 Let f : T → U and g : S → T be onto functions. Then f ◦ g is an onto function. Analysis of Proof The definition of onto uses nested quantifiers. Hypothesis: f : T → U and g : S → T are both onto functions. Conclusion: f ◦ g is onto. Core Proof Technique: Nested quantifiers. Preliminary Material: Let us recast the definition of onto for f ◦ g . To do this we need to be cognizant of the fact that f : T → U and g : S → T and f ◦ g : S → U . So the statement we need to prove is: For every y ∈ U there exists x ∈ S so that f (g (x)) = y . There are three instances of onto in the proposition. Two occur in the hypothesis and are associated with the functions f and g . The third occurs in the conclusion and is associated with the function f ◦ g . That is the one that interests us right now. The definition of onto begins with a universal qualifier. So we will use the Select Method applied to f ◦ g . Using our proof template we have the following. Proof in Progress 1. Let y ∈ U . 2. Consider the object x. We must construct the object x. 3. First, we show that x ∈ S . To be completed. 4. Now we show that f (g (x)) = y . To be completed. Constructing x seems difficult. We do not know what the sets S , T and U are and we have no idea what the functions f and g look like. But we have not made use of our hypotheses at all so let’s see if they can give us any ideas. Since f : T → U is onto, we know that for any u ∈ U , there exists a t ∈ T so that f (t) = u. 54 Chapter 7 Nested Quantifiers Since g : S → T is onto, we know that for any t ∈ T , there exists an s ∈ S so that g (s) = t. How does y fit in? Observe that y ∈ U . But f : T → U and is onto, so there exists a t ∈ T so that f (t ) = y . Since t ∈ T and g : S → T is onto, there exists an s ∈ S so that g (s ) = t . But what have we constructed? If we let x = s then we have an element that maps from S to T and then from T to U for which f (g (s )) = y . Let’s record these. Proof in Progress 1. Let y ∈ U . 2. Since f : T → U is onto, there exists a t ∈ T so that f (t ) = y . 3. Since t ∈ T and g : S → T is onto, there exists an s ∈ S so that g (s ) = t . 4. Hence, there exists s ∈ S so that f (g (s )) = f (t ) = y . 5. Hence, there exists x ∈ S so that f (g (x)) = y . Notice that our last two lines are essentially duplicates. When doing rough work, this is common. However, when writing up a proof, such duplications should be removed, consistent notation should be enforced and omitted steps should be included. In this case, the proof is almost done for us. Proof: Let y in U . Since f : T → U is onto, there exists a t ∈ T so that f (t ) = y . Since t ∈ T and g : S → T is onto, there exists an s ∈ S so that g (s ) = t . Hence, there exists s ∈ S so that f (g (s )) = f (t ) = y . 7.3 7.3.1 One-to-one or Injective Definition The definition of onto functions contained nested quantifiers which were different. The next definition uses nested quantifiers which are the same. Definition 7.3.1 One-to-one, Injective Let S and T be two sets. A function f : S → T is one-to-one (or injective) if and only if for every x1 ∈ S and every x2 ∈ S , f (x1 ) = f (x2 ) implies that x1 = x2 . Just as with onto functions, let’s parse the definition beginning with the universal quantifier “For every”. Recall that we must identify the quantifier, variable, domain and open sentence. Quantifier: Variable: Domain: Open sentence: ∀ x1 S for every x2 ∈ S , f (x1 ) = f (x2 ) implies that x1 = x2 The open sentence itself contains a quantifier! So we must again identify the four parts of this quantifier. Section 7.3 One-to-one or Injective Quantifier: Variable: Domain: Open sentence: 55 ∀ x2 S f (x1 ) = f (x2 ) implies that x1 = x2 It is important to note that the open sentence is an implication! We should be able to determine the structure of any proof that a function is one-to-one. The order of quantifiers is For all For all so we would expect the proof to be structured Select Method Select Method The Select Method selects a representative mathematical object within the appropriate domain, and shows that the object satisfies the corresponding open sentence. So a one-toone proof will look like this. Structure of a “One-to-one” Proof • Let x1 ∈ S . (This comes from the Select Method.) • Let x2 ∈ S . (This comes from the Select Method.) • Suppose that f (x1 ) = f (x2 ). (This is the hypothesis of the open sentence. Since we wish to show that the open sentence is true, we assume the hypothesis is true.) • Now we show that x1 = x2 . (This is the conclusion of the open sentence. Since we wish to show that the open sentence is true, we must show the conclusion is true.) 7.3.2 Reading Let’s work through an example. Notice how closely the proof follows the structure of a one-to-one proof. Proposition 3 Let m = 0 and b be fixed real numbers. The function f : R → R defined by f (x) = mx + b is one-to-one. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let x1 , x2 ∈ S . 2. Suppose that f (x1 ) = f (x2 ). 3. Now we show that x1 = x2 . 4. Since f (x1 ) = f (x2 ), mx1 + b = mx2 + b. 5. Subtracting b from both sides and dividing by m gives x1 = x2 as required. 56 Chapter 7 Nested Quantifiers Let’s perform an analysis of this proof. Analysis of Proof The definition of onto uses a nested quantifier. Hypothesis: m = 0 and b are fixed real numbers. f (x) = mx + b. Conclusion: f (x) is one-to-one. Core Proof Technique: Nested quantifiers. Preliminary Material: Let us remind ourselves of the definition of the defining property of one-to-one as it applies in this situation. For every x1 ∈ R and every x2 ∈ R, f (x1 ) = f (x2 ) implies that x1 = x2 . Sentence 1 Let x1 , x2 ∈ R. The author combines the first two sentences of the structure of a one-to-one proof into a single sentence. This works because both of the first two quantifiers in the definition are universal quantifiers and so the author uses the Select Method twice. That is, the author chooses elements (x1 andx2 ) in the domain (R). The author must now show that the open sentence is satisfied (f (x1 ) = f (x2 ) implies that x1 = x2 ). Sentences 2 and 3 Suppose that f (x1 ) = f (x2 ). Now we show that x1 = x2 . The open sentence that must be verified is an implication, and f (x1 ) = f (x2 ) is the hypothesis. To prove an implication, we assume the hypothesis and demonstrate that the conclusion, x1 = x2 , is true. Sentence 3 Since f (x1 ) = f (x2 ), mx1 + b = mx2 + b. This is just substitution. Sentence 4 Subtracting b from both sides and dividing by m gives x1 = x2 as required. Here the author confirms that the open sentence is satisfied. 7.3.3 Discovering Having read a proof, let’s discover one. Proposition 4 Let f : T → U and g : S → T be one-to-one functions. Then f ◦ g is a one-to-one function. Analysis of Proof The definition of one-to-one uses nested quantifiers. Hypothesis: f : T → U and g : S → T are both one-to-one functions. Conclusion: f ◦ g is one-to-one. Core Proof Technique: Nested quantifiers. Preliminary Material: Let us recast the definition of one-to-one for f ◦ g . For every x1 ∈ S and every x2 ∈ S , (f ◦ g )(x1 ) = (f ◦ g )(x2 ) implies that x1 = x2 . Section 7.4 Limits 57 There are three instances of one-to-one in the proposition. Two occur in the hypothesis and are associated with the functions f and g . The third occurs in the conclusion and is associated with the function f ◦ g . Let’s use the structure of a one-to-one proof as our starting point. Proof in Progress 1. Let x1 , x2 ∈ S . 2. Suppose that (f ◦ g )(x1 ) = (f ◦ g )(x2 ). 3. Now we show that x1 = x2 . 4. To be completed. 5. Hence, x1 = x2 as required. Since f and g are not specified, this may seem impossible. But let’s “follow our nose” and see what happens. Since (f ◦ g )(x1 ) = (f ◦ g )(x2 ), we know that f (g (x1 )) = f (g (x2 )). But since f is one-to-one, we know that g (x1 ) = g (x2 ). If this seems confusing, since f is one-to-one, f (y1 ) = f (y2 ) implies y1 = y2 . In this case, y1 = g (x1 ) and y1 = g (x1 ). Now back to g (x1 ) = g (x2 ). Since g is one-to-one, we know that x1 = x2 , which is exactly what we needed to show. A proof might look like the following. Proof: Let x1 , x2 ∈ S . Suppose that (f ◦ g )(x1 ) = (f ◦ g )(x2 ). Since (f ◦ g )(x1 ) = (f ◦ g )(x2 ), we know that f (g (x1 )) = f (g (x2 )). Since f is one-to-one, we know that g (x1 ) = g (x2 ). And since g is one-to-one, x1 = x2 as required. 7.4 7.4.1 Limits Definition Almost everyone who takes a calculus course encounters the notion of a limit. When we write lim f (x) = L x→a we informally mean that we can make the values of f (x) arbitrarily close to L by taking x sufficiently close to, but not equal to a. But formally we need to be more explicit about what “arbitrarily” and “sufficiently” mean. That leads to the infamous ε − δ definition of a limit. Definition 7.4.1 Limit The limit of f (x), as x approaches a, equals L means that for every real number ε > 0 there exists a real number δ > 0 such that 0 < |x − a| < δ ⇒ |f (x) − L| < ε Let’s carefully parse the definition beginning with the universal quantifier “For every”. Recall that we must identify the quantifier, variable, domain and open sentence. 58 Chapter 7 Quantifier: Variable: Domain: Open sentence: Nested Quantifiers ∀ ε real numbers > 0 there exists a real number δ > 0 such that 0 < |x − a| < δ ⇒ |f (x) − L| < ε The open sentence itself contains a quantifier, so we must again identify the four parts of the quantifier. Quantifier: Variable: Domain: Open sentence: ∃ δ real numbers > 0 0 < |x − a| < δ ⇒ |f (x) − L| < ε It is vitally important to observe that the open sentence is an implication. Because the existential quantifier is “nested” within the universal quantifier, this definition is another example of nested quantifiers. 7.4.2 Reading A Limit Proof Before we begin our example, we should be able to determine the structure of any limit proof. The order of quantifiers is For all ε there exists δ so we would expect the proof to be structured Select Method Construct Method The choice of δ will depend on the choice of ε and so δ will be a function of ε. The Construct Method identifies a mathematical object, shows that the object is within the domain, and that the object satisfies the open sentence. The open sentence is an implication with hypothesis 0 < |x − a| < δ (ε) and conclusion |f (x) − L| < ε. We assume that the hypothesis is true and show that the conclusion is true. So a limit proof will look like the following. Structure of a “Limit” Proof • Let ε > 0 be a real number. (This comes from the Select Method.) • Consider the real number δ (ε). (This comes from the Construct Method.) • First, we show that δ (ε) > 0. (This shows δ is within the domain.) • Now let 0 < |x − a| < δ (ε). (This is the hypothesis of the open sentence in the definition of limit.) • We show that |f (x) − L| < ε. (This is the conclusion of the open sentence.) The difficulty lies in finding a suitable choice of δ (ε). Let’s analyze a proof where someone else has made the choice of δ (ε) for us. Section 7.4 Limits Proposition 5 59 Let m = 0 be a real number. lim mx + b = ma + b x→a Proof: (For reference purposes, each sentence of the proof is written on a separate line.) 1. Let ε > 0 be a real number. 2. Consider the real number δ (ε) = 3. Since ε > 0 and |m| > 0, δ (ε) = ε . |m| ε > 0. |m| 4. Now 0 < |x − a| < δ (ε) ⇒ 0 < |x − a| < ε |m| ⇒ |m||x − a| < ε ⇒ |m(x − a)| < ε ⇒ |m(x − a) + (b − b)| < ε ⇒ |(mx + b) − (ma + b)| < ε ⇒ |f (x) − L| < ε as required. Analysis of Proof As usual, we begin with the hypothesis and the conclusion. Hypothesis: m = 0 is a real number. Conclusion: limx→a mx + b = ma + b. Core Proof Technique: Nested quantifiers. Preliminary Material: Definition of a limit. Notice how closely this proof follows the structure of a limit proof. Sentence 1 Let ε > 0 be a real number. The definition of limit begins with a universal quantifier so the first proof technique is the Select Method, just as in the structure of a limit proof. ε . |m| The next quantifier is an existential quantifier in the conclusion and so we use the Construct Method. This again follows the pattern of the proof structure. The conε structed object is the real number δ (ε) = . The author gives no indication why |m| that particular value was chosen or how it was derived. Sentence 2 Consider the real number δ (ε) = 60 Chapter 7 Nested Quantifiers ε > 0. |m| After an object is constructed, the Construct Method requires that the object be in the domain and that it satisfy the open sentence. Sentence 3 of the proof shows that δ is in the domain, the set of real numbers greater than zero. Sentence 3 Since ε > 0 and |m| > 0, δ (ε) = Sentence 4 Now . . . Sentence 4 demonstrates that δ satisfies the open sentence. The hypothesis of the open sentence is 0 < |x − a| < δ (ε) and the conclusion is |f (x) − L| < ε. The chain of reasoning begins with the hypothesis, and after arithmetic manipulation, arrives at the conclusion. Exercise 1 Justify each line of arithmetic in Sentence 4. 7.4.3 Discovering a Limit Proof We will prove Proposition 6 2 If f (x) = e−1/x then lim f (x) = 0. x→0 You might object that the function is not even defined at 0, which is true. But the definition of limx→a f (x) does not require f to be defined at a. As usual, we begin by explicitly identifying our hypothesis and conclusion. 2 Hypothesis: f (x) = e−1/x Conclusion: limx→0 f (x) = 0 This is a standard limit proof so we use our existing structure. Proof in Progress 1. Let ε > 0 be a real number. 2. Consider the real number δ (ε). 3. First, we show that δ (ε) > 0. 4. Now let 0 < |x| < δ (ε). (This is just 0 < |x − a| < δ (ε) with a = 0.) 2 2 5. We show that |e−1/x | < ε. (This is just |f (x) − L| < ε with f (x) = e−1/x and L = 0.) Section 7.4 Limits 61 The problem is: How do we construct a suitable δ ? Because ε is not numerically specified, our construction for δ will be a function of ε. Now is the time to go to scrap paper. Since we need 2 |e−1/x | < ε 2 we begin there and look for a way to get to 0 < |x| < δ (ε). e−1/x > 0 for all x so we do not need the absolute value signs. 1 <ε e1/x2 2 Now divide by ε (we are using the hypothesis that ε = 0) and multiply by e1/x (we are 2 using the fact that e1/x > 0) to get the following. 1 2 < e1/x ε Taking the natural log gives ln 1 ε < 1 x2 This is hopeful. We can invert the fractions to get x2 < 1 ln(1/ε) and since x2 > 0 we now have 0 < x2 < 1 ln(1/ε) Taking square roots gives 0 < |x| < 1 ln(1/ε) And this is precisely the form we want. Our constructed delta is δ= 1 ln(1/ε) This looks great. Unfortunately, we have made a dangerous assumption, that is ln(1/ε) > 0. This is only true when ε < 1. However, it is mathematical practice to consider ε as small, much smaller than one. We will adopt standard practice and ignore the case ε ≥ 1 though details could be given for it as well. We have already worked out the math so now we are in a position to write out the proof. Take a minute to read the proof. Proof: Let ε > 0. Since ε is small, we assume ε < 1. Consider δ = 1 ln(1/ε) . Since ε < 1, 1/ε > 1 which implies ln(1/ε) > 0 and so δ > 0. Now 0 < |x| < as required. 1 1 ⇒ 0 < x2 < ⇒ ln ln(1/ε) ln(1/ε) 1 ε < 1 1 2 2 ⇒ < e1/x ⇒ |e−1/x | < ε 2 x ε Chapter 8 Induction 8.1 Objectives The technique objective is: 1. Learn how to use sum and product notation, and recognize recurrence relations. 2. Learn how to use the Principle of Mathematical Induction, sometimes called Simple Induction. 3. Learn how to use the Principle of Strong Induction, usually called Strong Induction. 4. Define binomial coefficient. 5. Read a proof of the Binomial Theorem. 8.2 Notation A number of examples we will discuss use sum, product and recursive notation that you may not be familiar with. 8.2.1 Summation Notation The sum of the first ten perfect squares could be written as 12 + 22 + 32 + · · · + 102 In mathematics, a more compact and more helpful notation is used. 10 i2 i=1 62 Section 8.2 Notation Definition 8.2.1 63 The notation n Summation Notation xi i=m is called summation notation and it represents the sum xm + xm+1 + xm+2 + · · · + xn The summation symbol, , is the upper case Greek letter sigma. The letter i is the index of summation; the letter m is the lower bound of summation, and the letter n is the upper bound of summation. The expression i = m under the summation symbol means that the index i begins with an initial value of m and increments by i and stops when i = n. The index of summation is a dummy variable and any letter could be used in its place. Example 1 7 i2 = 32 + 42 + 52 + 62 + 72 i=3 3 sin(kπ ) = sin(0) + sin(π ) + sin(2π ) + sin(3π ) k=0 n i=1 11 1 1 = 1 + + + ··· + 2 i2 49 n This notation is often generalize to an arbitrary logical condition, and the sum runs over all values satisfying the condition. Example 2 For example: f (x) x∈ S is the sum of f (x) over all elements x in the set S . The expression d d|n,d>0 is the sum of all positive divisors of n. There are a number of rules that help us manipulate sums. Proposition 1 (Properties of Summation) 1. Multiplying by a constant n n cxi = c i=m xi where c is a constant i=m 64 Chapter 8 2. Adding two sums n n xi + i=m 3. Subtracting two sums n yi = i=m n (xi + yi ) i= m n xi − i=m Induction n (xi − yi ) yi = i=m i= m 4. Changing the bounds of the index of summation n n+k xi = i= m x i− k i=m+k The last property allows us to change the bounds of the index of summation, which is often useful when combining summation expressions. Question Panel here 8.2.2 Product Notation Just as summation notation using is an algebraic shorthand for a sum, product notation using is an algebraic shorthand for a product. Definition 8.2.2 The notation n Product Notation xi i=m is called product notation and it represents the product xm · xm+1 · xm+2 · · · · · xn The product symbol, , is the upper case Greek letter pi. The index i and the upper and lower bounds m and n behave just as they do for sums. Example 3 n 1− i=2 8.2.3 1 i2 = 1− 1 4 1− 1 9 1− 1 16 ··· 1 − 1 n2 Recurrence Relations You are accustomed to seeing mathematical expressions in one of two ways: iterative and closed form. For example, the sum of the first n integers can be expressed iteratively as 1 + 2 + 3 + ··· + n Section 8.3 Introduction to Induction 65 or in closed form as n(n + 1) 2 There is a third way. Definition 8.2.3 Recurrence Relation A recurrence relation is an equation that defines a sequence of number which is generated by one or more initial terms, and expressions involving prior terms. You are probably familiar with the Fibonacci sequence which is a recurrence relation. Example 4 (Fibonacci Sequence) The initial two terms are defined as f1 = 1 and f2 = 1. All subsequent terms are defined by the recurrence relation fn = fn−1 + fn−2 . The first eight terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 11, 19. Example 5 (Sum of First n Integers) We can define the sum of the first n terms recursively as f (1) = 1 and f (n) = f (n − 1) + n for n > 1 Question Panel here 8.3 Introduction to Induction Induction is a common and powerful technique and should be your first choice whenever you encounter a statement of the form For every integer n ≥ 1, P (n) is true. where P (n) is a statement that depends on n. Here are two examples of propositions in this form. Proposition 2 For every integer n ≥ 1 n i2 = i=1 n(n + 1)(2n + 1) . 6 Often the clause “For every integer n ≥ 1” is implied and does not actually appear in the proposition, as in the following version of the same theorem. 66 Chapter 8 Proposition 3 The sum of the first n perfect squares is Induction n(n+1)(2n+1) . 6 The second example uses sets, not equations. Proposition 4 Every set of size n has exactly 2n subsets. 8.4 Definition 8.4.1 Axiom Principle of Mathematical Induction An axiom of a mathematical system is a statement that is assumed to be true. No proof is given. From axioms we derive propositions and theorems. Sometimes axioms are described as self-evident, though many are not. Axioms are defining properties of mathematical systems. The Principle of Mathematical Induction is one such axiom. Axiom 1 Principle of Mathematical Induction (POMI) Let P (n) be a statement that depends on n ∈ N. If 1. P (1) is true, and 2. P (k ) is true implies P (k + 1) is true for all k ∈ N then P (n) is true for all n ∈ N. We use the Principle of Mathematical Induction to prove statements of the form For every integer n ≥ 1, P (n) is true. The structure of a proof by induction models the definition of induction. The three parts of the structure are as follows. Base Case Verify that P (1) is true. This is usually easy. You will often see the statement “It is easy to see that the statement is true for n = 1.” It is best to write this step out completely. Inductive Hypothesis Assume that P (k ) is true for some integer k ≥ 1. It is best to write out the statement P (k ). Inductive Conclusion Using the assumption that P (k ) is true, show that P (k + 1) is true. Again, it is best to write out the statement P (k + 1) before trying to prove it. Section 8.4 Principle of Mathematical Induction 8.4.1 67 Why Does Induction Work? The basic idea is simple. We show that P (1) is true. We then use P (1) to show that P (2) is true. And then we use P (2) to show that P (3) is true and continue indefinitely. That is P (1) ⇒ P (2) ⇒ P (3) ⇒ . . . ⇒ P (i) ⇒ P (i + 1) ⇒ . . . 8.4.2 Two Examples of Simple Induction Our first example is very typical and uses an equation containing the integer n. Proposition 5 n i2 = i=1 n(n + 1)(2n + 1) . 6 Proof: We begin by formally writing out our inductive statement n i2 = P (n) : i=1 n(n + 1)(2n + 1) . 6 Base Case We verify that P (1) is true where P (1) is the statement 1 i2 = P (1) : i=1 1(1 + 1)(2 × 1 + 1) . 6 As in most base cases involving equations, we can make our way from the left side of the equation to the right side of the equation with just a little algebra. 1 i2 = 12 = 1 = i=1 1(1 + 1)(2 × 1 + 1) . 6 Inductive Hypothesis We assume that the statement P (k ) is true for some integer k ≥ 1. k i2 = P (k ) : i=1 k (k + 1)(2k + 1) . 6 Inductive Conclusion Now show that the statement P (k + 1) is true. k+1 i2 = P (k + 1) : i=1 (k + 1)((k + 1) + 1)(2(k + 1) + 1) . 6 This is the difficult part. When working with equations, you can often start with the more complicated expression and decompose it into an instance of P (k ) with some 68 Chapter 8 Induction leftovers. That’s what we will do here. k+1 k i2 = i=1 i2 + (k + 1)2 partition into P (k ) and other i=1 k (k + 1)(2k + 1) + (k + 1)2 use the inductive hypothesis = 6 k (k + 1)(2k + 1) + 6(k + 1)2 = algebraic manipulation 6 (k + 1) 2k 2 + 7k + 6 = factor out k + 1, expand the rest 6 (k + 1)(k + 2)(2k + 3) = factor 6 (k + 1)((k + 1) + 1)(2(k + 1) + 1) = 6 The result is true for n = k + 1, and so holds for all n by POMI. Our next example does not have any equations. Proposition 6 Let Sn = {1, 2, 3, . . . , n}. Then Sn has 2n subsets. Let’s be very clear about what our statement P (n) is. P (n): Sn has 2n subsets. Now we can begin the proof. Proof: Base Case We verify that P (1) is true where P (1) is the statement P (1): S (1) has 2 subsets. We can enumerate all of the sets of S1 easily. They are { } and {1}, exactly two as required. Inductive Hypothesis We assume that the statement P (k ) is true for some integer k ≥ 1. P (k ): Sk has 2k subsets. Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): Sk+1 has 2k+1 subsets. The subsets of Sk+1 can be partitioned into two sets. The set A in which no subset contains the element k +1, and the complement of A, A, in which every subset contains the element k + 1. Now A is just the subsets of Sk and so, by the inductive hypothesis, has 2k subsets. A is composed of the subsets of Sk to which the element k +1 is added. So, again by our inductive hypothesis, there are 2k subsets of A. Since A and A are disjoint and together contain all of the subsets of Sk+1 , there must be 2k + 2k = 2k+1 subsets of Sk+1 . The result is true for n = k + 1, and so holds for all n by POMI. Section 8.5 Principle of Mathematical Induction 8.4.3 69 A Different Starting Point Some true statements cannot start with “for all integers n, n ≥ 1”. For example, “2n > n2 ” is false for n = 2, 3, and 4 but true for n ≥ 5. But the basic idea holds. If we can show that a statement is true for some base case n = b, and then show that P (b) ⇒ P (b + 1) ⇒ P (b + 2) ⇒ . . . ⇒ P (i) ⇒ P (i + 1) ⇒ . . . this is also induction. Perhaps this is not surprising because we can always recast a statement “For every integer n ≥ b, P (n)” as an equivalent statement “For every integer k ≥ 1, P (k )”. For example, For every integer n ≥ 5, 2n > n2 . is equivalent to For every integer k ≥ 1, 2k+4 > (k + 4)2 . In this case, we have just replaced n by k + 4 in the statement. The basic structure of induction is the same. To prove the statement For every integer n ≥ b, P (n) is true. the only changes we need to make are that our base case is P (b) rather than P (1), and that in our inductive hypothesis we assume P (k ) is true for k ≥ b rather than k ≥ 1. Here is an example. Proposition 7 For every integer n ≥ 5, 2n > n2 . As usual, let’s be very clear about what our statement P (n) is. P (n): 2n > n2 . Now we can begin the proof. Proof: Base Case We verify that P (5) is true where P (5) is the statement P (5): 25 > 52 This is just arithmetic. 25 = 32 > 25 = 52 Inductive Hypothesis We assume that the statement P (k ) is true for some integer k ≥ 5. P (k ): 2k > k 2 Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): 2k+1 > (k + 1)2 2k+1 = 2 × 2k > 2 × k 2 = k 2 + k 2 > k 2 + 2k + 1 = (k + 1)2 The result is true for n = k + 1, and so holds for all n by POMI. 70 Chapter 8 8.5 Induction Strong Induction Sometimes Simple Induction doesn’t work where it looks like it should. We then need to change our approach a bit. The following example is similar to examples that we’ve done earlier. Lets try to make Simple Induction work and see where things go wrong. Proposition 8 Let the sequence {xn } be defined by x1 = 0, x2 = 30 and xm = xm−1 + 6xm−2 for m ≥ 3. Then xn = 2 · 3n + 3 · (−2)n for n ≥ 1. This seems like a classic case for induction since the conclusion clearly depends on the integer n. Let’s begin with our statement P (n). P (n): xn = 2 · 3n + 3 · (−2)n . Now we can begin the proof. Proof: Base Case We verify that P (1) is true where P (1) is the statement P (1): x1 = 2 · 31 + 3 · (−2)1 . From the definition of the sequence x1 = 0. The right side of the statement P (1) evaluates to 0 so P (1) is true. Inductive Hypothesis We assume that the statement P (k ) is true for k ≥ 1. P (k ): xk = 2 · 3k + 3 · (−2)k . Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): xk+1 = 2 · 3k+1 + 3 · (−2)k+1 . xk+1 = xk + 6xk−1 k by the definition of the sequence k = 2 · 3 + 3 · (−2) + 6xk−1 by the Inductive Hypothesis Now two problems are exposed. The more obvious problem is what do we do with xk−1 ? The more subtle problem is whether we can even validly write the first line. When k + 1 = 2 we get x2 = x1 + 6 x0 and x0 is not even defined. The basic principle that earlier instances imply later instances is sound. We need to strengthen our notion of induction in two ways. First, we need to allow for more than one base case so that we avoid the problem of undefined terms. Second, we need to allow access to any of the statements P (1), P (2), P (3), ... , P (k ) when showing that P (k + 1) is true. This may seem like too strong an assumption but is, in fact, quite acceptable. This practice is based on the Principle of Strong Induction. Section 8.5 Strong Induction Axiom 2 71 Principle of Strong Induction (POSI) Let P (n) be a statement that depends on n ∈ N. If 1. P (1), P (2), . . . , P (b), are true, and 2. P (1), P (2), . . . , P (k ) are all true implies P (k + 1) is true for all k ∈ N, then P (n) is true for all n ∈ N. Just as before, there are three parts in a proof by strong induction. Base Cases Verify that P (1), P (2), . . . , P (b) are all true. This is usually easy. Inductive Hypothesis Assume that P (i) is true for i = 1, 2, 3, . . . , k , k ≥ b. This is sometimes written as Assume that P (1), P (2), . . . , P (k ) are true. Inductive Conclusion Using the assumption that P (1), P (2), . . . , P (k ) are true, show that P (k + 1) is true. As a rule of thumb, use Strong Induction when the general case depends on more than one previous case. Though we could use Strong Induction all the time, Simple Induction is often easier. Let’s return to our previous proposition. Proposition 9 Let the sequence {xn } be defined by x1 = 0, x2 = 30 and xm = xm−1 + 6xm−2 for m ≥ 3. Then xn = 2 · 3n + 3 · (−2)n for n ≥ 1. We will use Strong Induction. Recall our statement P (n). P (n): xn = 2 · 3n + 3 · (−2)n . Now we can begin the proof. Proof: Base Case We verify that P (1) and P (2) are true. P (1): x1 = 2 · 31 + 3 · (−2)1 . From the definition of the sequence x1 = 0. The right side of the statement P (1) evaluates to 0 so P (1) is true. P (2): x2 = 2 · 32 + 3 · (−2)2 . From the definition of the sequence x2 = 30. The right side of the statement P (2) evaluates to 30 so P (2) is true. 72 Chapter 8 Induction Inductive Hypothesis We assume that the statement P (i) is true for 1 ≤ i ≤ k , k ≥ 2. P (i): xi = 2 · 3i + 3 · (−2)i . Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): xk+1 = 2 · 3k+1 + 3 · (−2)k+1 . xk+1 = xk + 6xk−1 k by the definition of the sequence k = 2 · 3 + 3 · (−2) + 6(2 · 3 k −1 k−1 + 3 · (−2) ) by the Inductive Hypothesis = 3k−1 [2 · 3 + 6 · 2] + (−2)k−1 [3 · (−2) + 6 · 3] expand and factor = 18 · 3 k −1 k −1 + 12 · (−2) = 2 · 3k+1 + 3 · (−2)k+1 The result is true for n = k + 1, and so holds for all n by POSI. 8.5.1 Interesting Example A triomino is a tile of the form Proposition 10 A 2n × 2n grid of squares with one square removed can be covered by triominoes. As usual, we begin by explicitly stating P (n). P (n): A 2n × 2n grid of squares with one square removed can be covered by triominoes. We will use Simple Induction. Proof: Base Case We verify that P (1) is true. P (1): A 2 × 2 grid of squares with one square removed can be covered by triominoes. A 2 × 2 grid with one square removed looks like or or or . Each of these can be covered by one triomino. Inductive Hypothesis We assume that the statement P (i) is true for 1 ≤ i ≤ k . P (k ): A 2i × 2i grid of squares with one square removed can be covered by triominoes. Note that our hypothesis covers every possible position for the empty square within the grid. Section 8.6 Binomial Theorem 73 Inductive Conclusion We now show that the statement P (k + 1) is true. P (k + 1): A 2k+1 × 2k+1 grid of squares with one square removed can be covered by triominoes. Consider a 2k+1 × 2k+1 grid with any square removed. Split the 2k+1 × 2k+1 grid in half vertically and horizontally. The missing square occurs in one of the four 2k × 2k subgrids formed. We’ll start by placing one tile around the centre of the grid, not covering any of the 2k × 2k subgrids where the square is missing: We can now view the grid as being made up of four 2k × 2k subgrids, each with one square missing. The Inductive Hypothesis tells us that each of these can be covered by triominoes. Therefore, the whole 2k+1 × 2k+1 grid can be covered. The result is true for n = k + 1, and so holds for all n by POMI. 8.6 Definition 8.6.1 Binomial Binomial Theorem A binomial is the sum of two quantities, a + b for example. 74 Chapter 8 Induction You have probably encountered the following powers of a binomial. (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 The obvious question is: what is the expansion of (a + b)n for a positive integer n? The expansion of (a + b)n uses binomial coefficients. Definition 8.6.2 a b If 0 ≤ b ≤ a, then the binomial coefficient Binomial Coefficient a b where 0! is defined to be 1 so that a a = is defined by a! b!