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hw1solutions23b

# hw1solutions23b - Anna Medvedovsky [email protected] Math...

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Anna Medvedovsky Math 23b / Spring 2009 HW #1 solutions 1. A = { j 2 - j : j Z } and B = { k 2 + k : k Z ,k 0 } . Prove that A = B . First, I claim that A is a subset of B : If a A , then there exists j Z with j 2 - j = a . If j 1, then j - 1 Z 0 ; reexpressing a as a = ( j - 1) 2 + ( j - 1), we see that a B . If j 0, then - j Z 0 ; reexpressing a as a = ( - j ) 2 + ( - j ) we see that a is again in B . So a B regardless of j , and therefore A is a subset of B . Now I claim that B is a subset of A : If b B , then there exists k Z 0 with k 2 + k = b . But then b = ( k + 1) 2 - ( k + 1); since k + 1 Z , we see that b B . Since each set is a subset of the other, the two sets are equal, as desired. 2. A = { x 2 - y 2 : x,y Z } and B = { n Z : n even or n divisible by 4 } . To show: A = B . First, I claim that A is a subset of B . Indeed, if a A , then a has the form a = x 2 - y 2 = ( x + y )( x - y ) for some x and y in Z . If x and y have the same parity (both odd or both even) then x + y and x - y are both even numbers, so their product a = ( x + y )( x - y ) is divisible by 4 and

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hw1solutions23b - Anna Medvedovsky [email protected] Math...

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