/*! 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 21 (a) Prove there is no \(n \in \m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 21

(a) Prove there is no \(n \in \mathbb{N}\) such that \(0

Short Answer

Expert verified
There is no natural number between 0 and 1. Moreover, no natural number can be both even and odd.

Step by step solution

01

Proof 1 - The Well-Ordering Property

The well-ordering property of natural numbers states that any non-empty subset of natural numbers has a least element. Now suppose, for a contradiction, that there exists a natural number \(n\) between 0 and 1. This would mean that there is a subset of natural numbers between 0 and 1. This subset would include \(n\) as its least element. But no such \(n\) exists. We have no natural number that is greater than 0 and less than 1, contradicting the assumption.
02

Proof 2 - The Definition of Even and Odd Numbers

An even number is defined as any natural number that can be written in the form \(2k\), where \(k\) is an integer. An odd number is defined as any natural number that can be written in the form \(2k+1\), for an integer \(k\). Now suppose that we have a natural number \(n\) that is both even and odd. This would mean that \(n\) can be written in both forms simultaneously. This leads to a contradiction because \(2k\) and \(2k + 1\) are never equal for any integer \(k\). The assumption that \(n\) is both even and odd is false.

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

If \(I:=[a, b]\) and \(I^{\prime}:=\left\\{a^{\prime}, b^{\prime}\right]\) are closed intervals in \(\mathbb{R}\), show that \(I \subseteq I^{\prime}\) if and only if \(a^{\prime} \leq a\) and \(b \leq b^{\prime}\).

Let \(S_{3}=\\{1 / n: n \in \mathbb{N}\\} .\) Show that sup \(S_{3}=1\) and inf \(S_{3} \geq 0 .\) (It will follow from the Archimedean Property in Section \(2.4\) that inf \(\left.S_{3}=0 .\right)\)

Let \(S \subseteq \mathbb{R}\) be nonempty. Show that \(u \in \mathbb{R}\) is an upper bound of \(S\) if and only if the conditions \(t \in \mathbb{R}\) and \(t>u\) imply that \(t \notin S\).

If \(u>0\) is any real number and \(x

Use Mathematical Induction to show that if \(a \in \mathbb{R}\) and \(m, n, \in \mathbb{N}\), then \(a^{m+n}=a^{m} a^{n}\) and \(\left(a^{m}\right)^{n}=a^{m n} .\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

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.