(a − b)! = 1. This electronic assignment works through a proposed inductive proof of the following proposition. Proposition 11 (Binomial Theorem) If x and y are any numbers, and n ∈ N, then n (x + y )n = r=0 n n−r r xy r Example 6 3 (x + y )3 = r=0 3 3−r r xy r 3 3−0 0 3 3−1 1 3 3−2 2 3 3−3 3 x y+ x y+ x y+ xy 0 1 2 3 = = x3 + 3x2 y + 3xy 2 + y 3 Example 7 3 3 (2x − 3) = r=0 3 (2x)3−r (−3)r = 8x3 − 36x2 + 54x − 27 r Proof: Let P (n) be the statement: If x and y are any numbers, and n ∈ N, then n (x + y )n = r=0 n n−r r x y. r Base Case We verify that P (1) is true where P (1) is the statement Section 8.6 Binomial Theorem 75 P (1): If x and y are any numbers, then (x + y )1 = Since 1 r=0 1 1−r r x y= r 1 1 r=0 r x1−r y r . 1 1−0 0 1 1−1 1 x y+ x y = x + y = (x + y )1 0 1 the base case holds. Inductive Hypothesis We assume that the statement P (k ) is true for k ≥ 1. P (k ): If x and y are any numbers, then (x + y )k = k k r=0 r xk −r y r . Inductive Conclusion We now show that the statement P (k + 1) is true. P (k +1): If x and y are any numbers, then (x+y )k+1 = k+1 k+1 r=0 r xk+1−r y r . (x + y )k+1 = (x + y )(x + y )k (8.1) k k k k −r r xy r +y k k+1−r r x y r + = x(x + y ) + y (x + y ) k =x r=0 k = r=0 = k k+1 x + 0 k = xk+1 + r=1 k+1 = xk+1 + r=1 k+1 = r=0 k r=1 (8.2) k r=0 k r=0 k k−r r xy r (8.3) k k−r r+1 xy r (8.4) k k+1−r r x y+ r k k + r r−1 k −1 r=0 k k−r r+1 k k+1 xy + y r k xk+1−r y r + y k+1 k + 1 k+1−r r x y + y k+1 r k + 1 k+1−r r x y r The result is true for n = k + 1, and so holds for all n by POMI. (8.5) (8.6) (8.7) (8.8) Chapter 9 The Greatest Common Divisor 9.1 Objectives The content objectives are: 1. To discover a proof of the proposition GCD With Remainders. 2. Do an example of the Euclidean Algorithm. 3. Prove the GCD Characterization Theorem. 4. Compute gcds and certificates using the Extended Euclidean Algorithm. 9.2 Definition 9.2.1 Greatest Common Divisor Greatest Common Divisor Let a and b be integers, not both zero. An integer d > 0 is the greatest common divisor of a and b, written gcd(a, b), if and only if 1. d | a and d | b (this captures the common part of the definition), and 2. if c | a and c | b then c ≤ d (this captures the greatest part of the definition). Example 1 • gcd(24, 30) = 6 • gcd(17, 25) = 1 • gcd(−12, 0) = 12 • gcd(−12, −12) = 12 • gcd(0, 0) =?? 76 Section 9.2 Greatest Common Divisor Definition 9.2.2 gcd(0, 0) 77 For a = 0, the definition implies that gcd(a, 0) = |a| and gcd(a, a) = |a|. We define gcd(0, 0) as 0. This may sound counterintuitive, since all integers are divisors of 0, but it is consistent with gcd(a, 0) = |a| and gcd(a, a) = |a|. Let’s prove a seemingly unusual proposition about gcds. Proposition 1 (GCD With Remainders (GCD WR)) If a and b are integers not both zero, and q and r are integers such that a = qb + r, then gcd(a, b) = gcd(b, r). How would we discover a proof for this proposition? Let’s try the usual approach: identify the hypothesis and conclusion, and begin asking questions. Hypothesis: a, b, q and r are integers such that a = qb + r. Conclusion: gcd(a, b) = gcd(b, r) My first question typically starts with the conclusion and works backward. What is a suitable first question? How about “How do we show that two integers are equal?” There are lots of possible answers: show that their difference is zero, their ratio is one, each is less than or equal the other. However, here we are working with gcds rather than generic integers so perhaps a better question would be “How do we show that a number is a gcd?” The broad answer is relatively easy. Use the definition of gcd. After all, right now it is the only thing we have! A specific answer is less easy. Do we want to focus on gcd(a, b) or gcd(b, r)? Here is an easy way to do both. Let d = gcd(a, b). Then show that d = gcd(b, r). That gets us two statement in our proof. Proof in Progress 1. Let d = gcd(a, b). 2. To be completed. 3. Hence d = gcd(b, r). But how do we show that d = gcd(b, r)? Use the definition. Our proof can expand to Proof in Progress 1. Let d = gcd(a, b). 2. We will show (a) d | b and d | r, and (b) if c | b and c | r then c ≤ d. 3. To be completed. 4. Hence d = gcd(b, r). 78 Chapter 9 The Greatest Common Divisor For the first part of the definition, we ask “How do we show that one number divides another number?” Interestingly enough, there are two different answers - one for b and one for r, though that is not obvious. For b there is already a connection between d and b in the first sentence. Since d = gcd(a, b), we know from the definition of gcd that d | b. What about r? Using the definition of divisibility seems problematic. What propositions could we use? Transitivity of Divisibility (Proposition 1) doesn’t seem to apply. How about using the Divisibility of Integer Combinations (Proposition 1)? Proposition 2 (Divisibility of Integer Combinations) Let a, b and c be integers. If a | b and a | c, then a | (bx + cy ) for any x, y ∈ Z. Observe that r = a − qb. Since d | a and d | b, d divides any integer combination of a and b by the Divisibility of Integer Combinations. That is, d | (a(1) + b(−q )) so d | r. Let’s extend our proof in progress. Proof in Progress 1. Let d = gcd(a, b). 2. We will show (a) d | b and d | r, and (b) if c | b and c | r then c ≤ d. 3. Since d = gcd(a, b), we know from the definition of gcd that d | b. 4. Observe that r = a − qb. Since d | a and d | b, d | (a(1) + b(−q )) by the Divisibility of Integer Combinations, so d | r. 5. To be completed. 6. Hence d = gcd(b, r). That leaves us with the greatest part of greatest common divisor. This second part of the definition is itself an implication, so we assume that c | b and c | r and we must show c ≤ d. How do we show one number is less than or equal to another number? There doesn’t seem to be anything obvious but ask “Have I seen this anywhere before?”. Yes, we have. In the second part of the definition of gcd. But then you might ask “Isn’t that assuming what we have to prove?” Let’s be precise about what we are saying. We can use d for one inequality. Since d = gcd(a, b), for any c where c | a and c | b, c ≤ d. What we need to show is: if c | b and c | r then c ≤ d. These two statements are close, but not the same. If we assume that c | b and c | r, then c | (b(q ) + r(1)) by the Divisibility of Integer Combinations (again). Since a = qb + r, c | a. And now, since d = gcd(a, b) and c | a and c | b, c ≤ d as needed. Let’s add that to our proof in progress. Proof in Progress 1. Let d = gcd(a, b). Section 9.3 Greatest Common Divisor 79 2. We will show (a) d | b and d | r, and (b) if c | b and c | r then c ≤ d. 3. Since d = gcd(a, b), we know from the definition of gcd that d | b. 4. Observe that r = a − qb. Since d | a and d | b, d | (a(1) + b(−q )) by the Divisibility of Integer Combinations, so d | r. 5. Let c | b and c | r. Then c | (b(q ) + r(1)) by the Divisibility of Integer Combinations. Since a = qb + r, c | a. And now, since d = gcd(a, b) and c | a and c | b, c ≤ d by the second part of the definition of gcd. 6. Hence d = gcd(b, r). Having discovered a proof, we should now write the proof. Whenever you write, you should have an audience in mind. You actually have two audiences to keep in mind: your peers with whom you collaborate, and the markers. You do not need to specify each proof technique, since your peers and markers know all of them. It does help to provide an overall plan if you can. Also, proofs tend to work much more forwards than backwards because that helps to emphasize the notion of starting with hypotheses and ending with the conclusion. Here is one possible proof. Proof: Let d = gcd(a, b). We will use the definition of gcd to show that d = gcd(b, r). Since d = gcd(a, b), d | b. Observe that r = a − qb. Since d | a and d | b, d | (a − qb) by the Divisibility of Integer Combinations. Hence d | r, and d is a common divisor of b and r. Let c be a divisor of b and r. Since c | b and c | r, c | (qb + r) by the Divisibility of Integer Combinations. Now a = qb + r, so c | a. Because d = gcd(a, b) and c | a and c | b, c ≤ d. REMARK 1. If a = b = 0 this proposition is also true since the only possible choices for b and r are b = r = 0. 2. In general, there are many ways to work forwards and backwards. 3. The proof may records steps in a different order than their appearance in the discovery process. 4. Proofs are short and usually omit the discovery process. 5. Be sure that you can identify where each of your hypotheses was used in the proof. 80 Chapter 9 9.3 The Greatest Common Divisor Certificate of Correctess Suppose we wanted to compute gcd(1386, 322). We could factor both numbers, find their common factors and select the greatest. In general, this is very slow. Repeated use of GCD With Remainders allows us to efficiently compute gcds. For example, let’s compute gcd(1386, 322). Example 2 Since Since Since Since 1386 = 4 × 322 + 98, 322 = 3 × 98 + 28, 98 = 3 × 28 + 14, 28 = 2 × 14 + 0, gcd(1386, 322) = gcd(322, 98). gcd(322, 98) = gcd(98, 28). gcd(98, 28) = gcd(28, 14). gcd(28, 14) = gcd(14, 0). Since gcd(14, 0) = 14, the chain of equalities from the column on the right gives us gcd(1386, 322) = gcd(322, 98) = gcd(98, 28) = gcd(28, 14) = gcd(14, 0) = 14. This process is known as the Euclidean Algorithm. Exercise 1 Randomly pick two positive integers and compute their gcd using the Euclidean Algorithm. How do you know that you have the correct answer? Keep your work. You’ll need it soon. Because mistakes happen when performing arithmetic by hand, and mistakes happen when programming computers, it would be very useful if there were a way to certify that an answer is correct. Think of a certificate of correctness this way. You are a manager. You ask one of your staff to solve a problem. The staff member comes back with the proposed solution and a certificate of correctness that can be used to verify that the proposed solution is, in fact, correct. The certificate has two parts: a theorem which you have already proved and which relates to the problem in general, and data which relates to this specific problem. For example, here’s a proposition that allows us to produce a certificate for gcd(a, b). Proposition 3 (GCD Characterization Theorem (GCD CT)) If d is a positive common divisor of the integers a and b, and there exist integers x and y so that ax + by = d, then d = gcd(a, b). Our certificate would consist of this theorem along with integers x and y . If our proposed solution was d and d | a, d | b and ax + by = d, then we could conclude without doubt that d = gcd(a, b). In the example 2 above, the proposed gcd of 1386 and 322 is 14. Our certificate of correctness consists of the GCD Characterization Theorem and the integers d = 14, x = 10 and y = −43. Note that 14 | 1386 and 14 | 322 and 1386 × 10 + 322 × (−43) = 14, so we can conclude that 14 = gcd(1386, 322). Here is a proof of the GCD Characterization Theorem. Section 9.4 Certificate of Correctess 81 Proof: (For reference, each sentence of the proof is written on a separate line.) 1. We will show that d satisfies the definition of gcd(a, b). 2. From the hypotheses, d | a and d | b. 3. Now let c | a and c | b. 4. By the Divisibility of Integer Combinations (Proposition 1), c | (ax + by ) so c | d. 5. By the Bounds by Divisibility (Proposition 2), c ≤ d, and so d = gcd(a, b). Let’s do an analysis of the proof. Analysis of Proof As usual, we will begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: d is a positive common divisor of the integers a and b. There exist integers x and y so that ax + by = d. Conclusion: d = gcd(a, b) Core Proof Technique: Work forwards recognizing an existential quantifier in the hypothesis. Preliminary Material: Definition of gcd. An integer d > 0 is the gcd(a, b) if and only if 1. d | a and d | b, and 2. if c | a and c | b then c ≤ d. Sentence 1 We will show that d satisfies the definition of gcd(a, b). The author states the plan - always a good idea. The author is actually answering the question “How do I show that one number is the gcd of two other numbers?” Sentence 2 From the hypotheses, d | a and d | b. The author is working forwards from the hypothesis. This handles the first part of the definition of gcd. Sentence 3 Now let c | a and c | b. The second part of the definition of gcd is an implication with hypothesis c | a and c | b. The author must show c ≤ d. Sentence 4 By the Divisibility of Integer Combinations, c | (ax + by ) so c | d. This is where the author uses an existential quantifier in the hypothesis. The author assumes the existence of two integers x and y such that ax + by = d. The author does not state this explicitly. Having made this assumption, the author can use Sentence 3 to satisfy the hypotheses of Divisibility of Integer Combinations and so invoke the conclusion, that is, c | (ax + by ). Sentence 5 By the Bounds By Divisibility, c ≤ d, and so d = gcd(a, b). Having determined that c ≤ d, both parts of the definition of gcd are satisfied and so the author can conclude that d = gcd(a, b). Now the obvious questions is: “How do we find x and y ?” 82 Chapter 9 9.4 The Greatest Common Divisor The Extended Euclidean Algorithm (EEA) Given two positive integers, a and b, the EEA is an efficient way to compute not only d = gcd(a, b) but the data x and y for the certificate. Here is the algorithm and an example. Algorithm 1 Extended Euclidean Algorithm Require: a > b > 0 are integers. Ensure: The following conditions hold at the end of the algorithm. rn+1 = 0. rn = gcd(a, b). ri−2 = qi ri−1 + ri where 0 ≤ ri < ri−1 . In every row, axi + byi = ri . x = xn , y = yn is a solution to ax + by = gcd(a, b). {Initialize} Construct a table with four columns so that The columns are labelled x, y , r and q . The first row in the table is (1, 0, a, 0). The second row in the table is (0, 1, b, 0). {To produce the remaining rows (i ≥ 3)} repeat ri−2 qi ← ri−1 Rowi ← Rowi−2 − qi Rowi−1 until ri = 0 This may be easier to understand with an example. Let’s compute gcd(1386, 322) using the EEA. Since 1386 > 322 > 0 we can, in fact, legitimately use the EEA. Initially we get xy r q 1 0 1386 0 0 1 322 0 To generate the third row we must first compute q3 . Using the formula ri−2 ri−1 qi ← we get q3 = r1 r2 = 1386 =4 322 Now we use the formula Rowi ← Rowi−2 − qi Rowi−1 when i = 3 to get Row3 ← Row1 − q3 Row2 With q3 = 4 we get Row3 ← Row1 − 4 × Row2 Representing this in the table gives Row1 −4 × Row2 = Row3 xy r q 1 0 1386 0 01 322 0 1 −4 98 4 Section 9.4 The Extended Euclidean Algorithm (EEA) 83 In a similar fashion we get the fourth row. Row2 −3 × Row3 = Row4 x y r q 1 0 1386 0 0 1 322 0 1 −4 98 4 −3 13 28 3 The completely worked out example follows. x y r q 1 0 1386 0 0 1 322 0 1 −4 98 4 −3 13 28 3 10 −43 14 3 −23 99 0 2 We stop when the remainder is 0. The second last row provides the desired d, x and y . The gcd d is the entry in the r column, x is the entry in the x column and y is the entry in the y column. Hence, d = 14 (as before), and we can check the conditions of the GCD Characterization Theorem to certify correctness. Since 14 | 1386 and 14 | 322 and 1386 × 10 + 322 × (−43) = 14, we can conclude that 14 = gcd(1386, 322). If a or b is negative, apply the EEA to gcd(|a|, |b|) and then change the signs of x and y after the EEA is complete. If a < b, simply swap their places in the algorithm. This works because gcd(a, b) = gcd(b, a). We treat the EEA as a proposition where the preconditions are the hypotheses and the postconditions are the conclusions. Let’s record the algorithm in the form of a theorem. Proposition 4 (Extended Euclidean Algorithm (EEA)) If a and b are positive integers, then d = gcd(a, b) can be computed and there exist integers x and y so that ax + by = d. A proof of the correctness of the EEA is available in the appendix. Add to appendix. Exercise 2 A few minutes ago you computed the gcd of two numbers. Repeat that exercise using the EEA and verify that you can produce a certificate of correctness for your proposed gcd. Chapter 10 Properties Of GCDs 10.1 Objectives The technique objectives are: 1. To practice working with existential quantifiers. The content objectives are: 1. Discover a proof of Coprimeness and Divisibility. 2. Discover a proof of GCD Of One 3. Exercise: Discover a proof of Division by the GCD. 4. Exercise: Discover a proof of Primes and Divisibility. 10.2 Some Useful Propositions We begin with a proposition on coprimeness and divisibility. Definition 10.2.1 Two integers a and b are coprime if gcd(a, b) = 1. Coprime Proposition 1 (Coprimeness and Divisibility (CAD)) If a, b and c are integers and c | ab and gcd(a, c) = 1, then c | b. This proposition has two implicit existential quantifiers, one in the hypothesis and one in the conclusion. You might object and ask “Where?” It’s hidden - in the definition of divides. Recall the definition. An integer m divides an integer n if there exists an integer k so that n = km. We treat an existential quantifier in the hypothesis differently from an existential quantifier in the conclusion. Recall the following remarks from the chapter on quantifiers. 84 Section 10.2 Some Useful Propositions 85 REMARK When proving that “A implies B ” and A uses an existential quantifier, use the Object Method. 1. Identify the four parts of the quantified statement “there exists an x in the set S such that P (x) is true.” 2. Assume that a mathematical object x exists within the domain S so that the statement P (x) is true. 3. Make use of this information to generate another statement. When proving that “A implies B ” and B uses an existential quantifier, use the Construct Method. 1. Identify the four parts of the quantified statement. “there exists an x in the set S such that P (x) is true.” 2. Construct a mathematical object x. 3. Show that x ∈ S . 4. Show that P (x) is true. Let’s be clear about what “there exists an integer k so that b = kc”, the existential statement in the conclusion, means. Quantifier: Variable: Domain: Open sentence: ∃ k Z b = kc With all of this in mind, how do we go about discovering a proof for Coprimeness and Divisibility? As usual, we will begin by explicitly identifying the hypothesis, the conclusion, the core proof technique and any preliminary material we think we might need. Hypothesis: a, b and c are integers and c | ab and gcd(a, c) = 1. Conclusion: c | b. Core Proof Technique: We use the Object Method because of the existential quantifier in the hypothesis, and the Construct Method because of the existential quantifier in the conclusion. Preliminary Material: Definition of divides and gcd. When discovering proofs I prefer to start by working backwards from the conclusion. In this case, I would begin by asking “How do we show that one integer divides another?” We can answer with the definition of divisibility. We must construct an integer k so that b = ck . We will record this as follows. 86 Chapter 10 Properties Of GCDs Proof in Progress 1. To be completed. 2. Since b = kc, c | b. The problem is that it is not at all clear what k should be. Let’s work forwards from the hypothesis. Somehow we need an equation with a b alone on one side of the equality sign. We can’t start there but we can get an equation with a b. Since gcd(a, c) = 1, the EEA guarantees that we can find integers x and y so that ax + cy = 1. We could multiply this equation by b. Let’s record these forward statements. Proof in Progress 1. Since gcd(a, c) = 1, the EEA guarantees that we can find integers x and y so that ax + cy = 1 (1). 2. Multiplying (1) by b gives abx + cby = b (2). 3. To be completed. 4. Since b = kc, c | b. If we could factor the left hand side of (2), we’d be able to get a c and other stuff that we could treat as our k . But the first term has no c. Or maybe it does. Since c | ab there exists an integer h so that ch = ab. Substituting ch for ab in (2) gives chx + cby = b (3). We record this as Proof in Progress 1. Since gcd(a, c) = 1, the EEA guarantees that we can find integers x and y so that ax + cy = 1 (1). 2. Multiplying (1) by b gives abx + cby = b (2). 3. Since c | ab there exists an integer h so that ch = ab. Substituting ch for ab in (2) gives chx + cby = b (3). 4. To be completed. 5. Since b = kc, c | b. Now factor. Proof in Progress 1. Since gcd(a, c) = 1, the EEA guarantees that we can find integers x and y so that ax + cy = 1 (1). 2. Multiplying (1) by b gives abx + cby = b (2). 3. Since c | ab there exists an integer h so that ch = ab. Substituting ch for ab in (2) gives chx + cby = b (3). Section 10.2 Some Useful Propositions 87 4. This gives c(hx + by ) = b. 5. But then if we let k = hx + by we have an integer k so that ck = b. 6. Since b = kc, c | b. Here is a proof. Proof: By the Extended Euclidean Algorithm and the hypothesis gcd(a, c) = 1, there exist integers x and y so that ax + cy = 1. Multiplying by b gives abx + cby = b. Since c | ab there exists an integer h so that ch = ab. Substituting ch for ab gives chx + cby = b. Lastly, factoring produces (hx + by )c = b. Since hx + by is an integer, c | b. Let us consider more properties of the greatest common divisor. Proposition 2 (GCD Of One (GCD OO)) Let a and b be integers. Then gcd(a, b) = 1 if and only if there are integers x and y with ax + by = 1. This proposition has similar elements to the one we just proved, so it won’t be a surprise if we use similar reasoning. REMARK The important difference is that this statement is an “if and only if” statement. To prove A if and only if B we must prove two statements: 1. If A, then B . 2. If B , then A. We can restate the proposition as Proposition 3 (GCD Of One (GCD OO)) Let a and b be integers. 1. If gcd(a, b) = 1, then there are integers x and y with ax + by = 1. 2. If there are integers x and y with ax + by = 1, then gcd(a, b) = 1. In statement (1), there is an existential quantifier in the conclusion, so we would expect to use the Construction Method. The problem is “Where do we get x and y ?” In the previous proof, we used the EEA and it makes sense to use it here as well. By the EEA and the hypothesis gcd(a, b) = 1, there exist integers x and y so that ax + by = 1. 88 Chapter 10 Properties Of GCDs In statement (2), an existential quantifier occurs in the hypothesis so we can assume the existence of integers x and y so that ax + by = 1. Also, 1 | a and 1 | b. These are exactly the hypotheses of the GCD Characterization Theorem, so we can conclude that gcd(a, b) = 1. Here is a proof of the GCD Of One proposition. Proof: Since gcd(a, b) = 1, the EEA assures the existence of integers x and y so that ax + by = 1. Statement 1 is proved. Now, 1 | a and 1 | b. Also, by the hypothesis of Statement 2, there exist integers x and y so that ax + by = 1. These are exactly the hypotheses of the GCD Characterization Theorem, so we can conclude that gcd(a, b) = 1 and Statement 2 is proved. REMARK This proof illustrates the connection between the GCD Characterization Theorem and the Extended Euclidean Algorithm. Both assume integers a and b. The GCD Characterization Theorem starts with an integer d where d | a, d | b and integers x and y so that ax + by = d and concludes that d = gcd(a, b). The Extended Euclidean Algorithm computes a d so that d = gcd(a, b), hence it produces a d so that d | a and d | b, and also computes integers x and y so that ax + by = d. So, if we encounter a gcd in the conclusion, we can try the GCD Characterization Theorem. If we encounter a gcd in the hypothesis, we can try the Extended Euclidean Algorithm. Exercise 1 Proposition 4 Prove the following proposition. Compare your proof with the proof that follows. (Division by the GCD (DB GCD)) Let a and b be integers. If gcd(a, b) = d = 0, then gcd Proof: First, observe that gcd ab , dd ab , dd = 1. is meaningful. Since d | a and d | b, both b a and d d are integers. We will use the GCD Characterization Theorem. Since gcd(a, b) = d, the EEA assures the existence of integers x and y so that ax + by = d. Dividing by d gives a b x+ y =1 d d Since 1 divides both gcd ab , dd = 1. a b and , the GCD Characterization Theorem implies that d d Section 10.2 Some Useful Propositions Exercise 2 89 This exercise illustrates the use of Proof by Elimination and proves a very useful proposition that follows from Coprimeness and Divisibility. 1. Prove that A ⇒ B ∨ C ≡ (A ⇒ B ) ∨ (A ⇒ C ) ≡ ¬(A ⇒ B ) ∧ (A ⇒ C ) 2. A true statement of the form A ⇒ B ∨ C has two cases: A ⇒ B OR A ⇒ C . One way to prove a statement of the form A ⇒ B ∨ C is to show that one of the two cases must hold. This is often done by proving the statement ¬(A ⇒ B ) ∧ (A ⇒ C ) which is equivalent to proving “If the first case is not true, then the second case must be true.” This technique is called Proof By Elimination since one of the two cases is eliminated. With this in mind prove the proposition Primes and Divisibility below. Begin your proof with “Suppose p a. We must show” Proposition 5 (Primes and Divisibility (PAD)) If p is a prime and p | ab, then p | a or p | b. Chapter 11 Linear Diophantine Equations 11.1 Objectives The technique objectives are: 1. To practice working with universal quantifiers. 2. To practice working with subsets. The content objectives are: 1. Prove Coprimeness and Divisibility paying attention to the universal quantifier. 2. Define Diophantine equations. 3. Prove the Linear Diophantine Equation Theorem (Part 1) 4. Discover a proof to the Linear Diophantine Equation Theorem (Part 2). 5. Examples of the Linear Diophantine Equation Theorem. 11.2 The Select Method We have already proved Proposition 1 (Coprimeness and Divisibility (CAD)) If a, b and c are integers and c | ab and gcd(a, c) = 1, then c | b. Let’s restate the proposition using a universal quantifier. Proposition 2 (Coprimeness and Divisibility) For all integers a, b and c where c | ab and gcd(a, c) = 1, c | b. 90 Section 11.2 The Select Method 91 REMARK Whenever we encounter a universal quantifier, we use the Select Method. To prove a statement of the form For every x in the set S , P (x) is true. we should 1. Identify the four parts of the quantified statement. 2. Select a representative mathematical object x ∈ S . This cannot be a specific object. It has to be a placeholder so that our argument would work for any specific member of S . Note that if the the set S is empty, we proceed no further. The statement is vacuously true. 3. Show that P (x) is true. Let’s see how the Select Method is used in the following proof, most of which you have seen before. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let a, b and c be integers where c | ab and gcd(a, c) = 1. 2. By the Extended Euclidean Algorithm and gcd(a, c) = 1, there exist integers x and y so that ax + cy = 1. 3. Multiplying by b gives abx + cby = b. 4. Since c | ab there exists an integer h so that ch = ab. 5. Substituting ch for ab gives chx + cby = b. 6. Lastly, factoring produces c(hx + by ) = b. 7. Since hx + by is an integer, c | b. Here is an analysis. Analysis of Proof The statement begins with a quantifier so we should first identify the four parts of the quantified statement. Quantifier: Variable: Domain: Open sentence: ∀ a, b, c Z If c | ab and gcd(a, c) = 1, then c | b. 92 Chapter 11 Linear Diophantine Equations The open sentence is an implication so let’s identify the hypothesis and conclusion as well as the core proof technique and preliminary material. Hypothesis: c | ab and gcd(a, c) = 1. Conclusion: c | b. Core Proof Technique: We begin with the Select Method. We also use the Object Method because of the implicit existential quantifier in the hypothesis (c | ab) and the Construct Method because of the implicit existential quantifier in teh conclusion (c | b). Preliminary Material: Definitions of divide and gcd. Sentence 1 Let a, b and c be integers where c | ab and gcd(a, c) = 1. This follows exactly the plan of the Select Method. We begin by selecting representative objects in the domain. We end by showing that, for the chosen objects, the open sentence is true. This part appears in Sentence 7. Sentences 2 – 6 These appear just as they did in the original proof of Coprimeness and Divisibility (Proposition 1) Sentence 7 Since hx + by is an integer, c | b. The important part of this sentence is the demonstration that the open sentence is true. There are some important remarks to make here. • Just as in English, there is often more than one way to say the same thing. • Quantifiers are often implicit or hidden. • Condensed proofs typically do not illustrate the discovery process or explicitly identify techniques. 11.3 Linear Diophantine Equations In high school, you looked at linear equations that involved real numbers. We will look at linear equations involving only integers. Definition 11.3.1 Diophantine Equations Equations with integer co-efficients for which integer solutions are sought, are called Diophantine equations after the Greek mathematician, Diophantus of Alexandria, who studied such equations. Diophantine equations are called linear if each term in the equation is a constant or a constant times a single variable of degree 1. The simplest linear Diophantine equation is ax = b To emphasize, a, b ∈ Z and we want an x ∈ Z that solves ax = b. From the definition of divisibility, we know that this equation has an integer solution x if and only if a | b. Section 11.3 Linear Diophantine Equations 93 What about linear Diophantine equations with two variables? ax + by = c Theorem 3 (Linear Diophantine Equation Theorem, Part 1 (LDET 1)) Let gcd(a, b) = d. The linear Diophantine equation ax + by = c has a solution if and only if d | c. Before we study a proof of this theorem, let’s see how it works in practice. Example 1 Which of the following linear Diophantine equations has a solution? 1. 33x + 18y = 10 2. 33x + 18y = 15 Solution: 1. Since gcd(33, 18) = 3, and 3 does not divide 10, the first equation has no integer solutions. 2. Since gcd(33, 18) = 3, and 3 does divide 15, the second equation does have an integer solution. But how do we find a solution? Here are two simple steps that will allow us to find a solution. 1. Use the Extended Euclidean Algorithm to find d = gcd(a, b) and x1 and y1 where ax1 + by1 = d. 2. Multiply by k = c to get akx1 + bky1 = kd = c. A solution is x = kx1 and y = ky1 . d Returning to the exercise, the Extended Euclidean Algorithm gives x y rq 1 0 33 0 0 1 18 0 1 −1 15 1 −1 2 31 6 −11 0 5 hence 33 × −1 + 18 × 2 = 3 94 Chapter 11 Linear Diophantine Equations Multiplying by k = c/d = 15/3 = 5 gives 33 × −5 + 18 × 10 = 15 so one particular solution is x = −5 and y = 10. But are there more solutions? That’s where Part 2 of the Linear Diophantine Equation Theorem comes in and we will cover it later. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. First, suppose that the linear Diophantine equation ax + by = c has an integer solution x = x0 , y = y0 . That is, ax0 + by0 = c. 2. Since d = gcd(a, b), d | a and d | b. 3. But then, by the Divisibility of Integer Combinations, d | (ax0 + by0 ). That is d | c. 4. Conversely, suppose that d | c. 5. Then there exists an integer k such that c = kd. 6. Now, by the Extended Euclidean Algorithm, there exist integers x1 and y1 so that ax1 + by1 = d. 7. Multiplying this equation by k gives akx1 + bky1 = kd = c which, in turn, implies that x = kx1 and y = ky1 is a solution to ax + by = c. Let’s perform an analysis of this proof. Analysis of Proof This is an “if and only if” statement so we must prove two statements. 