# HW 4 - MATH 290 HOMEWORK 4 SOLUTIONS 1. For each natural...

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MATH 290 – HOMEWORK 4 – SOLUTIONS 1. For each natural number n , let M n = { k Z : n | k } . (In other words, for each natu- ral number n , M n = { ..., - 3 n, - 2 n, - n, 0 , n, 2 n, 3 n, . .. } .) Prove that [ n =1 M n = Z and \ n =1 M n = { 0 } . (Hint: It is not suﬃcient to just explain why the given identities are true, you have to supply an actual proof.) Solution: (a). To see that [ n =1 M n = Z we must show inclusion both ways. To see that [ n =1 M n Z note that each M n Z so that if x M n for some n , then x Z . To see that Z [ n =1 M n , note that since 1 divides every integer, Z M 1 (in fact Z = M 1 but we don’t need that here), and so Z M 1 [ n =1 M n . Therefore [ n =1 M n = Z . (b). To see that \ n =1 M n = { 0 } we again must show inclusion both ways. To see that { 0 } ⊆ \ n =1 M n , note that since every natural number divides 0, that 0 M n for all natural numbers n . Hence { 0 } ⊆ \ n =1 M n . To see that \ n =1 M n ⊆ { 0 } , let k be a nonzero integer. We know that if m is a natural number such that m | k then m ≤ | k | . Therefore letting the natural number

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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.

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HW 4 - MATH 290 HOMEWORK 4 SOLUTIONS 1. For each natural...

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