Calculus Once Again - David Santos

8/11/2019 Calculus Once Again - David Santos

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David A. SANTOS

[email protected]

January 2, 2010 Version
mailto:[email protected]:[email protected]


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Copyright 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modify this docu-

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the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A

copy of the license is included in the section entitled GNU Free Documentation License.


3/194

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Contents

GNU Free Documentation License iii

1. APPLICABILITY AND DEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

2. VERBATIM COPYING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

3. COPYING IN QUANTITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

4. MODIFICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

5. COMBINING DOCUMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

6. COLLECTIONS OF DOCUMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

7. AGGREGATION WITH INDEPENDENT WORKS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

8. TRANSLATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

9. TERMINATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

10. FUTURE REVISIONS OF THIS LICENSE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Preface viii

1 Preliminaries 1

1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Numerical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Injective and Surjective Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.2 Algebra of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.3 Inverse Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.4 Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Countability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 Groups and Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 Addition and Multiplication in R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.6 Order Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.6.1 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.7 Classical Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.7.1 Triangle Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.7.2 Bernoullis Inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.7.3 Rearrangement Inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.7.4 Arithmetic Mean-Geometric Mean Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.7.5 Cauchy-Bunyakovsky-Schwarz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.7.6 Minkowskis Inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.8 Completeness Axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.8.1 Greatest Integer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

v


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vi CONTENTS

2 TopologyofR 34

2.1 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2 Dense Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.3 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.4 Interior, Boundary, and Closure of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.5 Connected Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.6 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.7 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.8 Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.9 The Cantor Set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3 Sequences 47

3.1 Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.2 Convergence of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Classical Limits of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.4 Averages of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.5 Orders of Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.6 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.7 Topology of sequences. Limit Superior and Limit Inferior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4 Series 70

4.1 Convergence and Divergence of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.2 Convergence and Divergence of Series of Positive Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.3 Summation by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.4 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5 Real Functions of One Real Variable 85

5.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.3 Algebraic Operations with Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.4 Monotonicity and Inverse Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.5 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.5.1 Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.6 Classical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.6.1 Affine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.6.2 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.6.3 Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.6.4 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.6.5 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.6.6 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.6.7 Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.7 Continuity of Some Standard Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.7.1 Continuity Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97


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CONTENTS vii

5.7.2 Continuity of the Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.7.3 Continuity of the Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.8 Inequalities Obtained by Continuity Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.9 Intermediate Value Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.10 Variation of a Function and Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

5.11 Classical Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

6 Differentiable Functions 113

6.1 Derivative at a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.2 Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.3 Rolles Theorem and the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.4 Extrema. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.5 Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.6 Inequalities Obtained Through Differentiation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.7 Asymptotic Preponderance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

6.8 Asymptotic Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6.9 Asymptotic Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7 Integrable Functions 143

7.1 The Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1527.2 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

7.3 Riemann-Stieltjes Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

7.4 Eulers Summation Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

8 Sequences and Series of Functions 160

8.1 Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

8.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

8.3 Integrals and Derivatives of Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

8.4 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

8.5 Maclaurin Expansions to know by inspection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

8.6 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

8.7 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

8.8 Abels Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

A Answers and Hints 163

Answers and Hints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163


8/194

Preface

For many years I have been lucky enough to have students ask for more: more challenging problems, more illuminating

proofs to different theorems, a deeper look at various topics, etc. To those students I normally recommend the books in

the bibliography. Some of the same students have complained of not finding the books or wanting to buy them, but being

impecunious, not being able to afford to buy them. Hence I have decided to make this compilation.

Here we take a semi-rigorous tour through Calculus. We dont construct the real numbers, but we examine closer the

real number axioms and some of the basic theorems of Calculus. We also consider some Olympiad-level problems whose

solution can be obtained through Calculus.

The reader is assumed to be familiar with proofs using mathematical induction, proofs by contradiction, and the me-chanics of differentiation and integration.

David A. SANTOS

[email protected]

viii
mailto:[email protected]:[email protected]


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Chapter 1

Preliminaries

Why bother? We will use the language of set theory throughout these notes. There are various elementary

results that pop up in later proofs, among them, the De Morgan Laws and the Monotonicity Reversing of Com-

plementation Rule.

The concept of a functionlies at the core of mathematics. We will give a brief overview here of some basic

properties of functions.

1.1 Sets

This section contains some of the set notation to be used throughout these notes. The one-directional arrow= readsimplies and the two-directional arrow reads if and only if.

1 Definition We will accept the notion ofsetas a primitive notion, that is, a notion that cannot be defined in terms of more

elementary notions. By asetwe will understand a well-defined collection of objects, which we will call the elementsof the

set. If the elementxbelongs to the set Swe will writex S, and in the contrary case we will write x S.1 Thecardinalityofa set is the number of elements the set has. It can either be finite or infinite. We will denote the cardinality of the set Sby

card(S).

Some sets are used so often that merit special notation. We will denote by

N= {0,1,2,3,...}

the set of natural numbers, by

Z= {. . . ,3,2,1,0,1,2,3,... }2

byQ the set of rational numbers3, byR the real numbers, and byC the set of complex numbers. We will occasionally also use

Z= {. . . ,3,2,,0,, 2, 3, . . . }, etc.

We will also denote the empty set, that is, the set having no elements by.

2 Definition Theunionof two sets Aand Bis the set

AB= {x: (xA)or(x B)}.

This is read Aunion B. See figure 1.1.Theintersectionof two sets Aand Bis

AB= {x: (x A)and(x B)}.1 Georg Cantor(1845-1918), the creator of set theory, said A set is any collection into a whole of definite, distinguishable objects, calledelements, of our

intuition or thought.2Z for the German wordZhlenmeaning integer.

3Q for quotients.

1


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Sets

A B

Figure 1.1: AB

A B

Figure 1.2: AB

A B

Figure 1.3: A\ B

This is read AintersectionB. See figure 1.2.Theset differenceof two sets Aand Bis

A\ B= {x: (x A)and(x B)}.

