Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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1.13. Let x A . Then x = 2 k - 1 for some k Z . Let k 1 = k - 1. We then have 2 k 1 + 1 B , but also 2 k 1 + 1 = 2( k - 1) + 1 = 2 k - 1 = x , so x B . Hence A B . Conversely, let x B so that x = 2 m + 1 for some m Z and let m 1 = 2 m - 1. Then 2 m 1 - 1 A and x = 2 m +1 = 2( m +1) - 1 = 2 m 1 - 1 so x A . Thus B A . It follows that A = B . 1.14. We claim that [ a,b ] [ c,d ] = [ a,d ] - ( b,c ). We first show [ a,b ] [ c,d ] [ a,d ] - ( b,c ). Let x [ a,b ] [ c,d ]. Then a x b or c x d . In the first case, x d , since x b < d , and x 6 > b , since x b , so x [ a,d ] - ( b,c ). In the latter case, x a , since x c > a , and x 6 < c , since x c , so x [ a,d ] - ( b,c ). Thus [ a,b ] [ c,d ] [ a,d ] - ( b,c ) as desired. Conversely we show [ a,d ] - ( b,c ) [ a,b ] [ c,d ]. Let x [ a,d ] - ( b,c ). Then a x d and x / ( b,c ). Thus x b or x c . In the former case,
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HW1solns - 1.13 Let x A Then x = 2k 1 for some k Z Let k1 =...

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