Homework 6 Solution - MATH 74 HOMEWORK 6 (DUE WEDNESDAY...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 74 HOMEWORK 6 (DUE WEDNESDAY OCTOBER 17) 1. Let A = { a 2- b 2 : a, b Z } and B = { 2 k- 1 : k Z } . 1(a). No, A is not a subset of B . For example, 0 A because 0 = 0 2- 2 and Z . But 0 is not in B : if there were an integer k with 2 k- 1 = 0, then k would have to satisfy 2 k = 1; but the only real number solution this equation has is k = 1 2 which is not an integer. This contradiction shows that 0 6 B , and thus A 6 B . 1(b). Yes, B A . To see this, let x B be arbitrary. By definition of B there is k Z with x = 2 k- 1 . A little algebra shows us that then x = k 2- ( k- 1) 2 . Since k and k- 1 are both integers, this shows that x is in A (by the definition of A ). Since x was arbitrary this shows that B A . 1(c). A 6 = B because A is not a subset of B . 2(a). Yes, it is true that A C B C . To see this, let x A B be given. From the definition of A B we conclude that x A and also x C . From the fact that x A and the given information that...
View Full Document

This note was uploaded on 02/10/2010 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at University of California, Berkeley.

Ask a homework question - tutors are online