TITLE: For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$
QUESTION [6 upvotes]: I am unable to prove that For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$. Can anyone please help me here?
It is clear that there is at most one such real $y$. But how do I go about the existence of such a real?

REPLY [3 votes]: Assume that $y^n=z^n$ with $y,z>0$, $n\in \mathbb N$, $n>1$.
Then $$0=y^n-z^n=(y-z)(y^{n-1}+y^{n-2}z+y^{n-3}z^2+\ldots + yz^{n-2}+z^{n-1}).$$
The secod parentheses is a sum of $n-1$ positive summands, hence  nonzero. Therefore the other factor $y-z$ must be zero, i.e. $y=z$ and thus there can be at most one $n$th root of $x>0$.
For existence, $f\colon[0,x+1]\to\mathbb R$, $y\mapsto y^n$ is continuous and we have $f(0)=0<x$ and $f(x+1)=(x+1)^n=x^n+nx^{n-1}+\cdots +n x + 1>nx\ge x$, hence by IMV, there exists an $y$ with $f(y)=x$. 
You might in fact even try a very direct approach and show that $\{q\in\mathbb \mid q>0, q^n>x\}$ defines a Dedekind cut ...