Clarifying a specific step in proof of density of multiples of irrationals This question is about a specific step in the proof accepted as an answer here. I will use the same notation.
Half way through the proof, it is claimed that for all $m\in[0,n-1]$ there exists $k\in[1,M]$ such that
$$k(\{i\alpha\}-\{j\alpha\})\in\left[\frac{m}{n},\frac{m+1}{n}\right].$$
The reason given being that: (1) the length of $\left[\frac{m}{n},\frac{m+1}{n}\right]$ is $1/n$ and (2) for all $\ell\in\mathbb{N}$
$$|(\ell+1)(\{i\alpha\}-\{j\alpha\})-\ell(\{i\alpha\}-\{j\alpha\})|<1/n.$$
Can someone please clarify this step? For example, let $0\leq m\leq n-1$. How can I pick $k\in[1,M]$ such that
$$\frac{m}{n}\leq k(\{i\alpha\}-\{j\alpha\})\leq\frac{m+1}{n}?$$
I would like to know how (1) and (2) together prove the claim.
 A: I visualize the situation as follows. (You should sketch as you read.)
Look at $\quad0 < \{ i \alpha \} - \{ j \alpha \} < \dfrac{1}{n} $. We have a strictly positive real number $1/n.$ I think of this as my meter stick, or yard stick if you prefer. I see $n$ meter sticks lined up to produce $[0,1]$.
Then I take a new meter stick and split it in two, and keep one piece. This new
less-than-a-meter stick is my representation of $\{ i \alpha \} - \{ j \alpha \}$.
Call it a short stick.
We have $ M \ge 1 $ and such that
$\quad M (\{ i \alpha \} - \{ j \alpha \}) < 1. \quad \quad
(\spadesuit) $
In words this says $M$ short sticks lined up brings us less than $n$ meters. But if we lay $M+1$ short sticks, it will bring us past the $n$ meters (since $M$ was chosen maximal satisfying $M (\{ i \alpha \} - \{ j \alpha \}) \le 1$).
So we see $M+1$ short sticks stretching from zero, and $n$ meter sticks stretching not as far. We have $M$ positions at which two short sticks meet. The natural way to label them is $$1 (\{ i \alpha \} - \{ j \alpha \}),\; 2 (\{ i \alpha \} - \{ j \alpha \}), \;\dots,\; M (\{ i \alpha \} - \{ j \alpha \}).$$
Now, assume we are not able to find

integer $ k \in \{ 1,\ldots,M \} $ such that
$$
k (\{ i \alpha \} - \{ j \alpha \}) \in \! \left[ \frac{m}{n},\frac{m + 1}{n} \right].
$$

In our picture, this would mean that some meter stick has no meeting of two short sticks within it.
But since the $M+1$ short sticks cover the $n$ meter sticks, this can only mean that one of the short sticks is able to cover (by itself) one of the meter sticks, which contradicts the very first strict inequality.
