If the rational number
\[\frac{a}{b}\]
satisfies \[\| x- a\frac{a}{b} \| \lt \frac{1}{2b}\]
then \[\frac{a}{b}\]
is a convergent of the continued fraction representing \[x\]
.Proof
Suppose on the contrary that
\[\frac{a}{b}\]
is a rational number in its simplest form satisfying the above inequality and is not a convergent \[\frac{p_r}{q_r}\]
of the continued fraction. Let \[r\]
be the unique integer satisfying \[q_r \le b \le q_{r+1}\]
.
Using the Condition on Approximation to Infinite Continued Fraction,\[\| a_r x-p_r \| \le \| bx-a \| = b \| x- \frac{a}{b} \| \lt b(\frac{1}{2b^2}) =\frac{1}{2b}\]
Therefore
\[q_r \| x - \frac{p_r}{q_r} \| \lt \frac{1}{2b} \rightarrow \| x- \frac{p_r}{q_r} \| \lt \frac{1}{2q_rb}\]
.The Triangle Inequality gives
\[\| \frac{a}{b} - \frac{p_r}{q_r} \| \le \| x - \frac{p_r}{q_r} \| + \| \frac{a}{b} - x \| \lt \frac{1}{2q_rb}+ \frac{1}{2b^2}\]
On the other hand
\[\| \frac{a}{b} - \frac{p_r}{q_r} \| = \| \frac{aq_r-bp_r}{bq_r} \| \gt \frac{1}{q_rb}\]
.since
\[aq_r-bp_r\]
is a non zero integer.Hence
\[\frac{1}{q_rb} \lt \frac{1}{2q_rb} + \frac{1}{2b^2} \rightarrow q_r \gt b\]
, giving the contradiction and proving the theorem.