1. case: b = aq. Then a/b = 1/q.
Fibonacci is a mathematician! I really like this remark of Fibonacci. He
applies it immediately in the next case.
2. case: a = u + v + w + ... and all of the summands u, v, w are divisors of b.
If b = uu' = vv' = ww'= ..., then a/b = 1/u' + 1/v' + 1/w' + ... according to
case 1.
This case will be used in case 8. So one sees a little theory developing.
3. case: b = qa - 1. Then a/b = 1/q + 1/(qb).
4. case: b = q(a - 1) - 1. Then a/b = 1/b + (a - 1)/b and case 3.
5. case: b = 2u and b = v(a - 2) - 1. Then a/b = 1/u + (a - 2)/b and case 3.
6. case: b = 3u and b = v(a - 3) - 1. Similar to case 5.
7. case: b = aq + r. Then 1/(q + 1) < a/b < 1/q and
a/b = 1(q + 1) + (a - r)/((q + 1)b). Iterate! This works always.
8. case: Choose a natural number k. Determine q and r such that ak = qb + r
and r < b. Then a/b = q/k + r/(kb). Then try to decompose q and r in sums of
divisors of k. Apply 2.
So k should have many divisors. Fibonacci assumes always that b/2 < k < 2b.
He does not say, however, whether one can always find such a k that leads to
the desired decomposition of a/b. He uses the numbers 24, 36, 48 in his
examples. He mentions that the numbers 6, 8, 12, and 60 are also useful.
My comment. Case 8 works always. Let k := 2^t be such that b <= k <2b. It
follows from a/b < 1 and r < b that q < k and r < k. Develope q and r as
sums of powers of 2. Then case 2 works.
Best regards, Heinz Lueneburg