1. If the linear Diophantine equation ax + by = c has a solution, then d | c. 2. If d | c, then the linear Diophantine equation ax + by = c has a solution. Core Proof Technique: Both statements contain an existential quantifier in the hypothesis, so each will start with the Object Method. Though both statements also contain an existential quantifier in the conclusion, only one uses the Construction Method. The other uses a proposition we have already proved. Sentence 1 First, suppose that the linear Diophantine equation ax + by = c has an integer solution x = x0 , y = y0 . That is, ax0 + by0 = c. The author does not explicitly rephrase the “if and only if” as two statements. Rather, Sentence 1 indicates which of the two implicit statements will be proved by stating the hypothesis of Statement 1. Moreover, the first statement uses an existential quantifier in the hypothesis. The hypothesis of the first statement could be restated as Section 11.3 Linear Diophantine Equations 95 there exists an integer solution to the linear Diophantine equation The four parts are Quantifier: Variable: Domain: Open sentence: ∃ x0 , y0 Z ax0 + by0 = c. Since the existential quantifier occurs in the hypothesis, the author uses the Object Method. The author assumes the existence of the corresponding objects (x0 , y0 ) in a suitable domain (Z) and that these objects satisfy the related open sentence (ax0 + by0 = c). Sentence 2 Since d = gcd(a, b), d | a and d | b. This follows from the definition of gcd. Sentence 3 But then, by the Divisibility of Integer Combinations, d | (ax0 + by0 ). That is d | c. Since hypotheses of DIC (a, b and d are integers, and d | a and d | b) are satisfied, the author can invoke the conclusion of DIC (d | (ax0 + by0 )). And from Sentence 1, ax0 + by0 = c so d | c. Sentence 4 Conversely, suppose that d | c. The conversely indicates that the author is about to prove Statement 2. Recall that an “if and only if” always consists of a statement and its converse. The hypothesis of the converse is d | c. The definition of divides contains an existential quantifier and so, in Sentence 5, the authors uses the Object Method. The conclusion of Statement 2 contains an existential quantifier (there exists an integer solution to the linear Diophantine equation), so the author uses the Construction Method and builds a suitable solution. Here are the parts of the existential quantifier in the conclusion. Quantifier: Variable: Domain: Open sentence: ∃ x, y Z ax + by = c. Sentence 5 Then there exists an integer k such that c = kd. This is the Object Method and follows from the definition of divisibility. Sentence 6 Now, by the Extended Euclidean Algorithm, there exist integers x1 and y1 so that ax1 + by1 = d. This is prior knowledge. Sentence 7 Multiplying this equation by k gives akx1 + bky1 = kd = c which, in turn, implies that x = kx1 and y = ky1 is a solution to ax + by = c. This is where the solution is constructed, x = kx1 and y = ky1 , and where the open sentence is verified. The author does not explicitly check that kx1 and kx2 are integers, though we must when we analyse the proof. 96 Chapter 11 Linear Diophantine Equations LDET 1 tells us when solutions exist and how to construct a solution. It does not find all of the solutions. That happens next. Theorem 4 (Linear Diophantine Equation Theorem, Part 2, (LDET 2)) Let gcd(a, b) = d = 0. If x = x0 and y = y0 is one particular integer solution to the linear Diophantine equation ax + by = c, then the complete integer solution is b a x = x0 + n, y = y0 − n, ∀n ∈ Z. d d Before we discover a proof, let’s make sure we understand the statement. Example 2 Find all solutions to 33x + 18y = 15. Solution: Since gcd(33, 18) = 3, and 3 does divide 15, this equation does have integer solutions by the Linear Diophantine Equation Theorem, Part 1. If we can find one solution, we can use the Linear Diophantine Equation Theorem, Part 2 to find all solutions. Since we earlier found the solution x = −5 and y = 10 the complete solution is {(x, y ) | x = −5 + 6n, y = 10 − 11n, n ∈ Z} You we can check that these are solutions by substitution. Check: 33x + 18y = 33(−5 + 6n) + 18(10 − 11n) = −165 + 198n + 180 − 198n = 15 This check does not verify that we have found all solutions. It verifies that all of the pairs f integers we have fund are solutions. The expression “complete integer solution” in the statement of LDET 2 hides the use of sets. Let’s be explicit about those sets and what we need to do with them. There are, in fact, two sets in the conclusion, the set of solutions and the set of x and y pairs. We define them formally as follows. Complete solution Let S = {(x, y ) | x, y ∈ Z, ax + by = c} b Proposed solution Let T = {(x, y ) | x = x0 + d n, y = y0 − a n, ∀n ∈ Z} d The conclusion of LDET 2 is S = T . How do we show that two sets are equal? Two sets S and T are equal if and only if S ⊆ T and T ⊆ S . That is, at the risk of being repetitive, to establish that S = T we must show two things. 1. S ⊆ T and 2. T ⊆ S Section 11.3 Linear Diophantine Equations 97 Normally one of the two is easy and the other is harder. Suppose we want to show S ⊆ T . How do universal quantifiers figure in? Showing that S ⊆ T is equivalent to the following statement. Proposition 5 S ⊆ T if and only if, for every member s ∈ S , s ∈ T . If you prefer symbolic notation you could write ∀s ∈ S, s ∈ T or s ∈ S ⇒ s ∈ T . What are the components of the universal quantifier in Proposition 5? Quantifier: Variable: Domain: Open sentence: ∀ s S s∈T The Select Method works perfectly in these situations. As frequently as sets are used, they are usually implicit and our first task is to discern what sets exist and how they are used. Let’s return to the proof of LDET 2 where our sets are: Complete solution Let S = {(x, y ) | x, y ∈ Z, ax + by = c} b Proposed solution Let T = {(x, y ) | x = x0 + d n, y = y0 − a n, ∀n ∈ Z} d Let us discover a proof. We must keep in mind that we have two things to prove 1. S ⊆ T and 2. T ⊆ S In this case, item 2 is easier so we will do it first. How do we show that T ⊆ S ? We must show that x ∈ T ⇒ x ∈ S . We certainly don’t want to individually check every element of T so we choose a representative element of T , one that could be replaced by any element of T and the subsequent argument would hold. This is just the Select Method and it provides our first statement. b Let n0 ∈ Z. Then (x0 + d n0 , y0 − a n0 ) ∈ T . d To show that this element is in S we must show that the element satisfies the defining property of S , that is, the element is a solution. b ax + by = a x0 + n0 d a + b y0 − n 0 d ab ab = ax0 + by0 + n0 − n0 d d = ax0 + by0 =c And now we can conclude by hypothesis, x = x0 and y = y0 is an integer solution 98 Chapter 11 Linear Diophantine Equations b (x0 + d n0 , y0 − a n0 ) ∈ S . d To show that S ⊆ T we will need to recall the following proposition on Division by the GCD (Proposition 4). Proposition 6 (Division by the GCD) ab d, d Let a and b be integers. If gcd(a, b) = d = 0, then gcd =1 Let’s begin our analysis of S ⊆ T . How do we show that S ⊆ T ? We choose a representative element in S and show that it is in T , that is, that it satisfies the defining property of T . b Specifically, we must show that an arbitrary solution (x, y ) has the form (x0 + d n, y0 − a n). d Let (x, y ) be an arbitrary solution. Then (x, y ) ∈ S and we must show (x, y ) ∈ T . Let (x0 , y0 ) be a particular solution to the linear Diophantine equation ax + by = c. The existence of (x0 , y0 ) is assured by the hypothesis. Let’s do the obvious thing and substitute both solutions into the equation. ax + by = c ax0 + by0 = c Eliminating c and factoring gives a(x − x0 ) = −b(y − y0 ) b a We know that d = gcd(a, b) is a common factor of a and b so and are both integers. d d Dividing the previous equation by d gives a b (x − x0 ) = − (y − y0 ) d d Using Division by the GCD, gcd Coprimeness and Divisibility that ab d, d = 1. Since b d divides (11.1) a d (x b divides (x − x0 ) d By the definition of divisibility, there exists an n ∈ Z so that x − x0 = n b b ⇒ x = x0 + n d d b Also, substituting n d for x − x0 in Equation (11.1) yields a y = y0 − n d b So every solution is of the form (x, y ) = (x0 + d n0 , y0 − a n0 ) and so d − x0 ) we know from Section 11.3 Linear Diophantine Equations 99 (x, y ) ∈ T A very condensed proof of Linear Diophantine Equation Theorem, Part 2 might look like the following. Notice the lack of mention of sets. Theorem 7 (Linear Diophantine Equation Theorem, Part 2, (LDET 2)) Let gcd(a, b) = d = 0. If x = x0 and y = y0 is one particular integer solution to the linear Diophantine equation ax + by = c, then the complete integer solution is b a x = x0 + n, y = y0 − n, ∀n ∈ Z. d d b Proof: Substitution shows that integers of the form x = x0 + n d , y = y0 − a n, n ∈ Z are d solutions. Now, let (x, y ) be an arbitrary solution and let (x0 , y0 ) be a particular solution to the linear Diophantine equation ax + by = c. Then ax + by = c ax0 + by0 = c Eliminating c and factoring gives a(x − x0 ) = −b(y − y0 ) (1). Dividing by d and using b Division by the GCD and Coprimeness and Divisibility we have d | (x − x0 ). Hence, there b exists an n ∈ Z so that x = x0 + d n (2). Substituting (2) in (1) gives y = y0 − a n0 as d needed. Exercise 1 Find all solutions to 1. 35x + 21y = 28 2. 35x − 21y = 28 Chapter 12 Practice, Practice, Practice: Quantifiers and Sets 12.1 Objectives This class provides an opportunity to practice working with quantifiers and sets. 12.2 Exercise 1 Exercises For each of the following statements, identify each quantifier, its parts and your approach to a proof of the statement. 1. For every integer a, 2 | a(a + 1). 2. If n is an integer, then 8 | (52n + 7). 3. If there exist integer solutions to the Diophantine equation ax2 + by 2 = c, then gcd(a, b) | c. Exercise 2 For each of the following definitions, identify each quantifier, its parts and the proof techniques that you would use to prove that a specific object satisfies the definition. 1. Saying that the function f of one real variable is bounded above means that there is a real number y such that for every real number x, f (x) ≤ y . 2. Saying that a set of real numbers S is bounded means that there is a real number M > 0 such that for every element s ∈ S , |s| < M . 3. Saying that the function f of one real variable is continuous at the point x means that for every real number ε > 0 there is a real number δ > 0 such that, for all real numbers y with |x − y | < δ , |f (x) − f (y )| < . 100 Section 12.2 Exercises Exercise 3 101 Prove each of the following propositions. 1. Suppose a and b are fixed integers. Then {ax + by | x, y ∈ Z} = {n · gcd(a, b) | n ∈ Z}. 2. An integer p > 1 is called a prime if its only positive divisors are 1 and p; otherwise it is called composite. Let a and b be fixed integers. If p is a prime and p | ab, then p | a or p | b. Exercise 4 Solve the following problems. 1. Find the complete solution to 7x + 11y = 3. 2. Find the complete solution to 35x − 42y = 14. 3. Find the complete solution to 28x + 60y = 10. 4. For what value of c does 8x + 5y = c have exactly one solution where both x and y are strictly positive? Exercise 5 The proof of the following statement is incomplete. Identify two sets used in the statement and proof (they are used implicitly) and state why the proof is incomplete. Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. Then the quadratic equation ax2 + bx + c = 0 has the solution x= −b ± √ b2 − 4ac 2a Proof: To show that a particular value is a solution, it is enough to substitute that value into the equation and show that the equation is satisfied. Consider √ −b + b2 − 4ac x= 2a Substitution gives √ √ 2 −b + b2 − 4ac −b + b2 − 4ac a +b +c 2a 2a √ √ a(b2 − 2b b2 − 4ac + b2 − 4ac) −b2 + b b2 − 4ac = + +c 4a2 2 √ √a b2 − 2b b2 − 4ac + b2 − 4ac −2b2 + 2b b2 − 4ac 4ac = + + 4a 4a 4a =0 102 Chapter 12 Practice, Practice, Practice: Quantifiers and Sets Since a similar result holds for x= the proposition holds. −b − √ b2 − 4ac 2a Chapter 13 Congruence 13.1 Objectives The content objectives are: 1. Define a is congruent to b modulo m. 2. Read a proof of Congruence is an Equivalence Relation. 3. Discover the proof of Properties of Congruence. 4. Read the proof of Congruences and Division. 5. Read the proof of Congruent Iff Same Remainder. 6. Do examples. 13.2 Congruences 13.2.1 Definition of Congruences One of the difficulties in working out properties of divisibility is that we don’t have an “arithmetic” of divisibility. Wouldn’t it be nice if we could resolve problems about divisibility in much the same way that we usually do arithmetic: add, subtract, multiply and divide? Carl Friedrich Gauss (1777 - 1855) was the greatest mathematician of the last two centuries. In a landmark work, Disquisitiones Arithmeticae, published when Gauss was 23, he introduced congruences and provided a mechanism to treat divisibility with arithmetic. Definition 13.2.1 Congruent Let m be a fixed positive integer. If a, b ∈ Z we say that a is congruent to b modulo m, and write a ≡ b (mod m) if m | (a − b). If m (a − b), we write a ≡ b (mod m). 103 104 Chapter 13 Example 1 Congruence Verify each of the following 1. 20 ≡ 2 (mod 6) 2. 2 ≡ 20 (mod 6) 3. 20 ≡ 8 (mod 6) 4. −20 ≡ 4 (mod 6) 5. 24 ≡ 0 (mod 6) 6. 5 ≡ 3 (mod 7) Question Panel here REMARK One already useful trait of this definition is the number of equivalent ways we have to work with it. a ≡ b (mod m) ⇐⇒ m | (a − b) ⇐⇒ ∃k ∈ Z ⇐⇒ ∃k ∈ Z 13.3 a − b = km a = km + b Elementary Properties Another extraordinarily useful trait of this definition is that it behaves a lot like equality. Equality is an equivalence relation. That is, it has the following three properties: 1. reflexivity, a = a. 2. symmetry, If a = b then b = a. 3. transitivity, If a = b and b = c, then a = c. Most relationships that you can think of do not have these three properties. The relation greater than fails reflexivity. The relation divides fails symmetry. The non-mathematical relation is a parent of fails transitivity. Proposition 1 (Congruence Is An Equivalence Relation (CER)) Let a, b, c ∈ Z. Then 1. a ≡ a (mod m). Section 13.3 Elementary Properties 105 2. If a ≡ b (mod m), then b ≡ a (mod m). 3. If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m) These may seem obvious but as the earlier examples showed, many relations do not have these properties. So, a proof is needed. We will give a condensed proof for all of them, and then an analysis for part 3. Proof: We show each part in turn. 1. Because a − a = 0 and m | 0, the definition of congruence gives a ≡ a (mod m). 2. Since a ≡ b (mod m), m | (a − b) which in turn implies that there exists k ∈ Z so that km = a − b. But if km = a − b, then (−k )m = b − a and so m | (b − a). By the definition of congruence, b ≡ a (mod m). 3. Since a ≡ b (mod m), m | (a − b). Since b ≡ c (mod m), m | (b − c). Now, by the Divisibility of Integer Combinations, m | ((1)(a − b) + (1)(b − c)) so m | (a − c). By the definition of congruence, a ≡ c (mod m). Analysis of Proof We will prove part 3 of the proposition Congruence Is An Equivalence Relation. Hypothesis: a, b, c ∈ Z, a ≡ b (mod m) and b ≡ c (mod m). Conclusion: a ≡ c (mod m). Sentence 1 Since a ≡ b (mod m), m | (a − b). The author is working forward from the hypothesis using the definition of congruence. Sentence 2 Since b ≡ c (mod m), m | (b − c). The author is working forward from the hypothesis using the definition of congruence. Sentence 3 Now, by the Divisibility of Integer Combinations, m | ((1)(a − b) + (1)(b − c)) so m | (a − c). Here it is useful to keep in mind where the author is going. The question “How do I show that one number is congruent to another number?” has the answer, in this case, of showing that m | (a − c) so the author needs to find a way of generating a − c. And a − c follows nicely from an application of the Divisibility of Integer Combinations. Sentence 4 By the definition of congruence, a ≡ c (mod m). The author is working forward from m | (a − c) using the definition of congruence. Proposition 2 (Properties of Congruence (PG)) If a ≡ a (mod m) and b ≡ b (mod m), then 106 Chapter 13 Congruence 1. a + b ≡ a + b (mod m) 2. a − b ≡ a − b (mod m) 3. ab ≡ a b (mod m) We will discover a proof of the third part and leave the first two parts as exercises. As usual we begin by identifying the hypothesis and the conclusion. Hypothesis: a ≡ a (mod m) and b ≡ b (mod m) Conclusion: ab ≡ a b (mod m) Let’s consider the question “How do we show that two numbers are congruent to one another?” The obvious abstract answer is “Use the definition of congruent.” We may want to keep in mind, however, that there are several equivalent forms. a ≡ b (mod m) ⇐⇒ m | (a − b) ⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b It is not at all clear which is best or whether, in fact, several could work. Since the conclusion of part three involves the arithmetic operation of multiplication, and we don’t have multiplication properties for equivalence or divisibility, it makes sense to consider either the third or fourth of the equivalent forms. There isn’t much to separate them. I’ll choose the last form and see how it works. So, the answer to “How do we show that two numbers are congruent to one another?” in the notation of this proof is “We must find an integer k so that ab = km + a b . Let’s record that. Proof in Progress 1. To be completed. 2. Since there exists k so that ab = km + a b , ab ≡ a b (mod m). The problem is how to find k . There is no obvious way backwards here so let’s start working forward. The two hypotheses a ≡ a (mod m) and b ≡ b (mod m) can be restated in any of their equivalent forms. Since we have already decided that we would work backwards with the fourth form, it makes sense to use the same form working forwards. That gives two statements. Proof in Progress 1. Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1). 2. Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2). 3. To be completed. Section 13.3 Elementary Properties 107 4. Since there exists k so that ab = km + a b , ab ≡ a b (mod m). But now there seems to be a rather direct way to produce an ab and an a b which we want for the conclusion. Just multiply equations (1) and (2) together. Doing that produces ab = m2 jh + mjb + a mh + a b = (mjh + jb + a h)m + a b If we let k = mjh + jb + a h then k is an integer and satisfies the property we needed in the last line of the proof, that is ab = km + a b . Let’s record this. Proof in Progress 1. Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1). 2. Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2). 3. Multiplying (1) by (2) gives ab = m2 jh + mjb + a mh + a b = (mjh + jb + a h)m + a b . 4. Since there exists k so that ab = km + a b , ab ≡ a b (mod m). Lastly, we write a condensed proof. Note that the reader of the proof is expected to be familiar with the equivalent forms. Proof: Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1). Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2). Multiplying (1) by (2) gives ab = m2 jh + mjb + a mh + a b = (mjh + jb + a h)m + a b . Since mjh + jb + a h is an integer, ab ≡ a b (mod m). Exercise 1 Prove the remainder of the Properties of Congruence proposition. There are four arithmetic operations with integers, but analogues to only three have been given. It turns out that division is problematic. A statement of the form ab ≡ ab (mod m) ⇒ b ≡ b (mod m) seems natural enough, simply divide by a. This works with the integer equation ab = ab . But consider the case where m = 12, a = 6, b = 3 and b = 5. It is indeed true that 18 ≡ 30 (mod 12) and so 6×3≡6×5 (mod 12) but “dividing” by 6 gives the clearly false statement 3≡5 (mod 12). Division works only under the specific conditions of the next proposition. 108 Chapter 13 Proposition 3 Congruence (Congruences and Division (CD)) If ac ≡ bc (mod m) and gcd(c, m) = 1, then a ≡ b (mod m). Before we read the proof, let’s look at an example. Example 2 Examples of division in congruence relations. 1. 8 × 7 ≡ 17 × 7 (mod 3) ⇒ 8 ≡ 17 (mod 3) 2. For 6 × 3 ≡ 6 × 5 (mod 12), CD cannot be invoked. Why? Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Since ac ≡ bc (mod m), m | (ac − bc). That is, m | c(a − b). 2. By the proposition Coprimeness and Divisibility, m | (a − b). 3. Hence, by the definition of congruence a ≡ b (mod m). Exercise 2 Analyze the proof of the proposition on Congruences and Division. We now give one more statement that is equivalent to a ≡ b (mod m). Proposition 4 (Congruent Iff Same Remainder (CISR)) a ≡ b (mod m) if and only if a and b have the same remainder when divided by m. Because this proposition is an “if and only if” proposition, there are two parts to the proof: a statement and its converse. We can restate the proposition to make the two parts more explicit. Proposition 5 (Congruent Iff Same Remainder (CISR)) 1. If a ≡ b (mod m), then a and b have the same remainder when divided by m. 2. If a and b have the same remainder when divided by m, then a ≡ b (mod m). Section 13.3 Elementary Properties 109 In practice, the two statements are not usually written out separately. The authors assume that you do that whenever you read “if and only if”. Many “if and only if” proofs begin with some prefatory material that will help both parts of the proof. For example, they often introduce notation that will be used in both parts. Let’s look at a proof of the Congruent Iff Same Remainder proposition. Before we do an analysis, make sure that you can identify 1. prefatory material (if any exists) 2. the proof of a statement 3. the proof of the converse of the statement Proof: The Division Algorithm applied to a and m gives a = q1 m + r1 , where 0 ≤ r1 < m The Division Algorithm applied to b and m gives b = q2 m + r2 , where 0 ≤ r2 < m Subtracting the second equation from the first gives a − b = (q1 − q2 )m + (r1 − r2 ), where − m < r1 − r2 < m If a ≡ b (mod m), then m | (a − b) and there exists an integer h so that hm = a − b. Hence a − b = (q1 − q2 )m + (r1 − r2 ) ⇒ hm = (q1 − q2 )m + (r1 − r2 ) ⇒ r1 − r2 = m(h − q1 + q2 ) which implies m | (r1 − r2 ). But, −m < r1 − r2 < m so r1 − r2 = 0. Conversely, if a and b have the same remainder when divided by m, then r1 = r2 and a − b = (q1 − q2 )m so a ≡ b (mod m). The prefatory material is quoted below. The Division Algorithm applied to a and m gives a = q1 m + r1 , where 0 ≤ r1 < m The Division Algorithm applied to b and m gives b = q2 m + r2 , where 0 ≤ r2 < m Subtracting the second equation from the first gives a − b = (q1 − q2 )m + (r1 − r2 ), where − m < r1 − r2 < m The proof of Statement 1 is 110 Chapter 13 Congruence If a ≡ b (mod m), then m | (a − b) and there exists an integer h so that hm = a − b. Hence a−b = (q1 −q2 )m+(r1 −r2 ) ⇒ hm = (q1 −q2 )m+(r1 −r2 ) ⇒ r1 −r2 = m(h−q1 +q2 ) which implies m | (r1 − r2 ). But, −m < r1 − r2 < m so r1 − r2 = 0. The proof of the converse of Statement 1, Statement 2, is Conversely, if a and b have the same remainder when divided by m, then r1 = r2 and a − b = (q1 − q2 )m so a ≡ b (mod m). We will do an analysis of the proof of Statement 1. An analysis of the proof Statement 2 is left as an exercise. Analysis of Proof In many “if and only if” statements one direction is much easier than the other. In this particular case, we are starting with the harder of the two directions. Hypothesis: a ≡ b (mod m). Conclusion: a and b have the same remainder when divided by m. Sentence 1 If a ≡ b (mod m), then m | (a − b) and there exists an integer h so that hm = a − b. Here the author is working forwards using two definitions. The definition of congruence allows the author to assert that “If a ≡ b (mod m), then m | (a − b)”. The definition of divisibility allows the author to assert that “m | (a − b) [implies that] there exists an integer h so that hm = a − b.” Sentence 2 Hence a − b = (q1 − q2 )m + (r1 − r2 ) ⇒ hm = (q1 − q2 )m + (r1 − r2 ) ⇒ r1 − r2 = m(h − q1 + q2 ) which implies m | (r1 − r2 ). This is mostly arithmetic. The author begins with a − b = (q1 − q2 )m + (r1 − r2 ) from the prefatory paragraph, substitutes hm for a − b, isolates r1 − r2 and factors out an m from the remaining terms. Since h − q1 + q2 is an integer, the author deduces that m | (r1 − r2 ). Sentence 3 But, −m < r1 − r2 < m so r1 − r2 = 0. This part is not so obvious. The author is working with two pieces of information. The prefatory material provides −m < r1 − r2 < m. Sentence 2 provides m | (r1 − r2 ). Now, what are the possible values of r1 − r2 ? Certainly r1 − r2 can be zero but are there any other possible choices? If there were another choice it would be of the form mx with x = 0. But that would make r1 − r2 = xm > m or r1 − r2 = xm < −m both of which are impossible because −m < r1 − r2 < m. Hence, r1 − r2 = 0. The conclusion does not say r1 − r2 = 0. It says that a and b have the same remainder when divided by m. Since r1 and r2 are those remainders, and r1 − r2 = 0 ⇒ r1 = r2 , the author leaves it to the reader to deduce the conclusion. Section 13.3 Elementary Properties Exercise 3 111 Perform an analysis of the proof of Statement 2. REMARK The proposition Congruent Iff Same Remainder gives us another part to our chain of equivalent statements. a ≡ b (mod m) ⇐⇒ m | (a − b) ⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b ⇐⇒ a and b have the same remainder when divided by m The propositions covered in this lecture are surprisingly powerful. Consider the following example. Example 3 What is the remainder when 347 is divided by 7? Solution: You could attempt to compute 347 with your calculator but it might explode. Here is a simpler way. First, recognize that the remainder when 347 is divided by 7 is just 347 (mod 7). Now observe that 32 ≡ 2 (mod 7) and 33 ≡ 3 × 2 ≡ 6 ≡ −1 (mod 7). But then 347 ≡ 345 32 ≡ (33 )15 32 ≡ (−1)15 32 ≡ (−1)(2) ≡ −2 ≡ 5 (mod 7) Hence, the remainder when 347 is divided by 7 is 5. Chapter 14 Modular Arithmetic 14.1 Objectives The content objectives are: 1. Define the congruence class modulo m. 2. Construct Zm and perform modular arithmetic. Highlight the role of additive and multiplicative identities, and additive and multiplicative inverses. 3. State Fermat’s Little Theorem. 4. Read a proof to a corollary of Fermat’s Little Theorem. 5. Discover a proof to the Existence of Inverses in Zp . 14.2 Modular Arithmetic In this section we will see the creation of a number system which will likely be new to you. Definition 14.2.1 The congruence class modulo m of the integer a is the set of integers Congruence Class Example 1 [a] = {x ∈ Z | x ≡ a (mod m)} For example, when m = 4 [0] [1] [2] [3] = = = = {x ∈ Z | x ≡ 0 (mod m)} = {x ∈ Z | x ≡ 1 (mod m)} = {x ∈ Z | x ≡ 2 (mod m)} = {x ∈ Z | x ≡ 3 (mod m)} = {. . . , −8, −4, 0, 4, 8, . . .} {. . . , −7, −3, 1, 5, 9, . . .} {. . . , −6, −2, 2, 6, 10, . . .} {. . . , −5, −1, 3, 8, 11, . . .} 112 = = = = {4k | k ∈ Z} {4k + 1 | k ∈ Z} {4k + 2 | k ∈ Z} {4k + 3 | k ∈ Z} Section 14.2 Modular Arithmetic 113 REMARK Note that congruence classes have more than one representation. In the example above [0] = [4] = [8] and, in fact [0] has infinitely many representations. If this seems strange to you, remember that fractions are another example of where one number has infinitely many representations. For example 1/2 = 2/4 = 3/6 = · · · . Definition 14.2.2 We define Zm to be the set of m congruence classes Zm Zm = {[0], [1], [2], . . . , [m − 1]} and we define two operations on Zm , addition and multiplication, as follows: [a] + [b] = [a + b] [a] · [b] = [a · b] Though the definition of these operations may seem obvious there is a fair amount going on here. 1. Sets are being treated as individual “numbers”. Modular addition and multiplication are being performed on congruence classes which are sets. 2. The addition and multiplication symbols on the left of the equals signs are in Zm and those on the right are operations in the integers. 3. We are assuming that the operations are well-defined. That is, we are assuming that these operations make sense even when there are multiple representatives of a congruence class. REMARK Since [a] = {x ∈ Z | x ≡ a (mod m)} we can extend our list of equivalent statements to [a] = [b] in Zm ⇐⇒ a ≡ b (mod m) ⇐⇒ m | (a − b) ⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b ⇐⇒ a and b have the same remainder when divided by m Just as there were addition and multiplication tables in grade school for the integers, we have addition and multiplication tables in Zm . 114 Chapter 14 Example 2 Addition and multiplication tables in Z4 + [0] [1] [2] [3] Exercise 1 Modular Arithmetic [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] · [0] [1] [2] [3] [3] [3] [0] [1] [2] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1] Write out the addition and multiplication tables in Z5 14.2.1 [0] ∈ Zm By looking at the tables for Z4 and Z5 it seems that [0] ∈ Zm behaves just like 0 ∈ Z. In Z ∀a ∈ Z, a + 0 = a ∀a ∈ Z, a · 0 = 0 and in Zm ∀[a] ∈ Zm , [a] + [0] = [a] ∀[a] ∈ Zm , [a] · [0] = [0] This actually follows from our definition of addition and multiplication in Zm . ∀[a] ∈ Zm , [a] + [0] = [a + 0] = [a] ∀[a] ∈ Zm , [a] · [0] = [a · 0] = [0] 14.2.2 [1] ∈ Zm In a similar fashion, by looking at the multiplication tables for Z4 and Z5 it seems that [1] ∈ Zm behaves just like 1 ∈ Z. In Z ∀a ∈ Z, a · 1 = a and in Zm ∀[a] ∈ Zm , [a] · [1] = [a] This follows from our definition of multiplication in Zm . ∀[a] ∈ Zm , [a] · [1] = [a · 1] = [a] 14.2.3 Identities and Inverses in Zm Many of us think of subtraction and division as independent from the other arithmetic operations of addition and multiplication. In fact, subtraction is just addition of the inverse. Now, what’s an inverse? To answer that question we must first define an identity. Section 14.2 Modular Arithmetic Definition 14.2.3 Identity 115 Given a set and an operation, an identity is, informally, “something that does nothing”. More formally, given a set S and an operation designated by ◦, an identity is an element e ∈ S so that ∀a ∈ S, a ◦ e = a The element e has no effect. Having something that does nothing is extremely useful though parents might not say that of teenagers. Example 3 Here are examples of sets, operations and identities. • The set of integers with the operation of addition has the identity 0. • The set of rational numbers excluding 0 with the operation of multiplication has the identity 1. • The set of real valued functions with the operation of function composition has the identity f (x) = x. • The set of integers modulo m with the operation of modular addition has the identity [0]. Definition 14.2.4 The element b ∈ S is an inverse of a ∈ S if a ◦ b = b ◦ a = e. Inverse Example 4 Here are examples of inverses. • Under the operation of addition, the integer 3 has inverse −3 since 3 + (−3) = (−3) + 3 = 0. • Under the operation of multiplication, the rational number 34 43 4 · 3 = 3 · 4 = 1. 3 4 has inverse 4 3 since • Under the operation of function composition ln x has the inverse ex since ln(ex ) = eln x = x • Under the operation of modular addition, [3] has the inverse [−3] in Z7 since [3] + [−3] = [−3] + [3] = [0]. When the operation is addition, we usually denote the inverse by −a. Otherwise, we typically denote the inverse of a by a−1 . This does cause confusion. Many students interpret a−1 as the reciprocal. This works for real or rational multiplication but fails in other contexts like function composition. We will use −a to mean the inverse of a under addition and a−1 to mean the inverse under all other operations. 116 Chapter 14 14.2.4 Modular Arithmetic Subtraction in Zm Let’s return to Zm . The identity under addition in Zm is [0] since ∀[a] ∈ Zm , [a] + [0] = [a] Given any [a] ∈ Zm , [−a] exists and [a] + [−a] = [a − a] = [0] That is, every element [a] ∈ Zm has an additive inverse, [−a]. This allows us to define subtraction in Zm . Definition 14.2.5 We will define subtraction as addition of the inverse. Thus Subtraction [a] − [b] = [a] + [−b] = [a − b] 14.2.5 Division in Zm Division is related to multiplication in the same way that subtraction is related to addition. So first, we must identify the multiplicative identity in Zm . Since ∀[a] ∈ Zm , [a][1] = [a] we know that [1] is the identity under multiplication in Zm . Inverses are more problematic with multiplication. Looking at the multiplication table for Z5 we see that [2]−1 = [3] since [2][3] = [6] = [1]. But what is the inverse of [2] in Z4 ? It doesn’t exist! Looking at the row containing [2] in the multiplication table for Z4 we cannot find [1]. Unlike addition in Zm where every element has an additive inverse, it is not always the case that a non-zero element in Zm has a multiplicative inverse. We define division analogously to subtraction. Definition 14.2.6 Division Division by a ∈ Zm is defined as multiplication by the multiplicative inverse of a ∈ Zm , assuming that the multiplicative inverse exists. 14.3 Fermat’s Little Theorem Pierre de Fermat conjectured what we now call Fermat’s Last Theorem. He also proved a much smaller but extremely useful result called Fermat’s Little Theorem. Theorem 1 (Fermat’s Little Theorem (F T)) If p is a prime number that does not divide the integer a, then ap−1 ≡ 1 (mod p) Section 14.3 Fermat’s Little Theorem 117 A proof of Fermat’s Little Theorem is available in the Appendix. Add proof of F T for the appendix. We will examine two corollaries. Corollary 2 For any integer a and any prime p ap ≡ a (mod p) Proof: Let a ∈ Z and let p be a prime. If p a, then ap−1 ≡ 1 (mod p). Multiplying both sides of the equivalence by a gives ap ≡ a (mod p). If p | a, then a ≡ 0 (mod p) and ap ≡ 0 (mod p). Thus ap ≡ a (mod p). Let’s make sure we understand the proof. Analysis of Proof There are two important items to note: the use of nested quantifiers in the hypothesis and the use of cases in the proof. Hypothesis: a ∈ Z, p is a prime Conclusion: ap ≡ a (mod p) Core Proof Technique: Select Method Preliminary Material: Fermat’s Little Theorem Sentence 1 Let a ∈ Z and let p be a prime. The hypotheses contain two universal quantifiers, so we use the Select Method twice, once for integers and once for primes. Sentence 2 If p a, then ap−1 ≡ 1 (mod p). The author breaks up the proof into two parts depending on whether or not p divides a. The author will need two distinct cases because the approach differs based on the case. In the case where p does not divide a, the author uses F T. Sentence 3 Multiplying both sides of the equivalence by a gives ap ≡ a (mod p). This is just modular arithmetic. Sentence 4 If p | a, then a ≡ 0 (mod p) and ap ≡ 0 (mod p). Thus ap ≡ a (mod p). This is the second case where p does divide a. Both ap and a are congruent to zero mod p so they are congruent to each other. Corollary 3 (Existence of Inverses in Zp (INV Zp )) Let p be a prime number. If [a] is any non-zero element in Zp , then there exists an element [b] ∈ Zp so that [a] · [b] = [1] This corollary is equivalent to stating that every non-zero element of Zp has an inverse. Let’s discover a proof. As usual, we begin by identifying the hypothesis and the conclusion. Hypothesis: p is a prime number. [a] is any non-zero element in Zp . 118 Chapter 14 Modular Arithmetic Conclusion: There exists an element [b] ∈ Zp so that [a] · [b] = [1]. Three points are salient. First, the corollary only states that an inverse exists. It doesn’t tell us what the inverse is or how to compute the inverse. Second, there are three quantifiers. 1. Let p be a prime number is equivalent to For all primes p. Since this is an instance of a universal quantifier we would expect to use the Choose Method. 2. [a] is any non-zero element in Zp is another instance of a universal quantifier so we would expect to use the Choose Method again. 3. There is an existential quantifier in the conclusion so we would expect to use the Construct Method. Together these give us the following. Proof in Progress 1. Let p be a prime number. 2. Let [a] be a non-zero element in Zp . 3. Construct [b] as follows. 4. To be completed. The third salient point is that this statement is a corollary of Fermat’s Little Theorem. Now Fermat’s Little Theorem uses congruences, not congruence classes. But we could restate F T with congruence classes as Theorem 4 (Fermat’s Little Theorem (F T)) If p is a prime number that does not divide the integer a, then [ap−1 ] = [1] in Zp Now an analogy to real numbers provides the final step. In the reals ap−1 = a · ap−2 so why not let [b] = [ap−2 ]? This would give Proof in Progress 1. Let p be a prime number. 2. Let [a] be a non-zero element in Zp . 3. Consider [b] = [ap−2 ]. 4. To be completed. Now we can invoke Fermat’s Little Theorem but first we need to make sure the hypotheses are satisfied. Proof in Progress Section 14.3 Fermat’s Little Theorem 119 1. Let p be a prime number. 2. Let [a] be a non-zero element in Zp . 3. Consider [b] = [ap−2 ]. 4. Since [a] = [0] in Zp , p a and so by F T [a][b] = [a][ap−2 ] = [ap−1 ] = [1] A proof might look as follows. Proof: Let p be a prime number. Let [a] be a non-zero element in Zp . Consider [b] = [ap−2 ]. Since [a] = [0] in Zp , p a and so by F T [a][b] = [a][ap−2 ] = [ap−1 ] = [1] REMARK In summary, if p is a prime number and [a] is any non-zero element in Zp , then [a]−1 = [ap−2 ] Exercise 2 What is [3]−1 in Z7 ? Chapter 15 Linear Congruences 15.1 Objectives The content objectives are: 1. Define a linear congruence in the variable x. 2. State and prove the Linear Congruence Theorem. 3. Do examples. 15.2 The Problem One of the advantages of congruence over divisibility is that we have an “arithmetic” of congruence. This allows us to solve new kinds of “equations”. Definition 15.2.1 Linear Congruence A relation of the form ax ≡ c (mod m) is called a linear congruence in the variable x. A solution to such a linear congruence is an integer x0 so that ax0 ≡ c (mod m) The problem for this lecture is to determine when linear congruences have solutions and how to find them. Recalling our table of statements equivalent to a ≡ b (mod m) we see that ax0 ≡ c (mod m) if and only if there exists an integer y0 such that ax0 + my0 = c 120 Section 15.2 The Problem 121 REMARK Thus ax ≡ c (mod m) has a solution ⇐⇒ there exists an integer x0 such that ax0 ≡ c (mod m) ⇐⇒ there exists an integer y0 such that ax0 + my0 = c ⇐⇒ gcd(a, m) | c (by the Linear Diophantine Equation Theorem, Part 1) Moreover, the Linear Diophantine Equation Theorem, Part 2 tells us what the solutions to ax + by = c look like. Theorem 1 (Linear Diophantine Equation Theorem, Part 2, (LDET 2)) Let gcd(a, m) = d = 0. If x = x0 and y = y0 is one particular integer solution to the linear Diophantine equation ax + my = c, then the complete integer solution is x = x0 + m a n, y = y0 − n, ∀n ∈ Z. d d But then, if x0 ∈ Z is one solution to ax ≡ c (mod m) the complete solution will be m x ≡ x0 (mod ) where d = gcd(a, m) d Equivalently, m m m x ≡ x0 , x0 + , x0 + 2 , · · · , x0 + (d − 1) (mod m) d d d Take note that there are d = gcd(a, m) distinct solutions modulo m. We record this discussion as the following theorem. Theorem 2 (Linear Congruence Theorem, Version 1, (LCT 1)) Let gcd(a, m) = d = 0. The linear congruence ax ≡ c (mod m) has a solution if and only if d | c. Moreover, if x = x0 is one particular solution, then the complete solution is x ≡ x0 (mod m ) d or, equivalently, x ≡ x0 , x0 + m m m , x0 + 2 , · · · , x0 + (d − 1) d d d (mod m) 122 Chapter 15 Linear Congruences Another way of considering the same problem is to reframe it in Zm . Since [a] = {x ∈ Z | x ≡ a (mod m)} solving ax ≡ c (mod m) is equivalent to finding a congruence class [x0 ] ∈ Zm that solves [a][x] = [c] in Zm Thus Theorem 3 (Linear Congruence Theorem, Version 2, (LCT 2)) Let gcd(a, m) = d = 0. The equation [a][x] = [c] in Zm has a solution if and only if d | c. Moreover, if x = x0 is one particular solution, then the complete solution is [x0 ] , x0 + 15.3 m m m , x0 + 2 , · · · , x0 + (d − 1) d d d in Zm Extending Equivalencies Putting all of this together we have several views of the same problem. REMARK [a][x] = [c] has a solution in Zm ⇐⇒ ax ≡ c (mod m) has a solution ⇐⇒ there exists an integer x0 such that ax0 ≡ c (mod m) ⇐⇒ there exists an integer y0 such that ax0 + my0 = c ⇐⇒ gcd(a, m) | c Moreover, if x0 , y0 is a particular integer solution to ax + my = c then a m n, y = y0 − n, ∀n ∈ Z d d m ⇐⇒ the complete solution to ax ≡ c (mod m) is x ≡ x0 (mod ) d m m m ⇐⇒ the complete solution to ax ≡ c (mod m) is x ≡ x0 , x0 + , x0 + 2 , · · · , x0 + (d − 1) (mod m) d d d m m m ⇐⇒ the complete solution to [a][x] = [c] in Zm is [x0 ] , x0 + , x0 + 2 , · · · , x0 + (d − 1) in Z d d d the complete solution to ax + my = c is x = x0 + Section 15.4 Examples 15.4 Example 1 123 Examples If possible, solve the linear congruence 3x ≡ 5 (mod 6) Solution: Since gcd(3, 6) = 3 and 3 5, there is no solution to 3x ≡ 5 (mod 6) by the Linear Congruence Theorem, Version 1. Example 2 If possible, solve the linear congruence 4x ≡ 6 (mod 10) Solution: Since gcd(4, 10) = 2 and 2 | 6, we would expect to find two solutions to 4x ≡ 6 (mod 10). Since ten is a small modulus, we can simply test all possibilities modulo 10. x (mod 10) 4x (mod 10) 0 0 1 4 2 8 3 2 4 6 5 0 6 4 7 8 8 2 9 6 Hence, x ≡ 4 or 9 (mod 10). Example 3 If possible, solve the linear congruence 3x ≡ 5 (mod 76) Solution: Since gcd(3, 76) = 1 and 1 | 5, we would expect to find one solution to 3x ≡ 5 (mod 76). We could try all 76 possibilities but there is a more efficient way. Thinking of our list of equivalencies, solving 3x ≡ 5 (mod 76) is equivalent to solving 3x + 76y = 5 and that we know how to do that using the Extended Euclidean Algorithm. x y rq 1 0 76 0 0 1 30 1 −25 1 25 −3 −76 0 3 From the second last row, 76(1) + 3(−25) = 1, or to match up with the order of the original equation, 3(−25) + 76(1) = 1. Multiplying the equation by 5 gives 3(−125) + 76(5) = 5. Hence x ≡ −125 ≡ 27 (mod 76) We can check our work by substitution. 3 · 27 ≡ 81 ≡ 5 (mod 76). 124 Chapter 15 Example 4 Linear Congruences Find the inverse of [13] in Z29 . Solution: By definition, the inverse of [13] in Z29 is the congruence class [x] so that [13][x] = [1] in Z29 . Since gcd(13, 29) = 1, we know by the Linear Congruence Theorem, Version 2 that there is exactly one solution. We could try all 29 possibilities or recall that solving [13][x] = [1] in Z29 is equivalent to solving 13x + 29y = 1 and that we know how to do using the Extended Euclidean Algorithm. x y rq 1 0 29 0 0 1 13 0 1 −2 3 2 −4 9 14 13 −29 0 3 From the second last row, 29(−4) + 13(9) = 1, or to match up with the order of the original equation, 13(9) + 29(−4) = 1. Hence [13]−1 = [9] in Z29 We can check our work by substitution. [13][9] = [117] = [1] in Z29 . Chapter 16 Chinese Remainder Theorem 16.1 Objectives The content objectives are: 1. Do examples. 2. Discover a proof of the Chinese Remainder Theorem. 16.2 An Old Problem The following problem was posed, likely in the third century CE, by Sun Zi in his Mathematical Manual and republished in 1247 by Qin Jiushao in the Mathematical Treatise in Nine Sections. There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder is 2. What will be the number? The word problem asks us to find an integer n that simultaneously satisfies the following three linear congruences. n≡2 (mod 3) n≡3 (mod 5) n≡2 (mod 7) Before we solve this problem, we will begin with two simultaneous congruences whose moduli are coprime. 125 126 Chapter 16 16.3 Example 1 Chinese Remainder Theorem Chinese Remainder Theorem Solve n≡2 (mod 5) n≡9 (mod 11) Solution: The first congruence is equivalent to n = 5x + 2 where x ∈ Z (16.1) Substituting this into the second congruence we get 5x + 2 ≡ 9 (mod 11) ⇒ 5x ≡ 7 (mod 11) Have we seen anything like this before? Of course, this is just a linear congruence! Its solution is x ≡ 8 (mod 11) Now x ≡ 8 (mod 11) is equivalent to x = 11y + 8 where y ∈ Z (16.2) Substituting Equation 16.2 into Equation 16.1 gives the solution n = 5(11y + 8) + 2 = 55y + 42 for all y ∈ Z which is equivalent to n ≡ 42 (mod 55) We can check by substitution. If n = 55y + 42, then n ≡ 2 (mod 5) and n ≡ 9 (mod 11). Theorem 1 (Chinese Remainder Theorem (CRT)) If gcd(m1 , m2 ) = 1, then for any choice of integers a1 and a2 , there exists a solution to the simultaneous congruences n ≡ a1 (mod m1 ) n ≡ a2 (mod m2 ) Moreover, if n = n0 is one integer solution, then the complete solution is n ≡ n0 (mod m1 m2 ) Before we begin our discovery of a solution, let’s be clear that there are two things to prove. First, that a solution exists and second, what a complete solution looks like. With regards to the first part let’s identify, as usual, the hypothesis and the conclusion. Hypothesis: gcd(m1 , m2 ) = 1. Section 16.3 Chinese Remainder Theorem 127 Conclusion: For any choice of integers a1 and a2 , there exists a solution to the simultaneous congruences n ≡ a1 (mod m1 ) n ≡ a2 (mod m2 ) Since there is an existential quantifier in the conclusion, we have to construct a solution. There is nothing obvious from the statement of the theorem that will help us, but we have already solved such a problem once in Example 1. Perhaps we could mimic what we did there. From the first linear congruence The integer n satisfies n ≡ a1 (mod m1 ) if and only if n = a1 + m1 x for some x ∈ Z The next thing we did was substitute this expression into the second congruence. The number n satisfies the second congruence if and only if a1 + m 1 x ≡ a2 (mod m2 ) m 1 x ≡ a2 − a1 (mod m2 ) Have we seen anything like this before? Of course, this is just a linear congruence! Since gcd(m1 , m2 ) = 1, the Linear Congruence Theorem tells us that this congruence has a solution, say x = b and that the complete solution is x = b + m2 y for all y ∈ Z If we set y = 0 we get x = b and hence n = a1 + m1 b is one particular solution. Now let’s consider the second part, a complete solution. Following on what we have done above, an integer n satisfies the simultaneous congruences if and only if n = a1 + m 1 x = a1 + m1 (b + m2 y ) = (a1 + m1 b) + m1 m2 y for all y ∈ Z But these are the elements of exactly one congruence class modulo m1 m2 . Hence, if n = n0 is one solution, then the complete solution is n ≡ n0 Exercise 1 (mod m1 m2 ) Using the analysis above, write a proof for the Chinese Remainder Theorem. Question Panel 128 Chapter 16 Exercise 2 Solve n≡2 (mod 3) n≡3 (mod 5) n≡2 (mod 3) n≡3 (mod 5) n≡4 Exercise 3 Chinese Remainder Theorem (mod 11) Solve The exercise above makes it clear that we can solve more than two simultaneous linear congruences simply by solving pairs of linear congruences successively. We record this as Theorem 2 (Generalized Chinese Remainder Theorem (GCRT)) If m1 , m2 , . . . , mk ∈ Z and gcd(mi , mj ) = 1 whenever i = j , then for any choice of integers a1 , a2 , . . . , ak , there exists a solution to the simultaneous congruences n ≡ a1 (mod m1 ) n ≡ a2 . . . (mod m2 ) n ≡ ak (mod mk ) Moreover, if n = n0 is one integer solution, then the complete solution is n ≡ n0 (mod m1 m2 . . . mk ) Chapter 17 Practice, Practice, Practice: Congruences 17.1 Objectives The content objectives are: 1. Computational practice. 2. Preparing for RSA. 17.2 Linear and Polynomial Congruences Let’s recall how to solve linear congruences. Example 1 Solve 13x ≡ 1 (mod 60). Solution: Since gcd(13, 60) = 1 and 1 | 1 we would expect to find one congruence class as a solution to 13x ≡ 1 (mod 60). Now 13x ≡ 1 (mod 60) is equivalent to the linear Diophantine equation 13x+60y = 1 so we can use the EEA. (Note that we have interchanged the labels for x and y . The statement of the algorithm finds a solution to ax + by = gcd(a, b) where a > b. In this particular case, b = 60 > 13 = a. Because we have interchanged a and b, we also interchange x and y .) y x rq 1 0 60 0 0 1 13 0 1 −4 8 4 −1 5 51 2 −9 3 1 −3 14 21 5 −23 1 1 −13 60 02 Thus 13(−23)+60(5) = 1 and so x ≡ −23 ≡ 37 (mod 60) is a solution to 13x ≡ 1 (mod 60). 129 130 Chapter 17 Practice, Practice, Practice: Congruences Though we have efficient means to solve linear congruences, we have no equivalent means to solve polynomial congruences. Example 2 Solve x2 ≡ 1 (mod 8) by substitution. Your first reaction might be that there are zero, one or two solutions as there would be in the reals. Solution: x (mod 8) x2 (mod 8) 0 0 1 1 2 4 3 1 4 0 5 1 6 4 7 1 Hence, the solution is x ≡ 1, 3, 5 or 7 (mod 8). Example 3 Solve 36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2 (mod 5). Reduce terms and use Fermat’s Little Theorem or its corollaries before substitution. Solution: Since 36 ≡ 1 (mod 5) the term 36x47 reduces to x47 (mod 5). Since 5 ≡ 0 (mod 5) the term 5x9 reduces to 0 (mod 5). Thus, 36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2 (mod 5) reduces to x47 + x3 + x2 + x + 1 ≡ 2 (mod 5) By Fermat’s Little Theorem, x4 ≡ 1 (mod 5) and so x47 ≡ (x4 )11 x3 ≡ 111 x3 ≡ x3 (mod 5) and the polynomial congruence further reduces to x3 + x3 + x2 + x + 1 ≡ 2 (mod 5) or, more simply, 2x3 + x2 + x + 1 ≡ 2 x (mod 5) 2x3 + x2 + x + 1 (mod 5) (mod 5) 0 1 1 0 2 3 3 2 4 4 Hence, the only solution to 36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2 is x≡3 Example 4 Solve n3 ≡ 127 (mod 165). (mod 5) (mod 5) Section 17.2 Linear and Polynomial Congruences 131 Solution: We could try all 165 possibilities but perhaps there is another way. Observing that 165 = 3 × 5 × 11 and all three factors are relatively prime as pairs, maybe we could split the problem into three linear congruences and then apply the Chinese Remainder Theorem. Unfortunately, the polynomial is not linear. Let’s see what happens anyway. Since n3 ≡ 127 (mod 165), n3 ≡ 127 ≡ 1 (mod 3) 3 (mod 5) 3 (mod 11) n ≡ 127 ≡ 2 n ≡ 127 ≡ 6 Let’s consider each of the three congruences separately. In the case n3 ≡ 1 (mod 3) we can use a corollary to Fermat’s Little Theorem. Since n3 ≡ n (mod 3) by FlT, n3 ≡ 1 (mod 3) reduces to n ≡ 1 (mod 3) which is just the solution to the first congruence. For the case n3 ≡ 2 (mod 5) we will use a table. n (mod 5) n3 (mod 5) 0 0 1 1 2 3 3 2 4 4 6 7 7 2 The only solution to n3 ≡ 2 (mod 5) is n ≡ 3 (mod 5) For the case n3 ≡ 6 (mod 11) we will use a table. n (mod 11) n3 (mod 11) 0 0 1 1 2 8 3 5 4 9 5 4 8 6 9 3 10 10 The only solution to n3 ≡ 6 (mod 11) is n ≡ 8 (mod 11) Hence, a solution to n3 ≡ 127 (mod 165) can be found by solving the simultaneous linear congruences n≡1 (mod 3) n≡3 (mod 5) n≡8 (mod 11) Though these could be solved by eye (note that n ≡ 8 (mod 55) is a solution to the last two) we will solve these, for practice, by writing out and substituting equations. From n ≡ 1 (mod 3) we have n = 3x + 1 where x ∈ Z (17.1) Substituting into the second equation we get 3x + 1 ≡ 3 (mod 5) ⇒ 3x ≡ 2 (mod 5) ⇒ x ≡ 4 (mod 5) Now x ≡ 4 (mod 5) is equivalent to x = 5y + 4 where y ∈ Z (17.2) 132 Chapter 17 Practice, Practice, Practice: Congruences Substituting Equation 17.2 into Equation 17.1 gives the solution to the first two linear congruences. n = 3(5y + 4) + 1 = 15y + 13 for all y ∈ Z which is equivalent to n ≡ 13 (mod 15) n ≡ 13 (mod 15) Now we need to solve n≡8 (mod 11) From n ≡ 13 (mod 15) we have n = 15x + 13 where x ∈ Z (17.3) Substituting into the second equation we get 15x +13 ≡ 8 (mod 11) ⇒ 4x +2 ≡ 8 (mod 11) ⇒ 4x ≡ 6 (mod 11) ⇒ x ≡ 7 (mod 11) Now x ≡ 7 (mod 11) is equivalent to x = 11y + 7 where y ∈ Z (17.4) Substituting Equation 17.4 into Equation 17.3 gives the solution. n = 15(11y + 7) + 13 = 165y + 118 for all y ∈ Z which is equivalent to n ≡ 118 (mod 165) and which is the solution to the original problem n3 ≡ 127 (mod 165). Checking we have n2 ≡ 1182 ≡ 64 (mod 165) and n3 ≡ 118 × 64 ≡ 127 (mod 165). Example 5 Determine, with justification all solutions of the congruence equation x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 143) Solution: We could simply try all 143 distinct values modulo 143. However, computing numbers like 7061 might be problematic. Have we seen anything like this before? Note that 143 = 11 × 13. The previous question had a polynomial on the left and a composite modulus on the right so perhaps we could do now what we did in the previous exercise, break up the larger problem into several smaller problems. If x0 is a solution to x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 143) then x0 is also a solution to x61 + 26x41 + 11x25 + 5 ≡ 0 61 x + 26x 41 + 11x 25 +5≡0 (mod 11) (17.5) (mod 13) (17.6) Section 17.3 Linear and Polynomial Congruences 133 Let’s start with the polynomial congruence 17.5. x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 11) The most obvious thing to do is reduce each term modulo 11. This gives x61 + 4x41 + 5 ≡ 0 (mod 11) Since 11 is prime, as long as 11 x0 , we can use F T which implies that x10 ≡ 1 (mod 11) So then x61 ≡ x60 x1 ≡ (x10 )6 x1 ≡ 16 x1 ≡ x (mod 11) Similarly, x41 ≡ x (mod 11) and so the congruence reduces to x + 4x + 5 ≡ 0 (mod 11) or 5x ≡ −5 (mod 11) By Congruences and Division x ≡ −1 ≡ 10 (mod 11) We still have to deal with the possibility that 11 | x0 . If 11 did divide x0 , then 11 would be a solution to x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 11) Replacing x by 0 in the above equation, since 11 ≡ 0 (mod 11), gives 5 ≡ 0 (mod 11) which is false. So 11 x0 . Similarly, x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 13) reduces to Insert Question Panel. 12x + 5 ≡ 0 (mod 13) or, since 12 ≡ −1 (mod 13) −x ≡ −5 (mod 13) which has the solution x≡5 (mod 13) Note again that 13 x0 . But now we have two simultaneous linear congruences x ≡ 10 x≡5 (mod 11) (mod 13) Have you seen anything like this before? Insert Question Panel. 134 Chapter 17 17.3 Practice, Practice, Practice: Congruences Preparing for RSA This exercise will help us understand the implementation of the RSA scheme which we will look at next. In commercial practice the numbers chosen are large but here, choose numbers small enough to work with by hand. I will give an example. You follow along but use your own numbers. 1. Choose two distinct primes p and q and let n = pq . I will choose p = 7 and q = 11 so n = 77. 2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1). I will choose e = 13 which satisfies gcd(13, 60) = 1 and 1 < 13 < 60. 3. Solve ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1). In my case d = 37. Chapter 18 The RSA Scheme 18.1 Objectives The content objectives are: 1. Illustrate the use of RSA. 2. Prove that the message sent will be the message received. 18.2 Why Public Key Cryptography? In a private key cryptographic scheme, like the substitution cipher or Vigen`re cipher that e you have already learned about, participants share a common key. This raises the problem of how to distribute a large number of keys between users, especially if these keys need to be changed frequently. For example, there are almost 200 countries in the world. If Canada maintains an embassy in each country and allows Canadian embassies to communicate with one another, the embassies must exchange a common key between each pair of embassies. 00 That means there are 22 = 19, 900 keys to exchange. Worse yet, for security reasons, keys should be changed frequently and so 19, 900 keys might need to be exchanged daily. In a public key cryptographic scheme, keys are divided into two parts: a public encryption key which is shared in an open repository of some sort, and a private decryption key held secretly by each participant. For user A to send a private message to user B , A would look up B ’s public key, encrypt the message and send it to B . Since B is the only person who possesses the secret key required for decryption, only B can read the message. Such an arrangement solves the key distribution problem. The public keys do not need to be kept secret and only one per participant needs to be available. Thus, in our embassy example previously, only 200 keys need to be published. The possibility of public key cryptography was first published in 1976 in a paper by Diffie, Hellman and Merkle. The RSA scheme, named after its discoverers Rivest, Shamir and Adleman is an example of a commercially implemented public key scheme. RSA is now widely deployed. • Add list here. 135 136 Chapter 18 18.3 The RSA Scheme Implementing RSA In RSA, messages are integers. How does one get an integer from plaintext? In much the same way we did with a Vigen`re cipher, assign a number to each letter of the alphabet e and then concatenate the digits together. 18.3.1 Setting up RSA 1. Choose two large, distinct primes p and q and let n = pq . 2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1). 3. Solve ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1). 4. Publish the public encryption key (e, n). 5. Keep secure the private decryption key (d, n). 18.3.2 Sending a Message To send a message: 1. Look up the recipient’s public key (e, n). 2. Generate the integer message M so that 0 ≤ M < n. 3. Compute the ciphertext C as follows: Me ≡ C (mod n) where 0 ≤ C < n 4. Send C . 18.3.4 Example All of the computation in this part was done in Maple. Setting up RSA 1. Choose two large, distinct primes p and q and let n = pq . Let p be 9026694843 0929817462 4847943076 6619417461 5791443937, and let q be 7138718791 1693596343 0802517103 2405888327 6844736583 so n is 6443903609 8539423089 8003779070 0502485677 Section 18.3 Implementing RSA 18.3.3 137 Receiving a Message To decrypt a message: 1. Use your private key (d, n). 2. Compute the messagetext R from the ciphertext C as follows: Cd ≡ R (mod n) where 0 ≤ R < n 3. R is the original message. 1034536315 4526254586 6290164606 1990955188 1922989980 3977447271. 2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1). Now (p − 1)(q − 1) is 6443903609 8539423089 8003779070 0502485677 1034536313 8360840952 3666750800 6340495008 2897684191 1341266752. Choose e as 9573596212 0300597326 2950869579 7174556955 8757345310 2344121731. It is indeed the case that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1). 3. Solve ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1). Solving this LDE gives d as 5587652122 6351022927 9795248536 5522717791 7285682675 6100082011 1849030646 3274981250 2583120946 4072548779. 4. Publish the public encryption key (e, n). 5. Keep secure the private decryption key (d, n). Sending a Message To send a message: 1. Look up the recipient’s public key (e, n). 2. Generate the integer message M so that 0 ≤ M < n. We will let M = 3141592653. 3. Compute the ciphertext C as follows: Me ≡ C (mod n) where 0 ≤ C < n 138 Chapter 18 Computing gives C 4006696554 3080815610 2814019838 8509626485 8151054441 5245547382 5506759308 1333888622 4491394825 3742205367. 4. Send C . Receiving a Message To decrypt a message: 1. Use your private key key (d, n). 2. Compute the messagetext R from the ciphertext C as follows: Cd ≡ R 3. R is the original message. R = 3141592653. (mod n) where 0 ≤ R < n The RSA Scheme Section 18.4 Does M = R? 18.4 139 Does M = R? Are we confident that the message sent is the message received? Theorem 1 (RSA) If 1. p and q are distinct primes, 2. n = pq 3. e and d are positive integers such that ed ≡ 1 (mod (p − 1)(q − 1)), 4. 0 ≤ M < n 5. M e ≡ C (mod n) 6. C d ≡ R (mod n) where 0 ≤ R < n then R = M . The proof is long and can appear intimidating but, in fact, it is structurally straightforward if we break it into pieces. The proof is done in four parts. 1. Write R as a function of M , specifically R ≡ M M k(p−1)(q−1) (mod n) 2. Show that R ≡ M (mod p). We will do this in two cases: (i) p M and (ii) p | M . 3. Show that R ≡ M (mod q ). 4. Use the Chinese Remainder Theorem to deduce that R = M . Proof: First, we will show that R ≡ M M k(p−1)(q−1) (mod n) Since ed ≡ 1 (mod (p − 1)(q − 1)), there exists an integer k so that ed = 1 + k (p − 1)(q − 1) Now R ≡ Cd (mod n) ed ≡ (M ) ≡M ed (mod n) (mod n) ≡ M 1+k(p−1)(q−1) (mod n) k(p−1)(q −1) (mod n) ≡ MM 140 Chapter 18 Second, we will show that R ≡ M (mod p). Suppose that p Theorem, M p−1 ≡ 1 (mod p) The RSA Scheme M . By Fermat’s Little Hence M k(p−1)(q−1) ≡ (M p−1 )k(q−1) ≡ 1k(q−1) ≡1 (mod p) (mod p) (mod p) Multiplying both sides by M gives M M k(p−1)(q−1) ≡ M (mod p) Since R ≡ M M k(p−1)(q−1) (mod n) ⇒ R ≡ M M k(p−1)(q−1) (mod p) we have R≡M (mod p) Now suppose that p | M . But then M ≡ 0 (mod p) and so M M k(p−1)(q−1) ≡ 0 (mod p). That is, M M k(p−1)(q−1) ≡ M (mod p) Again, since R ≡ M M k(p−1)(q−1) (mod n) ⇒ R ≡ M M k(p−1)(q−1) (mod p) we have R≡M (mod p) In either case, we have R ≡ M (mod p). Third, we will show that R ≡ M (mod q ). But this similar to R ≡ M (mod p). Fourth and last, we will show that R = M . So far we have generated two linear congruences that have to be satisfied simultaneously. R≡M (mod p) R≡M (mod q ) Since gcd(p, q ) = 1 we can invoke the Chinese Remainder Theorem and conclude that R≡M (mod pq ) R≡M (mod n) Since pq = n we have Now, R and M are both integers congruent to each other modulo n, and both lie between 0 and n − 1, so R = M . 18.5 How Secure Is RSA? The basic idea behind RSA is that multiplying is easy and factoring is difficult. Hence it is easy to generate n, which is part of the key, and difficult to factor a large n, say 200 digits, into p and q which would make it easy to decrypt any message. Complete this. Chapter 19 Negation 19.1 Objectives The technique objectives are: 1. To learn how to negate systems. 2. To learn when to use counter-examples. 3. To practice finding counter-examples. 19.2 Negating Statements You will frequently encounter the negation of statement A. Definition 19.2.1 Negation The negation of the statement A is the statement NOT A. Because statements cannot be both true and false, exactly only one of A and NOT A can be true. In some instances, finding the negation of a statement is easy. For example A: f (x) has a real root. NOT A: f (x) does not have a real root. When A is already negated, a truth table tells us what to do. A ¬A ¬(¬A) TF T FT F Thus, ¬(¬A) = A. Two negatives are a positive, or equivalently, one NOT cancels another NOT. For example, A: 7 is not a divisor of 28. NOT A: 7 is a divisor of 28. 141 142 Chapter 19 Negation You have already seen DeMorgan’s Laws when we worked with truth tables. Proposition 1 (De Morgan’s Law’s (DML)) If A and B are statements, then 1. ¬(A ∨ B ) ≡ (¬A) ∧ (¬B ) 2. ¬(A ∧ B ) ≡ (¬A) ∨ (¬B ) REMARK Thus, there is a specific rule applied when negating a statement containing the word AND. A: B AND C NOT A: (NOT B ) OR (NOT C ) Note that the connecting word has changed from AND to OR and that each term in the expression has been negated. The brackets are not needed because NOT precedes OR in logical evaluation, but the brackets are useful to emphasize the change. Here is a specific example. For example, A: T is isosceles and it has perimeter 42. NOT A: T is not isosceles or it does not have perimeter 42. REMARK Similar to the conjunctive AND, a specific rule is applied when negating a statement containing the word OR. A: B OR C NOT A: (NOT B ) AND (NOT C ) Note that the connecting word has changed from OR to AND and, again, each term in the expression has been negated. As before, the brackets are not needed because NOT precedes AND in logical evaluation, but the brackets are useful to emphasize the change. Here is a specific example. For example, A: T is isosceles or it has perimeter 42. NOT A: T is not isosceles and it does not have perimeter 42. A: T is isosceles or it has perimeter 42. NOT A: T is not isosceles and it does not have perimeter 42. Section 19.3 Negating Statements with Quantifiers 19.3 143 Negating Statements with Quantifiers Negating statements that contains quantifiers is more complicated. We first observe that: • The negation of a universal statement results in an existential statement. • The negation of an existential statement results in a universal statement. REMARK A statement with an existential quantifier looks like There exists an x in the set S such that P (x) is true. Its negation is For every x in the set S , P (x) is false. REMARK A statement with a universal quantifier looks like For every x in the set S , P (x) is true. Its negation is There exists an x in the set S such that P (x) is false. REMARK To negate a statement using nested quantifiers, do the following. Step 1 Put the word NOT in front of the entire statement. Step 2 Move the NOT from left to right replacing quantifiers by their opposites and in each case place the NOT just before the open sentence. Repeat until there are no quantifiers to the right of NOT. Step 3 When all of the quantifiers are to the left of NOT, incorporate the NOT into the open sentence. Let’s do some examples. 144 Chapter 19 Negation Example 1 1. For every x ∈ S , f (x) = 0. (a) NOT [For every x ∈ S , f (x) = 0.] (b) There exists x ∈ S such that NOT [f (x) = 0]. (c) There exists x ∈ S such that f (x) = 0. 2. There exists x ∈ S such that f (x) = 0. (a) NOT [There exists x ∈ S such that f (x) = 0.] (b) For every x ∈ S , NOT [f (x) = 0]. (c) For every x ∈ S , f (x) = 0. 3. For every x ∈ S and for every f ∈ F , f (x) = 0. (a) NOT [For every x ∈ S and for every f ∈ F , f (x) = 0.] (b) There exists x ∈ S such that NOT [for every f ∈ F , f (x) = 0]. There exists x ∈ S and there exists f ∈ F such that NOT [f (x) = 0]. (c) There exists x ∈ S and there exists f ∈ F such that f (x) = 0. 4. There exists x ∈ S such that, for every f ∈ F , f (x) = 0. (a) NOT [There exists x ∈ S such that for every f ∈ F , f (x) = 0.] (b) For every x ∈ S , NOT [for every f ∈ F , f (x) = 0]. For every x ∈ S there exists a f ∈ F , NOT [f (x) = 0]. (c) For every x ∈ S there exists a f ∈ F , f (x) = 0. 