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Unformatted text preview: COT 3100: Applications of Discrete Structures February 16, 2010 Solution to Homework 5 Lecturer: Prof. My T. Thai TA: Thang N. Dinh 1 Section 2.2 Problem 16 Solution: a. If x is in A ∩ B , then perforce it is in A (by definition of intersection). b. If x is in A , then perforce it is in A ∪ B (by definition of union). c. If x is in A B , then perforce it is in A (by definition of difference). d. If x ∈ A then x / ∈ B A . Therefore there can be no elements in A ∩ ( B A ), so A ∩ ( B A ) = ∅ . e. The lefthand side consists precisely of those things that are either elements of A or else elements of B but not A , in other words, things that are elements of either A or B (or, of course, both). This is precisely the definition of the righthand side. Problem 17 Solution: This exercise asks for a proof of a generalization of one of De Morgan’s laws for sets from two sets to three. a. This proof is similar to the proof of the twoset property, given in Example 10. Suppose x ∈ A ∩ B ∩ C . Then x / ∈ A ∩ B ∩ C , which means that x fails to be in at least one of these three sets. In other words, x / ∈ A or x / ∈ B or x / ∈ C . This is equivalent to saying that x ∈ ¯ A or x ∈ ¯ B or x ∈ ¯ C . There for x ∈ ¯ A ∪ ¯ B ∪ ¯ C , as desired. Conversely if x ∈ ¯ A ∪ ¯ B ∪ ¯ C , then x ∈ ¯ A or x ∈ ¯ B or x ∈ ¯ C , so x cannot be in the intersection of A,B and C . Since x / ∈ A ∩ B ∩ C , we conclude that x ∈ A ∩ B ∩ C , as desired. b. The following membership table gives the desired equality, since columns five and nine are identical. Page 11 A B C A ∩ B ∩ C A ∩ B ∩ C ¯ A ¯ B ¯ C ¯ A ∪ ¯ B ∪ ¯ C 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Problem 18 Solution: a. Suppose that x ∈ A ∪ B . Then either x ∈ A or x ∈ B . In either case, certainly x ∈ A ∪ B ∪ C . This establishes the desired inclusion. b. Suppose that x ∈ A ∩ B ∩ C . Then x is in all three of these sets. In particular, it is in both A and B and therefore in A ∩ B , as desired. c. Suppose that x ∈ ( A B ) C . Then x is in A B but not in C . Since x ∈ A B , we know that x ∈ A (we also know that x / ∈ B , but that won’t be used here). Since we have established that x ∈ A but x / ∈ C , we have proved that x ∈ A C . d. To show that the set given on the lefthand side is empty, it suffices to assume that x is some element in that set and derive a contradiction, thereby showing that no such x exists. So suppose that x ∈ ( A C ) ∩ ( C B ). Then x ∈ A C and x ∈ C B . The first of these statements implies by definition that x / ∈ C , while the second implies that x ∈ C . This is impossible, so our proof by contradiction is complete....
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 Spring '08
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