$x^q < y^q \implies x < y$ proof $x^q &lt; y^q \implies x &lt; y$ , $q $ is a rational number. For a natural number it is quite easy. We can show it by induction. As $x&lt;y \leftrightarrow x*x &lt; y*x &lt; y*y$ and so on..But how could it be done for any rational number ? 
I guess it is only possible for $x,y &gt; 0$
Let $x&gt;0$, $y&gt;0$, $q&gt;0$. Suppose that $q=m/n$ with $m,n\in\mathbb N$. From the fact that $x^q&lt;y^q$ we get

\begin{align*}
0&lt;y^q-x^q&amp;=y^{m/n}-x^{m/n}=\left(y^{1/n}\right)^m-\left(x^{1/n}\right)^m\\
&amp;=\left(y^{1/n}-x^{1/n}\right)\cdot\left(y^{(m-1)/n}+y^{(m-2)/n}\cdot x^{1/n}+\cdots+y^{1/n}\cdot x^{(m-2)/n}+x^{(m-1)/n}\right)
\end{align*}

The expression in the last brackets is positive therefore 

$$
y^{1/n}-x^{1/n}&gt;0\quad\Rightarrow\quad y^{1/n}&gt;x^{1/n}\quad\Rightarrow\quad \underbrace{y^{1/n}\cdot y^{1/n}\cdots y^{1/n}}_{\text{$n$-times}}&gt;\underbrace{x^{1/n}\cdot x^{1/n}\cdots x^{1/n}}_{\text{$n$-times}}\quad\Rightarrow\quad y&gt;x.
$$
$q &gt;0, z&gt;0$.

$f(z):=z^{1/q}$ is strictly increasing, since

$f'(z)=(1/q)z^{(1/q-1)} &gt;0 $.

For $x^q&lt;y^q$ :

$f(x^q)=x&lt;f(y^q)=y$.

What happens if $q &lt;0$ ?