19.3.1 Counterexamples So far in the course, we have worked on proving that statements are true. How do we prove that a statement is false? In principle, this is relatively easy. To show that the statement A is false, we only need to prove that the statement NOT A is true. Suppose A is the statement: A: For every x ∈ [−π, π ], sin(x) = 0. This statement is very similar to our first example. NOT A is the statement NOT A: There exists x ∈ [−π, π ] such that sin(x) = 0. In this case, NOT A is easy to prove using our construction method. If I consider x = 0, I know that 0 ∈ [−π, π ] and sin(x) = 1 = 0. The number 0 is a counterexample. Definition 19.3.1 Counterexample In general, if we wish to prove that a universal statement A is false, we show that its negation, which is an existential statement, is true. The particular object which we use to show that the existential statement is true is called a counterexample of statement A. Section 19.3 Negating Statements with Quantifiers 145 The same idea arises when we want to show that a statement of the form “A implies B ” is false. It is enough to show a particular instance where A is true and B is false, or equivalently NOT B is true. For example, consider the following statement. Statement 2 S : If a, b and c are integers, and a | (bc), then a | b and a | c. The hypothesis is A: a, b and c are integers, and a | (bc) and the conclusion is B : a | b and a | c. To show that S is false, we must find a specific instance where A is true and B is false. To show that B is false we must show that NOT B is true. NOT B : a b or a c. Choosing a = 3, b = 6 and c = 7 we have an instance where the hypothesis A is true (since 3 | 42) and the conclusion B is false, equivalently, NOT B is true. The values a = 3, b = 6 and c = 7 are a counterexample for S . Chapter 20 Contradiction 20.1 Objectives The technique objectives are: 1. Learn how to read and discover proofs by contradiction. The content objectives are: 1. Read a proof of Prime Factorization. 2. Discover a proof of Infinitely Many Primes. 20.2 How To Use Contradiction We have mostly used the Direct Method to discover proofs, often in conjunction with one of the methods associated with quantifiers. There are times when this is difficult. A proof by contradiction provides a new method. REMARK Suppose that we wish to prove that the statement “A implies B ” is true. We assume that A is true. We must show that B is true. What would happen if B were true, but we assumed it was false and continued our reasoning based on the assumption that B was false? Since a mathematical statement cannot be both true and false, it seems likely we would eventually encounter a mathematically non-sensical statement. Then we would ask ourselves “How did we arrive at this nonsense?” and the answer would have to be that our assumption that B was false was wrong and B is, in fact, true. REMARK A proof by contradiction of the statement “A implies B ” structures proofs in exactly this way. Proceed as follows. 146 Section 20.2 How To Use Contradiction 147 1. Assume that A is true. 2. Assume that B is false, or equivalently, assume that NOT B is true. 3. Reason forward from A and NOT B to reach a contradiction. Unfortunately, it is not always clear what contradiction to find, or how to find it. What is more clear is when to use contradiction. 20.2.1 When To Use Contradiction The general rule of thumb is to use contradiction when the statement NOT B gives you useful information. There are typically two instances when this is useful. The first instance is when the statement B is one of only two alternatives. For example, if the conclusion B is the statement f (x) = 0 then the only two possibilities are f (x) = 0 and f (x) = 0. NOT B is the statement f (x) = 0 which could be useful to you. The second instance is when B contains a negation. As we saw earlier, NOT B eliminates the negation. 20.2.2 Reading a Proof by Contradiction Suppose we want to prove the following proposition. Proposition 1 (Prime Factorization (PF)) If n is an integer greater than 1, then n can be expressed as a product of primes. Example 1 The integers 2, 3, 5 and 7 are primes and each is a product unto itself, that is, it is a product consisting of one factor. The integers 4 = 2 × 2, 6 = 2 × 3 and 8 = 2 × 2 × 2 have been factored as products of primes. Here is a proof. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let N be the smallest integer, greater than 1, that cannot be written as a product of primes. 2. N is not itself a prime, so we can write N = rs where 1 < r ≤ s < N . 3. Since r and s are less than N , they can be written as a product of primes. 4. But then it follows that N = rs can be written as a product of primes, a contradiction. Analysis of Proof An interpretation of sentences 1 through 4 follows. 148 Chapter 20 Contradiction Sentence 1 Let N be the smallest integer, greater than 1, that cannot be written as a product of primes. The first sentence of a proof by contradiction usually gives the specific form of NOT B that the author is going to work with. In this case, the author identifies that this is a proof by contradiction by assuming the existence of an object which contradicts the conclusion, an integer N which cannot be written as a product of primes. Moreover, of all such candidates for N the author chooses the smallest one. Though it may not be obvious when first encountering the proof why the author would stipulate such a condition, it always has to do with something needed later in the argument. Once you know that this is a proof by contradiction, look ahead to find the contradiction. In this case, the contradiction appears in Sentence 4. Sentence 2 N is not itself a prime, so we can write N = rs where 1 < r ≤ s < N . If N were prime, then N by itself is a product of primes (with just one factor). But the author has assumed that N is not a product of primes, hence N is composite and can be written as the product of two non-trivial factors r and s. Sentence 3 Since r and s are less than N , they can be written as a product of primes. This sentence makes it clear why N needs to be the smallest integer that cannot be written as a product of primes. In order to generate the contradiction, r and s must be written as products of primes. If it were the case that N was not the smallest such integer, it might be the case that neither r nor s could be written as a product of primes. Sentence 4 But then it follows that N = rs can be written as a product of primes, a contradiction. Since both r and s can be written as a product of primes, the product rs = N can certainly be written as a product of primes. But this contradicts the assumption in Sentence 1 that N cannot be written as a product of primes. Since our reasoning is correct, it must be the case that our assumption that there is an integer which cannot be written as a product of primes is incorrect. That is, every integer can be written as a product of primes. 20.2.3 Discovering and Writing a Proof by Contradiction Discovering a proof by contradiction can be difficult and often requires several attempts at finding the path to a contradiction. Let’s see how we might discover a proof to a famous theorem recorded by Euclid. Proposition 2 (Infinitely Many Primes (INF P)) The number of primes is infinite. We should always be clear about our hypothesis and conclusion. There is no explicit hypothesis in this case and the conclusion is the statement Conclusion: The number of primes is infinite. Section 20.2 How To Use Contradiction 149 This statement contains a negation, infinite is an abbreviation of not finite, and so is a candidate for a proof by contradiction. Our first statement in a proof by contradiction is a negation of the conclusion so we have Proof in Progress 1. Assume that the number of primes is finite. (This is NOT B.) 2. To be completed. Now comes the tough part. What do we do from here? How do we generate a contradiction? Well, if the number of primes is finite, could we somehow use that assumption to find a “new” prime not in our finite list of primes? Our candidate should not have any of the finite primes as a factor. At this point, it sounds like we need to list our primes. Proof in Progress 1. Assume that the number of primes is finite. (This is NOT B.) 2. Label the finite number of primes p1 , p2 , p3 , . . . , pn . 3. To be completed. Now we have a way to express a candidate for a “new” prime. Proof in Progress 1. Assume that the number of primes is finite. (This is NOT B.) 2. Label the finite number of primes p1 , p2 , p3 , . . . , pn . 3. Consider the integer N = p1 p2 p3 · · · pn + 1. 4. To be completed. Clearly N is larger than any of the pi so, by the first sentence, N cannot be in the list of primes. Thus Proof in Progress 1. Assume that the number of primes is finite. (This is NOT B.) 2. Label the finite number of primes p1 , p2 , p3 , . . . , pn . 3. Consider the integer N = p1 p2 p3 · · · pn + 1. 4. Since N > pi ∀i, N is not a prime. 5. To be completed. And this is where we can find our contradiction. N has no non-trivial factors since dividing N by any of the pi leaves a remainder of 1. But that means N cannot be written as a product of primes, which contradicts the previous proposition. The contradiction in this proof arises from a result which is inconsistent with something else we have proved. Proof in Progress 150 Chapter 20 Contradiction 1. Assume that the number of primes is finite. (This is NOT B.) 2. Label the finite number of primes p1 , p2 , p3 , . . . , pn . 3. Consider the integer N = p1 p2 p3 · · · pn + 1. 4. Since N > pi ∀i, N is not a prime. 5. Since N = pi q + 1 for each of the primes pi , no pi is a factor of N . Hence N cannot be written as a product of primes, which contradicts our previous proposition. Putting all of the statements together gives the following proof. Proof: Assume that there are only a finite number of primes, say p1 , p2 , p3 , . . . , pn . Consider the integer N = p1 p2 p3 · · · pn + 1. Since N > pi ∀i, N is not a prime. But N = pi q + 1 for each of the primes pi , so no pi is a factor of N . Hence N cannot be written as a product of primes, which contradicts our previous proposition. Chapter 21 Contrapositive 21.1 Objectives The technique objectives are: 1. Define the contrapositive. 2. Read a proof using the contrapositive. 3. Discover and write a proof using the contrapositive. 21.2 The Contrapositive We begin with an exercise. Exercise 1 Definition 21.2.1 Use truth tables to show that A ⇒ B ≡ ¬B ⇒ ¬A. The statement ¬B ⇒ ¬A is called the contrapositive of A ⇒ B . Contrapositive The logical equivalence between a statement and its contrapositive gives us another proof technique. Instead of proving “A implies B ” we prove “ NOT B implies NOT A” using any of the existing techniques. 21.2.1 When To Use The Contrapositive 151 152 Chapter 21 Contrapositive REMARK This is very similar to contradiction. Use the contrapositive when the statement NOT A or the statement NOT B gives you useful information. This is most likely to occur when A or B contains a negation or is one of two possible choices. When both A and B contain negations, it is highly likely that using the contrapositive will be productive. 21.3 Reading a Proof That Uses the Contrapositive Consider the following proposition. Proposition 1 Suppose a is an integer. If 32 ((a2 + 3)(a2 + 7)) then a is even. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. We will prove the contrapositive. 2. If a is odd we can write a as 2k + 1 for some integer k . 3. Substitution gives (a2 + 3)(a2 + 7) = ((2k + 1)2 + 3)((2k + 1)2 + 7) = (4k 2 + 4k + 1 + 3)(4k 2 + 4k + 1 + 7) = (4k 2 + 4k + 4)(4k 2 + 4k + 8) = 4(k 2 + k + 1) × 4(k 2 + k + 2) = 16(k 2 + k + 1)(k 2 + k + 2) 4. Since one of k 2 + k + 1 or k 2 + k + 2 must be even, and the last line above shows that a factor of 16 already exists disjoint from (k 2 + k + 1)(k 2 + k + 2), (a2 + 3)(a2 + 7) must contain a factor of 32. That is 32 | ((a2 + 3)(a2 + 7)). Analysis of Proof Since the hypothesis of the proposition contains a negation, and the conclusion is one of two possible choices, it makes sense to consider the contrapositive. Sentence 1 We will prove the contrapositive. Not all authors will be so obliging as to state the proof technique up front. The provided proof would also be correct if this sentence was omitted. Correct, but less easy to understand. As usual, we begin by identifying the hypothesis and the conclusion. Hypothesis: A: 32 ((a2 + 3)(a2 + 7)). Conclusion: B : a is even. For the contrapositive Section 21.3 Reading a Proof That Uses the Contrapositive 153 Hypothesis: NOT B : a is even. Conclusion: NOT A: 32 | ((a2 + 3)(a2 + 7)) How would we know that the author was using the contrapositive if this sentence were omitted? The clause “If a is odd” is NOT B so the author is using one of only two proof techniques that begin this way, contradiction or contrapositive. Looking ahead to the last line, we see that the author concludes with NOT A, so this is a proof of the contrapositive. Had the author concluded with a contradiction, we would know that this is a proof by contradiction. Sentence 2 If a is odd we can write a as 2k + 1 for some integer k . This is the statement NOT B . Knowing from Sentence 1 that the author is using the contrapositive we would expect to see statements moving forward from the hypothesis of the contrapositive (a is even) or backwards from the conclusion of the contrapositive (32 | ((a2 + 3)(a2 + 7))). Sentence 3 Substitution gives (a2 + 3)(a2 + 7) = . . . = 16(k 2 + k + 1)(k 2 + k + 2). This is just arithmetic. Sentence 4 Since one of k 2 + k +1 or k 2 + k +2 must be even, and the last line above shows that a factor of 16 already exists disjoint from (k 2 + k + 1)(k 2 + k + 2), (a2 + 3)(a2 + 7) must contain a factor of 32. That is 32 | ((a2 + 3)(a2 + 7)). These sentences establish the conclusion of the contrapositive. Since the contrapositive is true, the original statement is true. 21.3.1 Discovering and Writing a Proof Using The Contrapositive The important observation here is that once you decide to use the contrapositive, all of your existing skills apply. The difficulty is in deciding whether or not to use the contrapositive. For our example, we will begin with a definition. Definition 21.3.1 Bounded Proposition 2 A set S of real numbers is bounded if there is a real number M > 0 such that, for all elements x ∈ S , |x| < M . Suppose that S and T are sets of real numbers with S ⊆ T . If S is not bounded, then T is not bounded. We should always be clear about our hypothesis and conclusion. Hypothesis: A: S is not bounded. Conclusion: B: T is not bounded. Since both the hypothesis and conclusion are negated, it makes sense to try to prove the contrapositive “If T is bounded, then S is bounded.” This gives us two statements in our proof. Proof in Progress 154 Chapter 21 Contrapositive 1. Suppose that T is bounded. (This is just NOT B .) 2. To be completed. 3. Hence, S is bounded. (This is just NOT A.) Working backwards from the conclusion we can ask “How do we show that S is bounded?” Using the definition of bounded, we can write Proof in Progress 1. Suppose that T is bounded. (This is just NOT B .) 2. To be completed. 3. For every x ∈ S , we have |x| < M . 4. Hence, S is bounded. (This is just NOT A.) Now the question becomes “Where can we find such an M ?” If we use the definition of bounded and work forward from the hypothesis we can write Proof in Progress 1. Suppose that T is bounded. (This is just NOT B .) 2. Since T is bounded, there exists a real number M > 0 such that, for all x ∈ T , |x| < M . 3. To be completed. 4. For every x ∈ S , we have |x| < M . 5. Hence, S is bounded. (This is just NOT A.) Next, we need to connect the two sets and show that the M of the set T is the same as the M of the set S . But we know Since x ∈ S and S ⊆ T , x ∈ T . Combining this with second sentence we have Since x ∈ S , x ∈ T and so |x| < M . Putting all of the statements together gives the following proof. Proof: We will prove the contrapositive. Suppose that T is bounded. Hence, there exists a real number M > 0 such that, for all x ∈ T , |x| < M . Let x ∈ S . Since S ⊆ T , x ∈ T and so |x| < M . But then S is bounded as required. Chapter 22 Uniqueness 22.1 Objectives The technique objective is: 1. Learn how to prove a statement about uniqueness in the conclusion. 22.2 Introduction You have already encountered statements that contain the adjective unique. Instead of the word “unique” you may see “one and only one” or “exactly one”. Prior to this course you have probably seen statements like the following. Example 1 1. Two lines in the plane which are not parallel will intersect in one and only one point. 2. There is a unique function f (x) such that f (x) = f (x). And earlier in this course you saw the Division Algorithm. Proposition 1 (Division Algorithm (DA)) If a and b are integers, and b > 0, then there exist unique integers q and r such that a = qb + r where 0 ≤ r < b. To prove a statement of the form If . . ., then there is a unique object x in the set S such that P (x) is true. 155 156 Chapter 22 Uniqueness there are basically two approaches. 1. Assume that there are two objects X and Y in the set S such that P (X ) and P (Y ) are true. Show that X = Y . 2. Assume that there are two distinct objects X and Y in the set S such that P (X ) and P (Y ) are true. Derive a contradiction. You can use whichever is easier in the circumstance. 22.3 Showing X = Y The method is as follows. 1. Assume that there are two objects X and Y in the set S such that P (X ) and P (Y ) are true. 2. Show that X = Y . For example, let us prove the following statement. Proposition 2 If a and b are integers with a = 0 and a | b, then there is a unique integer k so that b = ka. As usual, we begin by explicitly identifying the hypothesis and conclusion. Hypothesis: a and b are integers with a = 0 and a | b. Conclusion: There is a unique integer k so that b = ka. The appearance of “unique” in the conclusion tells us to use one of the two approaches described in the previous section. In this case, we will assume the existence of two integers k1 and k2 and show that k1 = k2 . But first, we need to show that at least one integer k exists, and this follows immediately from the definition of divisibility. Proof in Progress 1. Since a | b, at least one integer k exists so that b = ka. 2. Let k1 and k2 be integers such that b = k1 a and b = k2 a. (Note how closely this follows the standard pattern. k1 corresponds to X . k2 corresponds to Y . Both come from the set of integers and if P (x) is the statement “b = xa”, then P (X ) and P (Y ) are assumed to be true. 3. To be completed. 4. Hence, k1 = k2 . Section 22.4 Finding a Contradiction 157 The obvious thing to do is equate the two equations to get k1 a = k2 a Since a is not zero we can divide both sides by a to get k1 = k2 A proof might look like the following. Proof: Since a | b, at least one integer k exists so that b = ka. Now let k1 and k2 be integers such that b = k1 a and b = k2 a. But then k1 a = k2 a and dividing by a gives k1 = k2 . 22.4 Finding a Contradiction The method is as follows. 1. Assume that there are two distinct objects X and Y in the set S such that P (X ) and P (Y ) are true. 2. Derive a contradiction. For example, let us prove the following statement. Proposition 3 Suppose a solution to the simultaneous linear equations y = m1 x + b1 and y = m2 x + b2 exists. If m1 = m2 , then there is a unique solution to the simultaneous linear equations y = m1 x + b1 and y = m2 x + b2 . As usual, we begin by explicitly identifying the hypothesis and conclusion. Hypothesis: A solution to the simultaneous linear equations y = m1 x+b1 and y = m2 x+b2 exists. m1 = m2 . Conclusion: There is a unique solution to the simultaneous linear equations y = m1 x + b1 and y = m2 x + b2 . The appearance of “unique” in the conclusion tells us to use one of the two approaches described in the previous section. In this case, we will assume the existence of two distinct points of intersection and derive a conclusion. Proof in Progress 1. Suppose that y = m1 x + b1 and y = m2 x + b2 intersect in the distinct points (x1 , y1 ) and (x2 , y2 ). (Note again how closely this follows the standard pattern. (x1 , y1 ) corresponds to X . (x2 , y2 ) corresponds to Y . Both come from the set of ordered pairs and both satisfy the statement “are a solution to the simultaneous linear equations y = m1 x + b1 and y = m2 x + b2 .”.) 2. To be completed, hence a contradiction. 158 Chapter 22 Uniqueness But now if we substitute (x1 , y1 ) and (x2 , y2 ) into y = m1 x + b1 y1 = m1 x1 + b1 (22.1) y2 = m1 x2 + b1 (22.2) which implies that y1 − y2 = m1 (x1 − x2 ) Similarly, substituting (x1 , y1 ) and (x2 , y2 ) into y = m2 x + b2 gives y1 − y2 = m2 (x1 − x2 ) Equating the two expressions for y1 − y2 gives (m1 − m2 )(x1 − x2 ) = 0 Since m1 = m2 , m1 − m2 = 0 so x1 − x2 = 0. That is, x1 = x2 . But also, y1 − y2 = m1 (x1 − x2 ) and x1 − x2 = 0 imply y1 − y2 = 0 That is, y1 = y2 . But then the points (x1 , y1 ) and (x2 , y2 ) are not distinct, a contradiction. Exercise 1 Write a proof for the preceding proposition. 22.5 The Division Algorithm Suppose that in a proof of the Division Algorithm it has already been established that integers q and r exist and only uniqueness remains. A proposed proof of uniqueness follows. Proposition 4 (Division Algorithm) If a and b are integers and b > 0, then there exist unique integers q and r such that a = qb + r where 0 ≤ r < b Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Suppose that a = q1 b + r1 with 0 ≤ r1 < b. Also, suppose that a = q2 b + r2 with 0 ≤ r2 < b and r1 = r2 . 2. Without loss of generality, we can assume r1 < r2 . 3. Then 0 < r2 − r1 < b and 4. (q1 − q2 )b = r2 − r1 . 5. Hence b | (r2 − r1 ). Section 22.5 The Division Algorithm 159 6. By Bounds By Divisibility, b ≤ r2 − r1 which contradicts the fact that r2 − r1 < b. 7. Therefore, the assumption that r1 = r2 is false and in fact r1 = r2 . 8. But then (q1 − q2 )b = r2 − r1 implies q1 = q2 . Let’s make sure that we understand every line of the proof. Sentence 1 Suppose that a = q1 b + r1 with 0 ≤ r1 < b. Also, suppose that a = q2 b + r2 with 0 ≤ r2 < b and r1 = r2 . Since a statement about uniqueness appears in the conclusion, we would expect one of the two uniqueness methods to be used. In fact, both are used. The assertion of uniqueness applies to both q and r. Since the author writes r1 = r2 , that is, there are distinct values of r1 and r2 , we should look for a contradiction regarding r. But the author does not assume distinct values of q and so we would expect that the author will show q1 = q2 . Sentence 2 Without loss of generality, we can assume r1 < r2 . “Without loss of generality” is an expression that means the upcoming argument would hold identically if we made any other choice, so we will simply assume one of the possibilities. Sentence 3 Then 0 < r2 − r1 < b and This is a particularly important line. It comes, in part, from r1 < r2 by subtracting r1 from both sides (this gives 0 < r2 − r1 ) and by remembering that the largest possible value of r2 is b − 1 and the smallest possible value of r1 is 0, so the largest possible difference is b − 1, thus r2 − r1 < b Sentence 4 (q1 − q2 )b = r2 − r1 . This follows from equating a = q1 b + r1 and a = q2 b + r2 . Sentence 5 Hence b | (r2 − r1 ). This follows from the definition of divisibility. Sentence 6 By Bounds By Divisibility, b ≤ r2 − r1 which contradicts the fact that r2 − r1 < b. Note the importance of the strict inequality in the relation b ≤ r2 − r1 < b Sentence 7 Therefore, the assumption that r1 = r2 is false and in fact r1 = r2 . The contradiction we were looking for. The Division Algorithm states that both q and r are unique. So far, only the uniqueness of r has been established. Sentence 7 But then (q1 − q2 )b = r2 − r1 implies q1 = q2 . And this is where the uniqueness of q is established. Originally, the author assumed the existence of q1 and q2 and now has shown that they are, in fact, the same. Chapter 23 Introduction to Primes 23.1 Objectives The technique objectives are: 1. Practice with induction. 2. Practice with arguments of uniqueness. The content objectives are: 1. Recall the definition of prime and composite. 2. Discover a proof by induction of the Prime Factorization Theorem. 23.2 Introduction to Primes The second problem that the course focuses on is Fermat’s Last Theorem. Theorem 1 (Fermat’s Last Theorem (FLT)) If n ≥ 3, then there are no solutions to xn + y n = z n where x, y and z are positive integers. To make progress on this problem, we need to work with prime numbers. Recall our definition of prime number. Definition 23.2.1 An integer p > 1 is called a prime if its only divisors are 1 and p, and composite otherwise. Prime, Composite 160 Section 23.3 Induction Example 1 161 The integers 2, 3, 5 and 7 are primes. The integers 4 = 2 × 2, 6 = 2 × 3 and 8 = 2 × 2 × 2 are composite. Note, that by definition, 1 is not a prime. We have already proved three propositions about primes, one of which is a consequence of Coprimeness and Divisibility, and the other two were proved in the chapter on contradiction. Proposition 2 (Primes and Divisibility (PAD)) If p is a prime and p | ab, then p | a or p | b. Proposition 3 (Prime Factorization (PF)) If n is an integer greater than 1, then n can be written as a product of prime factors. Proposition 4 (Infinitely Many Primes (INF P)) The number of primes is infinite. We will prove Prime Factorization again, this time with induction. 23.3 Induction Recall how induction, Strong Induction in this case, works. Axiom 3 Principle of Strong Induction (POSI) Let P (n) be a statement that depends on n ∈ P. If 1. P (1), P (2), . . . , P (b), are true, and 2. P (1), P (2), . . . , P (k ) are all true implies P (k + 1) is true then P (n) is true for all n ∈ P. Recall the three parts in a proof by strong induction. Base Cases Verify that P (1), P (2), . . . , P (b) are all true. Inductive Hypothesis Assume that P (i) is true for i = 1, 2, 3, . . . , k where k ≥ b. Inductive Conclusion Using the assumption that P (1), P (2), . . . , P (k ) are true, show that P (k + 1) is true. 162 Chapter 23 Introduction to Primes We will use Strong Induction to prove Proposition 5 (Prime Factorization (PF)) If n is an integer greater than 1, then n can be expressed as a product of prime factors. First, we formulate our statement P (n) that relies on the integer n. P (n): n can be expressed as a product of prime factors. Now we can begin the proof. Proof: Base Case We verify P (2). Recall that the base case does not need to start at 1. P (2): 2 can be expressed as a product of prime factors. This is trivially true. Inductive Hypothesis We assume that P (i) is true for i = 2, 3, . . . , k where k ≥ 2. P (i): i can be expressed as a product of prime factors. Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): k + 1 can be expressed as a product of prime factors. If k + 1 is prime, then k + 1 by itself is a product of prime factors. It is a product with just one factor. In this case, P (k + 1) is true. If k + 1 is composite, then we can write k + 1 = rs where 1 < r ≤ s < k + 1. Since r and s are less than k + 1, they can be written as a product of prime factors by the inductive hypothesis. Hence, k + 1 is a product of prime factors and P (k + 1) is true in this case also. The result is true for n = k + 1, and so holds for all n by POSI. 23.4 Fundamental Theorem of Arithmetic In grade school you used prime numbers to write the prime factorization of any positive integer greater than one. You probably never worried about the possibility that there might be more than one way to do this. However, in some sets “prime” factorization is not unique. √ √ 5 Consider the set S = {a + b 5 | a, b ∈ Z}. In S , the number 4 = 4 + 0 √ can be factored √ √ √ in two different ways, 4 = 2 × 2 and 4 = ( 5 + 1)( 5 − 1). Moreover, 2, 5 + 1 and 5 − 1 are all prime numbers in S ! Since multiplication in the integers is commutative, the prime factorizations can be written in any order. For example 12 = 2 × 2 × 3 = 2 × 3 × 2 = 3 × 2 × 2. However, up to the order of the factors, the factorization of integers is unique. This property is so basic it is referred to as the Fundamental Theorem of Arithmetic. It is also referred to as the Unique Factorization Theorem. Section 23.4 Fundamental Theorem of Arithmetic Theorem 6 163 (Fundamental Theorem of Arithmetic or Unique Factorization Theorem (UFT)) If n > 1 is an integer, then n can be written as a product of prime factors and, apart from the order of factors, this factorization is unique. Observe that the conclusion contains two parts: 1. n can be written as a product of prime factors (which we proved earlier), and 2. apart from the order of factors, this factorization is unique. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. That n can be written as a product of prime factors follows from the proposition Prime Factorization. 2. Now suppose that n is factored into primes in two ways, n = p1 p2 . . . pk = q1 q2 . . . q (23.1) where all of the p’s and q ’s are primes. 3. Since p1 | n, p1 | q1 q2 . . . q . 4. By repeatedly applying the proposition Primes and Divisibility, p1 must divide one of the q ’s. If necessary, rearrange the q ’s so that p1 | q1 . 5. Since q1 is prime, and p1 > 1, it must be the case that p1 = q1 . 6. Dividing Equation 23.1 by p1 = q1 gives p2 p3 . . . pk = q2 q3 . . . q (23.2) 7. By continuing in this way, we see that each p must be paired off with one of the q s until there are no factors on either side. 8. Hence k = the same. and, apart from the order of the factors, the two expressions for n are Let’s perform an analysis of the proof. As usual, we begin with the hypothesis and the conclusion. Hypothesis: n is an integer, n > 1 Conclusion: There are two parts. 1. n can be written as a product of prime factors, and 2. apart from the order of factors, this factorization is unique. 164 Chapter 23 Introduction to Primes Core Proof Technique: Uniqueness Preliminary Material: Primes and Divisibility Sentence 1 That n can be written as a product of prime factors follows from the proposition Prime Factorization. The first of the two parts of the conclusion is just the conclusion of a previous proposition. Sentence 2 Now suppose that n is factored into primes in two ways, n = p1 p2 . . . pk = q1 q2 . . . q where all of the p’s and q ’s are primes. This is a classic use of the Uniqueness Method. We assume that there are two representations of the same object, and show that the two representations are, in fact, identical. One representation of n is the product p1 p2 . . . pk and the second representation is the product q1 q2 . . . q . Sentences 3 – 5 Since p1 | n, p1 | q1 q2 . . . q . By repeatedly applying the proposition Primes and Divisibility, p1 must divide one of the q ’s. If necessary, rearrange the q ’s so that p1 | q1 . Since q1 is prime, and p1 > 1, it must be the case that p1 = q1 . The author shows that the two representations of n are equal by showing that they have identical factors. Here, the author demonstrates that p1 = q1 . Sentences 6 – 7 Dividing Equation 23.1 by p1 = q1 gives p2 p3 . . . pk = q2 q3 . . . q By continuing in this way, we see that each p must be paired off with one of the q s until there are no factors on either side. This continues the author’s plan of showing that the two representations of n are equal by showing that they have identical factors. Sentence 8 Hence k = n are the same. and, apart from the order of the factors, the two expressions for This is a typical conclusion to the Uniqueness Method. The two representations of the same object are identical. 