Existence of Positive Root of Positive Real Number/Negative Exponent

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

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

Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.

Proof
Let $m = -n$.

Then $m > 0$.

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

Since $x \ge 0$:
 * $\dfrac 1 x \ge 0$

By Existence of Positive Root: Positive Exponent there is a $y > 0$ such that:
 * $g \left({y}\right) = \dfrac 1 x$

It follows from the definition of power that: