Existence of Root

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

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

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

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 12.11$