23.5 Finding a Prime Factor The previous proposition does not provide an algorithm for finding the prime factors of a positive integer n. The next proposition shows that we do not have to check all of the prime factors less than n, only those less than or equal to the square root of n. Proposition 7 (Finding a Prime Factor (FPF)) An integer n > 1 is either prime or contains a prime factor less than or equal to Let’s begin by identifying the hypothesis and the conclusion. √ n. Section 23.5 Finding a Prime Factor 165 Hypothesis: n is an integer and n > 1. Conclusion: n is either prime or contains a prime factor less than or equal to √ n. Before we see a proof, let’s do an example. Example 2 Is 73 a prime number? Solution: Using Finding a Prime Factor , we can check for divisibility by primes less than √ √ or equal to 73. Now 73 ≈ 8.544 so any possible prime factor must be less than or equal to 8. The only candidates to check are 2, 3, 5 and 7. Since none of these divide 73, 73 must be prime. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Suppose that n is not prime. 2. Let p be the smallest prime factor of n. 3. Since n is composite we can write n = ab where a and b are integers such that 1 < a, b < n. 4. Since p is the smallest prime factor, p ≤ a and p ≤ b and so n = ab ≥ p · p = p2 . That √ is p ≤ n. Analysis of Proof Since or appears in the conclusion, we will use Proof By Elimination. The equivalent statement that is proved is: If n is an integer greater than 1 and n is not prime, then n contains a prime √ factor less than or equal to n. The word “a” should alert us to the presence of an existential quantifier. We could reword the statement as If n is an integer greater than 1 and n is not prime, then there exists a √ prime factor of n which is less than or equal to n. This is the statement that will actually be proved. Hypothesis: n is an integer greater than 1 and n is not prime. Conclusion: There exists a prime factor of n which is less than or equal to √ n. Core Proof Technique: Construct Method Sentence 1 Suppose that n is not prime. This sentence tells that the author is going to use Proof by Elimination. Sentence 2 Let p be the smallest prime factor of n. The conclusion has an existential quantifier and so the author uses the Construct Method. The prime p will be the desired prime factor though it is not clear yet why “smallest” is important. The proposition on Prime Factorization guarantees us that a prime factor exists. 166 Chapter 23 Introduction to Primes Sentence 3 Since n is composite we can write n = ab where a and b are integers such that 1 < a, b < n. By the hypotheses of the restated proposition, n > 1 and n is not prime, so n is composite and can be factored. Sentence 4 Since p is the smallest prime factor, p ≤ a and p ≤ b and so n = ab ≥ p·p = p2 . √ That is p ≤ n. This is where “smallest” is used. The conclusion follows from arithmetic and the fact that p is the smallest prime factor. 23.6 Working With Prime Factorizations The next proposition, which we will state but not prove, gives us a means to list all of the divisors of a positive integer. A proof is available in the Appendix. Add proof. Proposition 8 (Divisors From Prime Factorization (DFPF)) If a > 1 is an integer and a = pα1 pα2 · · · pαk 12 k is the prime factorization of a into powers of distinct primes p1 , p2 , . . . , pk , then the positive divisors of a are integers of the form d = pd1 pd2 · · · pdk where 0 ≤ di ≤ αi for i = 1, 2, . . . , k 12 k Exercise 1 Using Divisors From Prime Factorization, list all of the positive factors of 45. Exercise 2 How many positive divisors are there to the integer a whose prime factorization is a = pα1 pα2 · · · pαk 12 k Proposition 9 (GCD From Prime Factorization (GCD PF)) If a = pα1 pα2 · · · pαk 12 k and b = pβ1 pβ2 · · · pβ 12 are the prime factorizations of a and b, where some of the exponents may be zero, then gcd(a, b) = pd1 pd2 · · · pdk where di = min{αi , βi } for i = 1, 2, . . . , k 12 k Though this method works well enough on small examples, it is much slower than the Extended Euclidean Algorithm for computing gcds. Section 23.6 Working With Prime Factorizations Exercise 3 Use GCD PF to compute gcd(33 51 74 131 , 52 77 131 232 ). Exercise 4 Use the definition of gcd to prove GCD From Prime Factorization. 167 Chapter 24 Introduction to Fermat’s Last Theorem 24.1 Objectives The content objectives are: 1. Provide an historical introduction. 2. Define gcd(x, y, z ), trivial solutions, Pythagorean triple and primitive Pythagorean triple. 3. State Extending Coprimeness. 4. Read a proof of Multiples of Pythagorean Triples. 5. Discover a proof of Relative Primeness of Pythagorean Triples. 6. Read a proof of Parity of Primitive Pythagorean Triples. 7. State Decomposition of n-th Powers. 24.2 History of Fermat’s Last Theorem Pierre de Fermat (1601 (?) – 1635) was a brilliant French mathematician. It was his habit to make notes in the margins of his books and one such note is famous. Fermat possessed a copy of Bachet’s translation of Diophantus’ Arithmetica. Problem II.8 of the Arithmetica reads Partition a given square into two squares. Diophantus did not require the squares to be integers so we might write Problem II.8 as For what positive rational numbers x, y and z is the equation x2 + y 2 = z 2 satisfied? 168 Section 24.2 History of Fermat’s Last Theorem 169 Adjacent to Problem II.8, and in the margin of his copy of Arithmetica, Fermat wrote (translated) It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvellous proof of this, which this margin is too narrow to contain. Fermat was asserting Theorem 1 (Fermat’s Last Theorem) If n ≥ 3, then xn + y n = z n has no solutions when x, y and z are positive integers. No proof was ever published by Fermat, or found among his notes after his death. It seems very unlikely that he did have a proof and it was not until Andrew Wiles’ publications in 1994 that the Theorem, conjecture really, was proved. Fermat did prove the case n = 4, as we shall do. First though, we will clarify our language. Clearly there are solutions to xn + y n = z n . One solution is x = y = z = 0, another solution is x = 0, y = z . Definition 24.2.1 Trivial We will say that any solution to xn + y n = z n for which at least one of x, y or z is zero, is trivial. So we restate Fermat’s Last Theorem as Theorem 2 (Fermat’s Last Theorem) If n ≥ 3, then xn + y n = z n has no non-trivial integer solutions. Our starting point will be a much more familiar problem. x2 + y 2 = z 2 (24.1) You will recognize this as the equation of the Pythagorean Theorem. Our task is to identify all positive integer solutions to 24.1. 170 Chapter 24 24.3 Introduction to Fermat’s Last Theorem Pythagorean Triples We begin with some definitions. Definition 24.3.1 A Pythagorean triple is a set of non-zero integers x, y and z such that x2 + y 2 = z 2 . Pythagorean Triple Equivalently, a Pythagorean triple is a non-trivial solution to x2 + y 2 = z 2 . Now we expand our definition of gcd. Definition 24.3.2 Greatest Common Divisor Let a, b and c be integers, not all zero. An integer d > 0 is the greatest common divisor of a, b and c, written gcd(a, b, c), if and only if 1. d | a, d | b and d | c (this captures the common part of the definition), and 2. if e | a and e | b and e | c then e ≤ d (this captures the greatest part of the definition). Definition 24.3.3 A Pythagorean triple is said to be primitive if gcd(x, y, z ) = 1. Primitive Triple Example 1 Both (6, 8, 10) and (3, 4, 5) are Pythagorean triples. However • (6, 8, 10) is not a primitive Pythagorean triple since gcd(6, 8, 10) = 2 = 1. • (3, 4, 5) is a primitive Pythagorean triple since gcd(6, 8, 10) = 1. We leave the proof of the following very useful lemma as an exercise. Lemma 3 (Extending Coprimeness (EC)) If x, y and z are integers, not all zero, and gcd(x, y ) = 1, then gcd(x, y, z ) = 1. Proposition 4 (Multiples of Pythagorean Triples (MPT)) Let d = gcd(x, y, z ). The three integers x, y and z are a Pythagorean triple if and only if x y z the three integers x1 = , y1 = and z1 = are a Pythagorean triple. d d d Example 2 (6, 8, 10) is a Pythagorean triple since 62 +82 = 102 . Since gcd(6, 8, 10) = 2, by the Multiples of Pythagorean Triples, (3, 4, 5) is a Pythagorean triple. Also, if (3, 4, 5) is a Pythagorean triple, then (3d, 4d, 5d) is a Pythagorean triple. This is a simple “if and only if” proof that can be proved using a chain of “if and only if” statements. Section 24.3 Pythagorean Triples 171 Proof: x, y and z are a Pythagorean triple ⇐⇒ x2 + y 2 = z 2 ⇐⇒ x2 y2 (defn of Pythagorean triple) z2 (divide by d2 ) + 2= 2 d2 d d 2 2 ⇐⇒ x2 + y1 = z1 1 ⇐⇒ x1 , y1 and z1 are a Pythagorean triple (substitution) (defn of Pythagorean triple) Take ten minutes to prove the following proposition and then compare your proof with the proof that follows. Proposition 5 (Relative Primeness of Pythagorean Triples (RPPT)) If x, y and z are a primitive Pythagorean triple, then gcd(x, y ) = gcd(x, z ) = gcd(y, z ) = 1. Proof: We will show that gcd(x, y ) = 1. The other pairs are similar. Suppose to the contrary that gcd(x, y ) = d > 1. Then there exists a prime p so that p | d. Since p | x and p | y , p | (x2 + y 2 ) by the Divisibility of Integer Combinations. Since x2 + y 2 = z 2 , p | z 2 and so p | z by Primes and Divisibility. But then gcd(x, y, z ) ≥ p > 1 which contradicts the hypothesis that x, y and z are a primitive Pythagorean triple. Let us walk through a proof of the following proposition. Proposition 6 (Parity of Primitive Pythagorean Triples (PPPT)) If x, y and z are a primitive Pythagorean triple, then one of the integers x or y is even and the other is odd. First, let’s check this against our experience. Example 3 (Parity of Primitive Pythagorean Triples) 1. In the primitive Pythagorean triple (3, 4, 5), 3 is odd and 4 is even. 2. The Pythagorean triple (6, 8, 10) has no odd elements, but the proposition does not apply to the Pythagorean triple (6, 8, 10) since it is not primitive. 3. In the primitive Pythagorean triple (8, 15, 17), 8 is even and 15 is odd. 172 Chapter 24 Introduction to Fermat’s Last Theorem Proof: (For reference, each sentence of the proof is written on a separate line.) 1. We will proceed by contradiction using two cases: x and y are both even, and x and y are both odd. 2. Consider the first case. Suppose that x and y are both even. 3. But then gcd(x, y ) = 2 = 1 which contradicts the Relative Primeness of Pythagorean Triples. 4. Consider the second case. Suppose that x and y are both odd. 5. This implies that x2 ≡ 1 (mod 4) and y 2 ≡ 1 (mod 4) which in turn implies z 2 = x2 + y 2 ≡ 2 (mod 4) 6. But this is impossible since the square of any integer can only be congruent to 0 or 1 modulo 4. 7. Since the two integers cannot both be even or odd, exactly one must be even and one must be odd. As usual, we will begin our analysis by identifying the hypothesis, conclusion, core proof techniques and preliminary material. Hypothesis: x, y and z are a primitive Pythagorean triple. Conclusion: One of the integers x or y is even and the other is odd. Core Proof Technique: There are only three possible cases: x and y are both even, x and y are both odd or x and y have opposite parity. The first two cases will be eliminated leaving the third as the only possible outcome. Each of the first two cases is dealt with using contradiction. The use of contradiction several times within a proof is common. Preliminary Material: primitive Pythagorean triple, congruences Let’s examine each collection of sentences. Sentence 1. We will proceed by contradiction eliminating two cases: x and y are both even, and x and y are both odd. The author indicates the plan of the proof, always a good idea. There are only three possible cases: x and y are both even, x and y are both odd or x and y have opposite parity. The author will disprove the first two cases using contradiction, hence by elimination leaving only opposite parity. Sentence 2. Consider the first case. Suppose that x and y are both even. This sentence begins the first of the two embedded proofs by contradiction. Section 24.3 Pythagorean Triples 173 Sentence 3. But then gcd(x, y ) = 2 = 1 which contradicts the Relative Primeness of Pythagorean Triples. To invoke the Relative Primeness of Pythagorean Triples we should make sure that the hypothesis of RPPT is satisfied. All that is required is that “x, y and z are a primitive Pythagorean triple”, which is assured from the hypothesis of the proposition we are proving. Sentence 4. Consider the second case. Suppose that x and y are both odd. This sentence begins the second of the two embedded proofs by contradiction. Sentence 5. This implies that x2 ≡ 1 (mod 4) and y 2 ≡ 1 (mod 4) which in turn implies z 2 = x2 + y 2 ≡ 2 (mod 4) Sentence 6. But this is impossible since the square of any integer can only be congruent to 0 or 1 modulo 4. This part of proof is quite different from the earlier part. Since any odd integer a can be written in the form 2t + 1, a2 has the form 4t2 + 4t + 1 which is congruent to 1 (mod 4). Thus z 2 = x2 + y 2 ≡ 1 + 1 ≡ 2 (mod 4). But how could this be? If z were odd, z 2 ≡ 1 (mod 4), and if z were even, z 2 ≡ 0 (mod 4). Sentence 7. Since the two integers cannot both be even or odd, exactly one must be even and one must be odd. Since the cases of x and y both even or x and y both odd have been eliminated, all that remains is that x and y have opposite parity. REMARK If x, y , z is a Pythagorean triple, we will assume as a convention that x is even and y is odd. Notice that this implies that z is odd. (Why?) We conclude with a small proposition that is very useful. The proof appears in the Appendix. Proposition 7 (Decomposition of n-th Powers (DNP)) If ab = cn and gcd(a, b) = 1, then there exist integers a1 and b1 so that a = an and b = bn . 1 1 Example 4 Consider 592, 704 which is just 843 . With n = 3, c = 84, a = 64 and b = 9261, the hypotheses of the proposition are satisfied. Hence, there exist integers a1 and b1 so that a = 64 = an and b = 9261 = bn . So a1 = 4 and b1 = 21, and 43 213 = 843 . 1 1 Notice that our choice of a and b satisfied gcd(a, b) = 1. With a = 8 = 23 and b = 74088 = 423 , even though ab = cn is still true, the proposition does not apply since gcd(a, b) = 1 Chapter 25 Characterization of Pythagorean Triples 25.1 Objectives The content objectives are: 1. State and prove the Characterization of Pythagorean Triples theorem. 2. Illustrate the theorem. 25.2 Pythagorean Triples We are now able to characterize all non-trivial, primitive Pythagorean triples. The proof in this section follows that done by David Burton in Elementary Number Theory, Seventh Edition. Theorem 1 (Characterization of Pythagorean Triples (CPT)) The complete set of non-trivial, primitive solutions to x2 + y 2 = z 2 is given by x = 2st y = s2 − t2 z = s2 + t2 for integers s > t > 0 such that gcd(s, t) = 1 and s ≡ t (mod 2). Let’s understand what the theorem is saying. Every choice of s and t satisfying integers s > t > 0 such that gcd(s, t) = 1 and s ≡ t (mod 2) should produce a Pythagorean triple and these are the only non-trivial, primitive Pythagorean triples. 174 Section 25.2 Pythagorean Triples 175 The table below lists some primitive Pythagorean triples arising from small values of s and t. s t 2 3 4 4 5 5 1 2 1 3 2 4 x 2st 4 12 8 24 20 40 y s2 − t2 3 5 15 7 21 9 z s2 + t2 5 13 17 25 29 41 Before we read the proof, let’s do some analysis. The expression “complete solution” should indicate to us that we are working with sets. So, the first step is to identify which sets are used and what their relationship is. One set is the collection of non-trivial, primitive Pythagorean triples and can be defined by S = {x, y, z ∈ N | x2 + y 2 = z 2 , gcd(x, y, z ) = 1, 2 | x} Note that the use of N is equivalent to non-trivial, that gcd(x, y, z ) = 1 is equivalent to primitive and 2 | x follows our convention that in a primitive Pythagorean triple, x is even and y and z are odd. The other set is the collection of triples determined by formula and can be defined by T = {x, y, z ∈ N | s, t ∈ N, x = 2st, y = s2 − t2 , z = s2 + t2 , s > t, gcd(s, t) = 1, s ≡ t (mod 2)} The Characterization of Pythagorean Triples theorem asserts that S = T . We would expect the proof to show that S = T by showing that S ⊆ T and T ⊆ S , though this is done implicitly. Here is the proof. Be sure to identify 1. where S and T appear in the proof, 2. where each of the elements that define set membership are satisfied, 3. where each of the elements that define set membership are used. Question Panel walk through of proof here. Proof: Let x, y and z be a non-trivial, primitive Pythagorean triple. Since x is even and y and z are odd, z − y and z + y are even. Suppose z − y = 2u and z + y = 2v . Then u= z−y z+y and v = 2 2 and v − u = y and u + v = z and the equation x2 + y 2 = z 2 may be rewritten as x2 = z 2 − y 2 = (z − y )(z + y ) 176 Chapter 25 Characterization of Pythagorean Triples Dividing the preceding equation by 4 gives x 2 2 = z−y 2 z+y 2 = uv We claim that u and v are relatively prime. Suppose they were not and gcd(u, v ) = d > 1. Then d | (v − u) and d | (u + v ). But v − u = y and u + v = z so d | y and d | z which contradicts the fact that y and z are relatively prime. Now we can use our proposition on Decomposing n-th Powers to conclude that u and v are perfect squares. Hence, for some positive integers s and t u = t2 v = s2 Using these values of u and v produces z = v + u = s2 + t2 y = v − u = s2 − t2 x2 = 4vu = 4s2 t2 ⇒ x = 2st We claim that gcd(s, t) = 1. If d > 1 were a common factor of s and t, d would be a common factor of y and z contradicting the fact that gcd(y, z ) = 1. Finally, if s and t are both even or both odd, then y and z are even, a contradiction. Hence, exactly one of s and t is odd, the other is even. Symbolically, s ≡ t (mod 2). Conversely, let the natural numbers s and t satisfy s > t, gcd(s, t) = 1, s ≡ t (mod 2). Using the provided formulas for x, y and z we have x2 + y 2 = (2st)2 + (s2 − t2 )2 = (s2 + t2 )2 = z 2 so x, y and z are a Pythagorean triple. To see that the triple is non-trivial, we must show that x, y and z are all positive. Since s, t > 0, x = 2st > 0 and z = s2 + t2 > 0. Since s > t, y = s2 − t2 > 0. To see that the triple is primitive, assume that gcd(x, y, z ) = d > 1 and let p be any prime divisor of d. Since one of s and t is odd and the other is even, z is odd. Since p | z , p = 2. From p | y and p | z , we know that p | (z + y ) and since z + y = 2s2 , p | 2s2 . Hence, p | s. Similarly, p | t. But then p is a common factor of s and t contradicting gcd(s, t) = 1. Since no such p can exist, gcd(x, y, z ) = 1 and x, y and z are a primitive triple. Chapter 26 Fermat’s Theorem for n = 4 26.1 Objectives The content objectives are 1. State and prove: The Diophantine equation x4 + y 4 = z 2 has no non-trivial solution. 2. State and prove: The Diophantine equation x4 + y 4 = z 4 has no non-trivial solution. 3. Show a reduction of FLT to If p is an odd prime, then the Diophantine equation xp + y p = z p has no non-trivial solution. 26.2 n=4 Having completely resolved the case of Pythagorean triples, we can now turn our attention to the one instance of FLT proved by Fermat. Actually, we will prove a slightly stronger result and the case n = 4 will follow as a corollary. The approach in this section mostly follows Elementary Number Theory, Seventh Edition by David Burton. Theorem 1 (FLT, Strong Version of n = 4) The Diophantine equation x4 + y 4 = z 2 has no non-trivial solution. The proof is demanding but it has a straightforward structure. 1. This is a proof by contradiction. It assumes the existence of a “minimal” solution x0 , y0 , z0 to x4 + y 4 = z 2 . 2. Using x0 , y0 , z0 the author constructs a non-trivial primitive Pythagorean triple. 3. Using the Characterization of Pythagorean Triples the author finds various algebraic expressions involving s and t. 4. The author uses these algebraic expressions to construct another non-trivial primitive Pythagorean triple. 177 178 Chapter 26 Fermat’s Theorem for n = 4 5. Lastly, the author uses this triple to construct a solution x1 , y1 , z1 to x4 + y 4 = z 2 which is “smaller” than x0 , y0 , z0 , hence a contradiction. Extensive Question Panel here. Proof: By way of contradiction, suppose there exists a positive integer solution to x4 + y 4 = z 2 . Of all such solutions, choose any one in which z is smallest. Call this solution x0 , y0 , z0 . Without loss of generality, we may also assume that gcd(x0 , y0 ) = 1. (Why?) This in turn implies that gcd(x0 , y0 , z0 ) = 1. (Why?) Since x0 , y0 , z0 is a solution we know 4 2 x4 + y0 = z0 0 which we can rewrite as x2 0 2 2 + y0 2 2 = z0 2 But that means that x2 , y0 and z0 are non-trivial primitive solutions of a2 + b2 = c2 so we 0 can make use of the Characterization of Pythagorean Triples. In particular, we know that 2 one of x2 and y0 is even. We can assume that x2 is even, hence x0 is even, and that there 0 0 exist integers s and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t (mod 2) satisfying x2 = 2st 0 2 y0 = s2 − t2 z0 = s2 + t2 Since s ≡ t (mod 2), exactly one of s and t are even. Suppose s is even and t is odd. Now 2 2 consider the equation y0 = s2 − t2 modulo 4. Because y0 is odd, 2 y0 = s2 − t2 ⇒ 1 ≡ 0 − 1 (mod 4) ⇒ 1 ≡ 3 (mod 4) which is impossible. Therefore s is odd and t is even so we write t = 2r. Then x2 = 2st ⇒ x2 = 4sr ⇒ 0 0 x0 2 2 = sr Now gcd(s, t) = 1 implies that gcd(s, r) = 1 (why?) and so we can use the proposition on Decomposing n-th Powers. Since (x0 /2)2 is a perfect square, s and r must be perfect 2 2 squares and we can write s = z1 and r = w1 for positive integers z1 and w1 . 2 Rewrite y0 = s2 − t2 as 2 t2 + y0 = s2 Because gcd(s, t) = 1 implies gcd(s, t, y0 ) = 1, the triple t, y0 , s is a primitive Pythagorean triple and we can use the Characterization of Pythagorean Triples again. With t even, the Characterization of Pythagorean Triples assures us of the existence of integers u and v so that u > v > 0 and gcd(u, v ) = 1 and u ≡ v (mod 2) satisfying t = 2uv y0 = y 2 − v 2 s = u2 + v 2 Now, observe that uv = t 2 = r = w1 2 Section 26.4 Reducing the Problem 179 and so by the proposition on Decomposing n-th Powers, u and v are perfect squares. Suppose 2 u = x2 and v = y1 where x1 and y1 are positive integers. But then 1 2 s = u2 + v 2 ⇒ z1 = x2 1 2 2 + y1 2 and so 2 4 z 1 = x 4 + y1 1 That is, x1 , y1 , z1 is a solution to x4 + y 4 = z 2 . Since z1 and t are positive 2 0 < z1 ≤ z1 = s ≤ s2 < s2 + t2 = z0 That is, z1 < z0 But recall that x0 , y0 , z0 is a solution to x4 + y 4 = z 2 with the smallest possible value of z . But x1 , y1 , z1 is a solution to x4 + y 4 = z 2 with a smaller value of z ! The case n = 4 of Fermat’s Last Theorem follows immediately. Corollary 2 The Diophantine equation x4 + y 4 = z 4 has no positive integer solution. 2 Proof: If x0 , y0 , z0 were a positive integer solution of x4 + y 4 = z 4 , then x0 , y0 , z0 would 4 + y 4 = z 2 , contradicting the previous theorem. be a positive integer solution to x 26.3 Reducing the Problem It is not necessary to consider every exponent of xn + y n = z n to prove Fermat’s Last Theorem. If n > 2, then n is either a power of 2 or divisible by an odd prime p. In the first case, n = 4k for some k ≥ 1 and the equation xn + y n = z n can be rewritten as xk 4 + yk 4 = zk 4 We have just seen that this equation has no positive integer solution. In the second case, n = pk for some k ≥ 1 and the equation xn + y n = z n can be rewritten as p p p xk + y k = z k If it could be shown that up + v p = wp has no solution, then there would be no solution of the form u = xk , v = y k , w = z k and so there would be no solution to xn + y n = z n . Therefore, Fermat’s Last Theorem reduces to Theorem 3 (Fermat’s Last Theorem – Reduced) If p is an odd prime, then the Diophantine equation xp + y p = z p has no non-trivial solutions. 180 Chapter 26 26.4 History Awaiting copyright permission. Fermat’s Theorem for n = 4 Chapter 27 Problems Related to FLT 27.1 Objectives 1. Read a proof of Squares From the Difference of Quartics 2. Read a proof of a proposition on the area of Pythagorean triangles. 27.2 x4 − y 4 = z 2 From x4 + y 4 = z 2 , we turn to a closely related Diophantine equation, x4 − y 4 = z 2 . Our proof is very similar to that of the Strong Version of FLT for n = 4. The approach in this section mostly follows Elementary Number Theory, Seventh Edition by David Burton. Proposition 1 (Squares From the Difference of Quartics (SFDQ)) The Diophantine equation x4 − y 4 = z 2 has no non-trivial solution. Proof: Suppose that there exists a non-trivial solution to x4 − y 4 = z 2 . Of all such solutions x0 , y0 , z0 , choose any one in which x0 is smallest. Choosing x0 as small as possible forces x0 to be odd. (Why?) We now show that we can also assume that gcd(x0 , y0 ) = 1. Suppose gcd(x0 , y0 ) = d > 1. 4 Then writing dx1 = x0 and dy1 = y0 and substituting into x4 − y 4 = z 2 we get d4 (x4 − y1 ) = 1 2 2 z0 . So d4 | z0 , hence d2 | z0 . Thus z0 = d2 z1 for some integer z1 . But then 4 2 d4 x4 − d4 y1 = d4 z1 1 so x1 , y1 , z1 is a non-trivial solution to x4 − y 4 = z 2 with 0 < x1 < x0 contradicting our choice of a minimal x0 . 4 2 If the equation x4 − y0 = z0 is written in the form 0 x2 0 2 2 − y0 then 2 2 z0 + y0 2 2 2 = z0 = x2 0 181 2 182 Chapter 27 Problems Related to FLT 2 and we see that z0 , y0 , x2 constitute a primitive Pythagorean triple. 0 From here there are two cases: y0 odd and y0 even. Consider the case where y0 is odd. The Characterization of Pythagorean Triples asserts that there exist integers s and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t (mod 2) satisfying z0 = 2st (this is forced from y0 odd) 2 y0 = s2 − t2 x2 = s2 + t2 0 Observe that 2 s4 − t4 = (s2 + t2 )(s2 − t2 ) = x2 y0 = (x0 y0 )2 0 so s, t, x0 y0 is a positive solution to x4 − y 4 = z 2 . But 0<s< s2 + t2 = x0 which contradicts the minimality of x0 so y0 cannot be odd. Now consider the case where y0 is even. The Characterization of Pythagorean Triples asserts that there exist integers s and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t (mod 2) satisfying 2 y0 = 2st 2 (this is forced from y0 even) 2 z0 = s − t x2 = s2 + t2 0 Because of the symmetry of expressions for s and t, we may assume that s is even and t is odd. Consider the relation 2 y0 = 2st Since gcd(s, t) = 1 and s is even, we know that gcd(2s, t) = 1. This allows us to invoke the proposition on Decomposing n-th Powers. That is, 2s and t are each squares of positive integers, say 2s = w2 and t = v 2 . Because w must be even, set w = 2u to get s = 2u2 . Therefore x2 = s2 + t2 = 4u4 + v 4 0 and so 2u2 , v 2 , x0 form a Pythagorean triple. Since gcd(2u2 , v 2 ) = gcd(s, t) = 1, gcd(2u2 , v 2 , x0 ) = 1 and so the Pythagorean triple is primitive. The Characterization of Pythagorean Triples asserts that there exist integers a and b so that a > b > 0 and gcd(a, b) = 1 and a ≡ b (mod 2) satisfying 2u2 = 2ab v 2 = a2 − b2 x0 = a2 + b2 Now 2u2 = 2ab implies u2 = ab which implies, by the proposition on Decomposing n-th Powers, that a and b are perfect squares. Say a = c2 and b = d2 . And here we use a pattern we have seen before. Since v 2 = a2 − b2 = c4 − d4 c, d, v is a positive integer solution to x4 − y 4 = z 2 . But √ 0 < c = a < a2 + b2 = x0 Section 27.2 x4 − y 4 = z 2 183 contradicting the minimality of x0 so y0 cannot be even. Since the integer y0 cannot be either odd r even, it must be the case that our assumption that there is a non-trivial solution is incorrect. This proposition has an unexpected use in a statement about the areas of Pythagorean triangles. Definition 27.2.1 A Pythagorean triangle is a right triangle whose sides are of integral length. Pythagorean Triangle In the margin of his copy of Diophantus’ Arithmetica, Fermat stated and proved a proposition equivalent to the following. Proposition 2 The area of a Pythagorean triangle can never be equal to a perfect square. Here, perfect square means the square of an integer. Proof: We will proceed by contradiction. Consider a Pythagorean triangle ABC where the hypotenuse has length z and the other two sides have lengths x and y , so that x2 + y 2 = z 2 The area of gives (27.1) ABC is (1/2)xy and if this were a square we could write (1/2)xy = u2 . This 2xy = 4u2 (27.2) Now Equation (27.1) plus Equation (27.2) gives x2 + y 2 + 2xy = z 2 + 4u2 ⇒ (x + y )2 = z 2 + 4u2 and Equation (27.1) minus Equation (27.2) gives x2 + y 2 − 2xy = z 2 − 4u2 ⇒ (x − y )2 = z 2 − 4u2 Now multiply these last two equations together to get (x + y )2 (x − y )2 = (z 2 + 4u2 )(z 2 − 4u2 ) ⇒ x2 − y 2 or x2 − y 2 2 2 = z 4 − 16u4 = z 4 − (2u)4 But we know by our proposition on the Squares From the Difference of Quartics that no non-trivial solution to this equation is possible, hence a contradiction. Chapter 28 Practice, Practice, Practice: Prime Numbers 28.1 Objectives This class provides an opportunity to practice working with primes. 28.2 Exercises 184 Chapter 29 Complex Numbers 29.1 Objectives The content objectives are : 1. N ⊂ Z ⊂ Q ⊂ R ⊂ C 2. Define: complex number, C, real part, imaginary part 3. Operations: complex addition, complex multiplication, equality of complex numbers 4. State and prove properties of complex numbers. 29.2 Different Equations Require Different Number Systems When humans first counted, we tallied. We literally made notches on sticks, stones and bones. Thus the natural numbers, N, were born. But it wouldn’t be long before the necessity of fractions became obvious. One animal to be shared by four people (we will assume uniformly) meant that we had to develop the notion of 1/4. Though it would not have been expressed this way, the equation 4x = 1 does not have a solution in N and so we would have had to extend our notion of numbers to include fractions, the rationals. Q= a | a, b ∈ Z, b = 0 b This is an overstatement historically, because recognition of zero and negative numbers which are permitted in Q were very slow to come. But even these new numbers would not help solve equations of the form x2 = 2 which would arise naturally from isosceles right angled triangles. For this, the notion of number had to be extended to include irrational numbers, which combined with the rationals, give us the real numbers. 185 186 Chapter 29 Complex Numbers Eventually, via Hindu and Islamic scholars, western mathematics began to recognize and accept both zero and negative numbers. Otherwise, equations like 3x = 5x or 2x + 4 = 0 have no solution. Thus, mathematicians recognized that N⊂Z⊂Q⊂R but even R was inadequate because equations of the form x2 + 1 = 0 had no real solutions. And so, our number system was extended again. 29.3 Definition 29.3.1 Complex Number Complex Numbers A complex number z in standard form is an expression of the form x + yi where x, y ∈ R. The set of all complex numbers is denoted by C = {x + yi | x, y ∈ R} Example 1 Some examples are 3 + 4i, 0 + 5i (usually written 5i), 7 − 0i (usually written 7) and 0 + 0i (usually written 0). Definition 29.3.2 For a complex number z = x + yi, the real number x is called the real part and is written (z ) and the real number y is called the imaginary part and is written (z ). Real Part, Imaginary Part So (3 + 4i) = 3 and (3 + 4i) = 4. If z is a complex number where (z ) = 0, we will treat z as a real number and we will not write the term containing i. For example, z = 3 + 0i will be treated as a real number and will be written z = 3. Thus R⊂C and so N⊂Z⊂Q⊂R⊂C One has to wonder how much further the number system needs to be extended! Definition 29.3.3 Equality The complex numbers z = x + yi and w = u + vi are equal if and only if x = u and y = v . Section 29.3 Complex Numbers Definition 29.3.4 187 Addition is defined as Addition (a + bi) + (c + di) = (a + c) + (b + d)i Example 2 (1 + 7i) + (2 − 3i) = (1 + 2) + (7 − 3)i = 3 + 4i Definition 29.3.5 Multiplication is defined as Multiplication (a + bi) · (c + di) = (ac − bd) + (ad + cb)i Example 3 (1 + 7i) · (2 − 3i) = (1 · 2 − 7 · (−3)) + (1 · (−3) + 7 · 2)i = 23 + 11i The multiplication symbol is usually omitted and we write zw or (a + bi)(c + di). Exercise 1 Let u = 3 + i and v = 2 − 7i. Compute 1. u + v 2. u − v 3. uv 4. u2 v 5. u3 6. Exercise 2 v u (write the solution in the form x + yi where x, y ∈ R) Compute 1. i4k for any non-negative integer k 2. i4k+1 for any non-negative integer k 3. i4k+2 for any non-negative integer k 4. i4k+3 for any non-negative integer k The usual properties of associativity, commutativity, identities, inverses and distributivity that we associate with rational and real numbers also apply to complex numbers. 