sample-soln1 - q and squaring both sides we have 3 q 2 = p...

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Problem 1 a,c Problem 2 a,d Problem 3 c,d Problem 4 a Problem 5 a,d Problem 6 a,c,d,e Problem 7 a,b,d,e (note that the u in part d) is supposed to be y ) Problem 8 Let x ( A B ) ( B A ). By definition of union, intersection and complement, this is equivalent to [( x A ) ∧ ¬ ( x B )] [( x B ) ∧ ¬ ( x A )]. Let p be x A and q be x B , then we can manipulate the expression as follows. ( p ∧ ¬ q ) ( q ∧ ¬ p ) [( p ∧ ¬ q ) q ] [( p ∧ ¬ q ) ∨ ¬ p ] [( p q ) ( ¬ q q )] [( p ∨ ¬ p ) ( ¬ q ∨ ¬ p )] [( p q ) T ] [ T ( ¬ q ∨ ¬ p )] ( p q ) ( ¬ q ∨ ¬ p ) ( p q ) ∧ ¬ ( q p ) Translating back to set notation this is equivalent to x ( A B ) ( A B ) = ( A B ) - ( A B ). Thus ( A B ) ( B A ) = ( A B ) - ( A B ). Problem 9 ( p ∧ ¬ q ) r ≡ ¬ ( p ∧ ¬ q ) r ≡ ¬ p q r ( ¬ p r ) q ≡ ¬ ( p ∧ ¬ r ) q ( p ∧ ¬ r ) q Problem 10 a) xL ( r,x ) ∨ ∃ x [( yL ( x,y )) N ( r,x ))] b) xL ( r,x ) ∧ ∃ yL ( y,r ) Problem 11 a) Suppose 3 is rational. Then 3 = p q for some integers p,q (with q 6 = 0) and p,q have no common factors except 1 (i.e. we reduce the fraction to its simplest terms). Multiplying by
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Unformatted text preview: q and squaring both sides we have 3 q 2 = p 2 . Thus, p 2 is divisible by 3. So p is divisible by 3. Substitute p = 3 k in to the equation we get 3 q 2 = 9 k 2 , or equivalently q 2 = 3 k 2 which implies q is divisible by 3. This contradicts our assumption that p,q have no common factors. Thus 3 is irrational. b) We will prove x is rational i y is rational, which is equivalent to the given statement. Assume x is rational, then x = p q (for integers p,q , q 6 = 0). But y = 4-x = 4 q-p q . Thus y is also rational. The opposite direction that y is rational implies x is rational can be proved similarly. c) x = 0 ,y = 2 is a counterexample. 1...
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This note was uploaded on 04/01/2008 for the course EECS 203 taught by Professor Yaoyunshi during the Winter '07 term at University of Michigan.

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