Prove that any positive number has an $n$th root

Solution 1:

So, given $a>0$, $M>0$, $M^n\ne a$, you want to construct $x$ such that either $M^n<x^n\le a$ or $M^n>x^n\ge a$.

Try $x=My$. Then you want to find $y$ such that $1<y^n\le N$ or $1>y^n\ge N$ where $N=a/M^n\ne1$.

Let's do the second case, where $N<1$ first. Let $N=1-t$, and take $y=1-t/n$. Bernoulli's inequality implies $$1>y^n=\left(1-\frac tn\right)^n\ge 1-t=N.$$

The first case, where $N>1$ reduces to the second. There is $y'$ with $1>y'^n\ge 1/N$ so take $y=1/y'$.

Solution 2:

One useful tool when you do not already posses all the machinery from calculus is usually Bernoulli inequality i.e. (under suitable assumptions on the domain of $x$, see the Wikipedia page):

$$(1+x)^n\ge 1+nx\\ (1-x)^n\ge 1-nx$$

In our case, the contradiction can be obtained quite easily: if $M^n<a$ then we could find a number $\delta$ such that $\frac{1}{M}>\left(\frac1{nM}\left(1-\frac{M^n}{a}\right)\right)>\delta>0$.

Considering

$$\left(\frac{1}{M}-\delta\right)^n=\frac{1}{M^n}\left(1-\delta M\right)^n>\frac{1}{M^n}\left(1-n\delta M\right)>\frac1a$$.

Thus $M^n<\left(\frac{1}{M}-\delta\right)^{-n}=\frac{M^n}{(1-\delta M)^n}<a$, and this contradicts the assumption that $M$ is the least upper bound.

Similarly, a contradiction is obtained considering $M^n>a$ by setting $0<\delta<\frac{M}{n}\left(1-\frac{a}{M^n}\right)$

Solution 3:

Rudin does this well. The identity you gave implies $$b^n -a^n <(b-a)n b^{n-1}$$ whenever $0<a<b$. Then assume $$M^n <x.$$ Let $h$ be such that $0<h<1$ and $$h< \frac{x-M^n}{n(M+1)^{n-1}}.$$ Ok, seems random, but then $$(M+h)^n -M^n < hn(M+h)^{n-1} <hn(M+1)^{n-1} <x-M^n.$$Therefore, $$M+h \in S.$$ But since $M+h>M$, we are contradicting the fact that $S$ is bounded above by $M$. So $$x\leq M^n.$$ Now assume $M^n >x$. Let $$k = \frac{M^n -x}{nM^{n-1}}.$$ Clearly $k<M$. If $r\geq M-k$ then $$M^n -r^n \leq M^n -(M-k)^n <knM^{n-1} =M^n-x.$$ Therefore, $r^n >x$ and $r\notin S$. Thus, $M-k$ is an upper bound of $S$. But wait a second, we claimed $M$ to be the least upper bound of $S$, not $M-k$. Hence, $$M^n=x.$$