188 Proposition 1 Chapter 29 Complex Numbers Let u, v, z ∈ C. Then 1. Associativity of addition: (u + v ) + z = u + (v + z ) 2. Commutativity of addition: u + v = v + u 3. Additive identity: 0 = 0 + 0i has the property that z + 0 = z 4. Additive inverses: If z = x + yi then there exists an additive inverse of z , written −z with the property that z +(−z ) = 0. The additive inverse of z = x + yi is −z = −x − yi. 5. Associativity of multiplication: (u · v ) · z = u · (v · z ) 6. Commutativity of multiplication: u · v = v · u 7. Multiplicative identity: 1 = 1 + 0i has the property that z · 1 = z whenever z = 0. 8. Multiplicative inverses: If z = x + yi = 0 then there exists a multiplicative inverse of z , written z −1 , with the property that z · z −1 = 1. The multiplicative inverse of z = x + yi is z −1 = xx−yi2 . 2 +y 9. Distributivity: z · (u + v ) = z · u + z · v We will only prove the eighth property. Proof: We only need to demonstrate that z −1 = xx−yi2 is well-defined and that z · z −1 = 1. 2 +y 2 + y 2 = 0 and so z −1 is well-defined. Now we simply use complex arithmetic. Since z = 0, x x + yi · x − yi x2 + xy − xy − y 2 i2 x2 + y 2 = =2 =1 x2 + y 2 x2 + y 2 x + y2 Chapter 30 Properties Of Complex Numbers 30.1 Objectives The content objectives are: 1. Define conjugate and modulus 2. State and prove several properties of complex numbers. 30.2 Definition 30.2.1 Conjugate The complex conjugate of z = x + yi is the complex number Conjugate z = x − yi The conjugate of z = 2 + 3i is z = 2 − 3i. Proposition 1 (Properties of Conjugates (PCJ)) If z and w are complex numbers, then 1. z + w = z + w 2. zw = z w 3. z = z 4. z + z = 2 (z ) 5. z − z = 2i (z ) Exercise 1 Prove each part of the Properties of Conjugates proposition. (Hint: begin with “Let z = x + yi and w = u + vi.”) 189 190 Chapter 30 Properties Of Complex Numbers Exercise 2 Prove: Let z ∈ C. The complex number z is a real number if and only if z = z . Exercise 3 Prove: Let z ∈ C and z = 0. The complex number z is purely imaginary ( (z ) = 0) if and only if z = −z . Example 1 For z = i define z+i z−i Prove that w is a real number if and only if z is zero or purely imaginary. w= Solution: W is real ⇐⇒ w = w z+i z−i ⇐⇒ = z−i z+i ⇐⇒ z z − 1 + (z + z )i = z z − 1 − (z + z )i ⇐⇒ z + z = 0 ⇐⇒ (z ) = 0 ⇐⇒ z is zero or purely imaginary 30.3 Definition 30.3.1 Modulus The modulus of the complex number z = x + yi is the non-negative real number Modulus Example 2 |z | = |x + yi| = The modulus of z = 2 − 5i is |z | = x2 + y 2 (22 ) + (−5)2 = √ 29. Given two real numbers, say x1 and x2 , we can write either x1 ≤ x2 or x2 ≤ x1 . However, given two complex numbers, z1 and z2 , we cannot meaningfully write z1 ≤ z2 or z2 ≤ z1 . But since the modulus of a complex number is a real number, we can meaningfully write |z1 | ≤ |z2 |. The modulus gives us a means to compare the magnitude of two complex numbers, but not compare the numbers themselves. If Proposition 2 (z ) = 0, then the modulus corresponds to the absolute values of real numbers. (Properties of Modulus (PM)) If z and w are complex numbers, then 1. |z | = 0 if and only if z = 0 Section 30.3 Modulus 191 2. |z | = |z | 3. zz = |z |2 4. |zw| = |z ||w| 5. |z + w| ≤ |z | + |w|. This is the triangle inequality. Exercise 4 Prove each of the parts of the Properties of Modulus proposition. Chapter 31 Graphical Representations of Complex Numbers 31.1 Objectives The content objectives are: 1. Define complex plane, polar coordinates, polar form. 2. Convert between Cartesian and polar form. 3. Multiplication in polar form. 31.2 31.2.1 Definition 31.2.1 Complex Plane The Complex Plane (x, y ) The notation z = x + yi suggests a non-algebraic representation. Each complex number z = x + yi can be thought of as a point (x, y ) in a plane with orthogonal axes. Label one axis the real axis and the other axis the imaginary axis. The complex number z = x + yi then corresponds to the point (x, y ) in the plane. This interpretation of the plane is called the complex plane or the Argand plane. 192 Section 31.4 Polar Representation 193 Figure 31.2.1: The Complex Plane Exercise 1 Plot the following points in the complex plane. 1. 4 + i 2. −2 + 3i 3. −2 − i 31.2.2 Modulus Recall that the modulus of the complex number z = x + yi is the non-negative real number |z | = |x + yi| = x2 + y 2 There are a couple of geometric points to note about the modulus of z = x + yi. The Pythagorean Theorem is enough to prove that |z | is the distance from the origin to z in the complex plane, and that the distance between z and w = u + vi is just |z − w| = (x − u)2 + (y − v )2 . Exercise 2 Sketch all of the points in the complex plane with modulus 1. 31.3 Polar Representation There is another way to represent points in a plane which is very useful when working with complex numbers. Instead of beginning with the origin and two orthogonal axes, we begin with the origin O and a polar axis which is a ray leaving from the origin. The point P (r, θ) is plotted so that the distance OP is r, and the counter clockwise angle of rotation from the polar axis, measured in radians, is θ. Note that this allows for multiple representations since (r, θ) identifies the same point as (r, θ + 2πk ) for any integer k . The obvious question is how to go from one to the other. 194 Chapter 31 Graphical Representations of Complex Numbers Figure 31.3.1: Polar Representation 31.4 Converting Between Representations Simple trigonometry allows us to convert between polar and Cartesian coordinates. Figure 31.4.1: Connecting Polar and Cartesian Representations Given the polar coordinates (r, θ), the corresponding Cartesian coordinates (x, y ) are x = r cos θ y = r sin θ Given the Cartesian coordinates (x, y ), the corresponding polar coordinates are determined by r= x2 + y 2 x r y sin θ = r cos θ = Exercise 3 For each of the following polar coordinates, plot the point and convert to Cartesian coordinates. 1. (1, 0) Section 31.4 Converting Between Representations 195 2. (2, π/2) 3. (3, π ) 4. (2, 7π/2) 5. (4, π/4) 6. (4, 4π/3) Exercise 4 For each of the following Cartesian coordinates, plot the point and convert to polar coordinates. 1. (1, 0) 2. (0, 1) 3. (−1, 0) 4. (0, −1) 5. (1, 1) 6. (−1, 1) √ 7. (2, −2 3) From our earlier description of conversions, we can write the complex number z = x + yi as z = r cos θ + ri sin θ = r(cos θ + i sin θ) Definition 31.4.1 The polar form of a complex number z is Polar Form z = r(cos θ + i sin θ) where r is the modulus of z and the angle θ is called an argument of z . The expression cos θ + i sin θ is frequently abbreviated to cis θ and so we write z = rcis θ. Example 1 The following are representations of complex numbers in both Cartesian and polar form. 1. 1 = cis 0 √ 2. −1 + i = 2cis 3π/4 √ 3. −1 − 3i = 2cis 4π/3 196 Chapter 31 Graphical Representations of Complex Numbers One of the advantages of polar representation is that multiplication becomes very straightforward. Proposition 1 (Polar Multiplication of Complex Numbers (PMCN)) If z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ) are two complex numbers in polar form, then z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) Example 2 √ √ ( 2cis 3π/4) · (2cis 4π/3) = 2 2cis 3π 4π + 4 3 √ = 2 2cis 25π 12 √ = 2 2cis Proof: z1 z2 = r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 ) = r1 r2 ((cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(cos θ1 sin θ2 + sin θ1 cos θ2 )) = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) π 12 Chapter 32 De Moivre’s Theorem 32.1 Objectives The content objectives are: 1. State and prove De Moivre’s Theorem and do examples. 2. Derive Euler’s Formula. 32.2 Theorem 1 De Movre’s Theorem (De Movre’s Theorem (DMT)) If θ ∈ R and n ∈ Z, then (cos θ + i sin θ)n = (cos nθ + i sin nθ) Example 1 √ √ Consider the complex number z = 1/ 2 + i/ 2 which, in polar form is z = cos π/4 + i sin π/4. By De Moivre’s Theorem, z 10 = (cos π/4 + i sin π/4)10 = cos 10π/4 + i sin 10π/4 = cos π/2 + i sin π/2 = i. Proof: We will prove DeMoivre’s Theorem using three cases. 1. n = 0 2. n > 0 3. n < 0 197 198 Chapter 32 De Moivre’s Theorem For the case n = 0, DeMoivre’s Theorem reduces to (cos θ + i sin θ)0 = (cos 0 + i sin 0). By convention z 0 = 1 so the left hand side of the equation is 1. Since cos 0 = 1 and sin 0 = 0, the right hand side also evaluates to 1. For the case n > 0 we will use induction. P (n): (cos θ + i sin θ)n = (cos nθ + i sin nθ). Base Case We verify that P (1) is true where P (1) is the statement P (1): (cos θ + i sin θ)n = (cos 1θ + i sin 1θ). This is trivially true. Inductive Hypothesis We assume that the statement P (k ) is true for some k ≥ 1. P (k ): (cos θ + i sin θ)k = (cos kθ + i sin kθ). Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): (cos θ + i sin θ)k+1 = (cos(k + 1)θ + i sin(k + 1)θ) (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k (cos θ + i sin θ) by separating out one factor = (cos kθ + i sin kθ)(cos θ + i sin θ) by the Inductive Hypothesis = (cos(k + 1)θ + i sin(k + 1)θ) Polar Multiplication of Complex Numbers Lastly, for the case n < 0 we will use complex arithmetic. Since n < 0, n = −m for some m ∈ P. (cos θ + i sin θ)n = (cos θ + i sin θ)−m 1 = (cos θ + i sin θ)m 1 = (cos mθ + i sin mθ) = cos mθ − i sin mθ = cos(−mθ) + i(sin(−mθ)) = cos nθ + i sin nθ Corollary 2 If z = r(cos θ + i sin θ) and n is an integer, z n = rn (cos nθ + i sin nθ) Section 32.3 Complex Exponentials 32.3 199 Complex Exponentials If you were asked to find a real-valued function y with the property that dy = ky and y = 1 when x = 0 dx for some constant k , you would choose y = ekx And if you were asked to find the derivative of f (θ) = cos θ + i sin θ where i was treated as any other constant you would almost certainly write df (θ) = − sin θ + i cos θ dθ but then and so Definition 32.3.1 df (θ) = − sin θ + i cos θ = i(cos θ + i sin θ) = if (θ) dθ df (θ) = if (θ) and f (θ) = 1 when θ = 0 dθ By analogy, we define the complex exponential function by Complex Exponential eiθ = cos θ + i sin θ As an exercise, prove the following. Proposition 3 (Properties of Complex Exponentials (PCE)) eiθ · eiφ = ei(θ+φ) eiθ n = einθ The polar form of a complex number z can now be written as z = reiθ where r = |z | and θ is an argument of z . Out of this arises one of the most stunning formulas in mathematics. Setting r = 1 and θ = π we get eiπ = cos π + i sin π = −1 + 0i = −1 That is eiπ + 1 = 0 Who would have believed that e, i, π, 1 and 0 would have such a wonderful connection! Chapter 33 Roots of Complex Numbers 33.1 Objectives The content objectives are: 1. State and prove the Complex n-the Roots Theorem and do examples. 33.2 Definition 33.2.1 Complex n-th Roots If a is a complex number, then the complex numbers that solve Complex Roots zn = a are called the complex n-th roots. De Moivre’s Theorem gives us a straightforward way to find complex n-th roots of a. Theorem 1 (Complex n-th Roots Theorem (CNRT)) If r(cos θ + i sin θ) is the polar form of a complex number a, then the solutions to z n = a are √ θ + 2kπ θ + 2kπ n r cos + i sin for k = 0, 1, 2, . . . , n − 1 n n √ The modulus n r is the unique non-negative n-th root of r. This theorem shows that any complex number, including the reals, has exactly n different complex n-th roots. Example 1 Find all the complex fourth roots of −16. Solution: We will use the Complex n-th Roots Theorem. First, we write −16 in polar form as −16 = 16(cos π + i sin π ) 200 Section 33.2 Complex n-th Roots 201 Using the Complex n-th Roots Theorem the solutions are √ 4 16 cos = 2 cos π + 2kπ π + 2kπ + i sin 4 4 π kπ π kπ + i sin + + 4 2 4 2 for k = 0, 1, 2, 3 for k = 0, 1, 2, 3 The four distinct roots are given below π π + i sin 4 4 3π + i sin When k = 1, z1 = 2 cos 4 5π + i sin When k = 2, z2 = 2 cos 4 7π + i sin When k = 3, z3 = 2 cos 4 When k = 0, z0 = 2 cos =2 3π 4 5π 4 7π 4 i 1 √ +√ 2 2 −1 =2 √ + 2 −1 =2 √ + 2 1 =2 √ + 2 = i √ 2 −i √ 2 −i √ 2 √ √ 2+i 2 √ √ =− 2+i 2 √ √ =− 2−i 2 = √ √ 2−i 2 Graphing these solutions is illuminating. Figure 33.2.1: The Fourth Roots of -16 Note that the solutions are uniformly distributed around a circle whose radius is √ n r. Proof: As usual, when showing that a complete solution exists we work with two sets: the set S of solutions and the set T of specific representation of the solution. We then show that S = T by mutual inclusion. Our two sets are S = {z ∈ C | z n = a} 202 Chapter 33 and T= √ n r θ + 2kπ n cos θ + 2kπ n + i sin Roots of Complex Numbers k = 0, 1, 2, . . . , n − 1 where a = r(cos θ + i sin θ). We begin by showing that T ⊆ S . Let t = √ n r cos θ + 2kπ n θ + 2kπ n n = r(cos(θ + 2kπ ) + i sin(θ + 2kπ )) √ n r tn = n be an element of T . Now cos = r(cos θ + i sin θ) De Moivre’s Theorem trigonometry =a Hence, t is a solution of z n = a, that is, t ∈ S . Now we show that S ⊆ T . Let w = s(cos φ + i sin φ) be an n-th root of a. Since a = r(cos θ + i sin θ) we have wn = a ⇐⇒ (s(cos φ + i sin φ))n = r(cos θ + i sin θ) ⇐⇒ sn (cos nφ + i sin nφ) = r(cos θ + i sin θ) De Moivre’s Theorem Now two complex numbers in polar form are equal if and only if their moduli are equal and their arguments differ by an integer multiple of 2π . So √ sn = r ⇒ s = n r and θ + 2πk n where k ∈ Z. Hence, the n-th roots of a are of the form nφ − θ = 2πk ⇒ φ = √ n r cos θ + 2kπ n + i sin θ + 2kπ n for k ∈ Z But this is k ∈ Z, not k = 0, 1, 2, . . . , n − 1. Since w is an n-th root of a, there exists an integer k0 so that w= √ n r cos θ + 2 k0 π n + i sin θ + 2k0 π n w= √ n r cos θ + 2 k1 π n + i sin θ + 2k1 π n If we can show that if and only if k0 ≡ k1 (mod n) whenever r = 0, then w ∈ T . Now k0 ≡ k1 ⇐⇒ k0 − k1 = n ⇐⇒ 2πk0 − 2πk1 = 2πn 2πk0 2πk1 ⇐⇒ − = 2π n n θ + 2πk0 θ + 2πk1 ⇐⇒ − = 2π n n (mod n) for some ∈Z for some ∈Z for some ∈Z for some ∈Z Section 33.3 More Examples 33.3 203 More Examples Exercise 1 An n-th root of unity is a complex number that solves z n = 1. Find all of the sixth roots of unity. Express them in standard form and graph them in the complex plane. Exercise 2 Find the square roots of −2i. Express them in standard form and graph them in the complex plane. Chapter 34 An Introduction to Polynomials 34.1 Objectives The content objectives are: 1. Define polynomial, coefficient, F[x], degree, zero polynomial, linear polynomial, quadratic polynomial, cubic polynomial, equal, sum, difference, product, quotient, remainder, divides and factor. 2. Define operations on polynomials. 3. State the Division Algorithm for Polynmials. 4. Do examples. 34.2 Polynomials Our number systems were developed in response to the need to find solutions to real polynomials. We are now able to solve all equations of the form a2 x2 + a1 x + a0 = 0 or xn − a0 = 0 whether the coefficients are real or complex. In fact, a great deal more is known. Let F be a field. Roughly speaking, a field is a set of numbers that allows addition, subtraction, multiplication and division. The rational numbers Q, the real numbers R, the complex numbers C and the integers modulo a prime p, Zp are all fields. The integers are not a field because we cannot divide 2 by 4 and get an integer. Since division is just multiplication by an inverse, Z6 is not a field since [3] has no inverse. 204 Section 34.3 Operations on Polynomials Definition 34.2.1 205 A polynomial in x over F is an expression of the form Polynomial an xn an−1 xn−1 + · · · + a1 x + a0 where all of the ai belong to F. The ai are called coefficients. We use F[x] to denote the set of polynomials in x with coefficients from F. Example 1 1. x2 + 7x − 1 ∈ R[x] 2. x3 − 7ix + (5 − 2i) ∈ C[x] 3. [3]x5 + [2]x3 + [6] ∈ Z7 [x] It is important to be clear about what field the coefficients come from. The polynomial x2 + 1 belongs to both R[x] and C[x] but the equation x2 + 1 = 0 has complex solutions but no real solutions. Definition 34.2.2 If an = 0 in the polynomial Degree et al an xn an−1 xn−1 + · · · + a1 x + a0 then the polynomial is said to have degree n. The zero polynomial has all of its coefficients zero and its degree is not defined. Polynomials of degree 1 are called linear polynomials, of degree 2, quadratic polynomials, and of degree 3 cubic polynomials. 34.3 Operations on Polynomials We very frequently use f (x) to denote an element of F[x] and write n f (x) = an xn an−1 xn−1 + · · · + a1 x + a0 = ai xi i=0 Let f (x), g (x) ∈ F[x] where n n n−1 f (x) = an x an−1 x ai xi + · · · + a1 x + a0 = i=0 n g (x) = bn xn bn−1 xn−1 + · · · + b1 x + b0 = bi xi i=0 Definition 34.3.1 Equal The polynomials f (x) and g (x) are equal if and only if ai = bi for all i. 206 Chapter 34 An Introduction to Polynomials Polynomials can be added, subtracted and multiplied as algebraic expressions exactly as you have done in high school. Definition 34.3.2 The sum of the polynomials f (x) and g (x) is defined as Sum max(n,m) (ai + bi )xi f (x) + g (x) = i=0 where any “missing” terms have coefficient zero. Example 2 1. In R[x], if f (x) = x2 + 7x − 1 and g (x) = 3x4 − x3 + 4x2 − x + 5 then f (x) + g (x) = 3x4 − x3 + 5x2 + 6x + 4. 2. In C[x], if f (x) = x3 − 7ix + (5 − 2i) and g (x) = (4 + 3i)x + (7 + 7i)x then f (x) + g (x) = x3 + (4 − 4i)x + (12 + 5i)x. 3. In Z7 [x], if f (x) = [3]x5 + [2]x3 + [6] and g (x) = [2]x4 + [5]x3 + [2]x2 + [4] then f (x) + g (x) = [3]x5 + [2]x4 + [2]x2 + [3]. Definition 34.3.3 The difference of the polynomials f (x) and g (x) is defined as Difference max(n,m) (ai − bi )xi f (x) − g (x) = i=0 where any “missing” terms have coefficient zero. Exercise 1 Find the difference of each of the pairs of polynomials given in Example 2. The definition of the product of two polynomials looks more complicated than it is. Definition 34.3.4 The product of the polynomials f (x) and g (x) is defined as Product m+n ci xi f (x) · g (x) = i=0 where i ci = a0 bi + a1 bi−1 + · · · + ai−1 b1 + ai b0 = aj bi−j j =0 Section 34.3 Operations on Polynomials Example 3 207 In R[x], if f (x) = x2 + 7x − 1 and g (x) = 3x + 2 then f (x) · g (x) = Add long multiplication example here. Now we run into the same issue we had with the integers, division. Though it makes sense to say that x − 3 divides x2 − 9 since x2 − 9 = (x − 3)(x + 3) what do we do when there is a remainder? Just as we had a division algorithm for integers, we have a division algorithm for polynomials. Proposition 1 (Division Algorithm for Polynomials (DAP)) If f (x) and g (x) are polynomials in F[x] and g (x) is not the zero polynomial, then there exist unique polynomials q (x) and r(x) in F[x] such that f (x) = q (x)g (x) + r(x) Definition 34.3.5 Quotient, Remainder where deg r(x) < deg g (x) or r(x) = 0 The polynomial q (x) is called the quotient polynomial. The polynomial r(x) is called the remainder polynomial. If r(x) = 0, we say that g (x) divides f (x) or f (x) is a factor of f (x) and we write g (x) | f (x). Add long division examples over several fields here. Exercise 2 For each f (x) and g (x), find the quotient and remainder polynomials. 1. Let f (x) and g (x) be the real polynomials f (x) = 2x4 + 6x3 − x + 4 and g (x) = x2 + 3. 2. Let f (z ) and g (z ) be the complex polynomials f (z ) = iz 3 + (2 + 4i)z 2 + (3 − i)z + (4 − 8i) and g (z ) = iz + (2 − 2i). Chapter 35 Factoring Polynomials 35.1 Objectives The content objectives are: 1. Define polynomial equation, solution and root. 2. State the Fundamental Theorem of Algebra 35.2 Definition 35.2.1 Polynomial Equation Polynomial Equations A polynomial equation is an equation of the form an xn an−1 xn−1 + · · · + a1 x + a0 = 0 which will often be written as f (x) = 0. An element c ∈ F is called a root or zero of the polynomial f (x) if f (c) = 0. That is, c is a solution of the polynomial equation f (x) = 0. The history of mathematics is replete with exciting and sometimes bizarre stories of mathematicians as they looked, in vain, for an algorithm that would find a root of an arbitrary polynomial. We can now prove that no such algorithm exists. It is known though, that a root exists for every complex polynomial. This was proved in 1799 by the brilliant mathematician Karl Friedrich Gauss. Theorem 1 (Fundamental Theorem of Algebra (FTA)) For all complex polynomials f (z ) with deg(f (z )) ≥ 1, there exists a z0 ∈ C so that f (z0 ) = 0. Ironically, we can prove a root exists, we just can’t construct one in general. The proof of this theorem is demanding and is left for later courses. We can use the Division Algorithm for Polynomials to help though. Recall 208 Section 35.2 Polynomial Equations Proposition 2 209 (Division Algorithm for Polynomials (DAP)) If f (x) and g (x) are polynomials in F[x] and g (x) is not the zero polynomial, then there exist unique polynomials q (x) and r(x) in F[x] such that f (x) = q (x)g (x) + r(x) Proposition 3 where deg r(x) < deg g (x) or r(x) = 0 (Remainder Theorem (RT)) The remainder when the polynomial f (x) is divided by (x − c) is f (c). Example 1 Find the remainder when f (z ) = 3z 12 − 8iz 5 + (4 + i)z 2 + z + 2 − 3i is divided by z + i. Solution: One could do the painful thing and carry out long division. Another possibility is to use the Remainder Theorem and compute f (−i). f (−i) = 3(−i)12 − 8i(−i)5 + (4 + i)(−i)2 + (−i) + 2 − 3i = 3 − 8i(−i) + (4 + i)(−1) − i + 2 − 3i = 3 − 8 − 4 − i − i + 2 − 3i = −7 − 5i The remainder is −7 − 5i. Proof: By the Division Algorithm for Polynomials, there exist unique polynomials q (x) and r(x) such that f (x) = q (x)(x − c) + r(x) where deg r(x) < 1 or r(x) = 0 Therefore, the remainder r(x) is a constant (which could be zero) which we will write as r0 . Hence f (x) = q (x)(x − c) + r0 Substituting x = c into this equation gives f (c) = r0 . Corollary 4 (Factor Theorem 1 (FT1)) The linear polynomial (x − c) is a factor of the polynomial f (x) if and only if f (c) = 0. Equivalently, Corollary 5 (Factor Theorem 2 (FT2)) The linear polynomial (x − c) is a factor of the polynomial f (x) if and only if c is a root of the polynomial f (x). 210 Chapter 35 Factoring Polynomials How do we go about actually factoring polynomials? In general, this is hard to do. There are no formulas for roots if the polynomial has degree five or more. But if the polynomial has integer coefficients, we have a good starting point. Theorem 6 (Rational Roots Theorem (RRT)) Let f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 be a polynomial with integer coefficients. If p is a rational root with gcd(p, q ) = 1, then p | a0 and q | an . q In order to find a rational root of f (x), we only need to examine a finite set of rational numbers, those whose numerator divides the constant term and those whose denominators divide the leading coefficient. Note that the theorem only suggests those rational numbers that might be roots. It does not guarantee that any of these numbers are roots. Example 2 If possible, find a rational root of f (x) = 2x4 + x3 + 6x + 3. Solution: We will use the Rational Roots Theorem. The divisors of 2 are ±1 and ±2. The divisors of 3 are ±1 and ±3. Hence, the candidates for rational roots are 1 3 ±1, ± , ±3, ± 2 2 Now test each of these candidates. x 1 −1 f (x) 12 −2 Thus, the only root is Proof: If 1 2 25 4 −1 2 0 3 −3 210 120 3 2 51 2 −3 2 3 4 −1 2. p is a root of f (x) then q an p q n + an−1 p q n−1 + · · · + a2 p q 2 + a1 p q + a0 = 0 Multiplying by q n gives an pn + an−1 pn−1 q + · · · + a2 p2 q n−2 + a1 pq n−1 + a0 q n = 0 and an pn = −q an−1 pn−1 + · · · + a2 p2 q n−3 + a1 pq n−2 a0 q n−1 Since all of the symbols in this equation are integers, both the right hand side and left hand side are integers. Since q divides the the right hand side, q divides the left hand side, that is q | an p n Since gcd(p, q ) = 1 we can repeatedly use the proposition on Coprimeness and Divisibility to show that q | an . In a similar way, we can show that p | a0 . Section 35.2 Polynomial Equations 211 Exercise 1 Is x + 1 a factor of x10 + 1, or x9 + 1. Can you make a statement about when x + 1 divides or does not divide x2n + 1, or x2n+1 + 1 for n a positive integer? Exercise 2 If x + (2 + i) is a root of f (x) = x4 + 4x3 + 2x2 − 12x − 15, factor f (x) into products of real polynomials and complex polynomials of lowest degree. Exercise 3 Prove that x = √ n p is irrational for any integer n > 1. Complex roots of real polynomials come in conjugate pairs. Chapter 36 The Shortest Path Problem 36.1 Objectives The technique objectives are: 1. Abstract from a map to a graph. 2. Formulate an algorithm. 3. Extend plausible uses. 36.2 The Problem Suppose you are living in downtown Toronto (the pink dot on the map) on a co-op work term and you want to escape the intense July heat by going to Sibbald Point Provincial Park (the blue dot on the map) to swim in Lake Simcoe. See Figure 36.2.1. You could take Highway 404 past the 401, past the 407 up to the end of Highway 404, and then take a minor road to Highway 48 and go north from there. But perhaps it would be better to take Lakeshore Drive to Highway 48 and go straight north. Your task is to find an algorithm, a strategy, to find the shortest route between downtown Toronto and Sibbald Point Provincial Park. 212 Section 36.2 The Problem 213 Figure 36.2.1: Sibbald Point Provincial Park 214 Chapter 36 36.3 The Shortest Path Problem Abstraction Let’s focus on what’s important in the problem. Looking at the map there is, for our purpose, lots of information that is not important: colours, parking locations, where the Green Belt is, towns not along the way. What is really important are locations where we might change directions, routes between those locations, and distances. We’ll highlight locations on the map as grey dots and connections between locations as solid teal lines. See Figure 36.3.1. Figure 36.3.1: Locations and Connections Section 36.5 Algorithm 215 But since we don’t need the rest of the detail, let’s omit it and include only locations, connections and distances. See Figure 36.3.2. 20 45 15 120 60 60 10 10 10 10 30 5 20 25 Figure 36.3.2: The Essentials 36.4 Algorithm With a partner, draw a random map and attempt to discover an algorithm that will find the shortest route from one location to another. [Note to instructor: solicit some algorithms and have the class assess whether the algorithm might work.] 36.5 Extensions This problem is set as minimal distances between two points. But perhaps instead of distance we could use time or cost. And instead of a person travelling we could have couriers delivering packages, or electrical signals carrying phone calls. In fact, there are surprising uses as well including managing cutting stock in steel mills and finding optimal schedules for construction projects. Chapter 37 Paths, Walks, Cycles and Trees 37.1 Objectives The technique objectives are: 1. Practice with by contradiction. 2. Practice with unqueness. The content objectives are: 1. Define graph, walk, path, cycle, and tree. 2. Construct diagrams corresponding to graphs. 3. Observe: Any walk can be decomposed into at most one path and a collection of cycles. 4. Prove: There is a unique path between every pair of vertices in a tree. 37.2 Definition 37.2.1 Graph The Basics A graph G is a pair (V, E ) where V is a finite, nonempty set, and E is a set of unordered pairs of elements of V . The elements of V are called vertices and the elements of E called edges. It is often very useful to represent a graph as a drawing where vertices correspond to points and edges correspond to lines between vertices. Graphs may be represented by more than one diagram as illustrated in Example 1. 216 Section 37.2 The Basics Example 1 217 Let G = (V, E ) where V = {1, 2, 3, 4, 5, 6, 7} and E = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 1}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}} . 7 2 1 6 7 3 1 5 2 3 4 5 6 4 Figure 37.2.1: Two representations of the same graph Definition 37.2.2 Adjacent, Incident If edge e = {u, v }, then we say that u and v are adjacent vertices, and that edge e is incident with vertices u and v . We can also say that the edge e joins u and v . Vertices adjacent to a vertex u are called neighbours of u. A graph is completely specified by the pairs of vertices that are adjacent, and the only function of a line in the diagram is to indicate that two vertices are adjacent. Definition 37.2.3 Walk A walk W is a non-empty sequence of edges W = {{v0 , v1 }, {v1 , v2 }, {v2 , v3 }, . . . , {vn−1 , vn }} . Since vi−1 and vi uniquely determine an edge e of a walk, we will usually just list the vertices. Thus W = v0 , v1 , v2 , v3 , . . . , vn−1 , vn . Definition 37.2.4 Path Definition 37.2.5 Cycle If v0 = s and vn = t in the walk W , we call W an st-walk. If no vertex in the walk is repeated, that is, if v0 , v1 , v2 , . . . , vn are all distinct, then W is called a path. If v0 = vn and v0 , v1 , v2 , . . . , vn−1 are all distinct, then W is called a cycle. 218 Chapter 37 2 1 6 Paths, Walks, Cycles and Trees 7 5 3 4 Figure 37.2.2: The bold lines indicate the walk W = 1, 6, 7, 3, 4, 7, 3, 2. 2 1 6 7 5 3 4 Figure 37.2.3: The bold lines indicate the path P = 1, 6, 7, 3, 2 2 1 6 7 5 3 4 Figure 37.2.4: The bold lines indicate the cycle C = 7, 3, 4, 7 Note that the walk W = 1, 6, 7, 3, 4, 7, 3, 2 can be decomposed into the path P = 1, 6, 7, 3, 2 and the cycle C = 7, 3, 4, 7. In fact, we can always perform this kind of decomposition for walks but before we state the appropriate theorem, we need to define a few more terms. Definition 37.2.6 Collection, Decomposed By a collection we mean a family of objects where repetition is allowed. Let W be an st-walk. If s = t, we say that W can be decomposed into a collection C of cycles if, for every edge e the number of times e occurs in W is the same as the number of times e occurs in cycles of C . If s = t, we say that W can be decomposed into an st-path P and a collection C of cycles if, for every edge e the number of times e occurs in W is the same as Section 37.3 Trees 219 the number of times e occurs in P and the cycles of C . We will state, but not prove, the following proposition. Proposition 1 (Walk Decomposition (WD)) Let W be an st-walk. 1. If s = t, then W can be decomposed into a non-empty collection of cycles. 2. If s = t and a vertex is not repeated in W , then W is a path. 3. If s = t and a vertex is repeated in W , then W can be decomposed into a path and a non-empty collection of cycles. You may wonder what the difference is between the definition of decomposition and the proposition Walk Decomposition. The definition allows for the possibility that some walks cannot be decomposed. The proposition states that all walks can be decomposed. Definition 37.2.7 Connected To say that a graph G is connected means that there is a path between any two vertices of G. We will assume for this course that all of our graphs are connected, though in general, that is not a safe assumption. 37.3 Trees A tree is a very special and incredibly useful kind of graph. Definition 37.3.1 A tree is a connected graph with no cycles. Tree 2 1 6 7 5 3 4 Figure 37.3.1: A tree We will prove several propositions about trees starting with this one. 220 Proposition 2 Chapter 37 Paths, Walks, Cycles and Trees (Unique Paths in Trees (UPT)) There is a unique path between every pair of vertices in a tree. We normally begin our proofs by explicitly identifying the hypothesis and the conclusion. Unique Paths in Trees is not in “If A, then B .” form, so let’s first restate it. Recall that the hypothesis is what we get to start with, and the conclusion is what we must show. We start with a tree. Call it T . We must show that there is a unique path between every pair of vertices in T . Hence, we could restate Unique Paths in Trees as Proposition 3 (Unique Paths in Trees (UPT)) If T is a tree, then there is a unique path between every pair of vertices in T . Working forwards and backwards to prove this proposition will be problematic. So, it’s time for a different technique, proof by contradiction. Normally, when we wish to prove that the statement “A implies B ” is true, we assume that A is true and show that B is true. What would happen if B were true, but we assumed it was false and continued our reasoning based on the assumption that B was false? Since a mathematical statement cannot be both true and false, it seems likely we would eventually encounter a mathematically non-sensical statement. Then we would ask ourselves “How did we arrive at this nonsense?” and the answer would have to be that our assumption that B was false was wrong and B is, in fact, true. Proofs by contradiction have the following structure. 