Hurwitz's Theorem (Number Theory)/Lemma 2

Lemma
Let $\xi$ be an irrational number.

Let there be $3$ consecutive convergents of the continued fraction to $\xi$.

Then at least one of them, $\dfrac p q$ say, satisfies:
 * $\size {\xi - \dfrac p q} < \dfrac 1 {\sqrt 5 \, q^2}$

Proof
By definition of simple infinite continued fraction, the partial quotients are strictly positive integers:


 * $\forall n \in \N_{>0}: a_n > 0$

Let $\dfrac {p_k} {q_k}$ be an arbitrary convergent to $\xi$.

Let:
 * $\dfrac {q_{n - 1} } {q_n} = b_n$

By definition of numerators and denominators of continued fraction:
 * $q_{n + 1} = a_{n + 1} q_n + q_{n - 1}$

Then:

From Difference between Adjacent Convergents of Simple Continued Fraction:

Therefore:

We aim to show that given $3$ consecutive convergents of the continued fraction to $\xi$, at least one of them satisfies:
 * $\size {\xi - \dfrac p q} < \dfrac 1 {\sqrt 5 \, q^2}$

that $3$ consecutive convergents do NOT satisfy the inequality.

Thus for arbitrary $n$:

Therefore:

Using the Quadratic Formula, we arrive at:

Therefore:
 * $\phi - 1 < b_{n + 1} < \phi$

Taking reciprocals:
 * $\phi > \dfrac 1 {b_{n + 1} } > \phi - 1$

Starting with $(4)$ above and replacing $n$ with $n - 1$, and replacing $n + 1$ with $n$, an identical argument can be made for $b_n$.

We now have:
 * $\phi - 1 < \dfrac 1 {b_{n + 1} } < \phi$

and
 * $\phi - 1 < b_n < \phi$

And from $(2)$ above:
 * $a_{n + 1} = \dfrac 1 {b_{n + 1} } - b_n$

Therefore, from $(2)$ and the two inequalities immediately above:
 * $a_{n + 1} = \dfrac 1 {b_{n + 1} } - b_n < 1$

But this contradicts our premise that the partial quotients are strictly positive integers.

Hence the result by Proof by Contradiction.