# Every positive real number has a unique positive n'th root

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1. Sep 12, 2016

### Math_QED

1. The problem statement, all variables and given/known data

Show, using the axiom of completeness of $\mathbb{R}$, that every positive real number has a unique n'th root that is a positive real number.

Or in symbols:

$n \in \mathbb{N_0}, a \in \mathbb{R^{+}} \Rightarrow \exists! x \in \mathbb{R^{+}}: x^n = a$

2. Relevant equations

Axiom of completeness:

Suppose $I_n = [x_n,y_n]$ is a sequence of closed intervals and they are nested such that:
$I_1 \supset I_2 \supset \dots \ I_n \dots$.

Then, there exists at least one number x, for which:

$x \in \bigcap\limits_{n=0}^{\infty} I_n$

3. The attempt at a solution

My textbook says that I can prove it as an application of the axiom of completeness. But it doesn't ask to proof this as an exercise, so I'm not sure whether I have covered enough to proof this problem.

So, I thought of a proof by contradiction. So, suppose that there are $2$ different numbers $y,z \in \mathbb{R}$, for which: $y^n = a$ and $z^n = a$. Then I wanted to take them as bounds in $[x_n, y_n]$, and somehow show that this interval only contains 1 element, but it seems that that's not the correct approach. Any hints in the right direction will be appreciated.

Last edited: Sep 12, 2016
2. Sep 12, 2016

### PeroK

The axiom of completeness is required for existence of an nth root. I would use the existence of $sup$ and $inf$ for any set.

For uniqueness, you could show that $x < y \ \Rightarrow x^2 < y^2$ and hence $x^n < y^n$.

3. Sep 12, 2016

### Math_QED

Sup and inf are defined later in the book so I was right that I did not cover enough yet. I will come back later to this thread. Thanks!

4. Sep 12, 2016

### Ray Vickson

You don't need to know about sup and inf to do it; using your completeness property is enough.

You are trying to show the existence of a root of the equation $y^n = x$, so you can just apply the so-called bisecting search method to do it. See, eg., https://en.wikipedia.org/wiki/Bisection_method
or http://www.sosmath.com/calculus/limcon/limcon07/limcon07.html . All you need to get started is the existence of a number $b > 0$ such that $b^n > x$.

5. Sep 12, 2016

### Math_QED

I'm sorry but the proof for that method seems to require some analysis (continuous function etc) and at this point I'm not allowed to use that.

6. Sep 12, 2016

### PeroK

The idea is that there are, for example, rationals to any number of decimal places, $k$, such that:

$r_1^n \le x \le r_2^n$

And $r_2 - r_1 = 10^{-k}$

7. Sep 12, 2016

### micromass

The absolute easiest proof is by using the intermediate value theorem (which is equivalent to completeness) on the function $f(x) = x^n - a$. But ok, you want to prove it from scratch without using any further analysis. Perhaps you could tell us what textbook you are using, because I don't know what properties you get to accept as true and which not.

8. Sep 12, 2016

### Ray Vickson

Are you worried that you cannot use the so-called "intermediate-value theorem"? Well, you do not need it.

You have $a_1 \leq a_2 \leq a_3 \leq \cdots \leq b_3 \leq b_2 \leq b_1$, with $b_k - a_k = b_0 2^{-k}$. Completeness gives
$$\bigcap_{j=1}^{\infty} [a_k, b_k] = y \in \mathbb{R}.$$
What can you say about the value of $y^n$, using only properties like $y_1 < y_2 \Rightarrow y_1^n < y_2^n$, etc.?

9. Sep 12, 2016

### Math_QED

I'm not really using a textbook but the syllabus from my brother. It's called Analysis 1 and it's in Dutch. I'll send you a link in PM.