<s>[INST] How to prove that: if $q= b+d$, then $p = a+c$? 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 to prove that:  if $q= b+d$, then $p = a+c$? Is there a simple way?
 [/INST] Quite an easy problem. while it is informed that all the variables in the problem is natural numbers so, $p = a+c$ is equivalent to $a+c-1 < p <a+c+1$, that is to say $\frac{a+c-1}{b+d}<\frac{c}{d}$ and $\frac{a+c+1}{b+d}>\frac{a}{b}$.in fact:
$$
ad - d<bc
$$
$$
ad+cd -d <bc+cd
$$
$$
\frac{a+c-1}{b+d}<\frac{c}{d}
$$
what's more,
$$
bc+b>ad
$$
$$
ab+bc+b>ab+ad
$$
$$
\frac{a+c+1}{b+d}>\frac{a}{b}
$$
</s>