Raising inequality to a power proof

Question

I have the to prove the following statements:

$$\forall a,b >0 \space\forall r > 0,r\in\mathbb{Q}:a<b \iff a^r<b^r $$ $$\forall a,0<a<1,\forall r,s\in\mathbb{Q},r<s: a^r>a^s$$

I think I managed to prove the first statement from left to right, by first reducing the problem to Natural powers first and then using induction. When I tried the proof from right to left, I started by assuming $a^r<b^r$. Then, by definition $\sqrt[q]{a^p}-\sqrt[q]{b^p}>0$ and I can multiply that with $\sqrt[q]{a^p}+\sqrt[q]{b^p}$ to get $a^p<b^p$. But here is where I seem to get stuck. I reduced the problem but I can't finish the proof. Any hints for that and the next inequality?

Answer 1

You can use calculus and show that for any positive $r$ we have that $f(x)=x^r$ is strictly increasing (for $x>0$). To that end, simply note that $f'(x)=rx^{r-1}>0$ for all $x>0$ (when $r>0$).