Uniqueness of Positive Root of Positive Real Number

Theorem
Let $x \in \R$ be a real number such that $x \ge 0$.

Let $n \in \Z$ be an integer such that $n \ne 0$.

Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.

Proof
Let the real function $f : \left[{0 \,.\,.\, \to}\right) \to \left[{0 \,.\,.\, \to}\right)$ be defined as:
 * $f \left({y}\right) = y^n$

First let $n > 0$.

By Identity Mapping is Order Isomorphism, the identity function $I_\R$ on $\left[{0 \,.\,.\, \to}\right)$ is strictly increasing.

We have that:
 * $f \left({y}\right) = \left({I_\R \left({y}\right) }\right)^n$

By Product of Positive Strictly Increasing Mappings is Strictly Increasing, $f$ is strictly increasing on $\left[{0 \,.\,.\, \to}\right)$.

From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
 * there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.

Now let $n < 0$.

Let $m = -n$.

Let $g$ be the real function defined on $\left[{0 \,.\,.\, \to}\right)$ defined by:
 * $g \left({y}\right) = y^m$

From the definition of power:
 * $g \left({y}\right) = \dfrac 1 {f \left({y}\right)}$

Hence $g \left({y}\right)$ is strictly decreasing.

From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
 * there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.