   Chapter 8.5, Problem 39ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Suppose that A is a totally ordered set. Use mathematical induction to prove that for any integer n ≥ 1 , every subset of A with n elements has both a least element and a greatest element.

To determine

To Prove:

For any integer n1 every subset of A with n elements has both a least element and a greatest element, using mathematical induction.

Explanation

Given information:

A is totally order set.

Concept used:

Mathematical induction

Proof:

Let A be a set that is totally ordered with respect to a relation . And let the property P(n) be the sentence “Every subset of A with n elements has both a least element and a greatest element”

Show that the property is true for n=1.

If A=ϕ then the property is true by default.

So assume that A has at least one element and suppose S={a1} is a subset of A with one element

Because is reflexive a1a1.

By the definition of least element and greatest element a1 ¡s both the least and greatest element of S.

Thus, the property is true for n=1.

Show that for all integers k1. If the property is true for n=k then it is true for n=k+1.

Let k be any integer with k1 and suppose that any subset of A with k elements has both a least element and a greatest element.

We must show that any subset of A with k+1 elements has both a least element and a greatest element. If A has fewer than k+1 elements, then the statement is true by default

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