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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.

user821980
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  • You (1) have a target of size $1/n$ and (2) as you increment $k$, you take steps of size less than $1/n$, so you can't step over the target -- at least one step (and possibly more than one) lands in the target. ((2) asserts that that the distance between the $\ell^\text{th}$ and $\ell+1^{\text{st}}$ step is less than $1/n$. So the procedure is start with $k = 1$, if you're not already done, keep incrementing $k$ until your small steps land you in the large target. – Eric Towers Oct 29 '21 at 21:25

1 Answers1

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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.

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