Dirichlet's Approximation Theorem

Theorem 1 (Dirichlet's Approximation Theorem). If $\alpha\in\mathbb{R}$, $t\in\mathbb{R}$, and $t\geq1$, then there exist $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ such that

\begin{align} \bigg\lvert \alpha-\frac{p}{q} \bigg\rvert<\frac{1}{qt}, \quad 0<q\leq t \end{align}

Proof of Theorem 1. Let $\{a\}$ and $[a]$ denote the fractional part and the integer part of $a$ respectively. We set $T=[t]+1$ and consider the $T+1$ numbers:

\begin{align} 0=\{\alpha\cdot0\},\{\alpha\cdot1\},\dots,\{\alpha(T-1)\},1 \end{align}

where

$$\{\alpha x\}=\alpha x-[\alpha x], \quad 0\leq \{\alpha x\}<1, \quad x=0,1,\dots,T-1$$

All of the points belong to the interval $0\leq y\leq 1$. We divide this interval into $T$ equal parts of the length $\frac{1}{T}$:

\begin{align} \frac{k}{T}\leq y< \frac{k+1}{T}, \quad k=0,1,\dots,T-2, \quad \frac{T-1}{T}\leq y\leq 1 \end{align}

Each of the points lies in one of the intervals. Since the number of points, namely $T+1$ points, is greater than the number of intervals, namely $T$ intervals, it follows that there must be an interval which contain two of the points. We now consider such an interval. There are two possible cases. 

    1. The interval is not the one at the extreme right. Suppose it contains the points $\{\alpha x_1\}$ and $\{\alpha x_2\}$, where $x_2>x_1$. Then we have

$$\lvert\{\alpha x_2\}-\{\alpha x_1\}\rvert=\lvert\alpha(x_2-x_1)-([\alpha x_2]-[\alpha x_1])\rvert<\frac{1}{T}<\frac{1}{t}$$

since the length of the half-open subinterval is $\frac{1}{T}$. We set $x_2-x_1=q$ and $[\alpha x_2]-[\alpha x_1]=p$. Obviously, $0<q\leq T-1\leq t$. We then have the inequalities

\begin{align} \lvert\alpha q-p\rvert<\frac{1}{t}, \quad 0<q\leq t \end{align}

from which Theorem 1 follows. 

    2. The interval is the extreme right interval. Suppose it contains the point $\{\alpha x_1\}$ as well as $1$, where $x_1\not=0$. Then

$$\lvert\{\alpha x_1\}-1\rvert=\lvert\alpha x_1-([\alpha x_1]+1)\rvert\leq \frac{1}{T}<\frac{1}{t}$$

since the length of the closed subinterval is $\frac{1}{T}$. We set $x_1=q$ and $[\alpha x_1]+1=p$. We then have $0<q\leq T-1\leq t$, and again we obtain the inequality in the first case, from which Theorem 1 follows. $\Box$

Theorem 2. For any irrational number $\alpha\in\mathbb{R}$, the inequality \begin{align} \bigg\lvert\alpha-\frac{p}{q}\bigg\rvert<\frac{1}{q^2} \end{align} has infinitely many solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$, and hence the set of denominators of the solutions to the inequality is unbounded. 

Proof of Theorem 2. By Dirichlet's Theorem, for any solution $(p,q)$, we have $$\bigg\lvert\alpha-\frac{p}{q}\bigg\rvert<\frac{1}{qt}\leq \frac{1}{q^2}$$ since $q\leq t$. Since $\lvert\alpha-\frac{p}{q}\rvert<\frac{1}{qt}$ has infinitely many solutions $(p,q)$ corresponding to different $t$. Consequently, the inequality above also has an infinite set of solutions $(p,q)\in\mathbb{Z}\times\mathbb{N}$. $\Box$