I'm studying Tao's Analysis I, in which the reader has to prove many of the theorems and lemmas.
$x$ and $y$ are nonnegative reals and $n$ is a positive integer.
Definition of nth root: $x^{1/n}=\sup\{z\in\mathbb{R}:z\geq0\text{ and }z^n\leq x\}$
I want to prove (a) $y=x^{1/n}\implies y^n=x$ and (b) $y^n=x\implies y=x^{1/n}$
My proof for part (b): Let $E=\{y\in\mathbb{R}:y\geq0\text{ and }y^n\leq x\}$. Suppose $y^n=x$. Then $y\in E$, and since $x^{1/n}$ is an upper bound of $E$, we have $x^{1/n}\geq y$. Now if we can show that $y$ is an upper bound of $E$, then $x^{1/n}\leq y$ since $x^{1/n}$ least upper bound, and hence $x^{1/n}=y$. Suppose for contradiction that $y$ is not an upper bound. Then there is $k\in E$ such that $k>y$. Then $k^n>y^n=x$, which implies $k\not\in E$, a contradiction.
Does this look correct?
As for part (a), I tried to derive a contradiction in the case that $y=x^{1/n}$ yet $y^n<x$, but I got stuck. Any ideas?
For (a), when $x=0$ we have $E=\{0\}$ and so clearly $y=0$ which satisfies $y^n=x$. The remainder applies to $x>0$.
Let $\varepsilon>0$. There exists $z\in E$ such that $y-\varepsilon<z$. Hence $(y-\varepsilon)^n<z^n\leq x$. The bound $(y-\varepsilon)^n\leq x$ with $\varepsilon$ being arbitrarily small implies $y^n\leq x$.
Similarly for every $\varepsilon>0$ there exists $z\in E$ such that $x-\varepsilon<z^n$. Thus $x-\varepsilon<z^n\leq y^n$. This shows that $x\leq y^n$.
The above relies on the fact that $x\mapsto x^n$ is increasing over $[0,\infty)$ and is also surjective.