jmc

We claim that, between any two fractions $a/b$ and $c/d$, if $bc-ad=1$, the fraction with smallest denominator between them is $\frac{a+c}{b+d}$. To prove this, we see that $$\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right)\ge \frac{1}{dq}+\frac{1}{bq},$$which reduces to $q\ge b+d$. We can easily find that $p=a+c$, giving an answer of $\boxed{7}$.