Simple questions about some inequalitites

[AxiomOfChoice]

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


(0 < x < 1) \Rightarrow \frac{1}{x^b} > \frac{1}{x^a}


and


(1 < x < \infty) \Rightarrow \frac{1}{x^a} > \frac{1}{x^b}


?????

[Landau]

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

Fact: Let 0<x<1. Then for all \alpha>0 we have 0<x^\alpha<1 and (hence) \frac{1}{x^\alpha}>1.

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

Together: 0<x^b<x^a<1. Conclusion:
0<\frac{1}{x^a}<\frac{1}{x^b}<1.

[AxiomOfChoice]

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

Fact: Let 0<x<1. Then for all \alpha>0 we have 0<x^\alpha<1 and (hence) \frac{1}{x^\alpha}>1.

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

Together: 0<x^b<x^a<1. Conclusion:
0<\frac{1}{x^a}<\frac{1}{x^b}<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.