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I have no idea of how to solve it. I would appreciate if someone gives me a hint, please.

Definitions Let $\,x^{1/n}:= sup\{\, y \in \mathbb{R}: y\ge0 \text{ and } y^n\le x\, \}$

Lemma: Let $x,y>0$ be positive reals, and let $n\ge 0$ be a positive integer.

(a) If $y = x^{1/n}$ then $y^n = x$

(b) Conversely, if $y^n = x$, then $y = x^{1/n}$

The big problems is I cannot use the binomial formula to prove it. I tried to used an argument by contradiction assume $y^n < x$ and so $y^n > x$ to get contradiction. But without the binomial formula I'm not sure of what use as estimator. The hint in the book is use denseness of $\mathbb{Q}$ in $\mathbb{R}$ and the basic properties of order in $\mathbb{R}$.

I thought to use something like this: Suppose that $y^n < x$ then there is a rational number such that $y^n < q < x$ and after this create a set $E_q$ which is bounded by $q$ find its supremum and try to get a contradiction and a similar argument to $x<y^n$. But that's not work very well. Could someone give me a hint, please?

I think I have the exercise (b)

Proof of (b): Suppose $y^n = x$ and $y > 0$. We set $E:=\{\, z \in \mathbb{R}: z\ge0 \text{ and } z^n\le x\, \}$. It follows that $E \not= \emptyset$ since $y\in E$. Now we need to show that is bounded above but that follows because $E$ is bounded by $\text{max} \{1,x\}$.

To prove the claim is sufficient to show that $y$ is the least upper bound of $E$.

First we have to show that $y$ is an upper bound for $E$. We may argue by contradiction, suppose that there is a $z\in E$ such that $z>y$ so $z^n>y^n=x$, i.e., $z\notin E$ a contradiction. Then for all $z\in E$ we must have $z\le y$.

Now to conclude the proof we need to show that $y$ is the least upper bound of $E$. Let $s$ be an upper bound for $E$ and suppose $s<y$. Then by the denseness of the rational numbers we have $s<q<y$.Then $q^n<y^n=x$ and $q>0$ so $\,q\in E$ contradicting that $s$ is an upper bound. Thus, the upper bound $s$ is greater than equal to $y$, thus $y$ is the least upper bound of $E$ and by definition $y= x^{1/n}$ as desired.

With the part (a) I'm not sure yet.

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Don't you mean $y\in \color{red}{\Bbb Q}$? – Git Gud Sep 24 '13 at 21:26
No, $y\in \mathbb{R}$. – Jose Antonio Sep 24 '13 at 21:28
So $1^1=\sup\left(\{y\in \Bbb R\colon y\ge 0 \land y^1\leq x\}\right)$? – Git Gud Sep 24 '13 at 21:32
I check the book and says:"Let $x > 0\,$ be a positive real, and let $n\ge 1\,$ be a positive integer. We define $x^{ 1/n}$, also known as the nth root of x, by the formula: $\,x^{1/n}:= sup\{\, y \in \mathbb{R}: y\ge0 \text{ and } y^n\le x\, \}$" – Jose Antonio Sep 24 '13 at 21:48
Ok, ok. I just find it odd. – Git Gud Sep 24 '13 at 21:49
up vote 1 down vote accepted

For problem (a) you can use this inequality, holding for $z>0$, $0<t\le 1$ and $n$ integer:

$$ (z+t)^n \le z^n + t((z+1)^n - z^n) $$

If $z^n<x$, then you can find $t$ such that $(z+t)^n<x$: just take $t$ such that $$ 0<t<\min\left\{\frac{x-z^n}{(z+1)^n-z^n},1\right\} $$ which surely exists. Therefore such a $z$ can't be the supremum of the set $\{y\ge0:y^n\le x\}$ and for the supremum $\xi$ of this set it must be $\xi^n=x$.

How do you find that inequality? It doesn't matter, because you can prove it by induction on $n$. It clearly holds for $n=0$. Suppose it holds for $n$; then \begin{align} (z+t)^{n+1} &=(z+t)^n(z+t)\\ &\le(z^n+t(z+1)^n-tz^n)(z+t)\\ &=z^{n+1}+tz(z+1)^n-tz^{n+1}+tz^n+t^2(z+1)^n-t^2z^n\\ &< z^{n+1}+tz(z+1)^n+t(z+1)^n-tz^{n+1}-t^2z^n\\ &=z^{n+1}+t(z+1)^{n+1}-tz^{n+1}-t^2z^n\\ &\le z^{n+1}+t(z+1)^{n+1}-tz^{n+1} \end{align}

The hypothesis $0<t\le 1$ is used to get the second $\le$, since $t^2\le t$.

Tell your teacher an elf suggested it. ;-) Of course one can also prove it with the binomial theorem: \begin{align} (z+t)^n &=z^n+\binom{n}{1}z^{n-1}t+\binom{n}{2}z^{n-2}t^2+\dots+\binom{n}{n-1}zt^{n-1}+t^n\\ &\le z^n+\binom{n}{1}z^{n-1}t+\binom{n}{2}z^{n-2}t+\dots+\binom{n}{n-1}zt+t\\ &=z^n+t((z+1)^n-z^n) \end{align} (which is how I first got it). The idea came from the same problem but with $n=2$, which is easier.

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Thank you so much the post is very helpful. – Jose Antonio Sep 27 '13 at 14:59

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