This is read Aset minus B. See figure1.3.

3 Definition Two sets Aand Bare disjointifAB=.

4 Example Write A Bas the disjoint union of three sets.

Solution: Observe that

AB= (A\ B) (AB) (B\A),

and that the sets on the dextral side are disjoint.

5 Definition AsubsetBof a setAis a subcollection ofA, and we denote this byB A. 4 This means thatx B= x A.

andAare always subsets of any setA.

Observe that

A= B (A B) and (BA).

We use this observation on the next theorem.

6T HEOREM(De Morgan Laws) LetA, B,Cbe sets. Then

A\ (BC) = (A\ B) (A\ C), A\ (BC) = (A\ B) (A\ C).

Proof: We have

x A\ (BC) x A and x (B or C) (x A) and ((x B) and (x C)) (x A and x B) and (x A and x C) (x A\ B) and (x A\ C)

x

(A\ B)

(A\ C).

Also,

x A\ (BC) x A and x (B and C) (x A) and ((x B) or (x C)) (x A and x B) or (xA and x C) (x A\ B) or (xA\ C) x (A\ B) (A\ C)

4There seems not to be an agreement here by authors. Some use the notation or instead of. Some see in the notation the exclusion of equality.In these notes, we will always use the notation , and if we wished to exclude equality we will write .

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Chapter 1

7T HEOREM(Monotonicity Reversing of Complementation) LetA, B,Xbe sets. Then

AB X\ BX\ A.

Proof: We haveAB (x A)= ( x B)

(x B)= (x A) (xX and x B)= ( xX and x A) X\ BX\ A.

8 Definition LetA1,A2, . . . ,An, be sets. TheCartesian Productof thesensets is defined and denoted by

A1 A2 An= {(a1, a2, . . . , an) : akAk},

that is, the set of all ordered n-tuples whose elements belong to the given sets.

In the particular case when all theAkare equal to a setA, we write

A1

A2

An=

An.

Ifa Aandb Awe write(a,b) A2.

9 Example The Cartesian product is not necessarily commutative. For example,(

2,1) RZ but(

2,1) ZR. SinceRZ has an element that ZR does not, RZ=ZR.

10 Example Prove that ifXX= Y Y thenX= Y.

Solution: LetxX. Then(x, x) XX, which gives(x, x) Y Y, soyY. HenceX Y.

Similarly, if y Y then(y,y) Y Y, which gives(y,y) XX, soyX. HenceY X.

ThusX Y andY X givesX= Y.

Homework

Problem 1.1.1 For a fixednN putAn= {nk: kN}.1. FindA2 A3.

2. Find

n=1An.

3. Find

n=1An.

Problem 1.1.2 Prove the following properties of the empty set:

A=, A=A.

Problem 1.1.3 Prove the following commutative laws:

AB=BA, AB=BA.

Problem 1.1.4 Prove by means of set inclusion the following dis-

tributive law:

(AB)C= (AC) (BC).

Problem 1.1.5 Prove the following associative laws:

A (BC) = (AB)C, A (BC) = (AB)C.

Problem 1.1.6 Prove that

AB=A A B.


AB=A BA.


A B= ACBC.


A B and C B= ACB.

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Numerical Functions

Problem 1.1.10 Prove the following distributive laws:

A (BC) = (AB) (AC), A (BC) = (AB) (AC).

Problem 1.1.11 Is there any difference between the sets,{}and{{}}? Explain.

Problem 1.1.12 Is the Cartesian product associative? Explain.

Problem 1.1.13 LetA,B, andCbe sets. Shew that

A (B\ C) = (AB) \ (AC).

Problem 1.1.14 Provethat a setwithNN elements has exactly2Nsubsets.

1.2 Numerical Functions

11 Definition By a(numerical) functionf :Domf Targetfwe mean the collection of the following ingredients:

anamefor the function. Usually we use the letter f.

a set of real number inputs called thedomainof the function. The domain offis denoted byDomfR.

aninput parameter, also calledindependent variableor dummy variable. We usually denote a typical input by the

letterx.

a set ofpossible real number outputs of the function, called the target setof the function. The target set offis denoted

byTargetfR. anassignment ruleor formula, assigning toevery inputa uniqueoutput. This assignment rule for fis usually de-

noted byx f(x). The output ofxunder fis also referred to as theimage ofxunderf, and is denoted byf(x).

The notation5

f : Dom

f Targetf

x f(x)read the function f, with domainDom

f, target setTarget

f, and assignment rule f mappingx to f(x) conveys all

the above ingredients.

Oftentimes we will only need to mention theassignment rule of a function, without mentioning its domain or target set.In such instances we will sloppily say the function f or more commonly, the functionx f(x), e.g., the square functionx x2.6

12 Definition TheimageImf

of a function fis its set of actual outputs. In other words,

Imf= {f(a) : a Domf}.

Observe that we always haveImf Targetf. For a setA, we also define

f(A) = {f(a) : a A}.

13THEOREM Letf :X Ybe a function and letAX, A X. Then

1. AA= f(A) f(A)2. f(AA) =f(A)f(A)

3. f(AA) f(A)f(A)

4. f(A) \f(A) f(A\A)

Proof:

5Notice the difference in the arrows. The straight arrow is used to mean that a certain set is associated with another set, whereas the arrow (readmaps to) is used to denote that an input becomes a certain output.

6This corresponds to the even sloppier American usage the functionf(x) = x2.

4


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Chapter 1

1. x A= x Aand hencef(x) f(A)= f(x) f(A)= f(A) f(A)2. SinceA AA andA AA, we have f(A) f(AA)and f(A) f(AA), by part (1) and thus

f(A) f(A) f(AA). Moreover, if yf(AA), thenx AAsuch thaty=f(x). Then eitherxAand sof(x) f(A)orx Aand sof x f(A). Either way, f(x) f(A)f(A)and

yf(AA)= y f(A)f(A)= f(AA) f(A) f(A).

