Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 12H / b

The convergents satisfyso $(p_n,q_n)$ and $(p_{n-1},q_{n-1})$ form an integer basis of ${\mathbb{Z}}^2$. Put ${\delta }_j=q_j\alpha -p_j$. Consecutive errors have opposite signs and $|{\delta }_{n-1}|>|{\delta }_n|$.

For integers $p,q$ with $0<q\le q_n$, writeIf $s>0$, the denominator bound forces $r\le 0$, while if $s<0$ positivity forces $r\ge 1$. Because ${\delta }_n$ and ${\delta }_{n-1}$ have opposite signs, in either case $s\ne 0$ impliesIf $s=0$, the denominator bound gives $r=1$ unless the fraction is zero or has the same value as $p_n/q_n$. Thus $p_n/q_n$ is a best approximation of the second kind. Dividing by $q\le q_n$ gives

Suppose $a_j\le A$ for every $j$. Given $b>0$, choose the first convergent denominator $q_m\ge b$. Then $q_{m-1}<b$ and the recurrence gives $q_m=a_m{q}_{m-1}+q_{m-2}<(A+1)b$. Part (i), applied at index $m$, gives $|b\alpha -a|\ge |q_m\alpha -p_m|$.

If ${\alpha }_{m+1}=[a_{m+1};a_{m+2},\dots ]$ is the complete quotient, the exact error formula isSince ${\alpha }_{m+1}<A+1$, this is greater than $1/((A+2)q_m)$ and therefore greater than $1/((A+1)(A+2)b)$. It follows that every rational $a/b$ satisfiesThus one may take $c=((A+1)(A+2){)}^{-1}$.

Consecutive convergents obey the standard error estimateIf the partial quotients are unbounded, then for every $c>0$ there are infinitely many indices with $a_{n+1}>1/c$; otherwise the whole sequence would be bounded after adjoining its finite initial segment. At each such index,The denominators of the infinite continued fraction tend to infinity, so these give infinitely many distinct rational numbers of the required kind.