/*! 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 17 17\. Write the negation of each ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 17

17\. Write the negation of each of the following true statements. For parts (a) and (b) the universe consists of all integers; for parts (c) and (d) the universe comprises all real numbers. a) For all integers \(n\), if \(n\) is not (exactly) divisible by 2 then \(n\) is odd. b) If \(k, m, n\) are any integers where \(k-m\) and \(m-n\) are odd, then \(k-n\) is even. c) If \(x\) is a real number where \(x^{2}>16\), then \(x<-4\) or \(x>4\) d) For all real numbers \(x\), if \(|x-3|<7\), then \(-4

Short Answer

Expert verified
Negation (a): There exists an integer \(n\) such that \(n\) is not divisible by 2 and \(n\) is not odd. Negation (b): There exist integers \(k, m, n\), such that \(k-m\) and \(m-n\) are odd, but \(k-n\) is not even. Negation (c): There exists a real number \(x\), such that \(x^{2}>16\) but \(x\geq-4\) and \(x\leq4\). Negation (d): There exists a real number \(x\), such that \(|x-3|<7\) but \(x\leq-4\) or \(x\geq10\).

Step by step solution

01

Negation of statement (a)

Statement (a): For all integers \(n\), if \(n\) is not divisible by 2 then \(n\) is odd. To form the negation, disprove the statement. Negation (a): There exists an integer \(n\) such that \(n\) is not divisible by 2 and \(n\) is not odd.
02

Negation of statement (b)

Statement (b): If \(k, m, n\) are any integers where \(k-m\) and \(m-n\) are odd, then \(k-n\) is even. To form the negation, disprove the statement. Negation (b): There exist integers \(k, m, n\), such that \(k-m\) and \(m-n\) are odd, but \(k-n\) is not even.
03

Negation of statement (c)

Statement (c): If \(x\) is a real number where \(x^{2}>16\), then \(x<-4\) or \(x>4\). To form the negation, disprove the statement. Negation (c): There exists a real number \(x\), such that \(x^{2}>16\) but \(x\geq-4\) and \(x\leq4\).
04

Negation of statement (d)

Statement (d): For all real numbers \(x\), if \(|x-3|<7\), then \(-4<x<10\). To form the negation, disprove the statement. Negation (d): There exists a real number \(x\), such that \(|x-3|<7\) but \(x\leq-4\) or \(x\geq10\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Negation of Statements
In mathematics, negating a statement involves flipping its truth value. This plays a crucial role in logical reasoning and proofs.
  • To negate the universal quantifier "for all," we replace it with "there exists."
  • The negation of a conditional statement, typically introduced by "if," involves changing the consequent 鈥� the part after "then".
For instance, given a statement like "For all integers \(n\), if \(P(n)\), then \(Q(n)\)," its negation would suggest that "there exists some integer \(n\) such that \(P(n)\) is true, but \(Q(n)\) is false." This method of negating statements helps to critically analyze statements by focusing on exceptions, or situations where things don't follow the expected rules.
Integer Properties
Integers, the set of whole numbers and their negatives (like -3, 0, 2), have distinct properties that make solving mathematical exercises interesting.
  • An integer's divisibility by 2 determines if it is "even" or "odd." An integer \(n\) is even if \(n \mod 2 = 0\), and odd if \(n \mod 2 = 1\).
  • When working with operations like addition, subtraction, and multiplication, the results often adhere to predictable patterns鈥攆or example, the sum of two odd integers always results in an even integer.
These properties are heavily used in creating logical conditions or statements that can be either true or false. Understanding these properties is fundamental in comprehending and negating logical statements that hold true across all integers.
Real Numbers
Real numbers include not just integers, but also fractions and irrational numbers (like \(\sqrt{2}\)). Understanding what falls into this category is key since they fill the number line completely.
  • Unlike integers, real numbers include values such as \(\frac{1}{3}\), \(\pi\), and roots, providing unparalleled flexibility in math.
  • They satisfy numerous orders and inequalities, such as \(x > y\) and \(x \leq z\), with clear graphical representations on the number line.
When considering logical statements involving real numbers, it鈥檚 important to visualize these inequalities and consider boundary cases. For instance, given a condition on \(x\) like \(x^2 > 16\), we need to determine ranges \(x\) could inhabit, significantly affecting the truth of associated conditional statements.
Conditional Statements
A conditional statement introduces a logical dependency between two statements, typically in the form "if \(P\), then \(Q\)."
  • The part \(P\) is known as the antecedent, while \(Q\) is the consequent.
  • For the statement to be true, whenever \(P\) is true, \(Q\) must be true as well.
To form the negation of such statements, we express that there is some instance where \(P\) is true, but \(Q\) is false. For instance, if our statement is "if \(x^2 > 16\), then \(x < -4\) or \(x > 4\)," its negation asserts that there is a real number \(x\) such that \(x^2 > 16\) and \(x\) does not satisfy \(x < -4\) or \(x > 4\). This approach demands understanding both the conditions and the ways they can break, which is a central exercise in logical evaluations.

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

15\. Define the connective "Nand" or "Not ... and ..." by \((p \uparrow q) \Leftrightarrow \neg(p \wedge q)\), for any statements \(p, q .\) Represent the following using only this connective. a) \(\neg p\) b) \(p \vee q\) c) \(p \wedge q\) d) \(p \rightarrow q\) e) \(p \leftrightarrow q\)

10\. Determine whether each of the following is true or false. Here \(p, q\) are arbitrary statements. a) An equivalent way to express the converse of " \(p\) is sufficient for \(q\) " is " \(p\) is necessary for \(q\)." b) An equivalent way to express the inverse of " \(p\) is necessary for \(q\) " is " \(\neg q\) is sufficient for \(\neg p\)." c) An equivalent way to express the contrapositive of \(" p\) is necessary for \(q\) " is " \(\neg q\) is necessary for \(\neg p\)."

1\. Determine whether each of the following sentences is a statement. a) In 2003 George W. Bush was the president of the United States, b) \(x+3\) is a positive integer. c) Fifteen is an even number. d) If Jennifer is late for the party, then her cousin Zachary will be quite angry. e) What time is it? f) As of June 30,2003 , Christine Marie Evert had won the French Open a record seven times.

26\. In calculus the definition of the limit \(L\) of a sequence of real numbers \(r_{1}, r_{2}, r_{3}, \ldots\) can be given as $$ \lim _{n \rightarrow \infty} r_{n}=L $$ if (and only if) for every \(\epsilon>0\) there exists a positive integer \(k\) so that for all integers \(n\), if \(n>k\) then \(\left|r_{n}-L\right|<\epsilon\). In symbolic form this can be expressed as $$ \lim _{n \rightarrow \infty} r_{n}=L \Leftrightarrow \forall \epsilon>0 \exists k>0 \forall n\left[(n>k) \rightarrow\left|r_{n}-L\right|<\epsilon\right] $$ Express \(\lim _{n \rightarrow \infty} r_{n} \neq L\) in symbolic form.

3\. Verify that each of the following is a logical implication by showing that it is impossible for the conclusion to have the truth yalue 0 while the hypothesis has the truth value 1 . a) \((p \wedge q) \rightarrow p\) b) \(p \rightarrow(p \vee q)\) c) \([(p \vee q) \wedge \neg p] \rightarrow q\) d) \([(p \rightarrow q) \wedge(r \rightarrow s) \wedge(p \vee r)] \rightarrow(q \vee s)\) e) \([(p \rightarrow q) \wedge(r \rightarrow s) \wedge(\neg q \vee \neg s)] \rightarrow(\neg p \vee \neg r)\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

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.