Hurwitz's Theorem (Number Theory)/Lemma 1

Lemma
Let $\xi$ be an irrational number.

Let $A \in \R$ be a real number strictly greater than $\sqrt 5$.

Then there may exist at most a finite number of relatively prime integers $p, q \in \Z$ such that:


 * $\left|{\xi - \dfrac p q}\right| < \dfrac 1 {A \, q^2}$

Proof
We will take as our example of such an irrational number:
 * $\xi = \dfrac {\sqrt 5 - 1} 2$

This is equal to $1 - \phi$, where $\phi$ is the Golden mean.

that there exist an infinite number of $p, q$ with $p \perp q$ such that:
 * $\left\lvert{\xi - \dfrac p q}\right\rvert < \dfrac 1 {A \, q^2}$

Then there exist an infinite number of $p, q$ with $p \perp q$ such that:
 * $\xi = \dfrac p q + \dfrac \delta {q^2}$

where:
 * $\left\lvert{\delta}\right\rvert < \dfrac 1 A < \dfrac 1 {\sqrt 5}$

Hence:

When $q$ is large, the of $(1)$ becomes less than $1$.

At the same time, the is always an integer.

Thus:
 * $p^2 + pq - q^2 = 0$

or:
 * $\left({2 p + q}\right)^2 = 5 q^2$

which would lead to:
 * $p = 2 q$

which contradicts the stipulation that $p$ and $q$ are coprime.

Hence by Proof by Contradiction there cannot be an infinite number of such $p, q$.

Hence the result.