Do Inequalities Change with Different Powers of X?

[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.