How to show density of 2^a 3^b Sounds like a nice homework problem, but this actually came up in preparing a lecture for a Music class; I want to show that if you try to build a set of notes where you can go up and down octaves and perfect fifths from any note (that is, the set of notes is closed over multiplication and division by 2 & 3, essentially), you wind up with an infinite number of keys on your piano.
Thinking about this, it seems "obvious" that this set--also describable as the rationals of the form $2^a 3^b$ where $a$ & $b$ can be positive or negative or zero--is dense in the reals. Proving it, though, is giving me fits. I feel like this should be easy.
The best I've got is to think about multiplication & division by 3 as addition & subtraction in the log realm, and then to argue that if there was a neighborhood of a point that contained no other point in the set, then there would have to be a nonzero greatest common divisor of $\log2$ and $\log3$, but that no such thing can exist, because then it would be the case that $\exists$ (nonzero) $y,z . 2^y = 3^z$, which is impossible.
Is there a simpler way to show this?

As Andre pointed out, $\alpha = \frac{\log 3}{ \log 2}$ is irrational. It is a theorem, looking for reference, that $m + n \alpha$ is dense in the reals. Multiply by $\log 2,$ we find $m \log 2 + n \log 3$ is dense in the reals, where $m,n$ are integers. so $e^{m \log 2 + n \log 3} = 2^m 3^n$ is dense in the positive reals. 
Alright, Chapter 3 in Diophantine Approximations by Ivan Niven.
On this site, For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$ among many others
