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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.

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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

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

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

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

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.

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

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