Show that nth root of a real number exists and it is unique

Question

I am studying real analysis with Real Mathematical Analysis (2nd. ed.) by Pugh and I am stuck in an exercise as follows:

Given $x>0$ and $n\in\mathbb{N}$, prove that there is a unique $y>0$ such that $y^n=x$. That is the $n^{\text{th}}$ root of $x$ exists and is unique. $[$Hint: Consider $$ y=\text{ l.u.b. } \{s\in\mathbb{R}:s^n\leq x\}$$ Then use Exercise 15 (shown below) to show that $y^n$ can be neither $<x$ nor $>x$.$]$

15. Given $y\in\mathbb{R}, n\in\mathbb{N}$ and $\epsilon>0$, show that for some $\delta>0$, if $u\in\mathbb{R}$ and $|u-y|<\delta$ then $|u^n-y^n|<\epsilon$. $[$Hint: Prove the inequality when $n=1, n=2$ and then do induction on $n$ using the identity

$$u^n-y^n=(u-y)(u^{n-1}+u^{n-2}y+\dots+y^{n-1})]$$

I cannot come up with an idea about how to start and complete this proof. I appreciate any kind of help.

Answer 1

Let $A=\{s\in\mathbb{R}:s^n\leq x\}$. Since $0\in A$ and all $a\in A\leq x+1$, the least upper bound $y$ for $A$ exists.

One of the following three cases must be true: $y^n<x, y^n=x, y^n>x$.

Case $y^n<x$:

• Let $\epsilon = x-y^n$.
• Then by exercise 15, there is a $\delta>0$ such that for some $u\in\mathbb{R}$, $|u-y|<\delta\implies|u^n-y^n|<\epsilon$.
• In particular, there is a $u'$ such that $y<u'<y+\delta$.
• This means that $u'^n-y^n<\epsilon=x-y^n \implies y^n<u'^n < x$, contradicting that $y$ is the least upper bound for $A$.

A similar argument can be made for $y^n > x$ so I'll omit it here.

Now that we showed $y^n=x$, we have to show that this $y$ is $unique$. This is trivial with the fact that least upper bounds are unique.

Note: I've seen a solution like this somewhere but I can't remember where.