I warmly welcome an approach on differencing between two continuous fractions Without applying the appropriate algebra to reduce the continuous fraction. I cannot find a way without completing the fraction into a single rational part, for example, given two prime integers whereby $\gcd(17,11)=1, \gcd(17,3)=1$ $$\frac{17}{11}=1+\frac{1}{1+\frac{1}{1+\frac{1}{5}}}, \frac{17}{3}=5+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}$$
When I take the difference of the two, for example, I denote $\frac{17}{11}=S, \frac{17}{3}=S^*$ then I get $$S^*-S=\frac{17}{3}-\frac{17}{11}=(5-1)+\left(\frac{1}{1+\frac{1}{1+\frac{1}{1}}}-\frac{1}{1+\frac{1}{1+\frac{1}{5}}}\right)$$
I have a partial method, however, I am no expert and would appreciate feedback on the approach, especially when taking the reciprocal and finding its quotient & remainder. Forexample:
$$S_1=\frac{p_1}{q_1}=a_0+\frac{r_1}{q_1}, S^*_1=\frac{p_1^*}{q_1^*}=a_0^*+\frac{r_1^*}{q_1^*} \\ S_1^*-S_1 = (a_0^*-a_0)+ \left(\frac{r_1^*}{q_1^*}-\frac{r_1}{q_1} \right)$$
Suppose I denote the fraction on the right as $A_1=\left(\frac{r_1^*}{q_1^*}-\frac{r_1}{q_1} \right)$, then taking the inverse, I get $$A_1 = \frac{1}{\left(\frac{q_1^*q_1}{r_1^*q_1-r_1q_1^*}\right)}$$
For positive integers $q_1,q^*_1 > 0$, then $\frac{1}{\left(\frac{r_1^*}{q_1^*}\right)=a_1^*+\frac{r_2^*}{q_2^*}}$, $\frac{1}{\left(\frac{r_1}{q_1}\right)=a_1+\frac{r_2}{q_2}}$, which I then find the next calculation to be $$\frac{1}{\left(\frac{q_1^*q_1}{r_1^*q_1-r_1q_1^*}\right)}=\frac{1}{\left(\frac{1}{\left(\frac{r_1^*q_1-r_1q_1^*}{q_1^*q_1}\right)}\right)}=\frac{1}{\left(\frac{1}{\left(a_1^*+\frac{r_2^*}{q_2^*}-a_1-\frac{r_2}{q_2}\right)}\right)}=\frac{1}{\left(\frac{1}{\left((a_1^*-a_1)+\frac{r_2^*}{q_2^*}-\frac{r_2}{q_2}\right)}\right)}$$
Then repeating the above algorithm recursively, although the approach is time-consuming and whether a faster method exists is strongly appreciated. Because I would have to apply the division algorithm separately for each rational remainder, then take the difference.
