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HW2_199_sol - HOMEWORK 2 DUE Fri Oct 7 NAME DIRECTIONS Turn...

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HOMEWORK 2 DUE: Fri., Oct. 7 NAME: DIRECTIONS: Turn in your homework as SINGLE-SIDED typed or handwritten pages. STAPLE your homework together. Do not use paper clips, folds, etc. STAPLE this page to the front of your homework. Be sure to write your name on your homework. Show all work, clearly and in order . You will lose point 0.5 points for each instruction not followed. Questions Points Score 1 1 2 1 3 1 4 2 5 3 6 1 7 1 Total 10 1
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HOMEWORK 2 DUE: Fri., Oct. 7 NAME: Problem 1: (1 point) Suppose A negationslash = and B negationslash = . Show that A × B = B × A iff A = B . Proof: Assume A negationslash = and B negationslash = . Part 1: Assume A × B = B × A . By definition of equality of sets, this means that every element of A × B is an element of B × A so there exist elements p , m A and elements q , l B such that x = ( p,q ) A × B and x = ( l,m ) B × A and ( p,q ) = ( l,m ). But by the definition of ordered pairs, that means p = l and q = m so for all p A , there exists an l B such that p = l hence A B . Similarly, B A . Part 2: Assume A = B . Let ( p,q ) A × B . Hence p A = B = p B and q B = A = q A . Hence ( p,q ) B × A as well by definition of Cartesian product, hence A × B ofB × A . Similarly, B × A A × B. Parts 1 and 2 imply that assuming A negationslash = and B negationslash = then A × B = B × A iff A = B . Q. E. D. Problem 2: (1 point) If A , B , and C are finite sets, show that # ( A B C ) = # A + # B + # C # ( A B ) # ( A C ) # ( B C ) + # ( A B C ) . Proof: Since this is equality of numbers, rather than sets, it suffices to show that for element x in the universal set X contributes the same number to both sides of the equation above. An element x A B C or x / A B C .
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