How to prove that: if $q= b+d$, then $p = a+c$?

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 to prove that: if $q= b+d$, then $p = a+c$? Is there a simple way?

Answer 1

Yes there is a simple way. Hint : show that $a+c+1$ is too large ($\frac{a+c+1}{b+d} > \frac{a}{b}$) and that $a+c-1$ is too small ($\frac{a+c-1}{b+d} < \frac{c}{d}$).

Answer 2

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, $$ bc+b>ad $$ $$ ab+bc+b>ab+ad $$ $$ \frac{a+c+1}{b+d}>\frac{a}{b} $$