sol1 - MATH 235 Algebra 1 Solutions to Assignment 1...

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MATH 235 – Algebra 1 Solutions to Assignment 1 September 18, 2007 Solution of 1. 1. If N > 1: N [ n =1 [ - n,n ] = [ - N,N ] . This is true because if x S N n =1 [ - n,n ], then x [ - n,n ] for some 1 6 n 6 N , and since [ - n,n ] [ - N,N ] we have x [ - N,N ]. To show the other inclusion, if x [ - N,N ] then x [ - n,n ] for n = N , so x S N n =1 [ - n,n ]. N \ n =1 [ - n,n ] = [ - 1 , 1] . This is true because if x T N n =1 [ - n,n ], then x [ - n,n ] for all 1 6 n 6 N , so, in particular, x [ - 1 , 1]. To show the other inequality, if x [ - 1 , 1], x [ - n,n ] for all 1 6 n 6 N (since [ - 1 , 1] [ - n,n ] for all such n ), so x T N n =1 [ - n,n ]. 2. [ n =1 [ n,n + 1] = [1 , ) . This is true because if x S n =1 [ n,n + 1], then x [ n,n + 1] for some n > 1, so x > n > 1, and thus x [1 , ). To show the other inclusion, if x [1 , ), and we let n be the greatest integer such that n 6 x , then x [ n,n + 1], and n > 1 (since x > 1), so x S n =1 [ n,n + 1]. [ n =1 ( n,n + 2) = (1 , ) . This is true because if x S n =1 ( n,n + 2), then x ( n,n + 2) for some n > 1, so x > n > 1, i.e. x (1 , ). To show the other inclusion, if x (1 , ), and we let n be the greatest integer such that n < x , then x ( n,n + 2), and n > 1 (since x > 1), so x S n =1 ( n,n + 2). 3. [ n =1 A n = A 1 = N . Clearly S n =1 A n N because every A i is, by definition, a subset of N . To show the other 1
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direction, note that every x N can be written as x n for n = 1, i.e. x A 1 , so x S n =1 A n . \ n =1 A n = { 0 , 1 } . Clearly 0 and 1 are in every A n since 0 = 0 n and 1 = 1 n for every n > 1. So { 0 , 1 } ⊂ T n =1 A n . We will prove the other direction by contradiction. Suppose that x T n =1 A n , x 6∈ { 0 , 1 } . Then, in particular,
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This note was uploaded on 04/18/2008 for the course MATH 235 taught by Professor Goren during the Fall '07 term at McGill.

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sol1 - MATH 235 Algebra 1 Solutions to Assignment 1...

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