Exercise 6.1.1 (Fraction immediately to the right of $1/2$ in the Farey sequence of order $n$)

Proof. To prove $b=b^{\prime }$, we use the symmetry

which preserves the Farey sequence.

More precisely, since $0\le \frac{a}{b}<\frac{1}{2}$ and $a\wedge b=1$, we obtain $\frac{1}{2}<\frac{b-a}{b}\le 1$, where $(b-a)\wedge b=1$, so $\frac{b-a}{b}$ is in the Farey sequence of order $n$. If some irreducible fraction $a^{″}∕b^{″}$ satisfies $1∕2<a^{″}∕b^{″}<(b-a)∕b$ and $b^{″}\le n$, then $a∕b<(b^{″}-a^{″})∕b^{″}<1∕2$: this is impossible, because $a∕b$ is the fraction immediately to the left of the fraction $1∕2$ in the Farey sequence of order $n$. Therefore $(b-a)∕b$ is the fraction immediately to the right of the fraction $1∕2$ in the Farey sequence of order $n$, so

This proves also the last assertion $a+a^{\prime }=b$.

Since

are two successive fractions in the Farey sequence of order $n$, by Theorem 6.1

thus $b=2a+1$, and

Since $a∕b$ is in the Farey sequence of order $n$, $b=2a+1\le n$, so

Moreover,

Since $a^{\prime }∕b^{\prime }$ is the fraction immediately to the right of the fraction $1∕2$ in the Farey sequence of order $n$, this shows that $2a+3>n$, so

By inequalities (2) and (3), $a\le \frac{n-1}{2}<a+1$, thus

In conclusion,

So $b$ is the greatest odd integer such that $b\le n$, and $a+a^{\prime }=b$. □