Problem_Set_02_Solutions

# Problem_Set_02_Solutions - ARS DIGITA UNIVERSITY MONTH 2:...

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ARS DIGITA UNIVERSITY MONTH 2: DISCRETE MATHEMATICS PROFESSOR SHAI SIMONSON PROBLEM SET 2 SOLUTIONS — SET, FUNCTIONS, BIG-O, RATES OF GROWTH (1) Prove by formal logic: (a) The complement of the union of two sets equals the intersection of the complements. Let x ( A B ). Then x is not in A , and x is not in B . Therefore x A B . Since x was an arbitrary element of ( A B ), we conclude that ( A B ) A B . Now, suppose that x A B . We can conclude that x A , and that x B . This impiles that x / A , and x / B , which implies x / ( A B ), which implies that x ( A B ). Again, since x was arbitrary, we conclude that ( A B ) A B . Two sets which are subsets of each other are equal (see below), so we can conclude that ( A B ) = A B (b) The complement of the interesection of two sets equals the union of the complements. Let x ( A B ). Then x cannot be in both A and B — at least one of the statements “ x A ”, “ x B ” is true. Therefore x A B , and ( A B ) A B . Now let x A B . Again, x cannot be in both A and B , so x / A B , so x ( A B ), and A B ( A B ). The two sets are equal. (c) ( B - A ) ( C - A ) = ( B C ) - A . Let x ( B - A ) ( C - A ). Then x ( B - A ) x ( C - A ). Assume without loss of generality that x ( B - A ). This implies that x B and x / A . Which implies that x ( B C ), and, combining this with the fact that x / A , we conclude that x ( B C ) - A , and ( B - A ) ( C - A ) ( B C ) - A . If x ( B C ) - A , then x / A , and x B x C . Therefore x ( B - A ) x ( C - A ), x ( B - A ) ( C - A ), and ( B C ) - A ( B - A ) ( C - A ), completing the proof. (d) If two sets are subsets of each other than they are equal. If two sets A and B are not equal, then there exists some element that is in one set that is not in the other. Without loss of generality, denote such an element x and assume that x A . Since x A and x / B , we conclude that A 6⊂ B . (2) Generalize De Morgan’s laws for n sets and prove the laws by induction. 1

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2 PROBLEM SET 2 The n -version generalization of De Morgan’s laws are as follows: n [ i =1 A i = n ± i =1 A i n ± i =1 A i = n [ i =1 A i To prove these laws by induction, we use n = 2 as our base cases: these are the traditional two-set De Morgan’s laws. Now, we assume that the laws hold for n sets, and show that they hold for n + 1 sets as well, using both our inductive hypothesis and the two-set De Morgan’s laws: n +1 [ i =1 A i = ( n [ i =1 A i ) A n +1 = ( n [ i =1 A i ) A n +1 = ( n ± i =1 A i ) A n +1 = n +1 ± i =1 A i n +1 ± i =1 A i = ( n ± i =1 A i ) A n +1 = ( n ± i =1 A i ) A n +1 = ( n [ i =1 A i ) A n +1 = n +1 [ i =1 A i (3) Prove by induction on the size of the set, that the power set P ( A ) has
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Problem_Set_02_Solutions - ARS DIGITA UNIVERSITY MONTH 2:...

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