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Unformatted text preview: Math 3040: Solutions for Homework 8. (1) For each of the following relations on the set of integers Z = { , 1 , 1 , 2 , 2 , 3 ... } determine: is it reflexive? Is it symmetric? Is it transitive? is it an equiv alence relation? Prove your claims. (a) ( a,b ) ∈ R iff 1 ≤ a b ≤ 1. (b) Let n be a positive integer. ( a,b ) ∈≡ n iff n divides a b . (c) ( a,b ) ∈ T iff ab is negative. (d) ( a,b ) ∈ S iff ab is nonnegative. Solution. (a) • Reflexive. Let a ∈ Z . Then a a = 0, so 1 < a a < 1. Therefore, aRa . • Symmetric. Let aRb . Then 1 < a b < 1, and so multiplying all three sides of the inequality by 1 gives 1 < b a < 1. Therefore bRa . • NOT Transitive. Counterexample: a = 0, b = 1, c = 2. 0 1 = 1 and 1 2 = 1, so aRb and bRc . However, 0 2 = 2, and so ( a,c ) 6∈ R . • R is NOT an equivalence relation. (b) • Reflexive. Let a ∈ Z . Then a a = 0, so n  0. Therefore, a ≡ n a . • Symmetric. Let a ≡ n b . Then n  a b , which means there is an m ∈ Z such that mn = a b . Then ( m ) n = b a , and so n  b a , and thus b ≡ n a . • Transitive. Let a ≡ n b and b ≡ n c . Then there are p,q ∈ Z such that pn = a b and qn = b c . Then ( p + q ) n = a b + b c = a c , and so n  a c , and thus a ≡ n...
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This note was uploaded on 01/25/2011 for the course MATH 3040 taught by Professor Kahn during the Spring '08 term at Cornell.
 Spring '08
 KAHN
 Integers

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