1. Assume that A is true. 2. Assume that B is false, or equivalently, assume that NOT B is true. 3. Reason forward from A and NOT B to reach a contradiction. We will prove Unique Paths in Trees by contradiction. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Suppose that u and v are any two distinct vertices of T . 2. Since T is connected, there is at least one path connecting u to v . 3. Now suppose that there are two distinct uv -paths, P1 = x0 , x1 , x2 , . . . , xn and P2 = y0 , y1 , y2 , . . . , ym . Thus u = x0 = y0 and v = xn = ym . 4. We can construct a walk W beginning with u and ending at u that consists of “walking” from u to v in P1 , then from v to u “backwards” in P2 . More specifically, W = x0 , x1 , x2 , . . . , xn , ym−1 , ym−2 , ym−3 , . . . , y0 . 5. By Part (1) of Proposition 1, W can be decomposed into a non-empty collection of cycles. Section 37.3 Trees 221 6. But then the tree T contains cycles, contradicting the definition of a tree. Analysis of Proof We will begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: T is a tree. Conclusion: There is a unique path between every pair of vertices in T . Core Proof Technique: Contradiction. Preliminary Material: Definition of tree. Sentence 1 Suppose that u and v are any two distinct vertices of T . The conclusion contains a universal quantifier, every. Let’s first identify the components of the universal quantifier. Objects: Universe of discourse: Certain property: Something happens: Vertices u and v Vertices of the tree T None specified. There is a unique path between u and v . Since we are using a universal quantifier in the conclusion of the proposition, the author uses Choose method. Sentence 2 Since T is connected, there is at least one path connecting u to v . Before the author can show that there is a unique path, the author must first show that a path exists. Sentence 3 Now suppose that there are two distinct uv -paths, P1 = x0 , x1 , x2 , . . . , xn and P2 = y0 , y1 , y2 , . . . , ym . Thus u = x0 = y0 and v = xn = ym . The author is negating the conclusion and so is going to use on of two techniques, Contradiction or Contrapositive. Since the author hasn’t indicated which, it is useful to look ahead in the proof to find out. The last sentence of the proof makes it clear that the author is using Contradiction. Sentence 4 We can construct a walk W beginning with u and ending at u that consists of “walking” from u to v in P1 , then from v to u “backwards” in P2 . More specifically, W = x0 , x1 , x2 , . . . , xn , ym−1 , ym−2 , ym−3 , . . . , y0 . Sentence 5 By Part (1) of Proposition 1, W can be decomposed into a non-empty collection of cycles. The difficult part in proofs by contradiction is finding a contradiction. In Sentence 4 the author constructs a walk and in Sentence 5 the author shows that the walk contains cycles. But cycles don’t exist in trees and so Sentence 6 But then the tree T contains cycles, contradicting the definition of a tree. This is also an example of working with uniqueness. Chapter 38 Trees 38.1 Objectives The technique objectives are: 1. Induction. The content objectives are: 1. Define degree. 2. Prove Two Vertices of Degree One. 3. Prove: Number of Vertices in a Tree. 38.2 Definition 38.2.1 Degree Proposition 1 Properties of Trees Let G be a graph. The number of edges incident with a vertex v is called the degree of v and is denoted by deg(v ). In Figure 38.2.1, vertex a has degree 3 and vertex b has degree 2. (Two Vertices of Degree One (TVDO)) If T is a tree with at least two vertices, then T has at least two vertices of degree one. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Find a longest path P = w0 w1 w2 . . . wn in T , say between nodes u = w0 and v = wn . 2. Since any edge in the tree constitutes a path, P must contain at least one edge so u = v. 3. Thus, the vertex wn−1 in the path is adjacent to v but distinct from v . 222 Section 38.2 Properties of Trees 223 i s h g f b e d c a s Figure 38.2.1: Graph corresponding to Toronto - Sibbald Point map 4. If deg(v ) > 1, there must be another vertex, w, distinct from wn−1 and adjacent to v . 5. If w is in P , then a cycle would exist but trees do not have cycles. Hence, w is not in P. 6. If w is not in P , then we could add edge {v, w} to P to get a path longer than P , contradicting the assumption that P is a longest path in T . 7. Hence, deg(v ) = 1. 8. Similarly, deg(u) = 1 and so two vertices of degree one exist in T . Analysis of Proof We will begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: T is a tree with at least two vertices. Conclusion: T has at least two vertices of degree one. Core Proof Technique: Construction and Contradiction (three times!). Preliminary Material: Definition of tree and of degree. 224 Chapter 38 Trees Sentence 1 Find a longest path P = w0 w1 w2 . . . wn in T , say between nodes u = w0 and v = wn . The conclusion contains an existential quantifier, has. Let’s first identify the components of the existential quantifier. Objects: Universe of discourse: Certain property: Something happens: Two vertices (unnamed) Vertices of the tree T None specified. Both vertices have degree 1. Since the proposition contains an existential quantifier in the conclusion, the author is using the Construct method. This sentence serves two purposes. First, it implicitly identifies the two objects that will be constructed, u and v . And second, it sets up the contradictions that will be needed later. Sentence 2 Since any edge in the tree constitutes a path, P must contain at least one edge so u = v . Given that the author intends to show that u and v are distinct vertices of degree one, the author must first establish that u = v . Also, the following argument will require that the path contain an edge. Sentence 3 Thus, the vertex wn−1 in the path is adjacent to v but distinct from v . The author is setting up the contradiction, though it is not at all clear from here how that contradiction will be displayed. Sentence 4 If deg(v ) > 1, there must be another vertex, w, distinct from wn−1 and adjacent to to v . From the analysis of the first sentence, the author intends to show that v has degree one. That means this sentence indicates the author is going to proceed by contradiction. Sentence 5 If w is in P , then a cycle would exist but trees do not have cycles. Hence, w is not in P . This is a miniature proof by contradiction of the statement “If deg(v ) > 1 and w is adjacent to v , then w is not in P .” Sentence 5 begins with the negation of the conclusion and finds a contradiction quickly. If w is in P , then the walk constructed by taking the subpath from w to v in P and adding the edge {v, w} yields a cycle, but trees do not contain cycles by definition. Sentence 6 If w is not in P , then we could add edge {v, w} to P to get a path longer than P , contradicting the assumption that P is a longest path in T . This is another miniature proof by contradiction, this time of the statement “If deg(v ) > 1 and w is adjacent to v , then w is in P .” Sentence 7 Hence, deg(v ) = 1. Assuming that deg(v ) > 1 leads to an adjacent vertex w being both in P and not in P , a contradiction. Since the author’s reasoning is correct, it must be the case that the assumption deg(v ) > 1 is false. Since T is connected, deg(v ) > 0 so deg(v ) = 1. Section 38.2 Properties of Trees 225 Sentence 8 Similarly, deg(u) = 1 and so two vertices of degree one exist in T . Similarly is a useful but dangerous word in proofs. If the conditions really are similar, then using “similarly” spares tedious effort in checking the details. However, if the conditions are not similar, the use of “similarly” could be masking a fatal error. In this case, the argument is identical when w1 replaces wn−1 . Proposition 2 (Number of Vertices in a Tree (NVT)) Let T = (V, E ) be a tree. Then |V | = |E | + 1. Since V is an integer, we could consider all trees with one vertex, two vertices, three vertices and so on, this seems like a perfect case for induction. Let’s be very clear about what our statement P (n) is. P (n): Let T = (V, E ) be a tree with n vertices. Then n = |E | + 1. Now we can begin the proof. Proof: Base Case We verify that P (1) is true where P (1) is the statement P (1): Let T = (V, E ) be a tree with one vertex. Then 1 = |E | + 1. This is equivalent to stating that |E | = 0. Since a tree with one vertex has no edges, the base case is true. Inductive Hypothesis We assume that the statement P (k ) is true for k ≥ 2. P (k ): Let T = (V, E ) be a tree with k vertices. Then k = |E | + 1. Inductive Conclusion Now show that the statement P (k + 1) is true. P (k + 1): Let T = (V, E ) be a tree with k + 1 vertices. Then k + 1 = |E | + 1. By Proposition 1, we know that there is at least one vertex of degree one in T . Let’s call such a vertex v . Since deg(v ) = 1, v is adjacent to only one vertex, say u. Deleting the vertex v and the edge {u, v } creates a new tree T where T has k vertices and |E | − 1 edges. By our Inductive Hypothesis therefore, k = (|E | − 1) + 1 ⇒ k = |E |. But T has one more vertex and more edge than T so k + 1 = |E | + 1 as required. The result is true for n = k +1, and so holds for all n by the Principle of Mathematical Induction. Chapter 39 Dijkstra’s Algorithm 39.1 Objectives The content objectives are: 1. Be able to execute Dijkstra’s Algorithm. 39.2 Dijkstra’s Algorithm [Note to instructors: How you start this depends on how you ended the class that introduced the Shortest Path Problem.] Let’s look at a formal expression for solving the shortest path problem. We can think of the “Require” statement as the pre-conditions to the algorithm, or the hypothesis to a proposition. In this case, we require a graph with non-negative weights on the edges, and a starting vertex s. We can think of the “Ensure” statement as the postconditions to the algorithm or the conclusion to a proposition. In this case, the algorithm should terminate with a tree of shortest paths rooted at s, and the distances of a shortest path from s to each node. Though our original problem talked about distances, the values we assign to the edges of the graph could also be time or capacity or costs. The convention is to call these values weights, which is why the function from the edges to the real numbers is named w. Let’s watch the algorithm in operation. Our example appears in Figure 39.2.1. The initialization steps of the algorithm set the distance to s at 0, and the provisional distances to all other vertices at infinity. By abuse of notation, we will treat infinity as a real number. We will record distances as numeric labels in blue near the vertices. The set V initially contains only s and E is empty. We will show the nodes in V as bold circles and the edges in E as bold lines. Note that at every stage of the algorithm, T = (V , E ) is a tree of shortest paths to the vertices in V . See Figure 39.2.2 Now the algorithm examines each edge with one vertex in V and one vertex not in V . If using the edge creates a shorter path to a vertex not in V , then the provisional distance to that vertex is updated. Figure 39.2.3 shows the results of the update. Edges and distances 226 Section 39.2 Dijkstra’s Algorithm 227 Algorithm 2 Dikstra’s Algorithm Require: G = (V, E ); w : E → R; w({u, v }) ≥ 0, ∀{u, v } ∈ E ; and a designated node s. Ensure: T = (V , E ) is a tree rooted at s of shortest paths from s to every other node; d : V → R gives the distance of a shortest path to v , ∀v ∈ V . {Initialize} d(s) ← 0 d(v ) ← ∞, ∀v ∈ V, v = s V ← {s} E ←∅ T ← (V , E ) repeat for every edge {u, v } ∈ E where u ∈ V and v ∈ V do if d(v ) < d(u) + w({u, v }) then d(v ) ← d(u) + w({u, v }) end if end for Choose a y ∈ V so that d(y ) = min{d(w) | w ∈ V } For the y just chosen, choose {x, y } ∈ E where x ∈ V and d(y ) = d(x) + w({x, y }) V ← V ∪ {y } E ← E ∪ {{x, y }} T ← (V , E ) until V = V 3 s 1 b 9 2 c 3 4 a 1 d Figure 39.2.1: Graph G with weights involved in the updates are shown in green. The infinite values previously assigned to vertices a, b and c have been crossed out. Continuing with the update, choose the vertex not in V with the smallest provisional distance. In this iteration, the choice is b. Add b to V and {s, b} to E . This update is shown in Figure 39.2.4. The nodes in V are shown as bold circles and the edges in E as bold lines. Note that T = (V , E ) is a tree of shortest paths to the vertices in V . We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, b}, V = V and so we continue. Again, the algorithm examines each edge with one vertex in V and one vertex not in V . If using the edge creates a shorter path to a vertex in V , then the provisional distance to that vertex is updated. Figure 39.2.5 shows the results of the update. Edges and distances involved in the updates are shown in green. Now choose the vertex not in V with the smallest provisional distance. In this iteration, 228 Chapter 39 ∞ 0 3 s 1 ∞ a 9 b Dijkstra’s Algorithm 2 c 3 4 ∞ 1 d ∞ Figure 39.2.2: After initialization 0 3 s 1 ∞1 9 b ∞3 a 2 ∞9 c 3 4 1 d ∞ Figure 39.2.3: First update of d 0 3 3 s 1 1 b a 9 2 c 3 4 9 1 d ∞ Figure 39.2.4: End of first iteration the choice is a. Add a to V and {s, a} to E . This update is shown in Figure 39.2.6. The nodes in V are shown as bold circles and the edges in E as bold lines. Again, note that T = (V , E ) is a tree of shortest paths to the vertices in V . We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, a, b}, V = V and so we Section 39.2 Dijkstra’s Algorithm 229 0 3 3 s 1 1 b a 9 2 c 3 4 94 1 d ∞5 Figure 39.2.5: Second update of d 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.6: End of second iteration continue. Again, the algorithm examines each edge with one vertex in V and one vertex not in V . If using the edge creates a shorter path to a vertex in V , then the provisional distance to that vertex is updated. In this iteration, no updates to provisional distances took place. Figure 39.2.7 shows the results of the update. Edges and distances involved in the updates are shown in green. 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.7: Third update of d 230 Chapter 39 Dijkstra’s Algorithm Now choose the vertex not in V with the smallest provisional distance. In this iteration, the choice is c. Add c to V and {b, c} to E . This update is shown in Figure 39.2.8. Again, note that T = (V , E ) is a tree of shortest paths to the vertices in V . 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.8: End of third iteration We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, a, b, c}, V = V and so we continue. Again, the algorithm examines each edge with one vertex in V and one vertex not in V . If using the edge creates a shorter path to a vertex in V , then the provisional distance to that vertex is updated. Figure 39.2.9 shows the results of the update. 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.9: Fourth update of d Now choose the vertex not in V with the smallest provisional distance. In this iteration, the choice is d. Add d to V . But now both and {b, d} and {c, d} match the condition to be added to E . Which one should be added or should both be added? It is only necessary to choose one, say {b, d}. This update is shown in Figure 40.3.1. Again, note that T = (V , E ) is a tree of shortest paths to the vertices in V . Now, finally V = V and the algorithm terminates. Exercise 1 (Note to instructors: you may wish to do another example before assigning Section 39.3 Certificate of Optimality 231 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.10: End of fourth iteration and termination of the algorithm this exercise.) Turn to a neighbour and create a random small graph, say of 6 vertices, and run Dijkstra’s algorithm on your graph. 39.3 Certificate of Optimality Based on our experiments when we began this section, the example we did together, and your own examples, it seems that we have lots of empirical evidence that Dijkstra’s algorithm works. But evidence is not a proof. Moreover, if a colleague were to provide us with a graph, edge weights and a proposed tree of shortest paths, it would be nice to have a certificate of optimality. Simply running the algorithm again might reproduce an existing error in the computer program that runs the algorithm. Let’s consider the two objects the algorithm is supposed to produce. 1. A tree of shortest paths rooted at s. 2. A function d : V → R which gives the distance of a shortest path to v , ∀v ∈ V . We won’t prove that Dijkstra’s algorithm produces these two objects, though we will certainly think about it. In the next couple of classes we will prove a theorem that allows us to certify that the output of Dijkstra’s algorithm is, in fact, correct. Let’s look at the algorithm more closely. Would we expect the algorithm to always produce a tree? That is, is T = (V , E ) a tree in every iteration? If there is some iteration where it is not a tree, then the end product will not be a tree because the algorithm only adds edges. The algorithm never deletes edges. The algorithm will have |V | − 1 iterations because we add a vertex to V at each iteration and V begins with s. We also add an edge at each iteration so we end up with |V | = |E | + 1. Proposition 2 is suggestive but not conclusive. It says that for a tree T = (V, E ), |V | = |E | + 1. It does not say that |V | = |E | + 1 implies that the graph (V, E ) is a tree. 232 Chapter 39 Dijkstra’s Algorithm Let’s consider the construction of T . A tree is defined as a connected graph with no cycles so let’s ask ourselves “Can the algorithm create a cycle in T ?” Suppose that it did and the cycle occurred when the edge {u, v } was added. That means both u and v already had to exist in V , but the edge that is added always contains a vertex not in V . Hence, no cycles exist in T . As for connectedness, this makes sense since, at each iteration an edge is added to an already connected graph constructed in the previous iteration. More problematic is guaranteeing that T is a tree of shortest paths. Let’s look at d more closely. Suppose {u, v } ∈ E and the path in E from s encounters u before it encounters v . Then d(u) = d(v ) + w({u, v }). That is not a surprise. That is how the algorithm adds edges to E . Now look at the exercise that you just completed. Examine any edge at all in E , say {x, y }. My guess is that you will see d(y ) ≤ d(x) + w({x, y }). This is what will help us generate a certificate of optimality. Chapter 40 Certificate of Optimality - Path 40.1 Objectives The content objectives are: 1. Define weight, distance potentials, feasible distance potentials, equality edges and tree of shortest paths. 2. Use a certificate of optimality to test that a proposed solution is optimal. 3. Prove A Path Shorter Than A Walk. 4. Prove Feasible Potentials. 5. Prove Certificate of Optimality for a Path. 6. Prove Shortest Paths Give Feasible Potentials. 7. Prove Shortest Path Optimality. 8. Prove Trees of Shortest Paths. 40.2 Certificate of Optimality Recall that a certificate consists of a theorem and data. If the data satisfy the hypothesis of the theorem, the theorem guarantees that the desired property holds. The data will be a tree T and a function d : V → R, exactly what is produced by Dijkstra’s algorithm. Our task is to find a theorem that will say “If the data satisfy a certain property, then 1. T is a tree of shortest paths rooted at s. 2. d : V → R gives the distance of a shortest path to v , ∀v ∈ V .” 233 234 Chapter 40 40.3 Certificate of Optimality - Path Weighted Graphs Suppose that G = (V, E ) is a connected graph with weights w : E → R. Let us also suppose that w({u, v }) ≥ 0, for every edge of E . Let W = v0 v1 v2 . . . vn be a walk in G. We define the weight of W to be the sum of the weights of all arcs in W . If the edge {u, v } occurs more than once in W , its weight is counted for each occurrence in W . More formally, n−1 w({vi , vi+1 }) w(W ) = i=0 We have been using this definition implicitly. The distance of a trip from downtown Toronto to Sibbald Point Provincial Park is the sum of distances of each part of the trip. Dijkstra’s algorithm also uses this definition implicitly. Proposition 1 (A Path Shorter Than A Walk (PSTW)) Let G = (V, E ) be a connected graph with non-negative real weights. Let W be an st-walk with s = t. Then there exists an st-path with w(P ) ≤ w(W ). Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Part 3 of Proposition 1 states that W can be decomposed into an st-path P and a collection of cycles C1 , C2 , . . . , Cr . 2. Now r w (W ) = w (P ) + w(Ci ). i=1 3. Since w(Ci ) ≥ 0 for all i = 1, 2, 3, . . . , r, w(P ) ≤ w(W ). Analysis of Proof We will begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: G = (V, E ) is a connected graph with non-negative real weights. W is an st-walk with s = t. Conclusion: There exists an st-path with w(P ) ≤ w(W ). Core Proof Technique: Construct method. Preliminary Material: Definitions related to weighted graphs. Sentence 1 Part 3 of Proposition 1 states that W can be decomposed into an st-path P and a collection of cycles C1 , C2 , . . . , Cr . The conclusion contains an existential quantifier so the author uses the Construct method. Let’s first identify the components of the existential quantifier. Section 40.3 Weighted Graphs 235 Quantifier: Variable: Domain: Open sentence: ∃ A path P All paths in G = (V, E ) w(P ) ≤ w(W ) The author must construct an st-path P and does so using Part 3 of Proposition 1. The author will now show that w(P ) ≤ w(W ). Sentence 2 Now r w (W ) = w (P ) + w(Ci ). i=1 This is the numeric implication of Proposition 1. Sentence 3 Since w(Ci ) ≥ 0 for all i = 1, 2, 3, . . . , r, w(P ) ≤ w(W ). This is arithmetic. The proof is very simple and relies very heavily on the fact that w(Ci ) ≥ 0 for all i = 1, 2, 3, . . . , r. What if the hypothesis “non-negative real weights” were simply “non-negative real weights”? Exercise 1 Show the necessity of “non-negative” in the hypothesis of Proposition 1. That is, find a counter-example to the statement: Let G = (V, E ) be a connected graph with non-negative real weights. Let W be an st-walk with s = t. Then there exists an st-path with w(P ) ≤ w(W ). You might argue that this is irrelevant because you never encounter negative distances. This may be true of distances, but this is not true of costs. Subsidies and rebates do, in fact, create negative cost edges in models. Let G = (V, E ) be a connected graph with non-negative weights w : E → R and d : V → R. The components of d are called distance potentials. We say that distance potentials are feasible when d(u) + w({u, v }) ≥ d(v ) for all uv ∈ E. Edges for which d(u) + w({u, v }) = d(v ) are called equality edges. Proposition 2 (Feasible Potentials (FP)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R, d : V → R be feasible distance potentials and W an st-walk. Then w(W ) ≥ d(t) − d(s). Moreover, w(W ) = d(t) − d(s) if and only if every arc of W is an equality edge. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Suppose W = v0 v1 v2 . . . vk where s = v0 and t = vk . 236 Chapter 40 Certificate of Optimality - Path 2. The feasible distance potentials satisfy d(v0 ) + w({v0 , v1 }) ≥ d(v1 ) d(v1 ) + w({v1 , v2 }) ≥ d(v2 ) d(v2 ) + w({v2 , v3 }) ≥ d(v3 ) . . . d(vk−1 ) + w({vk−1 , vk }) ≥ d(vK ) 3. Adding these inequalities together gives d(v0 ) + d(v1 ) + d(v2 ) + . . . + d(vk−1 ) + w({v0 , v1 }) + w({v1 , v2 }) + . . . + w({vk−1 , vk }) ≥d(v1 ) + d(v2 ) + d(v3 ) + . . . + d(vk ). 4. This simplifies to d(v0 ) + w(W ) ≥ d(vk ) or w(W ) ≥ d(t) − d(s). 5. Moreover, w(W ) ≥ d(t) − d(s) if and only if every inequality above holds with equality, that is, every edge in W is an equality edge. This is a straightforward proof so no analysis is provided. Theorem 3 (Certificate of Optimality for a Path (OPT P)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R and let s be a designated vertex and let P be an st-path. If there exist feasible distance potentials d : V → R such that every edge of P is an equality edge, then P is a shortest st-path. Before we examine the proof, let’s see how the theorem works as part of the certificate. Recall the tree and function d that resulted from our example of running Dijkstra’s algorithm. The dark edges indicate the tree and the blue labels adjacent to the vertices give d. Observe the sd-path P = sbd. All of the hypotheses of Theorem 3 are satisfied. G is a connected graph with non-negative weights. A vertex s has been designated. P = sbd is an sd-path. By examining each edge of G we can confirm that d are feasible distance potentials. By examining each edge of P we can confirm that every edge of P is an equality edge. Hence, by Theorem 3, P is a shortest sd-path. Now to the proof. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. By the first part of the conclusion of Proposition 2, every st-walk has weight at least w(t) − w(s). 2. By the second part of the conclusion of Proposition 2, w(P ) = w(t) − w(s). Section 40.3 Weighted Graphs 237 0 3 3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 40.3.1: Tree and d 3. Since the weight of every walk W is bounded below by w(t) − w(s), and P is a path that achieves that bound, P must be a shortest st-path. Analysis of Proof We will begin by explicitly identifying the hypothesis and the conclusion. Hypothesis: G = (V, E ) is a connected graph with non-negative weights w : E → R. s is a designated vertex and P is an st-path. There exist feasible distance potentials d : V → R such that every edge of P is an equality edge. Conclusion: P is a shortest st-path. Core Proof Technique: Forward-Backwards. Existential quantifiers occur in the hypothesis. Preliminary Material: Accumulated knowledge about weighted graphs. Sentence 1 By the first part of the conclusion of Proposition 2, every st-walk has weight at least w(t) − w(s). Since is is a form of the existential quantifier, the hypothesis “P is an st-path” allows the author to assume the existence of P . What the author must show is not that P exists, or that P is an st-path, but rather that P is a shortest st-path. The first sentence of the proof places an upper bound on w(P ). Sentence 2 By the second part of the conclusion of Proposition 2, w(P ) = w(t) − w(s). The hypotheses of the current theorem include “There exist feasible distance potentials d : V → R such that every edge of P is an equality edge.” The existential quantifier in this hypothesis allows the author to assume the existence of feasible distance potentials and equality edges. These are needed to invoke 2. Sentence 3 Since the weight of every walk W is bounded below by w(t) − w(s), and P is a path that achieves that bound, P must be a shortest st-path. Since no walk, and hence no path, can be shorter than w(t) − w(s), and w(P ) = w(t) − w(s), P must be a shortest st-path. 238 Chapter 40 Proposition 4 Certificate of Optimality - Path (Shortest Paths Give Feasible Potentials (SPGFP)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R and a designated node s. If d : V → R is defined as the length of a shortest path from s to v for all vertices in V , then d are feasible distance potentials. Proof: (For reference purposes, each sentence of the proof is written on a separate line.) 1. By contradiction, suppose that d are not feasible distance potentials. Then there exists {u, v } ∈ E such that d(u) + w({u, v }) < d(v ). 2. Let P be a shortest su-path. By the definition of d, w(P ) = d(u). 3. Consider the walk W constructed by appending the edge {u, v } to the path P . 4. By Proposition 1, there exists an sv -path P with w(P ) ≤ w(W ). 5. But w(W ) = w(P ) + w({u, v }) = d(u) + w({u, v }) < d(v ). 6. But then w(P ) < d(v ) so d(v ) cannot be the length of a shortest sv -path, a contradiction. Now we show that the converse of the certificate of optimality for paths also holds. Theorem 5 (Feasible Distance Potentials and Equality Edges) Let G = (V, E ) be a connected graph with non-negative weights w : E → R and let s be a designated vertex. If P is a shortest st-path, then there exist feasible distance potentials d : V → R such that every edge of P is an equality edge. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let d : V → R be defined as the length of a shortest path from s to v for all vertices in V . By 4, these are feasible distance potentials. 2. Hence, w(P ) = d(t) = d(t) − 0 = d(t) − d(s). 3. But then 2 implies that every edge of P is an equality edge. Together, the theorem on the optimality of paths (Theorem 3) and the existence of feasible distance potentials (Theorem 5) gives Theorem 6 (Shortest Path Optimality (SPO)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R and let s be a designated vertex. P is a shortest st-path if and only if there exist feasible distance potentials such that every edge of P is an equality edge. Section 40.4 Certificate of Optimality - Tree 40.4 239 Certificate of Optimality - Tree We have dealt so far with paths, but Dijkstra’s algorithm produces a tree, not a path. Fortunately, similar theorems hold. Theorem 7 (Trees of Shortest Paths (TSP)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R. Let s be a designated vertex and let T be a spanning tree rooted at s. If there exist feasible distance potentials such that every edge of T is an equality edge, then T is a tree of shortest paths rooted at s. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. Let us assume that there exist feasible distance potentials such that every edge of T is an equality edge. 2. For every node v in V , there is an st-path in T that satisfies the hypotheses of Theorem 3. 3. Hence, T is a tree of shortest paths rooted at s. Theorem 7 requires a spanning tree, feasible potentials and equality arcs. How do we know that these exist? Theorem 8 (Existence of Trees of Shortest Paths (ETSP)) Let G = (V, E ) be a connected graph with non-negative weights w : E → R and let s be a designated vertex. Then there exists a tree of shortest paths rooted at s. Proof: (For reference, each sentence of the proof is written on a separate line.) 1. For every node v ∈ V , let P (v ) be a shortest st-path in G and let d(v ) = w(P (v )). 2. Since d(v ) is the length of a shortest path to v , Proposition 4, tells us that d is a set of feasible distance potentials. 3. We know from Proposition 2 that every edge in a shortest sv -path is an equality arc. So, every edge of P (v ) is an equality edge for every v ∈ V . 4. Let E= P (v ). v ∈V 5. The edges of E contain a path consisting of equality arcs from s to every v ∈ V . Delete from E enough edges to produce a tree T . 6. But then 7 applies and T is a tree of shortest paths rooted at s. Chapter 41 Appendix Proposition 1 (Decomposing n-th Power (DNP)) If ab = cn and gcd(a, b) = 1, then there exist integers a1 and b1 so that a = an and b = bn . 1 1 Proof: Without loss of generality, we may assume that a > 1 and b > 1. If a = pk1 pk2 · · · pkr r 12 jj j b = q11 q22 · · · qss are the prime factorizations of a and b, then no px can occur among the qy otherwise the gcd(a, b) > 1. As a result, the prime factorization of ab is jj j ab = pk1 pk2 · · · pkr q11 q22 · · · qss r 12 Let us suppose that c can be factored into primes as c = ul1 ul2 · · · ult t 12 Then ab = cn can be written as jj j pk1 pk2 · · · pkr q11 q22 · · · qss = unl1 unl2 · · · unlt r t 12 1 2 This implies that each px and qy equals some uh and that the corresponding exponents are equal. That is kx = nlh (or jy = nlh ). This implies that all of the exponents of the px and qy are divisible by n. Thus, we can choose k /n k /n a = p1 1 p2 2 j /n j /n b = q11 q22 and a = an and b = bn as needed. 1 1 240 · · · pkr /n r j · · · qss /n ...
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This note was uploaded on 04/02/2012 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.

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