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 Morgans 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 wont 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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This note was uploaded on 01/22/2012 for the course COT 3100 taught by Professor Staff during the Spring '08 term at University of Florida.

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

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