Hence

f(AA) =f(A) f(A).3. Lety f(AA). Thenx AA such that f(x) = y. Thus we have bothx A= f(x) f(A)and

x A= f(x) f(A). Thereforef(x) f(A)f(A)and we conclude that f(AA) f(A) f(A).4. Letyf(A) \f(A). Thenyf(A)andyf(A). ThusxAsuch that f(x) = y. Sincey f(A), then

x A. Thereforex A\Aand finally, y f(A\A). This means that f(A) \f(A) f(A\A)as claimed.

1.2.1 Injective and Surjective Functions

14 Definition A function is injectiveor one-to-onewhenever two different values of its domain generate two different values

in its image. A function issurjectiveor ontoif every element of its target set is hit, that is, the target set is the same as the

image of the function. A function isbijectiveif it is both injective and surjective.

15 Example The function

a: R R

x x2

is neither injective nor surjective.

The function

b: R

0 ;+

x x2

is surjective but not injective.

The function

c:

0 ;+ R

x x2

is injective but not surjective.

The function

d:

0 ;+

0 ;+

x x2

is a bijection.

A bijection between two sets essentially tells us that the two sets have the same size. We will make this statement more

precise now for finite sets.

16THEOREM Letf :A Bbe a function, and letAand Bbe finite. Iff is injective, then card(A) card(B). Iff is surjectivethencard(B) card(A). Iffis bijective, thencard(A) = card(B).

Proof: Putn= card(A), A= {x1, x2, . . . , xn}andm= card(B), B= {y1,y2, . . . ,ym}.

Iffwere injective thenf(x1),f(x2),. . . ,f(xn)are all distinct, and among theyk. Hencen m.

If fwere surjective then eachykis hit, and for each, there is anxi with f(xi) = yk. Thus there are at leastmdifferent images, and son m.

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1.2.2 Algebra of Functions

17 Definition Let f : Domf Targetfand g : Domg Targetg. ThenDomf g= Domf Domgand the

sum (respectively, difference) function f+ g (respectively, f g) is given by

f g: Domf Domg Targetf g

x f(x) g(x) .

In other words, ifxbelongs both to the domain off andg, then

(f g)(x) =f(x) g(x).

18 Definition Let f : Domf Targetfand g: Domg Targetg. Then Dom f g = Domf Domgand the

product function f gis given by

f g: Dom

f Domg Targetf g

x f(x) g(x) .

In other words, ifxbelongs both to the domain off andg, then

(f g)(x) =f(x) g(x).

19 Definition Letg: Domg Targetgbe a function. Thesupportofg, denoted bysuppgis the set of elements inDomgwheregdoes not vanish, that issupp

g= {x Domg : g(x) = 0}.

20 Definition Let f : Domf Targetfand g: Domg Targetf. ThenDomf

g

= Domf suppgand the

quotient function f

gis given by

f

g:

Domfsuppg Targetf/g

x f(x)g(x)

.

In other words, ifxbelongs both to the domain off andgandg(x)=

0, then f

g(x)

= f(x)

g(x).

21 Definition Let f : Domf Targetf,g: Domg Targetg and let U= {x Domg : g(x) Domf}. We define

thecompositionfunction off andgas

f g: U Targetf g

x f(g(x)) . (1.1)

We readf gas f composed withg.

1.2.3 Inverse Image

22 Definition LetXand Ybe subsets ofR and let f :X Ybe a function. Let BY. Theinverse image ofBbyfis the set

f1(B) = {xX: f(x) B}.

IfB= {b} consists of only one element, we write, abusing notation, f1({b})= f1(b). It is clear that f1(Y)=X andf1() =.

23 Example Let

f : {2,1,0,1,3} {0,1,4,5,9}

x x2 .

Then f1({0, 1}) = {0,1, 1}, f1(1) = {1,1}, f1(5) =, f1(4) = 2, f1(0) = 0, etc. Notice that we have abused notationin all but the first example.

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Chapter 1

24THEOREM Letf :X Ybe a function and let BY, B Y. Then

1. BB= f1(B) f1(B)

2. f1(B B) = f1(B)f1(B)

3. f1(B B) = f1(B)f1(B)

4. f1(B) \f(B) =f1(B\ B)

Proof:

1. Assumexf1(B). Then there isy B B such that f(x) =y. Butyis also inB soxf1(B). Thusf1(B) f1(B).

2. SinceB B BandB B B, we havef1(B) f1(B B)and f1(B) f1(B B), by part (1).Thusf1(B)f1(B) f1(BB). Now, letxf1(BB). There isy BBsuch thatf(x) =y. Eithery BandsoyB= xf1(B) ory Band soy B= x f1(B). Either way, x f1(B)f1(B).Thusf1(B B) f1(B)f1(B). We conclude thatf1(BB) =f1(B) f1(B).

3. Letx

f1(B

B). Then

y

B

Bsuch thatf(x)

=y. Thus we have bothy

B

=x

f1(B)andy

B= xf1(B). Thereforex f1(B)f1(B) and we conclude thatf1(BB) f1(B)f1(B).Now, letx f1(B) f1(B). Thenx f1(B) andx f1(B). Then f(x) B and f(x) B. Thusf(x) B B and soxf1(B B). Hence f1(B) f1(B) f1(B B)also, and we conclude thatf1(B) f1(B) =f1(BB).

