Define $\mathscr L: \mathbb Q \mapsto \mathbb N$ as the minimal number of terms in the continued fraction of a rational number.
Example: the continued fraction of $\frac{5}{8}$ is $\left[0;1,1,1,2\right].$ Thus $\mathscr L(\frac{5}{8}) = 5.$
Assume we have two numbers $a,b \in \mathbb Q$ such that $$\mathscr L(a) = p \gg \mathscr L(b) = q.$$ (That is, $a$ has a much longer continued fraction than $b$.)
Assume also that if $a = \frac{m}{n}$ and $b=\frac{f}{g}$, then $m \gg f$ and $n \gg g.$ And finally that the maximal value in the continued fraction of $a$ is much bigger than in that of $b$.
Now set $c = a-b$.
Assume finally that $|c|<1.$
I would want to get some lower bound for $\mathscr L(c) := s.$
In the above case it would seem that number $a$ would have much more complex structure than $b$, so subtracting one from the other would likely keep the complexity of the result more like that of $a$ than that of $b$. Hence I would assume that in this case the bound could be set closer to $p$ than $q$.
But this is just a hunch based on some combination of naïve logic, wishful thinking and some computer experiments. So could someone provide a bound.
Ah, I should mention that I would like a "nontrivial" bound, so bigger than $1$ :)