When are two neighbouring fractions in Farey sequence are similarly ordered I am trying exercises from Tom M Apostol and I could not think about this problem in Chapter 5. 

Problem is - Two reduced fractions $a/b$ and $c/d$ are said to be similarly ordered if $(c-a)\times(d-b)\ge0$. Prove that any two neighboring fractions $\frac{a_i}{b_i}$ and $\frac{a_{i+1}}{b_{i+1}}$ are similarly ordered.

My attempt - I tried using result - for any two consecutive Farey fractions $a/b<c/d$, $bc-ad=1$ holds and then using definition of similarly ordered fractions. But it doesn't yields result when $b\neq d$.
Can someone please help.
 A: $a_{i + 1}b_i - a_ib_{i + 1} = 1$ is indeed the right identity to look at.
Note that, since $\frac{a_i}{b_i} < \frac{a_{i + 1}}{b_{i + 1}}$, the only way $(a_{i + 1} - a_i)(b_{i + 1} - b_i) \ge 0$ could possibly fail is if $a_{i + 1} \ge a_i + 1$ and $b_{i + 1} \le b_i - 1$. But then we'd have $$a_{i + 1}b_i - a_ib_{i + 1} \ge (a_i + 1)b_i - a_i(b_i - 1) \ge a_i + b_i > 1$$ which is a contradiction. 
A: The non-negative Farey fractions can be obtained by starting with $\frac01,\frac11,\frac21,\ldots$ (where the claim clearly holds) and then repeatedly inserting $\frac{a+c}{b+d}$ between adjacent fractions $\frac ab$ and $\frac cd$.
Near an inserted fraction, we have
$$ ((a+c)-a)\cdot ((b+d)-b)=cd\ge 0$$
and 
$$ (c-(a+c))\cdot (d-(b+d))=ab\ge0.$$
A: The equality $a_{i + 1}b_i - a_ib_{i + 1} = 1$ or the fact that Farey sequences can be generated by inserting $\frac{a+c}{b+d}$ are too powerfull and difficult facts for such an easy exercise. All the more this relation holds not only for consecutive Farey fractions: 
Let $\frac{a}{b}<\frac{c}{d}\le\frac{a+1}{b}$ then $(c-a)\times(d-b)\ge0$ obviously holds.
