Existence and 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 always exists a unique $$y \in \R: y \ge 0$$ such that $$y^n = x$$.

Hence the existence of the $n$th root of $$x = y^{1/n}$$.

Proof
Let $$n \in \Z: n > 0$$.

Consider the real function $$f$$ defined on the half open interval $$\left[{0 \,. \, . \, \infty}\right)$$ defined by $$f \left({y}\right) = y^n$$.

Since $$f \left({0}\right) = 0$$ and $$f \left({y}\right) \to + \infty$$ as $$y \to + \infty$$, it follows from the Continuity Property that the range of $$f$$ is also $$\left[{0 \,. \, . \, \infty}\right)$$.

Note that $$\forall y > 0: f^{\prime} \left({y}\right) = n y^{n-1} > y$$ from the power rule for derivatives: natural number index.

Thus from Derivative of Monotone Function‎, it follows that $$f$$ is strictly increasing on $$\left[{0 \,. \, . \, \infty}\right)$$.

The result follows from Strictly Monotone Mapping is Injective.

Hence the result has been shown to hold for $$n > 0$$.

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 $$g \left({y}\right) = \frac 1 {f \left({y}\right)}$$ and hence $$g \left({y}\right)$$ is strictly decreasing.

Again, the result follows from Strictly Monotone Mapping is Injective.

Hence the result has been shown to hold for $$m \in \Z: m < 0$$.