How prove that $q \geq b+d$ for $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$?

Question

Let $a,b,c,d,p$, and $q$ be natural numbers such that $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$.

How prove that $q \geq b+d$?

Answer 1

This is a well-known fact in the theory of Farey series. You have $aq>bp$ and $pd>qc$, so the integers $\delta_1=aq-bp$ and $\delta_2=dp-cq$ are both $\geq 1$. Now

$$ d\delta_1+b\delta_2=(ad-bc)q=q $$

It follows that $q\geq b+d$.