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

Proof
Let $f$ be the real function defined on the unbounded closed interval $\left[{0 \,.\,.\, \to}\right)$ defined by $f \left({y}\right) = y^n$.

Consider first the case of $n > 0$.

Since $x > 0$, $x + 1 > 1$.

By Strictly Positive Power of Element Greater than One Succeeds Element‎,

$f \left({ x + 1 }\right) \ge x + 1$.

Since $x + 1 > x$, it follows from transitivity that
 * $f \left({ x + 1 }\right) > x$.

Since $x \ge 0$,
 * $f \left({0}\right) \le x$.

By the Intermediate Value Theorem, there is a number $y$ such that $0 \le y \le x+1$ and $f\left({y}\right) = x$

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

Now consider the case of $n < 0$.

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 the case for $n > 0$ above, there is a $y > 0$ such that $g \left({y}\right) = \dfrac 1 x$.

It follows from the definition of power that

Hence the result has been shown to hold also for the case of $n < 0$.