 Solution

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