(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is from Wade's Intro to Analysis book; problem 5, Chapter 1.

Prove that if [tex]0 < a < 1[/tex] and [tex] b = 1 - \sqrt{1 - a} [/tex] then [tex]0 < b < a[/tex]

2. Relevant equations

The book hints to use this result: [tex]0 < a < 1[/tex] implies [tex]0 < a^{2} < a[/tex]

3. The attempt at a solution

First I substitute b into the equation to get:

[tex]0 < 1 - \sqrt{1 - a} < a < 1[/tex]

By the additive property I add [tex]-1[/tex] and get:

[tex]-1 < -\sqrt{1 - a} < a - 1 < 0[/tex]

Using the multiplicative property I multiply by [tex]-1[/tex] and get:

[tex]0 < 1 - a < \sqrt{1 - a} < 1[/tex]

That above bears resemblance to the hint equation but I don't know exactly where to go (and my inkling of an idea doesn't seem to prove the initial statement).

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# Homework Help: Easy Analysis Problem

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