# Is $\{\frac{m}{10^n}\mid m,n\in\mathbb Z,\ n\geq 0\}$ dense in $\mathbb R$?

The set $S$ of real numbers of the form $m/10^n$, $m,n$ integers and $n$ greater than or equal to $0$, is a dense subset of $\mathbb R$ or not??

I know dense means closure of $S$ in $\mathbb R$ is $\mathbb R$ then it will be dense. How to prove or disprove I have no idea.

For any $x\in\Bbb R$ and any $\epsilon>0$ you can pick an element $a$ out of this set such that $$|x-a|<\epsilon.$$ Why? If $\epsilon>0$, find $n$ large enough so that $1/10^n<\epsilon$. But $x$ must live in some interval of the form $[m/10^n,(m+1)/10^n]$, so..

Hint: Consider the decimal representation of real numbers.

Hint For any $x \in \mathbb R$ we have

$$x- \frac{1}{10^n}\leq \frac{\lfloor 10^n x \rfloor}{10^n} \leq x$$

That the closure of $S$ is $\mathbb{R}$ means that every ball in $\mathbb{R}$ contains at least one point of $S$ (infinitely many, in fact).

First notice that $S$ is closed under summation.

Then we can express any $x\in\mathbb{R}$ as $x=n.a_0a_1a_2...=n+\frac{a_0}{10}+\frac{a_0}{10²}+\dots$. Any partial sum of this expression will be in $S$, so we can approximate $x$ as much as we want with elements in $S$, which is equivalent to $S$ being dense.