Root is Strictly Increasing

Theorem
Let $x \in \R_{> 0}$.

Let $n \in \N$.

Let $f : \R_{> 0} \to \R$ be the real function defined as:
 * $f \left({x}\right) = \sqrt [n] x$

where $\sqrt [n] x$ denotes the $n$th root of $x$.

Then $f$ is strictly increasing.

Proof
Let $x, y \in \R$ such that $0 < x < y$.

that:
 * $\sqrt [n] x \ge \sqrt [n] y$

Now:

This contradicts the hypothesis that $x < y$.

Therefore, by Proof by Contradiction:
 * $0 < x < y \implies \sqrt [n] x \le \sqrt [n] y$

Hence the result by definition of strictly increasing real function.