Book of Proof - Amazon S3

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270 Solutions 19. Prove that {9 n : n ∈ Z} ⊆ {3 n : n ∈ Z}, but {9 n : n ∈ Z} ≠ {3 n : n ∈ Z}. Proof. Suppose a ∈ {9 n : n ∈ Z}. This means a = 9 n for some integer n ∈ Z. Thus a = 9 n = (3 2 ) n = 3 2n . This shows a is an integer power of 3, so a ∈ {3 n : n ∈ Z}. Therefore a ∈ {9 n : n ∈ Z} implies a ∈ {3 n : n ∈ Z}, so {9 n : n ∈ Z} ⊆ {3 n : n ∈ Z}. But notice {9 n : n ∈ Z} ≠ {3 n : n ∈ Z} as 3 ∈ {3 n : n ∈ Z}, but 3 ∉ {9 n : n ∈ Z}. ■ 21. Suppose A and B are sets. Prove A ⊆ B if and only if A − B = . Proof. First we will prove that if A ⊆ B, then A − B = . Contrapositive proof is used. Suppose that A − B ≠ . Thus there is an element a ∈ A − B, which means a ∈ A but a ∉ B. Since not every element of A is in B, we have A ⊈ B. Conversely, we will prove that if A − B = , then A ⊆ B. Again, contrapositive proof is used. Suppose A ⊈ B. This means that it is not the case that every element of A is an element of B, so there is an element a ∈ A with a ∉ B. Therefore we have a ∈ A − B, so A − B ≠ . ■ 23. For each a ∈ R, let A a = { (x, a(x 2 − 1)) ∈ R 2 : x ∈ R } . Prove that ⋂ a∈R A a = {(−1,0),(1,0))}. Proof. First we will show that {(−1,0),(1,0))} ⊆ ⋂ a∈R A a . Notice that for any a ∈ R, we have (−1,0) ∈ A a because A a contains the ordered pair (−1, a((−1) 2 −1) = (−1,0). Similarly (1,0) ∈ A a . Thus each element of {(−1,0),(1,0))} belongs to every set A a , so every element of ⋂ A a , so {(−1,0),(1,0))} ⊆ ⋂ A a . a∈R a∈R Now we will show ⋂ A a ⊆ {(−1,0),(1,0))}. Suppose (c, d) ∈ ⋂ A a . This means a∈R a∈R (c, d) is in every set A a . In particular (c, d) ∈ A 0 = { (x,0(x 2 − 1)) : x ∈ R } = {(x,0) : x ∈ R}. It follows that d = 0. Then also we have (c, d) = (c,0) ∈ A 1 = { (x,1(x 2 − 1)) : x ∈ R } = { (x, x 2 − 1) : x ∈ R } . Therefore (c,0) has the form (c, c 2 − 1), that is (c,0) = (c, c 2 − 1). From this we get c 2 − 1 = 0, so c = ±1. Therefore (c, d) = (1,0) or (c, d) = (−1,0), so (c, d) ∈ {(−1,0),(1,0))}. This completes the demonstration that (c, d) ∈ ⋂ A a a∈R implies (c, d) ∈ {(−1,0),(1,0))}, so it follows that ⋂ a∈R A a ⊆ {(−1,0),(1,0))}. Now it’s been shown that {(−1,0),(1,0))} ⊆ ⋂ A a and ⋂ A a ⊆ {(−1,0),(1,0))}, so it a∈R a∈R follows that ⋂ A a = {(−1,0),(1,0))}. ■ a∈R 25. Suppose A,B,C and D are sets. Prove that (A × B) ∪ (C × D) ⊆ (A ∪ C) × (B ∪ D). Proof. Suppose (a, b) ∈ (A × B) ∪ (C × D). By definition of union, this means (a, b) ∈ (A × B) or (a, b) ∈ (C × D). We examine these two cases individually. Case 1. Suppose (a, b) ∈ (A × B). By definition of ×, it follows that a ∈ A and b ∈ B. From this, it follows from the definition of ∪ that a ∈ A ∪ C and b ∈ B ∪ D. Again from the definition of ×, we get (a, b) ∈ (A ∪ C) × (B ∪ D).
