How do we prove $x < y$ implies $x^{\alpha} < y^{\alpha}$ for $\alpha > 0$, $\alpha \in \mathbb{Q}$?
Particularly, the case when $x$ and $y$ are between $0$ and $1$.
I wonder if there is a first principles explanation that does not rely on defining $e^x$, $\log(x)$.
EDIT: Given that natural number exponent is monotone, show that natural root is monotone.
Starting with: $x < y$
$x^{\frac{1}{n}} \stackrel{?}{=} y^{\frac{1}{n}}$
Using the fact that natural exponent is monotone, $$(x^{\frac{1}{n}})^n \stackrel{?}{=} (y^{\frac{1}{n}})^n$$
Since $x < y$, the above relationship propagates back to the root case, and we get: $x^{\frac{1}{n}} < y^{\frac{1}{n}}$
The fractional exponent is a compsition of an integer power and an integer root, so you just need to show both of those are monotone. And the integer root is the inverse of the integer power, so this reduces to showing that integer powers are monotone.