existence of $n$th root

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. ∎

Title existence of $n$th root ExistenceOfNthRoot 2013-03-22 15:52:15 2013-03-22 15:52:15 Wkbj79 (1863) Wkbj79 (1863) 21 Wkbj79 (1863) Theorem msc 26C10 msc 26A06 msc 12D99 ExistenceOfNthRoot