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:


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

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

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

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

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

where:
 * $A > \sqrt 5$

therefore,
 * $\dfrac 1 A < \dfrac 1 {\sqrt 5}$

and
 * $0 < \size \delta < \dfrac 1 A < \dfrac 1 {\sqrt 5}$

Hence:

Over the interval ${-\dfrac 1 {\sqrt 5} } < \delta < \dfrac 1 {\sqrt 5}$

The of $(1)$ takes on values of $\dfrac 1 {5q^2} - 1 < \dfrac {\delta^2} {q^2} - \delta \sqrt 5 < \dfrac 1 {5q^2} + 1$

At the same time, the is always an integer.

Thus, for the equlity to hold infinitely many times, it must hold at zero:

Which leads 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.