Exercise 6.1.5 (Conditions for two fractions of the Farey sequence to be adjacent)

Proof. Since $\mathit{ad}-\mathit{bc}=1>0$, and $d>0,b>0$, we obtain $c\wedge d=1,a\wedge b=1$, where $n=\max <!--\; FUNCTION\; APPLICATION\; -->(c,d)$, thus $c\le n$ and $d\le n$, and

so that the fractions $c∕d$ and $a∕b$ are two fractions in the Farey sequence of order $n$.

Consider any fraction $x∕y$ $(y>0)$ between $c∕d$ and $a∕b$ so that

By hypothesis, $\mathit{ad}-\mathit{bc}=1$, therefore

By (1), $\mathit{ay}-\mathit{bx}>0$ and $\mathit{dx}-\mathit{cy}>0$, therefore

where $\mathit{bdy}>0$, thus

Since $n=\max <!--\; FUNCTION\; APPLICATION\; -->(b,d)$, and $b\ge 1,d\ge 1$, we obtain $b+d\ge n+1$, so

This shows that no fraction $x∕y$ between $c∕d$ and $a∕b$ is in the Farey sequence of order $n$. Hence $c∕d<a∕b$ are adjacents fractions in this sequence.

In conclusion, if $\mathit{ad}-\mathit{bc}=1$, $b>0$, $d>0$ and $n=\max <!--\; FUNCTION\; APPLICATION\; -->(b,d)$, then $c∕d$ and $a∕b$ are adjacent fractions in the Farey sequence of order $n$. □