Proof of the uniqueness of $n^{th}$ root

Question

I am reading Rudin's Principles of Mathematical Analysis and am trying to work through the proof of Theorem 1.21 (in my edition). I have reproduced this below for completeness.

1.21 Theorem For every real $x>0$ and every integer $n>0$ there is one and only one real positive $y$ such that $y^n = x$.

First he proves that there is only one such $y$. Then he proves that given the set $E$, defined as being the set of all $t$ such that $t^n<x$, is bounded and nonempty and thus has a supremum in $\mathbb{R}$.

He then claims that $y=\sup E$. In order to prove that $y^n = x$ he goes on to prove that both $y^n<x$ and $y^n>x$ both lead to contradictions. I get the reasoning of the argument as above, but when it comes to proving these inequalities lead to contradictions I have difficulties justifying his steps.

Namely, he introduces the identity $$b^n-a^n = (b-a)(b^{n-1} + b^{n-2}a + ... + a^{n-1})$$ which is fine. But he then says that

$$b^n-a^n < (b-a)nb^{n-1}$$ for $0<a<b$. This step I do not quite follow. I can't see why it is necessarily true. That aside, I struggle with the next step even more. He then sets up the first inequality:

Assume $y^n<x$ Choose $h$ so that $0<h<1$ and $h<\frac{x-y^n}{n(y+1)^{n-1}}.$

Then using the identity above and setting $a=y$ and $b=y+h$, we quite clearly$^1$ get that $(y+h)^n < x$ and so by definition $y+h \in E$, but this is a contradiction. The step I am struggling to justify is why we can just introduce this convenient $h$ in order to get what we want. Am I being stupid? Perhaps I am just not used to this type of argument.

1. $(y+h)^n - y^n < hn(y+h)^{n-1} < hn(y+1)^{n-1}<x-y^n$

Answer 1

For the inequality $b^n-a^n<(b-a)nb^{n-1}$ (which is only strict for $n>1$, obviously):

Note that $b>a>0$, $b^{k}\geq a^k$, so $b^{m-k}a^k\leq b^m$ for any $m\geq k\geq 0$ (with strict inequalities unless $b=1$ and $k=0$, or $m=k=0$). Then in particular, $$\underbrace{b^{n-1}+b^{n-2}a+\ldots +a^{n-1}}_{n\text{ terms}}\leq \underbrace{b^{n-1}+b^{n-1}+\ldots+b^{n-1}}_{n\text{ terms}}=nb^{n-1}$$ with strict inequality for $n>1$. In particular, for $n>1$ $$b^n-a^n=(b-a)(b^{n-1}+\ldots+a^{n-1})<(b-a)nb^{n-1}.$$

For the choice of $h$, we have the assumption that $x>y^n$, $y>0$, and $n>0$. Then $x-y^n>0$ and $n(y+1)^n>0$, so $\frac{x-y^n}{n(y+1)^n}>0$. Then by the Archimedian Property of real numbers, there is a smaller positive number $\frac{x-y^n}{n(y+1)^n}>h'>0$. Finally, either $h'<1$ and we set $h=h'$; or $h'>1$, in which case we can pick $h$ to be any number smaller than 1 (say, $h=.1$).

Answer 2

We have $b^{n-2}a<b^{n-2}b=b^{n-1}$, $b^{n-3}a^2<b^{n-3}b^2=b^{n-1}$, etc. Hence $b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\dots)<(b-a)nb^{n-1}$ (because there are $n$ terms in the parentheses).

As to $h$, the fraction $(x-y^n)/(n(y+1)^{n-1})$ is positive, and you can take $$ h=\frac12\min\{1,(x-y^n)/(n(y+1)^{n-1})\}. $$