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Unformatted text preview: MATH 290 – HOMEWORK 6 – SOLUTIONS 1. Define the relation R on the integers Z by xRy if and only if either x  y or y  x . (a) Prove that the relation R is not transitive. (Hint: To show it is not transitive, you must find three integers x , y , and z such that xRy and yRz but x 6 Rz .) (b) For each n ∈ Z , define A n to be the set of integers related to n , that is, A n = { x ∈ Z : xRn } . Since R is symmetric we also have A n = { x ∈ Z : nRx } . Describe explicitly the sets A 20 and A . This description can be done by listing the set, or by giving a precise description of the set in words. (c) Prove that \ n ∈ Z A n = { 1 , , 1 } . (Hint: To prove the inclusion “ ⊆ ” you might want to try proof by contrapositive, that is, show that if m / ∈ { 1 , , 1 } then m / ∈ T n ∈ Z A n .) Solution: (a). Let x = 5, y = 10, and z = 2. Since 5  10, xRy and since 2  10, yRz . But since 2 does not divide 5 and 5 does not divide 2, 2 6 R 5 and hence R is not transitive. (b). The set A 20 consists of all multiples of 20, that is, all m ∈ Z such that 20  m , together with the factors of 20 and their negatives, that is, all m ∈ Z such that m  20....
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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.
 Fall '08
 Alligood,K
 Advanced Math, Integers

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