real numbers of the form $\frac{m}{10^n} $ with $m,n \in \mathbb{Z} $ and $n \geq 0$ is dense in $\mathbb{R}$ . [duplicate]

Question

Problem :

Verify if the statement if true of false -

The set $S$ of all real numbers of the form $\frac{m}{10^n} $ with $m,n \in \mathbb{Z} $ and $n \geq 0$ is dense in $\mathbb{R}$ .

I think this true .

Reason : $S$ is actually a subgroup of $\mathbb{R}$ with respect to addition .

Now i know the theorem that any subgroup of $\mathbb{R}$ is either cyclic or dense .

If $S$ were cyclic then there non-zero $m$ and an $n\geq 0$ such that $\frac{m}{10^n} $ generates $S$ . So there is a non-zero $i\in \mathbb{Z} $ such that $0= \frac{m}{10^n}+..\frac{m}{10^n}$ ( $i$ times adiition )$ = \frac {mi}{10^n} $ which is not possible because neither $m$ or $i$ is $0$ .So $S$ can be cyclic . Hence it is dense .

Answer 1

Take $\varepsilon=\frac{1}{10^n}$ and receive $\frac{\lfloor 10^na\rfloor}{10^n}\in (a-\varepsilon,a+\varepsilon)$

because $\lfloor 10^na\rfloor\leq10^na<\lfloor10^na\rfloor+1$

Answer 2

We have to show that $\forall r \in \mathbb{R}$ and $\forall \varepsilon>0, \exists m,n \in \mathbb{Z}: \left|r-\frac{m}{10^n}\right|<\varepsilon$.

From $$m=\left \lfloor r10^n \right \rfloor \leq r10^n < \left \lfloor r10^n \right \rfloor +1=m+1 \Rightarrow \frac{m}{10^n} \leq r < \frac{m}{10^n}+\frac{1}{10^n}$$ we have $$\left|r-\frac{m}{10^n}\right|<\frac{1}{10^n}$$ By "adjusting" $n>0$ we can find one such that $\frac{1}{10^n} < \varepsilon$.