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FinalReview-Solns

# FinalReview-Solns - MATH 127 Final Review Solutions Friday...

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MATH 127: Final Review Solutions Friday, April 29, 2011 1. Note that x 3 + y 3 x 2 y + xy 2 iff x 3 + y 3 - x 2 y - xy 2 0. We will show the latter. x 3 + y 3 - x 2 y - xy 2 = x 2 ( x - y ) + y 2 ( y - x ) = ( x 2 - y 2 )( x - y ) = ( x + y )( x - y ) 2 0 (since x + y 0 and ( x - y ) 2 0) Thus x 3 + y 3 x 2 y + xy 2 as desired. 2. We assume X B and we want to show A B A - X B - X ( ) Assume A B and let a A - X . Then a A and a 6∈ X . Since A B we know a B and a 6∈ X . Thus a B - X and hence A - X B - X ( ) Assume A - X B - X and let a A . Case 1: If a 6∈ X then a A - X and hence a B - X which implies a B Case 2: If a X then a B since X B Thus A B as desired. 3. We want to use double containment to show ( A - B ) ( B - A ) = ( A B ) - ( A B ) ( ): Let x ( A - B ) ( B - A ). Then x A - B or x B - A . WLOG, let x A - B . Then x A and x 6∈ B . Hence x A B (since x A ) and x 6∈ A B (since x A but x 6∈ B ). Thus x ( A B ) - ( A B ) ( ): Let x ( A B ) - ( A B . Thus x A B and x 6∈ A B . This implies x A or x B but not in both. Thus x A - B or x B - A . Hence x ( A - B ) ( B - A ) 4. Proof by induction. Base: n = 2 1 · 2 = 1 · 2 · 3 3 as desired IH: Assume the statement holds for n , we want to show it holds for n + 1 1 · 2 + · · · ( n - 1) n + n ( n + 1) = ( n - 1) n ( n + 1) 3 + n ( n + 1) (By IH) = ( n - 1) n ( n + 1) + 3 n ( n + 1) 3 = n ( n + 1)( n - 1 + 3) 3 = n ( n + 1)( n + 2) 3 as desired

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FinalReview-Solns - MATH 127 Final Review Solutions Friday...

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