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Revision difference : Dirichlet's approximation theorem
Version 3 Version 2
\PMlinkescapeword{unit} \PMlinkescapeword{unit}
Theorem (Dirichlet, c. 1840): For any real number $\theta$ and any integer Theorem (Dirichlet, c. 1840): For any real number $\theta$ and any integer
$n\ge 1$, there exist integers $n\ge 1$, there exist integers
$a$ and $b$ such that $1 \le a \le n$ and $a$ and $b$ such that $1 \le a \le n$ and
$ \arrowvert a\theta-b \arrowvert \le \frac{1}{n+1}$. $ \arrowvert a\theta-b \arrowvert \le \frac{1}{n+1}$.
Proof: We may assume $n\ge 2$. Proof: We can suppose $n\ge 2$.
For each integer $a$ in the interval $[1,n]$, write For each integer $a$ in the interval $[1,n]$, write
$r_a = a\theta - [a\theta] \in [0,1)$, where $[x]$ denotes $r_a = a\theta - [a\theta] \in [0,1)$. Since the $n+2$
the greatest integer less than $x$. Since the $n+2$
numbers $0, r_a, 1$ all lie in the same unit interval, some two numbers $0, r_a, 1$ all lie in the same unit interval, some two
of them differ (in absolute value) by at most $\frac{1}{n+1}$. of them differ (in absolute value) by at most $\frac{1}{n+1}$.
If $0$ or $1$ is in any such pair, then the other element of the If $0$ or $1$ is in any such pair, then the other element of the
pair is one of the $r_a$, and we are done. pair is one of the $r_a$, and we are done.
If not, then $0 \le r_k - r_l \le \frac{1}{n+1}$ for some distinct $k$ If not, then $0 \le r_k - r_l \le \frac{1}{n+1}$ for some distinct $k$
and $l$. If $k>l$ we have $r_k - r_l = r_{k-l}$, since each side is in and $l$. If $k>l$ we have $r_k - r_l = r_{k-l}$, since each side is in
$[0,1)$ and the difference between them is an integer. Similarly, $[0,1)$ and the difference between them is an integer. Similarly,
if $k<l$, we have $1-(r_k - r_l) = r_{l-k}$. So, with $a=k-l$ or if $k<l$, we have $1-(r_k - r_l) = r_{l-k}$. So, with $a=k-l$ or
$a=l-k$ respectively, we get $a=l-k$ respectively, we get
\[ \arrowvert r_a - c \arrowvert \le \frac{1}{n+1} \] \[ \arrowvert r_a - c \arrowvert \le \frac{1}{n+1} \]
where $c$ is $0$ or $1$, and the result follows. where $c$ is $0$ or $1$, and the result follows.
It is clear that we can add the condition $\gcd(a,b)=1$ to the conclusion. It is clear that we can add the condition $\gcd(a,b)=1$ to the conclusion.
The same statement, but with the weaker conclusion The same statement, but with the weaker conclusion
$ \arrowvert a\theta-b \arrowvert < \frac{1}{n}$, $ \arrowvert a\theta-b \arrowvert < \frac{1}{n}$,
admits a slightly shorter proof, and is sometimes also referred to admits a slightly shorter proof, and is sometimes also referred to
as the Dirichlet approximation theorem. (It was that shorter proof as the Dirichlet approximation theorem. (It was that shorter proof
which made the ``pigeonhole principle'' famous.) Also, the theorem which made the ``pigeonhole principle'' famous.) Also, the theorem
is sometimes restricted to irrational values of $\theta$, with the is sometimes restricted to irrational values of $\theta$, with the
(nominally stronger) conclusion (nominally stronger) conclusion
$\arrowvert a\theta-b \arrowvert < \frac{1}{n+1}$. $\arrowvert a\theta-b \arrowvert < \frac{1}{n+1}$.