4. Letx f1(B) \f1(B). Thenx f1(B) andx f1(B). Thus f(x) Band f(x) B. Thus f(x) B\ Band thereforexf1(B\ B), giving f1(B) \f1(B) f1(B\ B). Now, letxf1(B\ B). Thenf(x) B\ B, which means that f(x) B but f(x) B. Thusx f1(B) butx f1(B), which givesxf1(B) \f1(B)and sof1(B\ B) f1(B) \f1(B). This establishes the desired equality.

25THEOREM Letf :X

Ybe a function. Let A

BX

Y. Then

1. A (f1 f)(A)

2. (f f1)(B)B

Proof: We have

1. Letx A. Theny Ysuch thaty=f(x). Thusy f(A). Thereforex f1(f(A)).2. y (f f1)(B). Thenx f1(B)such thatf(x) =y. Thusx f1(y). Hencef(x) B. Thereforey B.

1.2.4 Inverse Function

26 Definition LetA BR2. A functionF: A Bis said to beinvertibleif there exists a function F1 (called theinverseofF) such that FF1 = IdBand F1 F= IdA. Here Id Sis the identity on the set Sfunction with ruleId S(x) = x.

The central question is now: given a functionF: A B, when isF1 : BAa function? The answer is given in the nexttheorem.

27THEOREM LetABR2. A functionf :A Bis invertible if and only if it is a bijection. That is, f1 : BAis a functionif and only iffis bijective.

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Numerical Functions

Proof: Assume first thatfis invertible. Then there is a functionf1 : BAsuch that

f f1 = Id B and f1 f= IdA. (1.2)

Let us prove thatf is injective and surjective. Lets, tbe in the domain offand such thatf(s) = f(t). Applyingf1 to both sides of this equality we get(f1 f)(s) = (f1 f)(t). By the definition of inverse function,(f1 f)(s) = sand(f1 f)(t) = t. Thuss= t.Hencef(s) =f(t)= s= timplying thatf is injective. To prove thatf is surjective we must shew that for everyb

f(A)

a

Asuch thatf(a)

=b.We takea

=f1(b) (observe that

f1(b) A). Then f(a) = f(f1(b)) = (f f1)(b) = bby definition of inverse function. This shews that f issurjective. We conclude that iff is invertible then it is also a bijection.

Assume now that fis a bijection. For everyb Bthere exists a uniqueasuch thatf(a) = b. This makes the ruleg: BAgiven byg(b) = aa function. It is clear thatg f= IdAand f g= Id B.We may thus take f1 = g.This concludes the proof.

Homework

Problem 1.2.1 Find all functions with domain{a,b}and target set

{c,d}.

Problem 1.2.2 Let A, B be finite sets with card (A)= n andcard (B) = m. Provethat

The number of functions fromAtoBismn.

Ifn m, the number of injective functions from A toB ism(m1)(m2) (mn+1). Ifn> mthere are no injectivefunctions fromAtoB.

Problem 1.2.3 LetAandBbe two finite sets withcard (A) = nandcard (B) = m. Ifn< mprove that there are no surjections fromAtoB. Ifn mprove that the number of surjective functions fromAtoBis

mnm1(m1)n+m

2(m2)nm

3(m3)n++(1)m1 m

m1(1)n.Problem 1.2.4 Leth: RR be given byh(1x) = 2x. Findh(3x).

Problem 1.2.5 Consider the polynomial

(1 x2 + x4)2003 = a0 +a1x+a2x2 + + a8012x8012.

Find

a0

a0 + a1 +a2 + + a8012 a0 a1 +a2 a3 + a8011 +a8012 a0 + a2 +a4 + + a8010 +a8012 a1 + a3 + + a8009 +a8011

Problem 1.2.6 Letf :RR, be a function such thatx]0;+[,

[f(x3 +1)]

x = 5,find the value of

f

27+y3

y3

27y

fory]0;+[.

Problem 1.2.7 Let f satisfy f(n+ 1)= (1)n+1n 2f(n),n 1Iff(1) = f(1001)find

f(1)+f(2) +f(3) + +f(1000).

Problem 1.2.8 Iff(a)f(b) = f(a+b) a,bR andf(x) > 0 xR, findf(0). Also, findf(a)andf(2a)in terms off(a).

Problem 1.2.9 Prove thatf :R\ {1} R\ {1}

x x1x+1

is a bijection

andfindf1.

Problem 1.2.10 Letf[1] (x) =f(x) = x+1,f[n+1] =f f[n],n 1.Find a closed formula forf[n]

Problem 1.2.11 Let f, g :

0 ;1 R be functions. Demonstrate

that there exist(a,b)

0 ; 12

such that1

4f(a)+g(b)ab.

Problem 1.2.12 Demonstrate that there is no function f : R \

{1/2} R such that

xR\ {1/2}= f(x)f

x12x1

= x2 +x+1

Problem 1.2.13 Find all functionsf :R\ {1,0} R such that

xR\ {1,0}= f(x)+f 1x+1 = 3x+2.Problem 1.2.14 Let f[1] (x)= f(x)=2x,f[n+1] = f f[n],n 1.Find a closed formula forf[n]

Problem 1.2.15 Find all functionsg: RR that satisfyg(x+y) +g(xy) = 2x2 +2y2.

Problem 1.2.16 Find all the functions f : R R that satisfyf(x y) =y f(x).

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Chapter 1

Problem 1.2.17 Find all functionsf :R\ {0} Rfor which

f(x)+2f

1

x

= x.

Problem 1.2.18 Find all functionsf :R\ {1} R such that

(f(x))2

f

1x1+x= 64x.

Problem 1.2.19 Letf[1] = fbe given byf(x) = 11x. Find

(i)f[2] (x) = (ff)(x),(ii)f[3] (x) = (fff)(x), and(iii)f[69] = (ff ff)

69compositions with itself

(x).

Problem 1.2.20 Let f : A

B andg :B

C be functions. Shew

that(i) ifgf is injective, thenf is injective. (ii) if gf is surjective,thengis surjective.

