Best Rational Approximation to an Irrational Number
\[x\]
has infinite continued fraction \[[ 0,2,4,6, 8, 10 ]\]
, Find the best rational approximation \[\frac{a}{b}\]
to \[x\]
with \[b \lt 100\]
.The first few convergents of the continued fraction are
\[0, \frac{1}{2}, \; \frac{1}{2 + \frac{1}{4}}= \frac{4}{9}, \; \frac{1}{2 + \frac{1}{4+ \frac{1}{6}}} = \frac{25}{56}, \; \; \frac{1}{2 + \frac{1}{4+ \frac{1}{6+ \frac{1}{8}}}} = \frac{204}{457} \]
.The convergent
\[\frac{35}{56}\]
satisfies \[\| x- \frac{25}{56} \| \lt \frac{1}{56 \times 457}\]
.Suppose there is some better rational approximation
\[\frac{a}{b}\]
with denominator less than 100. Then\[\begin{equation} \begin{aligned} \| \frac{a}{b} - x \| & \gt \| \frac{a}{b} - \frac{25}{56} \| \\ & \gt \| x- \frac{25}{56} \| \\ & \gt \frac{1}{56 \times 100} - \frac{1}{56 \times 457} \\ &= \frac{337}{56 \times 100 \times 457} \\ & \gt \frac{1}{56 \times 457} \end{aligned} \end{equation}\]
so
\[\frac{25}{56}\]
is a better approximation than \[\frac{a}{b}\]
. 
