# Is the set $A = \left\lbrace \frac{2^k}{3^l}: k,l\in\mathbb{Z}\right\rbrace$ dense?

Define $$A = \left\lbrace\frac{2^k}{3^l}: k,l\in\mathbb{Z}\right\rbrace.$$

Is the set $A$ dense in $\mathbb{R}^+$?

A canonical example of a dense set is $\left\lbrace\frac{n}{2^k}: n,k\in \mathbb{Z}\right\rbrace$. For the set $A$, I guess it's not dense. To show that, I have to show that for some $x\in\mathbb{R}^+$, no element of $A$ falls into the interval $(x-\varepsilon,x+\varepsilon)$ for some $\varepsilon>0$.

• – Watson Aug 19 '16 at 21:45

This is the image under $\exp$ of the group $\ln (2)\Bbb Z+\ln(3)\Bbb Z$, which is a dense subgroup of $\Bbb R$, since $\frac{\ln(3)}{\ln(2)}=\ln_2(3)$ is irrational by a standard argument. Thus it is dense in $\Bbb R_+$.