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This post is related to the question What is an easy way to prove the equality $r > s > 0$ implies $x^r > x^s$? which is essentially the same question reduced to the unit interval. In particular I was wondering if there was a simple proof for the following inequality:

Let $x \in (0,1)$ and $r,s \in \mathbb{R}$

How do you show the inequality $r > s > 0$ implies $x^r < x^s$?

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    Could you edit your title to ask the same thing as the question? [0, 1] isn't the same as (0, 1).2011-10-04

2 Answers 2

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$r-s>0$ and $\log x<0$ since $0< x<1$

Therefore $(r-s)\log x<0\implies \log x^{r-s}<0\implies x^{r-s}<1\implies x^r

For $x=0,1$ it doesn't happens

Alternative Solution:

Let $f(x)=a^x$ where $0 then f'(x)=(\ln a)a^x<0 for all $x$ since $0.

Therefore $f(p)>f(q)$ if $pa^s$ if $p

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You need to take $x\in(0,1)$, not in $[0,1]$: it isn’t true if $x=0$ or $x=1$. Use Gerry Myerson’s argument from the previous question: the only change is that if $0, $\log x$ is negative, so $r>s$ implies that $r\log x < s\log x$.