(In reply to From N/D to n
That works for me.
The only thing that remains to be proved is that Jer's reverse algorithm must terminate, and that is very clear, as every subtraction makes the (numerator + denominator) smaller, the reciprocal leaves the (numerator + denominator) unchanged, and we cannot cannot do two reciprocal operations in a row.
It therefore follows that each rational number appears in the sequence. And the fact that at each step there is only one option available (as the fraction must either be greater or less than 1) means that a given rational number can appear in the sequence only once.