Proof of Condition on Accuracy of Convergents of Continued Fractions

Theorem
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  
\[n \gt 2\]
.

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