1.3 Countability

28 Definition A set Xis countable if either it is finite or if there is a bijection f : X N, that is, the set X has as manyelements as N.

Any countable set can be thus enumerated a sequence

x1, x2, x3, . . . .

Thus the strictly positive integers can be enumerated as customarily:

1,2,3,....

Another possible enumeration7 is the following

3, 5, 7, 9, . .. , , 2 3, 2 5, 2 7, 2 9, .. . , 22 3,22 5, 22 7, 22 9,. .. , .. .24, 23,22,2,1,

that is, we start with the odd integers in increasing order, then 2times the odd integers,22 times the odd integers, etc., and

at the end we put the powers of2in decreasing order.

29LEMMA Any subsetX N is countable.

Proof: IfXis finite, then there is nothing to prove. If Xis infinite, we can arrange the elements ofX increasing

order, say,

x1 < x2 < x3


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Groups and Fields

Proof: One can take, as a bijection between the two sets, for example,f :ZN,

f(x) =

2x+ 1 ifx 02x ifx< 0.

32THEOREM Q is countable.

Proof: Considerf :QN givenfa

b

= 2|a|3b51+signum(a),

wherea

bis in least terms, andb> 0. By the uniqueness of the prime factorisation of an integer, fis an injection.

The above theorem means that there as many rational numbers as natural numbers. Thus the rationals can be enumer-ated as

q1, q2, q3, . . . ,

33THEOREM(Cantors Diagonal Argument) R is uncountable.

Proof: AssumeR were countable so that its complete set of elements may be enumerated, say, as in the list

r1 = n1.d11d12d13 . . .r2 = n2.d21d22d13 . . .r3 = n3.d31d32d33 . . . ,

where we have used decimal notation. Define the new realr= 0.d1d2d3 . . . bydi= 0ifdi i= 0 anddi= 1 ifdi i= 0. This is real number (as it is a decimal), but it differs fromriin theith decimal place. It follows that thelist is incomplete and the reals are uncountable.

34THEOREM The interval 1 ; 1 is uncountable.Proof: Observe that the mapf :

1 ; 1

R given byf(x) = tan x

2 is a bijection.

Homework

Problem 1.3.1 Prove that there as many numbers in[0;1]as in any

interval[a;b]witha< b.Problem 1.3.2 Prove that there as many numbers in

;+

as

in

0 ;+

.

1.4 Groups and Fields

Here we observe the rules of the game for the operations of addition and multiplication inR.

35 Definition LetS,Tbe sets. Abinary operationis a function

: S S T(a,b) (a,b) .

We usually use the infix notation a b rather than the prefix notation(a,b). IfS=Tthen we say that the binaryoperation isinternalorclosedand ifS= Tthen we say that it is external.

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Chapter 1

36 Example Ordinary addition is a closed binary operationon thesetsN, Z,Q,R. Ordinary subtraction is a binary operation

on these sets. It is not closed on N, since for example1 2 = 1 N, but it is closed in the remaining sets.

37 Example The operation : RRR given bya b= 1 + a b, where is the ordinary multiplication of real numbers iscommutative but not associative. To see commutativity we have

a b= 1+ ab= 1+ ba= b a.

Now,

1 (1 2) = 1 (1+ 1 2) = 1 (3) = 1+ 1 3 = 4, but (1 1) 2 = (1+1 1) 2 = 22 = 1+2 2 = 5,

so the operation is not associative.

38 Definition Let Gbe a non-empty set and be a binary operation on GG. Then G, is called agroupif the followingaxioms hold:

G1: is closed, that is,(a, b) G2, a b G,

G2:

is associative, that is,

(a,b,c) G3, a (b c) = (a b) c,

G3: Ghas an identity element, that is

e Gsuch that a G, e a= a e= a,

G4: Every element ofGis invertible, that is

a G, a1 Gsuch thata a1 = a1 a= e.

From now on, we drop the sign and rather use juxtaposition for the underlying binary operation in a given group.Thus we will say a groupG rather than the more precise a groupG,.

39 Definition A group Gis abelianif its binary operation is commutative, that is, (a, b) G2, a b= b a.

40 Example Z,+, Q,+, R,+, C,+ are all abelian groups under addition. The identity element is 0 and the inverse ofais a.

41 Example Q \ {0}, , R \ {0}, , C \ {0}, are all abelian groups under multiplication. The identity element is1 and theinverse ofais

1

a.

42 Example Z \ {0}, is not a group. For example the element 2does not have a multiplicative inverse.

43 Example Let V4 = {e, a,b,c} and define by the table below. e a b c e e a b c

a a e c b

b b c e a

c c b a e

It is an easy exercise to check that V4is an abelian group, called theKlein Viergruppe.

44THEOREM Let Gbe a group. Then

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Groups and Fields

1. There is only one identity element, the identity element is unique.

2. The inverse of each element is unique.

3.(a, b) G2 we have(ab)1 = b1 a1.

Proof:

1. Leteande be identity elements. Sinceeis an identity, e= ee. Sincee is an identity,e = ee. This givese= ee = e.

2. Letbandbbe inverses ofa. Thene= abandba= e. This gives

b= eb= (ba)b= b(ab) = be= b.

3. We have

(ab)(b1a1) = a(bb1)a1 = a(e)a1 = aa1 = e.Thusb1a1 works as a right inverse fora b. A similar calculation shews also that it works as a left inverse.Since inverses are unique, we must have

(ab)1

=b1a1.

This completes the proof.

45 Definition LetnZ and let Gbe a group. Ifa G, we define

a0 = e,

a|n| = a a a|n|as

,

and

a|n| = a1 a1 a1

|n|a1s

.

If(m,n) Z2, then by associativity(an)(am) = (am)(an) = am+n.

