existence of $n$th rootExistenceOfRoot2006-04-25 08:26:112007-04-21 02:24:39Theoremnth root\usepackage{amssymb}
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\newtheorem*{thm*}{Theorem}\begin{thm*}
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 $u^n=a$.
\end{thm*}
\begin{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.
\end{proof}