# Help understanding steps to prove Dirichlet's approximation theorem

In my elementary number theory textbook, there is a problem that is meant to help understand the proof of Dirichlet's rational approximation theorem. There are two parts to the problem: the first part asks:

Let $$x$$ be an irrational number, and let $$n$$ be a positive integer. For each integer $$b$$ between 1 and $$n$$, let $$r_b = bx - \lfloor bx \rfloor$$. The set $$S = \{0, r_1, r_2, \ldots, r_n, 1\}$$ contains $$n + 2$$ real numbers between 0 and 1. Prove that there are elements $$s, t \in S$$ such that $$0 < t - s \leq \frac{1}{n + 1}.$$

This I have proved and am convinced is true. However, the next part asks to prove that there exist integers $$a, b$$, such that $$1 \leq b \leq n$$ and $$|bx - a| \leq \frac{1}{n + 1}.$$ Hint: consider the previous part. Three cases are needed: $$s = 0, t = r_b$$, or $$s= r_b, t = 1$$, or $$s = r_b, t = r_e$$.

The first case is quite simple. We have that $$r_b = bx - \lfloor bx \rfloor$$, by definition. Rearranging, we obtain $$bx = r_b + \lfloor bx \rfloor$$. Let $$a = \lfloor bx \rfloor$$ which gives us $$|bx - \lfloor bx \rfloor| = |bx - a| = |r_b| \leq t - s \leq \frac{1}{n + 1}$$

since $$s = 0, t = r_b \in S$$. But the second and third case is where I get stuck. I try taking the same approach for the second case (where $$s = r_b$$ and $$t = 1$$), but it seems to not work. For instance, we have again that $$r_b = bx - \lfloor bx \rfloor \implies bx = r_b - \lfloor bx \rfloor$$. Let $$a = \lfloor bx \rfloor$$ which gives us

$$|bx - \lfloor bx \rfloor| = |bx - a| = |r_b| = |s| \leq t - s \leq \frac{1}{n + 1}$$

but this isn't always true? I am stuck here and any help would be appreciated.

For the second case, you're not appropriately using the absolute value in the expression

$$\lvert bx - a \rvert \le \frac{1}{n+1}$$

You're correct to use $$r_b = bx - \lfloor bx \rfloor$$, with this then becoming $$bx = r_b + \lfloor bx \rfloor$$. However, rather than $$a = \lfloor bx \rfloor$$, use $$a = \lfloor bx \rfloor + 1$$ instead to get

\begin{aligned} \lvert bx - a \rvert & = \lvert r_b + \lfloor bx \rfloor - (\lfloor bx \rfloor + 1) \rvert \\ & = \lvert r_b - 1 \rvert \\ & = \lvert 1 - r_b \rvert \\ & = t - s \\ & \le \frac{1}{n+1} \end{aligned}

Proceed similarly with the third case of $$s = r_d, t = r_e$$ (I used $$d$$ as the first index instead of $$b$$ to help avoid confusion). First, if $$e \gt d$$, then set $$b = e - d$$ and $$a = \lfloor ex \rfloor - \lfloor dx \rfloor$$ to get

\begin{aligned} \lvert bx - a \rvert & = \lvert (e - d)x - (\lfloor ex \rfloor - \lfloor dx \rfloor) \rvert \\ & = \lvert (ex - \lfloor ex \rfloor) - (dx - \lfloor dx \rfloor) \rvert \\ & = \lvert r_e - r_d \rvert \\ & = t - s \\ & \le \frac{1}{n+1} \end{aligned}

For the case where $$e \lt d$$, switch the other values around (i.e., $$b = d - e$$ and $$a = \lfloor dx \rfloor - \lfloor ex \rfloor$$), with using the absolute value meaning the same result as above will occur.

• Great answer, providing two ways to solve the problem. +1 Feb 21, 2023 at 0:23