# For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$

I am unable to prove that For every real $x>0$ and every integer $n>0$, there is one and only one real $y>0$ such that $y^n=x$. Can anyone please help me here?

It is clear that there is at most one such real $y$. But how do I go about the existence of such a real?

• This is false for $n$ even, for instance for $n=2$, $x=9$ you find both $y=-3$ and $y=3$. Before you proving, you should scrutinize the statement to be proved. Dec 10, 2012 at 17:48
• @MarcvanLeeuwen, sorry-I understood the problem.I have edited the question. Dec 10, 2012 at 17:52

What you are asking for is the existence of $$n$$-th root. This is true in $$\mathbb{R}$$ but not in $$\mathbb{Q}$$ (for example $$\not \exists q\in \mathbb{Q}:q^2=2$$). Any proof of this will require the completeness of the real line (Least Upper Bound Property).

LUB property: Every subset of $$\mathbb{R}$$ bounded above must have a supremum.

Here is a typical proof of this theorem only using the LUB property and the Binomial Theorem.

Let $$x>1$$ and $$S=\left\{ a>0: a^n. Obviously $$1\in S$$ and so $$S\neq \emptyset$$.

Since $$x>1\Rightarrow x^n>x$$ we have that $$S$$ is bounded above by $$x$$. Therefore, by the Least Upper Bound Property, $$\exists \sup S=r\in \mathbb{R}$$.

We shall prove that $$r^n=x$$.

Suppose that $$r^n>x$$ and let $$$$\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{r^n-x}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}$$$$ Then $$0<\epsilon <1$$ and so $$\epsilon ^n<1$$.

Therefore, by the Binomial Theorem, $$\begin{gather}(r-\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}(-\epsilon )^k}=r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}(-1)^{k-1}\ge \\ r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}> r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r-\epsilon)^n> r^n-\dfrac{r^n-x}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}=r^n-r^n+x=x\Rightarrow \left( r-\epsilon \right)^n>x\end{gather}$$ which is a contradiction since $$r=\sup S$$ and $$r-\epsilon

Suppose that $$r^n and let $$$$\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{x-r^n}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}$$$$ Then, $$0<\epsilon <1$$ and so $$\epsilon ^n<1$$. Therefore, by the Binomial Theorem, $$\begin{gather}(r+\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}\epsilon^k}=r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}<\\ r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r+\epsilon)^n< r^n+\dfrac{x-r^n}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}= r^n+x-r^n=x\Rightarrow (r+\epsilon)^n and so $$\sup S=r which is a contradiction.

Therefore, $$r^n=x$$ if $$x>1$$

Let $$0. Then $$\frac{1}{x}>1\Rightarrow \exists r>0:r^n=\frac{1}{x}\Rightarrow \exists r'=\frac{1}{r}>0:r'^n=\frac{1}{r}^n=\frac{1}{r^n}=x$$.

If $$x=1$$ then $$r=1$$.

EDIT: Motivation as per request: As I said this is statement is not true if we replace $$\mathbb{R}$$ with $$\mathbb{Q}$$. What sets $$\mathbb{R}$$ and $$\mathbb{Q}$$ apart is the completeness of $$\mathbb{R}$$. The LUB property therefore must in some way be used. This is why we define $$S$$. Because the LUB is an existensial theorem, it shows the existence of a supremum but not its value, it is often used in proofs by contradiction.

So assuming $$r^n>x$$ we need to arrive to a contradiction. Remember $$r$$ is a very special number, the supremum of $$S$$. If we could show that $$\exists m\in \mathbb{R}$$ so that $$m and is an upper bound of $$S$$ or $$m>r$$ and $$m\in S$$ then we are done. This is what we do with $$m=r-\epsilon$$ in the first case and with $$m=r+\epsilon$$ in the second.

It all boils down to finding an $$\epsilon>0$$ so that $$r-\epsilon$$ is an upper bound of $$S$$, that is $$(r-\epsilon)^n>x$$. We can make this $$\epsilon$$ as small as we want. I shall choose an $$\epsilon$$ for $$n=2$$. $$$$(r-\epsilon )^2=r^2-2r\epsilon+\epsilon^2>r^2-2r\epsilon-\epsilon$$$$ Remember we want $$(r-\epsilon )^2>x$$ and so it suffices $$$$r^2-2r\epsilon-\epsilon>x\Leftrightarrow \epsilon<\frac{r^2-x}{1+2r}$$$$

• That was monstrously quick-3 minutes! :O Either way,I need to read it thoroughly. Dec 10, 2012 at 17:47
• I have a written a file in my computer of all the proofs I like and this was one of them... I think you can understand why I was "monstrously quick". Dec 10, 2012 at 17:52
• Nice one.But it still seemed not so obvious.What is the motivation behind it? Dec 10, 2012 at 18:54
• @RichardNash Let me edit that in the answer. Dec 10, 2012 at 18:56
• @RichardNash I am afraid I can't add any more details now. I hope my edit has filled the void my $\epsilon$ s created. Dec 10, 2012 at 19:26

Assume that $y^n=z^n$ with $y,z>0$, $n\in \mathbb N$, $n>1$. Then $$0=y^n-z^n=(y-z)(y^{n-1}+y^{n-2}z+y^{n-3}z^2+\ldots + yz^{n-2}+z^{n-1}).$$ The secod parentheses is a sum of $n-1$ positive summands, hence nonzero. Therefore the other factor $y-z$ must be zero, i.e. $y=z$ and thus there can be at most one $n$th root of $x>0$.

For existence, $f\colon[0,x+1]\to\mathbb R$, $y\mapsto y^n$ is continuous and we have $f(0)=0<x$ and $f(x+1)=(x+1)^n=x^n+nx^{n-1}+\cdots +n x + 1>nx\ge x$, hence by IMV, there exists an $y$ with $f(y)=x$. You might in fact even try a very direct approach and show that $\{q\in\mathbb \mid q>0, q^n>x\}$ defines a Dedekind cut ...

• I am not comfortable with cuts.Can you please include an answer that uses them? Dec 10, 2012 at 18:07

Hint: Intermediate Value Theorem.

Remark: For $n$ even, there will be two real $y$. So we need to say that for every positive $x$, there is a unique positive $y$ such that $y^n=x$.