Problem suggested by the Pythagorean musica scale Is the set of fractions $3^m/2^n$ dense in the interval [1,2]?
If this can help: for each m there is exactly one n in the interval ($n$=integer part of $m\gamma$ , where $\gamma=\log 3/\log 2$)  
 A: Yes, the set of such fractions is dense in $[1, 2]$. The proof below is an elaboration of Henning Makholm's answer.


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*The fractions $\dfrac{3^m}{2^n}$ being dense in $[1, 2]$ is equivalent to the logarithms of those fractions being dense in $[\log 1, \log 2]$.
Proof: This is because the inverse of the logarithm function is continuous: a neighbourhood of $x \in [1, 2]$ contains the image of a neighbourhood of $\log x \in [\log 1, \log 2]$.
In concrete terms, if you want to show that for a particular $x$, for every $\epsilon$ there exists $(m, n)$ such that $|x - 3^m/2^n| < \epsilon$, it's enough to show that for every $\delta$ there exists $(m, n)$ such that $|\log x - \log(3^m/2^n)| < \delta$, as for sufficiently small $\delta$ (possibly depending on $x$ and $\epsilon$), we have $|\log x - \log(3^m/2^n)| < \delta \implies |x - 3^m/2^n| < \epsilon$.

*The logarithm of $\dfrac{3^m}{2^n}$ is $m\log 3 - n\log 2$. We want to prove that the set of such numbers is dense $[0, \log 2]$, or, equivalently, dividing throughout by $\log 2$ (which again we can do because this operation has a continuous inverse, say), we want to prove that the set of numbers of the form $m\frac{\log 3}{\log 2} - n$ is dense in $[0, 1]$.
As you observed, for any integer $m$, there is a unique integer $n = \left\lfloor \frac{m\log 3}{\log 2} \right\rfloor$ such that $m \log 3 - n\log 2$ lies in $[0, \log 2]$, or equivalently, such that $m \frac{\log 3}{\log 2} - n$ lies in $[0, 1]$). So $m \frac{\log 3}{\log 2} - n$ is precisely the fractional part of $m\frac{\log 3}{\log 2}$. The question now is to prove that the fractional parts of $m\frac{\log 3}{\log 2}$ are dense in $[0, 1]$.

*There is a well-known result that if $\alpha$ is irrational, then the set of fractional parts of $m\alpha$ as $m$ ranges over the natural numbers, is dense in $[0, 1]$. This you can see proved on this site, here, here, here and even on Wikipedia. Basically, you can use the pigeonhole principle: if you want to show that the fractional part of some $m\frac{\log 3}{\log 2}$ is within some distance $\frac{1}{q}$ of some number, consider more than $q$ such numbers; two of the fractional parts lie in the same $\frac{1}{q}$ "bucket"; by considering their difference and a suitable multiple, you can put the fractional part into the bucket you want.

*The number $\dfrac{\log 3}{\log 2}$ is irrational, because of $\dfrac{\log 3}{\log 2} = \dfrac{p}{q}$, then it means that $3^q = 2^p$, which is impossible for nonzero $q$. This concludes the proof.
A: Yes.
The set of multiples of $\log 3$ is dense in $\mathbb R$ modulo $\log 2$, because $\frac{\log 3}{\log 2}$ is irrational. If it was rational there'd be a nontrivial power of 2 that was also a power of 3, which contradicts the fundamental theorem of arithmetic.
