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 $f\colon \left[{0 \,.\,.\, \to}\right) \to \left[{0 \,.\,.\, \to}\right)$ be defined by
 * $f\left({y}\right) = y^n$.

First consider the case of $n > 0$.

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

Note that $f\left({y}\right) = \left({\operatorname{id} \left({y}\right) }\right) ^ n$.

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

The result follows from Strictly Monotone Mapping with Totally Ordered Domain is Injective.

Now consider the case of $n < 0$.

Now let $m = -n$.

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

It follows from the definition of power that $g \left({y}\right) = \dfrac 1 {f \left({y}\right)}$ and hence $g \left({y}\right)$ is strictly decreasing.

Again, the result follows from Strictly Monotone Mapping with Totally Ordered Domain is Injective.