Do Inequalities Change with Different Powers of X?

[AxiomOfChoice]

If a and b are positive and a < b, do we have

<br /> (0 &lt; x &lt; 1) \Rightarrow \frac{1}{x^b} &gt; \frac{1}{x^a}<br />

and

<br /> (1 &lt; x &lt; \infty) \Rightarrow \frac{1}{x^a} &gt; \frac{1}{x^b}<br />

?

 

[Landau]

I think you should be able to prove these yourself. Here's the first one:

Fact: Let 0&lt;x&lt;1. Then for all \alpha&gt;0 we have 0&lt;x^\alpha&lt;1 and (hence) \frac{1}{x^\alpha}&gt;1.

In particular:
* 0&lt;x^a&lt;1 and 0&lt;x^b&lt;1;
* \frac{x^a}{x^b}=\frac{1}{x^{b-a}}&gt;1, hence x^a&gt;x^b.

Together: 0&lt;x^b&lt;x^a&lt;1. Conclusion:
0&lt;\frac{1}{x^a}&lt;\frac{1}{x^b}&lt;1.

 

[AxiomOfChoice]

I think you should be able to prove these yourself. Here's the first one:

Fact: Let 0&lt;x&lt;1. Then for all \alpha&gt;0 we have 0&lt;x^\alpha&lt;1 and (hence) \frac{1}{x^\alpha}&gt;1.

In particular:
* 0&lt;x^a&lt;1 and 0&lt;x^b&lt;1;
* \frac{x^a}{x^b}=\frac{1}{x^{b-a}}&gt;1, hence x^a&gt;x^b.

Together: 0&lt;x^b&lt;x^a&lt;1. Conclusion:
0&lt;\frac{1}{x^a}&lt;\frac{1}{x^b}&lt;1.

Thanks for your help, Landau. This is one of those questions that I answered for myself when I was typing it up, but I thought I'd go ahead and post it anyway to make sure I wasn't crazy.