Exercise 6.1.4 (Bounds for the distances of consecutive fractions in the Farey sequence)

Proof. (of Lemma). If $a∕b<a^{\prime }∕b$ are two consecutive terms of the Farey sequence, then by Theorem 6.1 (and the remark at the end of section 6.1), $b{a}^{\prime }-\mathit{ab}=1$, thus $b(a^{\prime }-a)=1(b>0)$, thus $b=1$ and $a^{\prime }=a+1$. The two consecutive terms are $a$ and $a+1$, hence $n=1$. (There is an alternative proof in Hardy & Wright.) □

Proof. (of 6.1.4). Consider two consecutive terms $a∕b<a^{\prime }∕b^{\prime }$ in the (complete) Farey sequence of order $n$, so that $b\le n$ and $b^{\prime }\le n$. By the Lemma, $b\ne b^{\prime }$, and by Theorem 6.1,

Then

• Since $b\ne b^{\prime }$, $b{b}^{\prime }\le n(n-1)$, where $n>1$. Therefore by (2),

  $$\begin{array}{lll}\hfill \frac{a^{\prime }}{b^{\prime }}-\frac{a}{b}\ge \frac{1}{n(n-1)}.& & \hfill \end{array}$$

  Moreover, by Problem 5, $a∕b=1∕n$ and $a^{\prime }∕b^{\prime }=1∕(n-1)$ are consecutive fractions in the Farey sequence of order $n$, and $a^{\prime }∕b^{\prime }-a∕b=1∕(n(n-1))$. Hence

  $$\min <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{a^{\prime }}{b^{\prime }}-\frac{a}{b}\right)=\frac{1}{n(n-1)}.$$

• Since $\frac{a}{b}<\frac{a^{\prime }}{b^{\prime }}$ are consecutive fractions in the Farey sequence of order $n$, and $\frac{a}{b}<\frac{a+a^{\prime }}{b+b^{\prime }}<\frac{a^{\prime }}{b^{\prime }}$, we deduce that $\frac{a+a^{\prime }}{b+b^{\prime }}$ is not in the Farey sequence of order $n$. Therefore $b+b^{\prime }>n$ (otherwise the reduced fraction of $\frac{a+a^{\prime }}{b+b^{\prime }}$ would be in the Farey sequence of order $n$). So

  $$b+b^{\prime }\ge n+1.$$

  Then

  $$b{b}^{\prime }-n\ge b(n+1-b)-n=(n-b)(b-1)\ge 0,$$

  thus

  $$b{b}^{\prime }\ge n.$$

  By (2),

  $$\begin{array}{lll}\hfill \frac{a^{\prime }}{b^{\prime }}-\frac{a}{b}\le \frac{1}{n}.& & \hfill \end{array}$$

  Moreover, by Problem 5, $a∕b=0∕1$ and $a^{\prime }∕b^{\prime }=1∕n$ are consecutive fractions in the Farey sequence of order $n$, and $a^{\prime }∕b^{\prime }-a∕b=1∕n$. Hence

  $$\max <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{a^{\prime }}{b^{\prime }}-\frac{a}{b}\right)=\frac{1}{n}.$$