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# HW5_Sol - COT 3100 Applications of Discrete Structures...

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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 left-hand 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 right-hand 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 two-set 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 1-1 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 left-hand 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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HW5_Sol - COT 3100 Applications of Discrete Structures...

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