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
First consider the case of $n > 0$.

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

The result follows from Strictly Monotone Mapping is Injective.

Now consider the case of $n < 0$.

Now let $m = -n$.

Let $g$ be the real function defined on $\left[{0 \,.\,.\, +\infty}\right)$ 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 is Injective.