46 Definition LetFbe a set having at least two elements 0F and1F (0F= 1F) together with two binary operations (fieldmultiplication) and+ (field addition). AfieldF, ,+ is a triplet such thatF,+ is an abelian group with identity0F ,F\ {0F}, is an abelian group with identity1Fand the operations and + satisfy

a (b+ c) = (ab) + (ac),

that is, field multiplication distributes over field addition.

We will continue our practice of denoting multiplication by juxtaposition, hence the sign will be dropped.

47 Example Q, ,+, R, ,+, and C, ,+ are all fields. The multiplicative identity in each case is 1 and the additive identityis0.

Homework

12


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Chapter 1

Problem 1.4.1 Is the set of real irrationalnumbers closed under ad-

dition? Under multiplication?

Problem 1.4.2 Let

S= {xZ : (a,b) Z2, x= a3 +b3 +c3 3ab c}.

Prove thatSis closed under multiplication, that is, ifx

Sandy

S

thenx yS.

Problem 1.4.3 (Putnam, 1971) LetSbe a set and let be a binaryoperation onSsatisfying the two laws

(x S)(x x= x),

and

((x,y, z) S3)((xy) z= (y z) x).Shew that is commutative.

Problem 1.4.4 (Putnam, 1972) LetSbe a set andlet be a binaryoperation ofSsatisfying the laws(x,y) S2

x (xy) =y, (1.3)

(y x) x=y. (1.4)Shew that is commutative, but not necessarily associative.

Problem 1.4.5 OnQ]1;1[define the binary operation by

ab= a+b1+ ab ,

where juxtaposition means ordinary multiplication and+ is the or-dinary addition of real numbers. Prove thatQ]1;1[, is anabelian group by following these steps.

1. Prove that is a closed binary operation onQ]1;1[.

2. Prove that is both commutative and associative.

3. Find an elementeQ]1;1[ suchthat(aQ]1;1[)(ea= a).

4. Giveneas above and an arbitrary element a Q] 1;1[,solve the equationab= eforb.

Problem 1.4.6 LetGbe a group satisfying(a G)

a2 = e.

Prove thatGis an abelian group.

Problem 1.4.7 LetGbe a group where((a,b) G2)

((ab)3 = a3b3) and ((ab)5 = a5b5).

Shew thatGis abelian.

Problem 1.4.8 Suppose that in a groupGthere exists a pair(a,b) G2 satisfying

(ab)k = akbk

for three consecutive integersk= i,i+1,i+2.Prove thatab= ba.

1.5 Addition and Multiplication inR

Since R is a field, it satisfies the following list of axioms, which we list for future reference.

48 Axiom (Arithmetical Axioms of R)R, ,+that is, the set of real numbers endowed with multiplication and addition+is a field. This entails that + and verify the following properties.

R1: + and are closed binary operations, that is,(a, b) R2, a+bR, a bR,

R2: + and are associative binary operations, that is,(a,b,c) R3, a+ (b+c) = (a+b)+ c, a (bc) = (a b) c

R3: + and are commutative binary operations, that is,(a,b) R2, a+ b= b+ a, ab= b a,

R4: R has an additive identity element 0, and a multiplicative identity element 1, with 0 = 1, such thataR, 0+ a= a+ 0 = a, 1 a= a 1 = a,

R5: Every element ofR has an additive inverse, and every element ofR\ {0} has a multiplicative inverse, that is,

aR, (a) R such thata+ (a) = (a) + a= 0,bR\ {0}, b1 R\ {0} such that bb1 = b1 b= 1,

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Addition and Multiplication inR

R6: + and satisfy the following distributive law:(a, b, c, ) R3, a (b+c) = a b+ a c.

Since + and are associative inR, we may write a suma1 + a2 + + anor a producta1a2 anof real numbers withoutrisking ambiguity. We often use the following shortcut notation.

49 Definition For real numbers aiwe define

a1 + a2 + + an=n

k=1ak and a1a2 an=

nk=1

ak.

By convention

kak= 0and

k

ak= 1.

50THEOREM(Lagranges Identity) Letak, bkbe real numbers. Then n

k=1akbk

2=

nk=1

a2k

n

k=1b2k

1k


23/194

Chapter 1

Using Pascals Identity we obtain Pascals Triangle.

00

1

0

1

1

2

0

2

1

2

2

3

0

3

1

3

2

3

3

4

0

4

1

4

2

4

3

4

4

50 51 52 53 54 55...

......

......

...

When the numerical values are substituted, the triangle then looks like this.

1

1 1

1 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 1...

......

......

...

We see from Pascals Triangle that binomial coefficients are symmetric. This symmetry is easily justified by the identityn

k

=

n

nk

. We also notice that the binomial coefficients tend to increase until they reach the middle, and that then they

decrease symmetrically.

53THEOREM(Binomial Theorem) FornN,

(x+y)n =n

k=0

n

k

xkynk.

Proof: The theorem is obvious forn= 0 (defining(x+y)0 =1), n=1 (as(x+y)1 = x+y), andn= 2 (as

(x+y)2 = x2 + 2x y+y2). Assumen 3. The induction hypothesis is that(x+y)n =n

k=0

n

k

xkynk.Then we

15


24/194


have(x+y)n+1 = (x+y)(x+y)n

= (x+y)

nk=0

n

k

xkynk

=n

k=0

n

k

xk+1ynk+

nk=0

n

k

xkynk+1

= xn+1 +n1k=0nkxk+1ynk+

nk=1nkxkynk+1 +yn+1

= xn+1 +n

k=1

n

k1

xkynk+1 +

nk=1

n

k

xkynk+1 +yn+1

= xn+1 +n

k=1

n

k 1

+

n

k

xkynk+1 +yn+1

= xn+1 +n

k=1

n+ 1

k

xkynk+1 +yn+1

=n+1k=0

n+ 1

k

xkynk+1,

proving the theorem.

