/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Let the universe for the variabl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 26

Let the universe for the variables in the following statements consist of all real numbers. In each case negate and simplify the given statement. a) \(\forall x \forall y[(x>y) \rightarrow(x-y>0)]\) b) \(\forall x \forall y\left[\left[(x>0) \wedge\left(y=\log _{10} x\right)\right] \rightarrow\left(x=10^{y}\right)\right]\) c) \(\forall x \forall y[(x0) \wedge(y>0))] \rightarrow[\exists z(x z>y)]\)

Short Answer

Expert verified
Negation results: a) \(\exists x \forall y[(x > y) \land (x - y \leq 0)]\), b) \(\exists x \exists y\left[(x > 0) \wedge (y = \log_{10} x) \land (x \neq 10^y)\right]\), c) \(\exists x \exists y \forall z [(x < y) \land (x \geq z \lor z \geq y)]\), d) \(\exists x \exists y[(|x|=|y|) \land (y \neq x \land y \neq -x)]\), e) \(\forall x \forall y \forall z [(x > 0 \land y > 0) \land xz \leq y]\)

Step by step solution

01

Negate and Simplify a)

To negate \(\forall x \forall y[(x> y) \rightarrow(x-y>0)]\), use the rule that \(\neg (A \rightarrow B)\) is equivalent to \(A \land \neg B\). Now this becomes \(\neg \forall x \exists y[(x > y) \land \neg (x - y > 0)]\), which further simplifies to \(\exists x \forall y[(x > y) \land (x - y \leq 0)]\).
02

Negate and Simplify b)

\(\forall x \forall y\left[\left[(x>0) \wedge\left(y=\log _{10} x\right)\right] \rightarrow\left(x=10^{y}\right)\right]\) becomes \(\exists x \exists y\left[\left[(x>0) \wedge\left(y=\log _{10} x\right)\right] \land \neg (x=10^{y})\right]\). Taking \(\neg (x=10^{y})\) into account, the statement further simplifies to \(\exists x \exists y [(x > 0) \land (y = \log_{10} x) \land (x \neq 10^y)]\).
03

Negate and Simplify c)

\(\forall x \forall y[(x<y) \rightarrow \exists z(x<z<y)]\) when negated becomes \(\exists x \exists y [(x < y) \land \neg \exists z(x < z < y)]\). This translates to \(\exists x \exists y [(x < y) \land \forall z \neg (x < z < y)]\), finally simplifying to \(\exists x \exists y \forall z [(x < y) \land (x \geq z \lor z \geq y)]\).
04

Negate and Simplify d)

The statement \(\forall x \forall y[(|x|=|y|) \rightarrow(y=\pm x)]\) negates to \(\exists x \exists y[(|x|=|y|) \land \neg (y = x \lor y = -x)]\). This simplifies to \(\exists x \exists y[(|x|=|y|) \land (y \neq x \land y \neq -x)]\). The result can be interpreted as there existing some \(x\) and \(y\) such that their magnitudes are the same, but \(y\) is neither \(x\) nor \(-x\).
05

Negate and Simplify e)

Negating \([\forall x \forall y((x>0) \wedge(y>0))] \rightarrow[\exists z(x z>y)]\) we get \(\forall x \forall y [(x > 0 \land y > 0) \land \neg \exists z (xz > y)]\). This further translates into \(\forall x \forall y \forall z [(x > 0 \land y > 0) \land xz \leq y]\). The result indicates that for all real numbers \(x\), \(y\) and \(z\), if both \(x\) and \(y\) are positive then \(xz\) is less than or equal to \(y\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each of the following, fill in the blank with the word converse, inverse, or contrapositive so that the result is a true statement. a) The converse of the inverse of \(p \rightarrow q\) is the of \(p \rightarrow q\). b) The converse of the inverse of \(p \rightarrow q\) is the of \(q \rightarrow p\). c) The inverse of the converse of \(p \rightarrow q\) is the of \(p \rightarrow q\). d) The inverse of the converse of \(p \rightarrow q\) is the of \(q \rightarrow p\). e) The converse of the contrapositive of \(p \rightarrow q\) is the of \(p \rightarrow q\). f) The converse of the contrapositive of \(p \rightarrow q\) is the of \(q \rightarrow p\). g) The inverse of the contrapositive of \(p \rightarrow q\) is the of \(p \rightarrow q\)

For each of the following statements provide an indirect proof [as in part (2) of Theorem 2.4] by stating and proving the contrapositive of the given statement. a) For all integers \(k\) and \(l\), if \(k l\) is odd, then \(k, l\) are both odd. b) For all integers \(k\) and \(l\), if \(k+l\) is even, then \(k\) and \(l\) are both even or both odd.

Let the universe for the variables in the following statements consist of all real numbers. In each case negate and simplify the given statement. a) \(\forall x \forall y[(x>y) \rightarrow(x-y>0)]\) b) \(\forall x \forall y\left[\left[(x>0) \wedge\left(y=\log _{10} x\right)\right] \rightarrow\left(x=10^{y}\right)\right]\) c) \(\forall x \forall y[(x0) \wedge(y>0))] \rightarrow[\exists z(x z>y)]\)

Let \(n\) be a positive integer greater than 1 . We call \(n\) prime if the only positive integers that (exactly) divide \(n\) are 1 and \(n\) itself. For example the first seven primes are \(2,3,5\), \(7,11,13\), and 17. (We shall learn more about primes in Chapter 4.) Use the method of exhaustion to show that every even integer in the universe \(4,6,8, \ldots, 36,38\) can be written as the sum of two primes.

Rewrite each of the following statements as an implication in the If-Then form. a) Practicing her serve daily is a sufficient condition for Darci to have a good chance of winning the tennis tournament. b) Fix my air conditioner or I won't pay the rent. c) Mary will be allowed on Larry's motorcycle only if she wears her heimet.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.