College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.4 - Page 751: 32

Answer

See below.

Work Step by Step

Proofs using mathematical induction consists of two steps: 1) The base case: here we prove that the statement holds for the first natural number. 2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$. Hence here: 1) For $n=1: 0\lt x^1=x\lt1$ , thus this is true by the condition. 2) Assume for $n=k: 0\lt x^k\lt1$. Then for $n=k+1:0\lt x^{k+1}=x\cdot x^k\lt1.$ This statement is true because we multiplied the value in the middle of the inequality of the inductive hypothesis by a number larger than $0$ (thus it stayed positive) but by a number less than $1$, thus it became smaller thus it is still less than $1$. Thus we proved what we wanted to.
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