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

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

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

Then:

It is sufficient to prove that:
 * $(1): \quad a'_i + b_i \le 5$

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

Suppose $(1)$ is true for $i = n - 1$ and $i = n$.

Then:
 * $a'_{n - 1} = a_{n - 1} + \dfrac 1 {a'_n}$

and:


 * $\dfrac 1 {b_n} = \dfrac {q_{n - 1} } {q_{n - 2} } = a_{n - 1} + b_{n - 1}$

Hence:
 * $\dfrac 1 {a'_n} + \dfrac 1 {b_n} = a'_{n - 1} + b_{n - 1} \le \sqrt 5$

and:
 * $1 = a'_n \dfrac 1 {a'_n} \le \paren {\sqrt 5 - b_n} \paren {\sqrt 5 - \dfrac 1 {b_n} }$

or:
 * $b_n + \dfrac 1 {b_n} \le \sqrt 5$

As $b_n$ is rational, the equality cannot happen.

We also have that $b_n < 1$.

Thus:
 * ${b_n}^2 -b_n \sqrt 5 + 1 < 0$


 * $\paren {\dfrac {\sqrt 5} 2 - b_n}^2 < \dfrac 1 4$

and so:
 * $b_n > \dfrac {\sqrt 5 - 1} 2$

If $(1)$ were also true for $i = n + 1$, it could be proved similarly that:
 * $b_{n + 1} > \dfrac {\sqrt 5 - 1} 2$

and we would then be able to substitute $n + 1$ for $n$ in the above equations, to get:
 * $a_n = \dfrac 1 {b_{n + 1} } - b_n < \dfrac {\sqrt 5 + 1} 2 - \dfrac {\sqrt 5 - 1} 2 = 1$

from which a contradiction is apparent.