271 Case 2. Suppose (a, b) ∈ (C × D). By definition of ×, it follows that a ∈ C and b ∈ D. From this, it follows from the definition of ∪ that a ∈ A ∪ C and b ∈ B ∪ D. Again from the definition of ×, we get (a, b) ∈ (A ∪ C) × (B ∪ D). In either case, we obtained (a, b) ∈ (A ∪ C) × (B ∪ D), so we’ve proved that (a, b) ∈ (A × B) ∪ (C × D) implies (a, b) ∈ (A ∪ C) × (B ∪ D). Therefore (A × B) ∪ (C × D) ⊆ (A ∪ C) × (B ∪ D). ■ 27. Prove {12a + 4b : a, b ∈ Z} = {4c : c ∈ Z}. Proof. First we show {12a + 4b : a, b ∈ Z} ⊆ {4c : c ∈ Z}. Suppose x ∈ {12a + 4b : a, b ∈ Z}. Then x = 12a + 4b for some integers a and b. From this we get x = 4(3a + b), so x = 4c where c is the integer 3a+b. Consequently x ∈ {4c : c ∈ Z}. This establishes that {12a + 4b : a, b ∈ Z} ⊆ {4c : c ∈ Z}. Next we show {4c : c ∈ Z} ⊆ {12a + 4b : a, b ∈ Z}. Suppose x ∈ {4c : c ∈ Z}. Then x = 4c for some c ∈ Z. Thus x = (12 + 4(−2))c = 12c + 4(−2c), and since c and −2c are integers we have x ∈ {12a + 4b : a, b ∈ Z}. This proves that {12a + 4b : a, b ∈ Z} = {4c : c ∈ Z}. ■ 29. Suppose A ≠ . Prove that A × B ⊆ A × C, if and only if B ⊆ C. Proof. First we will prove that if A×B ⊆ A×C, then B ⊆ C. Using contrapositive, suppose that B ⊈ C. This means there is an element c ∈ C with c ∉ B. Since A ≠ , there exists an element a ∈ A. Now consider the ordered pair (a, c). Note that (a, c) ∈ A × C, but (a, c) ∉ A × B. This means A × B ⊈ A × C. Conversely, we will now show that if B ⊆ C, then A × B ⊆ A × C. We use direct proof. Suppose B ⊆ C. Assume that (a, b) ∈ A × B. This means a ∈ A and b ∈ B. But, as B ⊆ C, we also have b ∈ C. From a ∈ A and b ∈ C, we get (a, b) ∈ A × C. We’ve now shown (a, b) ∈ A × B implies (a, b) ∈ A × C, so A × B ⊆ A × C. ■ 31. Suppose B ≠ and A × B ⊆ B × C. Prove A ⊆ C. Proof. Suppose B ≠ and A × B ⊆ B × C. In what follows, we show that A ⊆ C. Let x ∈ A. Because B is not empty, it contains some element b. Observe that (x, b) ∈ A × B. But as A × B ⊆ B × C, we also have (x, b) ∈ B × C, so, in particular, x ∈ B. As x ∈ A and x ∈ B, we have (x, x) ∈ A × B. But as A × B ⊆ B × C, it follows that (x, x) ∈ B × C. This implies x ∈ C. Now we’ve shown x ∈ A implies x ∈ C, so A ⊆ C. ■ Chapter 9 Exercises 1. If x, y ∈ R, then |x + y| = |x| + |y|. This is false. Disproof: Here is a counterexample: Let x = 1 and y = −1. Then |x + y| = 0 and |x| + |y| = 2, so it’s not true that |x + y| = |x| + |y|.
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Book of Proof Richard Hammack Virgi
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To my students
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v II How to Prove Conditional State
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Preface In writing this book I have
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ix The book is organized into four
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Part I Fundamentals
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4 Sets Some sets are so significant
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6 Sets These last three examples hi
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8 Sets 1.2 The Cartesian Product Gi
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10 Sets We can also take Cartesian
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12 Sets This idea of “making” a
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14 Sets This is a subset C ⊆ R 2
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16 Sets y y y x x x (a) (b) (c) Fig
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18 Sets B A ∪ B A ∩ B A − B A
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20 Sets Example 1.7 If P is the set
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22 Sets We can also think of A ∩
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24 Sets 1.8 Indexed Sets When a mat
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26 Sets Example 1.11 Here the sets
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28 Sets 1.9 Sets that Are Number Sy
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30 Sets Russell’s paradox arises
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32 Logic In proving theorems, we ap
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34 Logic R(f , g) : The function f
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36 Logic 2.2 And, Or, Not The word
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38 Logic To conclude this section,
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40 Logic You can think of P ⇒ Q a
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42 Logic that it’s impossible tha
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44 Logic P if and only if Q. P is a
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46 Logic Notice that when we plug i
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48 Logic There are two pairs of log
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50 Logic Likewise, a statement such
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52 Logic 2.8 More on Conditional St
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54 Logic Example 2.9 Consider Goldb
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56 Logic Example 2.10 Consider nega
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58 Logic Example 2.13 Negate the fo
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60 Logic 2.12 An Important Note It
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62 Counting Occasionally we may get
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64 Counting To answer this question
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66 Counting we can choose any one o
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68 Counting 3.2 Factorials In worki
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70 Counting We close this section w
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72 Counting Definition 3.2 If n and
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74 Counting Fact 3.3 ( ( ) n n! n I
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76 Counting 3.4 Pascal’s Triangle
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78 Counting coefficients 1 2 1. Sim
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80 Counting We seek the number of 3
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82 Counting 5. How many 7-digit bin
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CHAPTER 4 Direct Proof It is time t
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Definitions 87 4.2 Definitions A pr
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Definitions 89 it necessary to defi
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Direct Proof 91 So the setup for di
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Direct Proof 93 Here is another exa
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Direct Proof 95 This proposition te
