Show that $\mathbb{Q} \cap B(x,r) \neq \varnothing$ Let $x$ be an irrational number and let $r>0$. Consider the open ball $B(x,r) \subset \mathbb{R}$. Show that $\mathbb{Q} \cap B(x,r) \neq \varnothing$.

My thoughts: I was thinking that maybe you could engineer a sequence that consists of rational numbers which converges to $x$ - perhaps something along the lines of n times the integer part of x divided by n:
$$\bigg\{ \frac{\lfloor nx \rfloor}{n}\bigg\}_{n=1}^{\infty}$$
But how do I show formally that this in fact converges to x - I'm not that familiar with working with integer parts and limits. If I can show that I can just pick a number in the sequence so I have rational number closer to x (say $\epsilon$ close) than $r$ and I'd be done.
Or maybe I am way off or there is an easier way of showing it, any feedback is appreciated!
 A: We have
$$\tag 1 \text{For any } s \in \Bbb R, \; \lfloor s \rfloor \le s$$
$$\tag 2 \text{For any } s \in \Bbb R, \; s \lt \lfloor s \rfloor + 1$$
Let the irrational (or not) number $x$ and integer integer $n \gt 0$ be given.
Using $\text{(1)}$, we can write
$\tag 3  \frac{\lfloor nx \rfloor}{n} \le x$
Using $\text{(2)}$ and employing algebra, we can write
$\tag 4  \frac{\lfloor nx \rfloor}{n} \gt x -\frac{1}{n}$
Combining $\text{(3)}$ and $\text{(4)}$,
$$\tag 5  x -\frac{1}{n} \lt \frac{\lfloor nx \rfloor}{n} \le x $$
Using the squeeze theorem, the sequence $\bigg\{ \frac{\lfloor nx \rfloor}{n}\bigg\}_{n=1}^{\infty}$ converges to $x$.
A: Within the reals, B(x,r) = (x-r, x+r).
So the problem becomes finding a rational in (a,b) for all a < b.
That can be done by use of the Archimedean property of numbers. 
A: In the real set, $B(x,r) = (x-r, x+r)$. So the problem becomes: prove that in $(a,b)$ for all $a < b$ there is a ractional numbers.
Let's suppose $0\leq a$.
To do that we will use that for all $\epsilon>0$ there is an $n\in \mathbb{N}$, this could be simply prove thinking at the $\lim_{n\to +\infty} \epsilon\cdot n=+\infty$ for a fixed $\epsilon$.
So now we know that exist $n$ such that $n\cdot (b-a)>1$ so $\frac{1}{n}<b-a$.
Now let's consider the set $$A=({m\in\mathbb{N}:\frac{m}{n}\leq a})$$ it is 
limited above, infact $n\cdot$ a is bigger then any element of A, so admit sup. $L=supA$ and by the definiction of sup we have that for any $1>\epsilon>0$ exist $u\in A$ such that $L-\epsilon \leq u \leq L$ so $u\leq L \leq u+\epsilon \leq u+1$, so we have that $u$ is a element of $A$ and $u+1$ not, in fact is bigger that L. $$\frac{u}{n}\leq a \leq \frac{u+1}{n}=\frac{u}{n}+\frac{1}{n}<a+(b-a)=b$$
So the rational number $y=\frac{u+1}{n}$ satisfies the thesis. 
If $a<0<B$ the number $0$ satisfies the thesis.
If $b\leq0$ we have already test the existence of $y$ such that $-b<y<-a$ so $-y$ satisfies the thesis.
