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Fall 2007 - Wallach's Class - Exam 1

# Fall 2007 - Wallach's Class - Exam 1 - Solutions for...

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Unformatted text preview: Solutions for Midterm (Math 109) November 8, 2007 1. P Q not (P and Q) P = ) (not Q) (not P) or (not Q) (not Q) = ) (not P) T T F F F T T F T T T F F T T T T T F F T T T T The contrapositive of P = ) Q is (not Q) = ) (not P). 2. Claim : If A and B are &nite sets with B & A then j A ¡ B j = j A j¡j B j . Proof : First suppose A ¡ B = ; . Let a 2 A . Since A ¡ B = ; we have that a 2 B . Therefore A & B . Combining this with our hypothesis that B & A we get that A = B . Then in particular j A j = j B j , or in other words j A j¡j B j = 0 . And by assumption j A ¡ B j = 0 since the cardinality of the empty set is zero. Therefore j A ¡ B j = j A j ¡ j B j . Now suppose A ¡ B 6 = ; . First note that if B = ; then the claim is trivial. So suppose also that B 6 = ; . Let A ¡ B have cardinality n and B have cardinality m , where n and m are natural numbers. Then there exist bijections f : N n ¡! A ¡ B g : N m ¡! B We want to show that the cardinality of A is n + m , and to do that we have to...
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Fall 2007 - Wallach's Class - Exam 1 - Solutions for...

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