Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 6 - Set Theory - Exercise Set 6.1 - Page 350: 7

Answer

A = {x ∈ Z | x = 6a + 4 for some integer a}, B = {y ∈ Z | y = 18b − 2 for some integer b}, and C = {z ∈ Z | z = 18c + 16 for some integer c} a. False, A ⊈ B. Contradiction: Suppose that x is a particular but arbitrarily chosen element of A and B. By definition of A, there is an integer a such that x = 6a + 4. By definition of B, there is an integer b such that x = 18b – 2. By substitution, it follows that: 6a + 4 = 18b – 2 b = (6a + 4 + 2)/18 = (6a + 6)/18 = (a + 1)/3 Since b = (a + 1)/3, then b is not an integer since it is a quotient. It follows that b is not an integer and b is an integer, which is a contradiction. Therefore, the supposition is false and A ⊈ B. Counterexample: There are elements in A that are not in B. For example, if a = 3 is a particular but arbitrarily chosen element in A, then x = 6(3) + 4 = 22. If x = 22 is an element in B, then: 22 = 18b – 2 b = (22 + 2)/18 = 24/18 = 6*4/(6*3) = 4/3 b = 4/3 Since b = 4/3, b is not an integer since it is a quotient. It follows that b is not an integer and b is an integer, which is a contradiction. Therefore, the supposition is false and A ⊈ B. b. True, B⊆A Proof: Suppose that y is a particular but arbitrarily chosen element of B. By definition of B, there is an integer b such that y = 18b – 2. [18b – 2 = 6a + 4 a = (18b – 6)/6 = 3b – 1] Let a = 3b – 1. Then a is an integer since it is a sum of integers. By substitution, 6a + 4 = 6(3b – 1) + 4 = 18b – 6 + 4 = 18b – 2. Therefore, by definition of A, x is an element of A. c. True, B = C ↔ B⊆C ^ C⊆B Proof that B⊆C: Suppose that y is a particular but arbitrarily chosen element of B. By definition of B, there is an integer b such that y = 18b – 2. [18b – 2 = 18c + 16 c = (18b – 18)/18 = b – 1] Let c = b - 1. Then c is an integer since it is a sum of integers. By substituting c into the definition of C, 18c + 16 = 18(b – 1) + 16 = 18b – 18 + 16 = 18b – 2 = x. Therefore, by definition of C, x is an element of C. Proof that C⊆B: Suppose that z is a particular but arbitrarily chosen element of C. By definition of C, there is an integer c such that z = 18c + 16. [18c + 16 = 18b - 2 b = (18c + 18)/18 = c + 1] Let b = c + 1. Then b is an integer since it is a sum of integers. By substituting b into the definition of B, 18b – 2 = 18(c + 1) – 2 = 18c + 18 – 2 = 18c + 16 = x. Therefore, by definition of B, x is an element of B.

Work Step by Step

A = {x ∈ Z | x = 6a + 4 for some integer a}, B = {y ∈ Z | y = 18b − 2 for some integer b}, and C = {z ∈ Z | z = 18c + 16 for some integer c} a. False, A ⊈ B. Contradiction: Suppose that x is a particular but arbitrarily chosen element of A and B. By definition of A, there is an integer a such that x = 6a + 4. By definition of B, there is an integer b such that x = 18b – 2. By substitution, it follows that: 6a + 4 = 18b – 2 b = (6a + 4 + 2)/18 = (6a + 6)/18 = (a + 1)/3 Since b = (a + 1)/3, then b is not an integer since it is a quotient. It follows that b is not an integer and b is an integer, which is a contradiction. Therefore, the supposition is false and A ⊈ B. Counterexample: There are elements in A that are not in B. For example, if a = 3 is a particular but arbitrarily chosen element in A, then x = 6(3) + 4 = 22. If x = 22 is an element in B, then: 22 = 18b – 2 b = (22 + 2)/18 = 24/18 = 6*4/(6*3) = 4/3 b = 4/3 Since b = 4/3, b is not an integer since it is a quotient. It follows that b is not an integer and b is an integer, which is a contradiction. Therefore, the supposition is false and A ⊈ B. b. True, B⊆A Proof: Suppose that y is a particular but arbitrarily chosen element of B. By definition of B, there is an integer b such that y = 18b – 2. [18b – 2 = 6a + 4 a = (18b – 6)/6 = 3b – 1] Let a = 3b – 1. Then a is an integer since it is a sum of integers. By substitution, 6a + 4 = 6(3b – 1) + 4 = 18b – 6 + 4 = 18b – 2. Therefore, by definition of A, x is an element of A. c. True, B = C ↔ B⊆C ^ C⊆B Proof that B⊆C: Suppose that y is a particular but arbitrarily chosen element of B. By definition of B, there is an integer b such that y = 18b – 2. [18b – 2 = 18c + 16 c = (18b – 18)/18 = b – 1] Let c = b - 1. Then c is an integer since it is a sum of integers. By substituting c into the definition of C, 18c + 16 = 18(b – 1) + 16 = 18b – 18 + 16 = 18b – 2 = x. Therefore, by definition of C, x is an element of C. Proof that C⊆B: Suppose that z is a particular but arbitrarily chosen element of C. By definition of C, there is an integer c such that z = 18c + 16. [18c + 16 = 18b - 2 b = (18c + 18)/18 = c + 1] Let b = c + 1. Then b is an integer since it is a sum of integers. By substituting b into the definition of B, 18b – 2 = 18(c + 1) – 2 = 18c + 18 – 2 = 18c + 16 = x. Therefore, by definition of B, x is an element of B.
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