For every irrational $r$ does there exist a sequence in $\mathbb{Q}$ that converges to $r$? Let $r\in\mathbb{R}\setminus\mathbb{Q}$, how can I construct a sequence $(x_n)\subseteq\mathbb{Q}$ such that $\lim_{n\rightarrow\infty}x_n=r$?
I don't want a specific example of an irrational number such as $e$ where $e=\lim_{n\rightarrow\infty}x_n$ with $x_n=(1+1/n)^n$, but rather I want a generalization (if there is any).
 A: From lulu's comment, consider the sequence
$$x_n=\frac{\lfloor r\cdot 10^n\rfloor}{10^n}\quad \mbox{for $n\geq 0$}$$
A: $\mathbb Q$ is dense everywhere. It implies that for every $\varepsilon >0$ and $x\in\mathbb R$, there are infinitely many rationals in the interval $(x-\varepsilon, x+\varepsilon)$.  
So we obtain a sequence by fixing for every $n\in\mathbb N$ an element $x_n\in (x-\frac{1}{n},x+\frac{1}{n})\cap\mathbb Q$.
Put simply, for every $n\in\mathbb N$ we "pick" an element $x_n\in\mathbb Q$ whose distance from $x$ is less than $\frac{1}{n}$.
A: We have 
$$\lfloor 2^nr\rfloor \le 2^nr <  \lfloor 2^nr\rfloor +1\Longleftrightarrow  \frac{\lfloor 2^nr\rfloor}{2^n}\rfloor \le r \le  \frac{\lfloor 2^nr\rfloor}{2^n}+\frac{1}{2^n}\Longleftrightarrow  0\le r-\frac{\lfloor 2^nr\rfloor}{2^n}\le  \frac{1}{2^n}\to 0$$
That is $$\frac{\lfloor 2^nr\rfloor}{2^n}\to r,~~~\forall r\in\mathbb R$$
 but $$\lfloor 2^nr\rfloor \in\mathbb Z,~~~\forall r\in\mathbb R,\forall n\in\mathbb N $$
ie 
$$\frac{\lfloor 2^nr\rfloor}{2^n}\in \mathbb Q~~~\forall r\in\mathbb R,\forall n\in\mathbb N $$