54LEMMA IfaR,a= 1and nN \ {0}, then

1+ a+ a2 + an1 = 1 an

1 a .

Proof: For, putS= 1+ a+ a2 + + an1.ThenaS= a+ a2 + + an1 + an. Thus

S aS= (1+ a+ a2 + + an1) (a+ a2 + + an1 + an) = 1 an,

and from(1 a)S= S aS= 1 an we obtain the result.

55THEOREM Let nbe a strictly positive integer. Then

yn

xn

= (y x)(yn

1

+yn

2

x+ +y xn

2

+ xn

1

).

Proof: By making the substitutiona= xy

in Lemma54we see that

1+ xy

+

x

y

2+ +

x

y

n1=

1

xy

n1 xy

we obtain 1 x

y

1+ x

y+

x

y

2+ +

x

y

n1= 1

x

y

n,

or equivalently, 1x

y

1+x

y+x

2

y2+ +x

n1

yn1

= 1x

n

yn.

Multiplying byyn both sides,

y

1x

y

yn1

1+x

y+x

2

y2+ +x

n1

yn1

=yn

1x

n

yn

,

which is

yn xn = (y x)(yn1 +yn2 x+ +y xn2 + xn1),yielding the result.

16


25/194

Chapter 1

56THEOREM 1+2+ +n=n(n+1)2

.

First Proof: Observe that

k2 (k1)2 = 2k 1.

From this

1

2

02

= 2 1 122 12 = 2 2 132 22 = 2 3 1...

......

n2 (n1)2 = 2 n 1

Adding both columns,

n2 02 = 2(1+2+3+ +n) n.

Solving for the sum,

1+2+3+ + n= n2/2+ n/2 =n(n+1)2

.

Second Proof: We may utilise Gauss trick: If

An= 1+2+3+ +n

then

An= n+ (n 1) + +1.

Adding these two quantities,

An = 1 + 2 + + n

An = n + (n1) + + 12An = (n+ 1) + (n+1) + + (n+ 1)

= n(n+ 1),

since there arensummands. This givesAn=n(n+ 1)

2 , that is,

1+2+ +n= n(n+ 1)2

.

Applying Gausss trick to the general arithmetic sum

(a) + (a+d)+ (a+ 2d) + + (a+ (n1)d)

we obtain

(a)+ (a+ d) + (a+2d)+ +(a+ (n 1)d) = n(2a+ (n1)d)2

(1.5)

57THEOREM 12 + 22 +32 + +n2 = n(n+ 1)(2n+ 1)6

.

Proof: Observe that

k3 (k1)3 = 3k2 3k+ 1.

17


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Hence

13 03 = 3 12 3 1+ 123 13 = 3 22 3 2+ 133 23 = 3 32 3 3+ 1...

......

n3

(n

1)3

= 3

n2

3

n

+1

Adding both columns,

n3 03 = 3(12 + 22 +32 + +n2) 3(1+ 2+3+ +n)+ n.

From the preceding example1 +2+3+ + n= n2/2+n/2 = n(n+ 1)2

so

n3 03 = 3(12 + 22 + 32 + +n2) 32

n(n+ 1)+ n.

Solving for the sum,

1

2

+ 22

+ 32

+ +n2

=n3

3 +1

2 n(n+1) n

3 .

After simplifying we obtain

12 +22 +32 + +n2 = n(n+ 1)(2n+ 1)6

.

Homework

Problem 1.5.1 Prove that forn 1,

2n =n

k=0

nk

; 0 =

nk=0

(1)kn

k

, 2n1 =

0knkeven

nk

=

1knkodd

nk

.

Problem 1.5.2 Given that1002004008016032 has a prime factor

p> 250000,find it.

Problem 1.5.3 Prove that(a+b+c)2 = a2 +b2 +c2 +2ab+2bc+2ca.

Problem 1.5.4 Leta,b,cbe real numbers. Prove that

a3 +b3 +c3 3ab c= (a+b+c)(a2 +b2 +c2 abbcca).

Problem 1.5.5 Prove thatn

k

= n

k

n1k1

.

Problem 1.5.6 Prove thatn

k

= n

k n1

k1

n2k2

.


nk=1

kn

k

pk(1p)nk = np.


nk=2

k(k1)

n

k

pk(1p)nk = n(n1)p2.

Problem 1.5.9 Demonstrate that

nk=0

(knp)2

n

k

pk(1p)nk = np(1p).

Problem 1.5.10 LetxR\ {1}and letnN\ {0}. Prove thatn

k=0

2k

x2k +1

= 1x1

2n+1

x2n+1 +1

.

Problem 1.5.11 Consider the nk k-tuples (a1, a2, . . . ,ak) which

can be formed by taking ai {1,2,..., n}, repetitions allowed.Demonstrate that

ai{1,2,...,n}min(a1, a2, . . . ,ak) = 1k+2k+ +nk.

18


27/194

Chapter 1

1.6 Order Axioms

Vocabulary Alert!We will call a numberxpositiveifx 0andstrictly positiveifx> 0. Similarly, we will call

a numberynegativeify 0andstrictly negativeify< 0. This usage differs from most Anglo-American books,who prefer such terms asnon-negative andnon-positive.

We assumeR endowed with a relation >which satisfies the following axioms.

58 Axiom (Trichotomy Law) (x,y) R2 exactly one of the following holds:

x> y, x=y, or y> x.

59 Axiom (Transitivity of Order)(x,y, z) R3,

if x>y and y> z then x> z.

60 Axiom (Preservation of Inequalities by Addition) (x,y, z) R3,

if x>y then x+ z>y+ z.

61 Axiom (Preservation of Inequalities by Positive Factors) (x,y, z) R3,

if x>y and z> 0 then xz>y z.

x x.x ymeans that eithery> xory= x, etc.

