Density of the set $S=\{m/2^n| n\in\mathbb{N}, m\in\mathbb{Z}\}$ on $\mathbb{R}$? Let $S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}$, is  $S$ a dense set on $\mathbb{R}$?

Yes, is it, given open interval $(a,b)$ (suppose $a$ and $b$ positives) you can find $n\in\mathbb{N}$ such that $1/2^n<|b-a|$. Then consider the set:
$$X=\{k\in \mathbb{N}; k/2^n > b\}$$
This is a subset of $\mathbb{N}$, for well ordering principe $X$ has a least element $k_0$ then is enought taking $(k_0-1)/2^n\in(a,b)$.
The same is if $a$, $b$ or both are negatives (because $(a,b)$ is bounded).
