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22+PS3+Solutions

# 22+PS3+Solutions - 1 Handout#22 CS103 Robert Plummer...

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1 Handout #22 CS103 April 18, 2011 Robert Plummer Problem Set #3 Solutions 1. Suppose A and B are sets. Prove that (A B) - A = B - (A B). PROOF: We will prove equality by showing that LHS RHS and RHS LHS. [We'll put some whitespace in the first half of the proof to show how it improves readability.] Suppose x ((A B) – A). Then (x A x B) x A by the definition of union and set difference. Since x A and x A cannot both be true, we have x B x A. If x A, then x A B, so we have x B x (A B), which means x (B – (A B)) by the definition of set difference. So LHS RHS. For the second half of the proof, suppose x (B – (A B)). Then x B x (A B) by the definition of set difference. Thus x B (x A x B) by definition of intersection, and x B x A since x B and x B cannot both be true. Since x B, x A B, so we have x (A B) x A and x ((A B) – A) ) by the definition of set difference. So RHS LHS. 2. Consider the following three conditions: (i) A B (ii) A C (iii) A B - C = Can there exist sets A, B, and C that satisfy all three conditions? If so, provide an example. If not, provide a proof to that effect. Example: A = {1, 2}, B = {2, 3}, C = {2, 4}. Or, any non-empty sets where A = B = C. 3. If A and B are sets, is it possible that A B and A B? Give an example or prove that this is not possible. Example: A = {1, 2}, B = {1, 2, {1, 2}} 4. Suppose that A and B are sets. Prove that B A = A if and only if A B. [Indenting can improve readability, and placing a formula on a separate line makes it convenient to give it a number for later reference. Lower case Roman numerals are often used for this purpose.] PROOF. First the forward direction ( ): Suppose that (B A) = A (i) and suppose that x A. Then by (i), x B A., so x B. Thus A B. Now the backward direction ( ): Suppose that A B (ii) and suppose that x B A. Then x A. Now suppose x A. Then x B by (ii), and x B A. Thus B A = A.

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