Regarding similarly ordered fractions in farey sequences This question is from Apostol modular functions and Dirichlet series in number theory. 
It is related to this problem - 
When are two neighbouring fractions in Farey sequence are similarly ordered
Apostol in 2 nd part of this exercise asks to prove that any two 2nd neighborhoods $\frac {a_i }{ b_i} $ and $\frac { a_{i+2 }} {b_{i +2}} $ are similarly ordered. 

My attempt - There exists 2 cases if the fractions are not similarly ordered. 

Case1 -$a_{i+2}$ < $a_i$ and $b_{i+2}$ > $b_i$  . It is easy to obtain contradiction in this case and I obtained it. 

But in Case2 - $a_{i+2} $>  $a_i$ and $b_{i+2} $< $b_i$
   I cannot obtain  any contradiction. 

Can somebody please help. 
 A: If $\frac{a}{b} < \frac{c}{d}$ ($a,b,c,d$ positive integers [$a = 0$ is also admissible]) and $d \geqslant b$, then it immediately follows that
$$c > d\cdot \frac{a}{b} = \frac{d}{b}\cdot a \geqslant a\,,$$
so the fractions are similarly ordered. This holds generally, no Farey sequence properties required. (You already know this part, but it's here for completeness.)
So let's look at the case $b_{i+2} < b_i$ for the second neighbours in a Farey sequence. We want to deduce $a_{i+2} \leqslant a_i$. Suppose it weren't so. Then
$$\frac{a_{i+2}}{b_{i+2}} > \frac{a_{i+2} - 1}{b_{i+2}} \geqslant \frac{a_i}{b_{i+2}} \geqslant \frac{a_i}{b_i-1} > \frac{a_i}{b_i}$$
and since the two fractions are second neighbours the three fractions in the middle must all be equal, that is, we must have $a_{i+2} = a_i + 1$ and $b_{i+2} = b_i - 1$.
But we also know that the fraction between the two is their mediant,
$$\frac{a_{i+1}}{b_{i+1}} = \frac{a_{i} + a_{i+2}}{b_i + b_{i+2}}\,,$$
and that means we must have
$$\frac{a_{i}}{b_{i}-1} = \frac{2a_{i} + 1}{2b_{i} - 1} \iff \frac{2b_{i}-1}{b_{i}-1} = \frac{2a_{i}+1}{a_{i}} \iff \frac{1}{b_i-1} = \frac{1}{a_i} \iff a_i = b_i - 1\,.$$
This however means
$$\frac{a_{i+2}}{b_{i+2}} = \frac{a_i+1}{b_i-1} = \frac{b_i}{b_i-1} > 1\,,$$
which contradicts the assumption that $0 < \frac{a_{i+2}}{b_{i+2}} \leqslant 1$.
Therefore $\frac{a_i}{b_i}$ and $\frac{a_{i+2}}{b_{i+2}}$ must be similarly ordered also when $b_{i+2} < b_i$.
