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This isn't for a class, I was just wondering if I would be able to work out a proof for something like this myself for fun, and wanted to verify that my methods are correct. Basically, what I'm trying to prove, in terms of music theory is: Prove that it is impossible to stack a number of pure 5ths in just intonation, and to end up with a perfectly tuned octave, or multiple of octaves. Or more formally...

Let $ (\frac{3}{2})^m = (\frac{1}{2})^n $, where n and m are positive integers. Show that there exists no positive integers m and n such that the equation is true.

Assume the equation is true:

$ (\frac{3}{2})^m = (\frac{1}{2})^n $

$ (\frac{3}{2})^m = \frac{1}{2^n} $

$ (\frac{3}{2})^m = 2^{-n} $

$ \log{_2}[(\frac{3}{2})^m] = -n $

$ n = -\log{_2}[(\frac{3}{2})^m] $

$ n = -m\log{_2}[\frac{3}{2}] $

$ n \not\in Z^{+} $

$ \therefore $ by contradiction, there exists no n, m $\in Z^{+}$ such that $ (\frac{3}{2})^m = (\frac{1}{2})^n $.

Just for reference, showing a more elegant proof of the same thing would also be appreciated (maybe from using other proven theorems in mathematics).

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Looks fine to me. More simply, we instantly see that no matter what $n, m$ we choose, the two fractions will be reduced and thus they won't be equal. – Amateur Apr 15 '14 at 22:32
If you want both $m$ and $n$ to be positive integers, you should probably be looking at the equation $(3/2)^m=2^n$ rather than $(3/2)^m=(1/2)^n$... – Micah Apr 15 '14 at 22:52
up vote 4 down vote accepted

First of all, as Micah wrote in a comment, you should be looking at $(\frac32)^m=2^n$. Your version, with $(\frac12)^n$ on the right side, would amount to trying to go down a whole number ($n$) of octaves by going up a whole number of major fifths. Of course, that's impossible, just because of the discrepancy between down and up. That discrepancy shows up as the sign difference in your final equation where $m$ and $n$ can't both be positive because $-\log_2(\frac32)$ is negative.

So, if the contradiction at the end of your proof is just because of the sign issue, then it stems from the down-vs.-up issue, and all you've really proved is that going down by octaves will never match going up by major fifths.

On the other hand, if the contradiction you had in mind is based on $\log_2(\frac32)$ being irrational, then I have to agree with Micah that this irrationality, though true, needs a proof, which would be essentially the proof given by kike0001.

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Ahh, right. I think the issue is that informally in the tuning community I'm used to notating octaves as $\frac{1}{2}$, but I suppose mathematically it does make sense that it really should be $\frac{2}{1}$. – Sintrastes Apr 15 '14 at 23:12
If you use $\frac12$ for octaves, then to be consistent you should use $\frac23$ for major fifths. That way, in both cases, you'd have the frequency of the lower note over the frequency of the higher note. – Andreas Blass Apr 15 '14 at 23:17

Continuing with the idea, of @sanath:

The second step asserts that $3^m=2^{m-n}$. Here, $2$ and $3$ are prime numbers, so this equation then implies that there is a natural number with two different prime factorizations. However, this contradicts of the Fundamental theorem of Arithmetic. Therefore, $3^m = 2^{m-n}$ has no solutions for $m,n\in\mathbb{N}$.

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That would have been my proof, but you got there first! (I would have pointed out that @SanathDevalpurkar’s proof was perfectly fine, but that this argument is much more basic.) – Lubin Apr 15 '14 at 22:53
@WChargin, thank for the new edition, in the proof. – AsdrubalBeltran Apr 16 '14 at 1:33

Your proof is correct, assuming that you already know that $\log_2(3/2)$ is an irrational number. However, I suspect that in practice it ends up being circular. The easiest way to show that logarithms of rational numbers are irrational is to show that equations like your original one don't have solutions!

Because of this issue, I think number-theoretic proofs like this one are superior.

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Yes, your proof is correct.

Here's another proof: $$\left(\dfrac{3}{2}\right)^m=\left(\dfrac{1}{2}\right)^n\\ \implies 3^m=2^{m-n}\\ \implies m-n=m\log_23\not\in\mathbb{Z}^+\text{ as $m$ is an integer.}\\ \implies n=m(1-\log_23)\not\in\mathbb{Z}^+\text{ as $m$ is an integer.}\\ \implies \{n,m\}\not\subset\mathbb{Z}^+$$ Contradiction!

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