Exercise 6.2.6 (Consecutive terms of the Farey sequence of order $n$)

Question

If an irrational number $	heta$ lies between two consecutive terms $a/b$ and $c/d$ of the Farey sequence of order $n$, prove that at least one of the following inequalities holds:
$$|	heta - a/b| < 1/(2b^2),\qquad |	heta - c/d| < 1/(2d^2).$$

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that
$$\frac{a}{b} \leq 	heta \leq \frac{c}{d},$$
where $a/b$ and $c/d$ are two consecutive terms of the Farey sequence of order $n$.

If both inequalities
$$\left|	heta - \frac{a}{b}ight| < \frac{1}{2b^2},\qquad \left|	heta - \frac{c}{d}ight| < \frac{1}{2d^2}$$
were wrong, then
$$	heta - \frac{a}{b} = \left|	heta - \frac{a}{b}ight| \geq \frac{1}{2b^2},\qquad  \frac{c}{d} - 	heta = \left|	heta - \frac{c}{d}ight| \geq \frac{1}{2d^2}.$$
Therefore, using $bc -ad = 1$ (by Theorem 6.1),
\begin{align*}
\frac{1}{bd} &=  \frac{c}{d} - \frac{a}{b}\
 &= \left(	heta - \frac{a}{b} ight)+ \left(\frac{c}{d} - 	heta ight)\
&\geq \frac{1}{2b^2} +  \frac{1}{2d^2}.
\end{align*}
Hence
\begin{align*}
0&\geq \frac{1}{b^2} + \frac{1}{d^2} - \frac{2}{bd} = \left( \frac{1}{b} - \frac{1}{d} ight)^2, 
\end{align*}
so $b = d$. This is impossible by the Lemma in the proof of Problem 6.1.4: no consecutive fractions of the Farey sequence of order $n >1$ have the same denominator.

This contradiction shows that at least one of the following inequalities holds:
$$\left|	heta - \frac{a}{b}ight| < \frac{1}{2b^2},\qquad \left|	heta - \frac{c}{d}ight| < \frac{1}{2d^2}.$$
\end{proof}

Exercise 6.2.6 (Consecutive terms of the Farey sequence of order $n$)

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Exercise 6.2.6 (Consecutive terms of the Farey sequence of order $n$)

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