Analysis - Powers of real numbers

[gazzo]

Hey everyone!

1. Let $x \in (0, \infty)$

a) Prove that $(x^{\frac{1}{n}})^m = (x^{\frac{1}{q}})^p$ for all $m,p \in \mathbb{Z}$ such that $mq = np$

Because of the previous we can define $x^r \in \mathbb{R}$ for all $r \in \mathbb{Q}$ by $x^r = (x^{\frac{1}{n}})^m$ if $r = \frac{m}{n}$ with $m \in \mathbb{Z}$ and [iitex]n \in \mathbb{N}[/itex]

b) Prove that $x^{r+s} = {x^r}{x^s}$ and $(x^r)^s = x^{rs}$ for all $r,s \in \mathbb{Q}$

This question came up in my analysis assignment, we're studying continuity and differentiability at the moment so I'm unsure of my answer! It seems too short :(

For a), it seems to follow immediately,

If $mq = np$ then $m = \frac{np}{q}$ and so we have
$(x^{\frac{1}{n}})^m = (x^{\frac{1}{n}})^{\frac{np}{q}} = ((x^{\frac{1}{n}})^n)^{\frac{p}{q}} = (x^{\frac{p}{q}}) = x^{\frac{1}{n}} = (x^{\frac{1}{q}})^p$ as required.

And b) is similar (there are too many exponents to latexify before my stuff gets stolen upstairs in the study floors).
I went about it proving if $a \in \mathbb{R}$ and $m,n \in \mathbb{N}$ then $a^{m+n} = {a^m}{a^n}$ and $(a^m)^n = a^{mn}$ by induction. (Can you use induction over m and n by doing it twice fixing m the first time and inducting over n and the second time fixing n?).

Or it's probably better to first prove, if $m\in \mathbb{Z}, n \in \mathbb{N}, x \in (0,\infty)$ then $x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}$.
If $x \in (0,\infty)$ and $m,n \in \mathbb{Z}$ then $(x^m)^n = x^{mn} = (x^n)^m$. Now let $y:=x^{\frac{m}{n}} = (x^{\frac{1}{n}})^m > 0$. So $y^n = ((x^{\frac{1}{n}})^m)^n = ((x^{\frac{1}{n}})^n)^m = x^m$. Therefore it follows that $y = (x^m)^{\frac{1}{n}}$.

If we proved it for the naturals, how can we prove it for the rationals? It seems to follow naturally.

These analysis proofs seem so empty, although subtle. I have huge trouble sponging it all together into a valid proof.

Thanks a lot!
-Gareth

 

[lippyka]

For part b) you can use the fact that for any real number x and any rational numbers r and s, we have x^{r+s} = {x^r}{x^s}. This is because we can rewrite r+s as a fraction m/n, with m and n integers, and then use the result you proved in part a).