Disjoint sets of rationals both dense in $\mathbb{R}$ There exists a pair of disjoint subsets of $\mathbb{Q}$ such that both are dense in $\mathbb{R}$.
True or false?
The statement is true but how can I find an example of such disjoint subsets of $\mathbb{Q}$?
 A: Let $x \in \mathbb{R}$. Now, 
$$
0 < 2^nx - [2^nx] < 1
$$
and thus 
$$
0 < x - \frac{[2^nx]}{2^n} <  \frac{1}{2^n}
$$
This says that $D = \{\frac{m}{2^n}\}_{m \in \mathbb{Z}, n \in \mathbb{N}}$ is dense in $\mathbb{R}$. Now, let's consider a subset of this set, 
$$
A = \{\frac{m}{2^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 2 \not |n \}
$$
If $A$ were dense in $D$, it would be dense in $\mathbb{R}$. For this, we only have to show that $D \setminus A$ can be approximated by elements of $A$. Let $\frac{m}{2^n} \in D \setminus A$ and write $m = 2^ks$ with $2 \not | s$. It can't be that $k \leq n$, because that would mean $\frac{m}{2^n} = \frac{s}{2^{n-k}} \in A$. Therefore, $k \geq n$ and thus $\frac{m}{2^n} = 2^{k-n}s \in \mathbb{Z}$. This means that it is sufficient to show that $A$ can approximate integers: if $a \in \mathbb{Z}$ and $\varepsilon > 0$, then taking $\frac{1}{2^n} < \varepsilon$ we have that 
$$
\left|a - \frac{2^na + 1}{2^n}\right| = \frac{1}{2^n} < \varepsilon
$$
and $\frac{2^na + 1}{2^n} \in A$ because $2 \not | \  2^na+1$. In conclusion, we have shown that $A$ is dense in $D$ and since $D$ is dense in the reals, so is $A$. With an identical construction (which I encourage you to write out) one can show that the same holds for
$$
B = \{\frac{m}{3^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 3 \not |n \}
$$
To conclude, then, we just have to observe that $A \cap B = \emptyset$. In effect, if 
$$
\frac{m}{2^n} = \frac{j}{3^k}
$$
for some $n,m,j,k$, then $3^km = 2^nj$ and thus $2 | 2^nj = 3^km$. But since $2$ and $3$ are coprime, this would imply $2 | m$ which is absurd.
A: HINT.-It is enough to show an example in $I=[0,1]$. All rational is periodic after a finite number of digits
$$r=0.a_1a_2....a_m[b_1b_2....b_n]\in I$$ It is not hard to prove that $A$ and $B$ below are disjoint and dense in $I$
$$A=\{r\text{ such that }  b_n=1\}\\B=\{r\text{ such that }  b_n\ne1\}$$
