Theorem   If $a \in \mathbb{R}$ with $a>0$ and $n$ is a positive integer, then there exists a unique positive real number $u$ such that $u^n=a$ .

Proof. The statement is clearly true for $n=1$ (let $u=a$ ). Thus, it will be assumed that $n>1$ .

Define $p \colon \mathbb{R} \to \mathbb{R}$ by $p(x)=x^n-a$ . Note that a positive real root of $p(x)$ corresponds to a positive real number $u$ such that $u^n=a$ .

If $a=1$ , then $p(1)=1^n-1=0$ , in which case the existence of $u$ has been established.

Note that $p(x)$ is a polynomial function and thus is continuous. If $a<1$ , then $p(1)=1^n-a>1-1=0$ . If $a>1$ , then $p(a)=a^n-a=a(a^{n-1}-1)>0$ . Note also that $p(0)=0^n-a=-a<0$ . Thus, if $a \neq 1$ , then the intermediate value theorem can be applied to yield the existence of $u$ .

For uniqueness, note that the function $p(x)$ is strictly increasing on the interval $(0, \infty)$ . It follows that $u$ as described in the statement of the theorem exists uniquely.