Proof of Condition on Accuracy of Convergents of Continued Fractions
Out of any two consecutive convergents
\[\frac{p_n}{q_n}, \; \frac{p_{n+1}}{q_{n+1}}\]
of a continued fraction for a number \[x\]
one of them must satisfy the inequality \[\| x - \frac{p}{q} \|\]
where \[\frac{p}{q} \| \lt \frac{1}{2q^2}\]
is one of the continued fractions.The sign of
\[x - \frac{p_n}{q_n}\]
alternates so that \[x - \frac{p_n}{q_n}=-(x- \frac{p_{n+1}}{q_{n+1}})\]
. Hence\[\begin{equation} \begin{aligned} \| (x- \frac{p_{n+1}}{q_{n+1}})+ (x- \frac{p_n}{q_n} ) \| &= \| \frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} \| \\ &= \| \frac{p_{n+1}q_n - p_nq_{n+1}}{q_{n+1}q_n} \| \\ &= \frac{1}{q_{n+1}q_n} \| \end{aligned} \end{equation}\]
.If neither convergent satisfies the property then
\[\frac{1}{q_{n+1}q_n} \ge \frac{1}{2q_{n+1}^2} + \frac{1}{2q_n^2} \rightarrow (q_{n+1}=q_n)^2 \le 0\]
which is impossible since \[q_{n+1} \neq q_n\]
for \[n \gt 2\]
. 
