Math proof sought

I originally posted this message in tuning, but Paul Erlich suggested
I should repost here. All comments welcome.

From: "paulerlich" <paul@s...>
Date: Sat Jan 5, 2002 8:23 pm
Subject: Re: Mathematical proof sought

--- In tuning@y..., "tunerguy2002" <tunerguy2002@y...> wrote:
> Hi.
> I get a great deal of insight about tuning systems from the
> observation that:
>
> (3/2)^12 != (2/1)^7
>
> that is, twelve fifths does not exactly equal seven octaves. In
> fact, it seems that there are no M and N such that
>
> (3/2)^M != (2/1)^N
>
> but I can't prove it. Other "near misses" occur at, for example,
> M=53, N=31.
>
> Is anyone aware of a proof of the general case? All leads
> appreciated.

This is a very simple consequence of the Fundamental Theorem of
Arithmetic. I'm sure Gene can give you the most concise proof of
this. You should ask the question at

tuning-math@yahoogroups.com

--- In tuning-math@y..., "tunerguy2002" <tunerguy2002@y...> wrote:

> > Is anyone aware of a proof of the general case? All leads
> > appreciated.
>
> This is a very simple consequence of the Fundamental Theorem of
> Arithmetic. I'm sure Gene can give you the most concise proof of
> this. You should ask the question at

It seems to me Paul has basically given the proof, which is to cite the FTA. If you want the details, the FTA says that any positive rational number has a *unique* representation as 2^e1 * 3^e2 * ...,
where the exponents ep are integers, all but a finite number being zero. If you have positive rational numbers a and b, such that for some prime p the exponent of p in the product representation of a, which is called vp(a), the "valuation" at p of a, is not zero whereas vp(b)=0, then vp(a^n) = n vp(a) > 0, but vp(b^m) = m*0 = 0; since they are not equal in terms of the exponent of p, they cannot be equal by the FTA. I think this covers the situation you had in mind; and in the form I give here, can be generalized to situations where the Fundamental Theorem itself does not apply.