Unformatted text preview: MATH 74 HOMEWORK 6 (DUE WEDNESDAY OCTOBER 17) 1. Let A = { a 2 b 2 : a, b ∈ Z } and B = { 2 k 1 : k ∈ Z } . 1(a). No, A is not a subset of B . For example, 0 ∈ A because 0 = 0 2 2 and ∈ Z . But 0 is not in B : if there were an integer k with 2 k 1 = 0, then k would have to satisfy 2 k = 1; but the only real number solution this equation has is k = 1 2 which is not an integer. This contradiction shows that 0 6∈ B , and thus A 6⊆ B . 1(b). Yes, B ⊆ A . To see this, let x ∈ B be arbitrary. By definition of B there is k ∈ Z with x = 2 k 1 . A little algebra shows us that then x = k 2 ( k 1) 2 . Since k and k 1 are both integers, this shows that x is in A (by the definition of A ). Since x was arbitrary this shows that B ⊆ A . 1(c). A 6 = B because A is not a subset of B . 2(a). Yes, it is true that A ∩ C ⊆ B ∩ C . To see this, let x ∈ A ∩ B be given. From the definition of A ∩ B we conclude that x ∈ A and also x ∈ C . From the fact that x ∈ A and the given information that...
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 Fall '07
 COURTNEY
 Math, Empty set, Natural number, Cartesian product

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