Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 3 - The Logic of Quantified Statements - Exercise Set 3.3 - Page 129: 12


a. Negation: ∃x in D such that ∀y in E, x + y = 1. The negation is true. When x = −2, the only number y with the property that x + y = 1 is y = 3, and 3 is not in E. b. Negation: ∀x in D, ∃y in E such that x + y = −y. The negation is true and the original statement is false. To see that the original statement is false, take any x in D and choose y to be any number in E with y $\neq$ − x/2 . Then 2y = −x, and adding x and subtracting y from both sides gives x + y = −y. c.Negation: $\exists x $ in D such that $\forall y$ in E, xy < y. The original statement is true (which also means the negation is false). To see that the original is true, examine each of the possible 5 values of x in the domain and in each case select a y that makes xy $\geq$ y true. See details in the step by step description below. d. Negation: $\forall x $ in D, $\exists y$ in E such that x>y. The original statement is true. Let x=-2. Then for all the elements y in E, -2 $\leq$ y. This also means the negation is false.

Work Step by Step

c. To see that $\forall x$ in D, $\exists y$ in E such that xy $\geq$ y, examine all possible values of x and find a y in each case that makes the statement true. x = -2, choose y=-1, then 2 $\geq$ -2 x=-1, choose y=-1, then 1 $\geq$ -1 x=0, choose y=0, then 0 $\geq$ 0 x=1, choose y=2, then 2 $\geq$ 1 x=2, choose y=1, then 2 $\geq$ 2
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.