This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: CS237. Practice Problems Set 1: Sets, Basic Counting. Not Graded. Please complete by Thursday Jan 28, 2011. January 27, 2011 Reading. Schaums Chapter 1, 2.1. L&L Counting I. Exercise 1. Let A,B U . Prove the following. (Hint, use proofs by contradiction.): A B iff A B c = . Solution: In ( ) direction we can use the proof by contradiction. Let us assume that A B and A B c 6 = . From the latter assumption, it follows that there exists an element x A that is not in B . But, if x A then x B because of A B . Contradiction. For the other direction, we again use proof by contradiction. Assume that A B is false. Then we must have some x A , so that x / B . But then, A B c is not empty as assumed. Contradiction. A B iff A c B = U . Solution: The ( ) direction: By contradiction. Let us assume that A B and A c B 6 = U . From the latter assumption, it follows that there exists an element x U that is not in A c and not in B . That means that x A . But, if x A then x B because of A B . Contradiction. For the other direction, by contradiction: Assume that A B is false, so that for some x A , x / B . But then, we must have x U so A c B = U implies that either x A c which is not the case, or x B which is a contradiction. A B iff B c A c ....
View Full Document
This note was uploaded on 04/14/2011 for the course CS 237 taught by Professor Goldberg during the Spring '11 term at BU.
- Spring '11