62THEOREM The square of any real number is positive, that is, a R, a2 0. In fact, ifa= 0thena2 > 0.

Proof: Ifa= 0, then02 = 0and there is nothing to prove. Assume now thata= 0. By trichotomy, eithera> 0ora< 0. Assume first thata> 0. Applying Axiom61withx= z= aandy= 0we have

aa

>a0

=a2

>0,

so the theorem is proved ifa> 0.

Ifa< 0thena> 0and we apply the result just obtained:

a> 0= (a)2 > 0= 1 a2 > 0= a2 > 0,

so the result is true regardless the sign ofa.

Theorem62will prove to be extremely powerful and will be the basis for many of the classical inequalities that follow.

63THEOREM If(x,y) R2,x

>y

x

y

>0.

Proof: This is a direct consequence of Axiom60upon takingz= y.

64THEOREM If(x,y, a, b) R4,x> y and a b= x+ a>y+b.

Proof: We have

x>y= x+ a>y+ a, y+ ay+b,by Axiom60and so by Axiom59x+ a>y+ b.

19


28/194

Order Axioms

65THEOREM If(x,y, a, b) R4,x> y> 0 and a b> 0= x a> yb.

Proof: Indeed

x>y= x a>y a, y ayb,by Axiom61and so by Axiom59x a>yb.

66THEOREM 1 > 0.

Proof: By definition of R being a field0 = 1. Assume that1 < 0then12 > 0by Theorem62. But12 = 1and so1 > 0, a contradiction to our original assumption.

67THEOREM x> 0= x< 0 and x1 > 0.

Proof: Indeed, 1 < 0since1 = 0and assuming1 > 0would give0 = 1+1 > 1, which contradicts Theorem66.Thus

x= 1 x< 0.Similarly, assumingx1

1would give1 = x x1 > 1 1 = 1, a contradiction.

1.6.1 Absolute Value

69 Definition (The Signum (Sign) Function) Letxbe a real number. We define signum (x) =

1 if x< 0,0 if x= 0,

+1 if x> 0.

70LEMMA The signum function is multiplicative, that is, if(x,y) R2

thensignumxy= signum (x) signumy.Proof: Immediate from the definition of signum.

71 Definition (Absolute Value) LetxR. Theabsolute valueofxis defined and denoted by

|x| = signum (x) x.

72THEOREM LetxR. Then

1. |x| =

x ifx< 0,x ifx 0.

2. |x| 0,3. |x| = max(x,x),

4. |x| = |x|,

5.|x| x |x|.

6.

x2 = |x|

7. |x|2 = |x2| = x2

8. x= signum (x) |x|

20


29/194

Chapter 1

Proof: These are immediate from the definition of |x|.

73THEOREM ((x,y) R2), x y = |x| y .Proof: We have

x y = signumx yx y= signum (x) x signumyy= |x| y ,where we have used Lemma70. 74THEOREM Lett 0. Then

|x| t t x t.

Proof: Either|x| = xor |x| = x. If|x| = x,|x| t x t t 0 x t.

If|x| = x,|x| t x t t x 0 t.

75THEOREM If(x,y) R2, max(x,y) = x+y+xy

2 andmin(x,y) = x+y

xy2

.

Proof: Observe thatmax(x,y) + min(x,y) = x+y, since one of these quantities must be the maximum and theother the minimum, or else, they are both equal.

Now, eitherxy= xy, and soxy, meaning thatmax(x,y)min(x,y) = xy, orxy= (xy) =yx,

which means thaty xand somax(x,y)min(x,y) =yx. In either case we getmax(x,y)min(x,y) =xy.

Solving now the system of equations

max(x,y)+ min(x,y) = x+ymax(x,y)

min(x,y)

= xy ,formax(x,y)andmin(x,y)gives the result.Homework

Problem 1.6.1 Letx,ybe real numbers. Then

0 x


30/194

Classical Inequalities

Proof: From5in Theorem72, by addition,

|a| a |a|to

|b| b |b|we obtain

(|a|+ |b|) a+b (|a|+ |b|),

whence the theorem follows by applying Theorem74.

By induction, we obtain the following generalisation to nterms.

77C OROLLARY Letx1, x2, . . . , xnbe real numbers. Then

|x1 + x2 + + xn| |x1|+ |x2|+ + |xn| .

Proof: We apply Theorem76n 1times|x1 + x2 + + xn| |x1|+ |x2 + xn1 + xn|

|x1|+ |x2|+ |x3 + xn1 + xn|...

|x1|+ |x2|+ +|xn1 + xn| |x1|+ |x2|+ +|xn1|+ |xn| .

78C OROLLARY Let(a, b) R2. Then||a| |b|| |ab| . (1.7)

Proof: We have

|a| = |a b+b| |ab|+ |b|,giving

|a| |b| |a b|.Similarly,

|b| = |b a+ a| |b a|+ |a| = |ab|+ |a|,gives

|b| |a| |a b| = |a b| |a| |b| .Thus

|a b| |a| |b| |a b| ,and we now apply Theorem74.

79THEOREM Let bi> 0for1 i n. Then

mina1b1 ,a2b2 , . . . ,anbn a1 + a2 + + anb1 + b2 + +bn maxa1b1 ,a2b2 , . . . ,anbn .Proof: For everyk, 1 k n,

min

a1

b1,a2

b2, . . . ,

an

bn

ak

bk max

a1

b1,a2

b2, . . . ,

an

bn

= bkmin

a1

b1,a2

b2, . . . ,

an

bn

ak bkmax

a1

b1,a2

b2, . . . ,

an

bn

.

Adding all these inequalities for1 k n,

(b1 + b2 + +bn)min

a1

b1,a2

b2, . . . ,

an

bn

a1 + a2 + + an (b1 + b2 + +bn)max

a1

b1,a2

b2, . . . ,

an

bn

,

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