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

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

and from Numerators and Denominators of Continued Fraction we have:
 * $q_{n + 1} =a_{n+1} q_n + q_{n - 1}$

Therefore:

Therefore:

From Difference between Adjacent Convergents of Simple Continued Fraction, we know that:

And from Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent, we know one of the following two inequalities must be true:

Therefore:

Therefore:

It is now sufficient to prove that:
 * $(1): \quad a_{i+1} + b_i \lt \sqrt 5$

cannot be true for all of $n - 1$, $n$ and $n + 1$ of $i$.

To begin, from the definition of a Simple Infinite Continued Fraction, we know that the partial quotients are strictly positive integers ($a_n > 0$ for $n >0$).

that $(1)$ is true for $i = n - 1$, $i = n$ and $i = n + 1$:

Suppose that:

And since $q_n$ is strictly increasing as n increases, we know that:
 * $\forall i$, $0 \lt b_i \lt 1$

From which we deduce that $a_n$, $a_{n + 1}$ and $a_{n + 2}$ must be equal to either $1$ or $2$ for the inequalities above to hold.

From our supposition above, we have:

and also from above:

Therefore:

And we now have:

Which can only hold true if $a_{n + 1} = a_{n + 2} = 1$ since $\dfrac 1 {\sqrt 5 - 2 } \gt \sqrt 5$

To finish off the proof:

But this implies that $b_n \gt \sqrt 5 - 1$ which contradicts the fact that $\forall i$, $0 \lt b_i \lt 1$

This contradicts our premise that $a_{i+1} + b_i \lt \sqrt 5$ was true for all of $n - 1$, $n$ and $n + 1$ of $i$.

Hence the result by Proof by Contradiction.