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:

$\map f x = \sqrt [n] x$

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

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

Aiming for a contradiction, suppose that:

$\sqrt [n] x \ge \sqrt [n] y$

We have:

$\ds \sqrt [n] x$

$\ge$

$\ds \sqrt [n] y$

$\ds \leadsto \ \ $

$\ds \paren {\sqrt [n] x}^n$

$\ge$

$\ds \paren {\sqrt [n] y}^n$

$\ds \leadsto \ \ $

$\ds x$

$\ge$

$\ds y$

Definition of Root of Number

This contradicts the hypothesis that $x < y$.

$0 < x < y \implies \sqrt [n] x \le \sqrt [n] y$

Hence the result by definition of strictly increasing real function.

$\blacksquare$