Hurwitz's theorem is an extension of Minkowski's Theorem and deals with rational approximations to irrational numbers. The theorem states:
For every irrational number $\alpha$, there are infinitely many coprime integers $p$ and $q$ such that:
$$\left|\alpha – \frac{p}{q} \right| < \frac{1}{\sqrt{5}q^2} $$
It turns out that the $\sqrt{5}$ term is sharp. My question is why does the $\sqrt{5}$ term appear. What properties of $\sqrt{5}$ enable this to be a sharp bound? I am looking for an intuitive understanding.
Best Answer
As Terry mentions in the comments, the reason for the $\sqrt{5}$ is that the limiting case, the golden ratio, forces it. There is a very neat explanation of all of this in the classic number theory book by Hardy and Wright, pages 209 to 212. I give a brief sketch of the ideas.
As Hardy and Wright put it, "from the point of view of rational approximation, the simplest numbers are the worst. The "simplest" of all irrationals, from this point of view, is the number $\phi$."
The reason for this is that if we consider the best approximation for a given $\alpha$,
$$\left|\alpha - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} < \frac{1}{a_{n+1}q^2_n}$$
it is best when $a_{n+1}$ is large. But in the case of $\phi$, every $a_{n+1}$ is as small as possible.
The idea is to simply see what happens when we approximate $\phi$. It roughly goes like this:
$$\left|\phi - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} \sim \frac{1}{q^2_n}\frac{1}{1+2\phi}=\frac{1}{q^2_n\sqrt{5}}$$
This follows easily by contradiction. There are no infinitely many $p$, $q$ such that
$$\alpha=\frac{p}{q}+\frac{\delta}{q^2}$$ and $$|\delta|<\frac{1}{\sqrt{5}}$$
Now any proof of the theorem should look convincing enough, knowing where the $\sqrt{5}$ it presupposes comes from.
EDIT. I include for completeness a nice alternative proof brought up by Marty and Halbort in the comments.
L. R. Ford, "Fractions" (Amer. Math. Monthly, Vol 45, No 9 (Nov 1938))