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Treating Similar Cases 97 Propositi
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Treating Similar Cases 99 19. Suppo
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Contrapositive Proof 101 Since P
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Congruence of Integers 103 Proposit
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Mathematical Writing 105 Propositio
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Mathematical Writing 107 Since a |
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CHAPTER 6 Proof by Contradiction We
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Proving Statements with Contradicti
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Proving Conditional Statements by C
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Some Words of Advice 115 for intege
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Part III More on Proof
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120 Proving Non-Conditional Stateme
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130 Proofs Involving Sets Generally
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132 Proofs Involving Sets In practi
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134 Proofs Involving Sets Example 8
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136 Proofs Involving Sets Thus, sin
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138 Proofs Involving Sets Though it
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140 Proofs Involving Sets If we add
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142 Proofs Involving Sets so that d
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CHAPTER 9 Disproof E ver since Chap
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146 Disproof deciding whether a sta
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148 Disproof Example 9.2 Conjecture
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150 Disproof 9.3 Disproof by Contra
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CHAPTER 10 Mathematical Induction T
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154 Mathematical Induction This pic
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156 Mathematical Induction To round
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158 Mathematical Induction Proposit
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160 Mathematical Induction Strong i
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164 Mathematical Induction We next
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168 Mathematical Induction 13. For
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Part IV Relations, Functions and Ca
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174 Relations The set R encodes the
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176 Relations Exercises for Section
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178 Relations Example 11.7 Here A =
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180 Relations Example 11.8 Prove th
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182 Relations 11.2 Equivalence Rela
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184 Relations We close this section
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186 Relations 11.3 Equivalence Clas
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188 Relations Notationally, the uni
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190 Relations In a similar vein, [2
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192 Relations Exercises for Section
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CHAPTER 12 Functions You know from
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196 Functions value a to the output
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198 Functions To answer this, first
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200 Functions For more concrete exa
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202 Functions To see that g is surj
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204 Functions Although the underlyi
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206 Functions 12.4 Composition You
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208 Functions Proof. First suppose
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210 Functions A B a 1 a 1 b 2 b 2 c
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212 Functions We can check our work
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214 Functions If you continue your
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216 Cardinality of Sets On the othe
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218 Cardinality of Sets There is a
Page 230 and 231: 220 Cardinality of Sets And saying
Page 232 and 233: 222 Cardinality of Sets Here is a s
Page 234 and 235: 224 Cardinality of Sets Beginning a
Page 236 and 237: 226 Cardinality of Sets Exercises f
Page 238 and 239: 228 Cardinality of Sets f : A → P
Page 240 and 241: 230 Cardinality of Sets 13.4 The Ca
Page 242 and 243: 232 Cardinality of Sets and the whi
Page 244 and 245: 234 Cardinality of Sets Here are so
Page 246 and 247: 236 Cardinality of Sets On the face
Page 248 and 249: Solutions Chapter 1 Exercises Secti
Page 250 and 251: 240 Solutions Sketch the following
Page 252 and 253: 242 Solutions 5. Sketch the sets X
Page 254 and 255: 244 Solutions Section 1.8 1. Suppos
Page 256 and 257: 246 Solutions Section 2.4 Without c
Page 258 and 259: 248 Solutions 7. P ⇒ Q = (P∧
Page 260 and 261: 250 Solutions Section 2.10 Negate t
Page 262 and 263: 252 Solutions (b) How many such cod
Page 264 and 265: 254 Solutions 5. How many 7-digit b
Page 266 and 267: 256 Solutions 15. If n ∈ Z, then
Page 268 and 269: 258 Solutions 7. Proposition Suppos
Page 270 and 271: 260 Solutions 27. If a ≡ 0 (mod 4
Page 272 and 273: 262 Solutions 7. If a, b ∈ Z, the
Page 274 and 275: 264 Solutions This expresses a 3 +
Page 276 and 277: 266 Solutions 19. If n ∈ N, then
Page 278 and 279: 268 Solutions Chapter 8 Exercises 1
Page 282 and 283: 272 Solutions 3. If n ∈ Z and n 5
Page 284 and 285: 274 Solutions 25. For all a, b, c
Page 286 and 287: 276 Solutions 5. If n ∈ N, then 2
Page 288 and 289: 278 Solutions (2) Now assume the st
Page 290 and 291: 280 Solutions 23. Use induction to
Page 292 and 293: 282 Solutions Thus, with n + 1 line
Page 294 and 295: 284 Solutions a b a b a b a b a b a
Page 296 and 297: 286 Solutions Therefore 3x − 5z i
Page 298 and 299: 288 Solutions Chapter 12 Exercises
Page 300 and 301: 290 Solutions 13. Consider the func
Page 302 and 303: 292 Solutions Section 12.5 Exercise
Page 304 and 305: 294 Solutions 13. Let f : A → B b
Page 306 and 307: 296 Solutions 11. Partition N into
Page 308 and 309: 298 Solutions 7. Prove or disprove:
Page 310 and 311: 300 Index notation, 197 one-to-one,
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