Theorem: Given $\alpha \in \mathbb{R}$ and $n \in \mathbb{Z}^+$, there exist $a,b \in \mathbb{Z}$ where $1 \le b \le n$ and $ \left|\alpha - \frac{a}{b}\right| \le \frac{1}{bn} .$Proof: We operate on the $Q$ in $P \implies Q$. Note that after multiplying $Q$ by $b$, we find $ \left|\alpha - \frac{a}{b}\right| \le \frac{1}{bn} \iff \left|b\alpha - a \right| \le \frac{1}{n} $What do we mean at this point? Well, given $\alpha$, we want to guarantee that we can keep multiplying it $1\alpha, 2\alpha, 3\alpha, \ldots$ and guarantee that the fractional part is at at most $\frac{1}{n}$. But, we must guarantee that we can find such a coefficient of $\alpha$ within the set $\{1,2, \ldots, n\}$. Consider $\mathbb{R} / \mathbb{Z} = [0,1)$ and specifically a partition of this set into $n$ smaller sets: $\left[0,\frac{1}{n}\right) \cup \left[\frac{1}{n}, \frac{2}{n}\right) \cup \ldots \cup \left[\frac{n-1}{n},1\right).$ Now, consider the numbers $0, \alpha, 2\alpha, 3\alpha, \ldots, n\alpha$. By the pigeonhole principle, there exists an interval $\left[\frac{k}{n},\frac{k+1}{n}\right)$ with $0 \leq k <n$ for which there exists distinct $i\alpha, j\alpha$ where $0 \leq j < i \leq n$. Then, $(i-j)\alpha \in \left[0,\frac{1}{n}\right)$. We know that $0 <i-j \leq n$ so take $b := i-j$ and we are done.