Two rational numbers $\frac{a}{b}$ < $\frac{c}{d}$ will be called neighbors if $\frac{c}{d}$ - $\frac{a}{b}$ = $\frac{bc-ad}{bd}$ = $\frac{1}{bd}$.
Suppose $\frac{a}{b}$ and $\frac{c}{d}$ are neighbors in this sense and $\frac{m}{n}$ is a rational number such that:
$\frac{a}{b}$ < $\frac{m}{n}$ < $\frac{c}{d}$
Prove that $n \geq b + d$
wlog $a=0$ and $b=1$ because you can subtract $\frac{a}{b}$ from each $\frac{c}{d}$ and from $\frac{m}{n}$. So we get $0<\frac{m}{n}<\frac{c}{d}$ and $\frac{c}{d}-0=\frac{1}{d}$. Thus $0<\frac{m}{n}<\frac{1}{d}$ which implies $n\geq d+1=d+b$.
Suppose that $n < b + d$. Let $r = b + d - n > 0$. Let $s = a + c - m$. Then
$$ \frac{a}{b} < \frac{m}{n} = \frac{a + c - s}{b + d - r} < \frac{c}{d}$$
implies that
$$ ab + ad - ar < ab + bc - bs,$$ and $$ ad + cd - ds < bc + cd - cr.$$
Thus $$ bs - ar < bc - ad = 1,$$ $$ cr - ds < bc - ad = 1.$$
$a,b,c,d, s, r$ are integers, so we have $ bs - ar \leqslant 0$ and $ cr - ds \leqslant 0, $
which means $\frac{s}{r} \leqslant \frac{a}{b} $ and $\frac{c}{d} \leqslant \frac{s}{r}$ and this causes a contradiction since in the condition we can deduce that $\frac{a}{b} < \frac{c}{d}$.
