
The RMP 2/n table and the Liber Abaci's 7th distinction
Posted:
Aug 9, 2009 3:26 PM


Dear Forum members:
It has been a small number of years since Wikipedia and a blog reported an outline of these situations per:
http://en.wikipedia.org/wiki/Liber_Abaci
http://liberabaci.blogspot.com/
"7. Seventh method (distinction)
a. 4/9  1/13 = 3/(13 x 49)
= (1/319 0/637 1/617 1/319 1/13), not elegant
b. 4/49  1/14 = 7/(14 x 49)
= (1/2 0/49 1/14), elegant
c. 4/49 = 1/7 x (4/7) = 1/7 x (4/7  1/2 = 1/14)
= (1/2 0/49 1/14), alternate elegant
A vivid parallel is available in Ahmes' shorthand. In RMP 31 and RMP 36, two rational numbers 28/97 and 30/53 were not converted in unit fraction series in their present forms. Ahmes applied a 2/n table rule,
letting n/p = 2/p + (n 2)/p
to achieve a fast conversion to unit fraction series, the 2/n table method was adapted by Fibonacci into subtraction context by apparently using only 1/p, as mentioned in Fibonacci's 4th distinction:
20/53 = 19/53 + 1/53
with (19/53  1/3) = (3+ 1)/159 = 1/53 + 1/159
Ahmes' 2850 year older arithmetic considered the aliquot parts of 56, and 4 for 28/97 and 30 and 2 for 30/53 and writing:
28/97 = 2/97 + 26/97 = 2/97*(56/56) + 26/97*(4/4) = (97 + 8 + 7 ) + (97 + 4 + 2 + 1)/388
30/53 = 2/53 + 28/53 = 2/53 *(30/30) + 28/53*(2/2) = (53 + 5 + 2)/1590 + (53 + 2 + 1)/106
Fibonacci also considered simpler aliquot parts (divisors) properties of closely related first partition denominators.
Sigler did not discuss 30/53, a vulgar fraction that can not be solved by Fibonacci's 29/53 + 1/53 rule. There is no first partition that solves 29/53. Try 1/2, means
29/53  1/2 =(53 + 2 + 2 + 1)/106 is needed,
thereby breaking a repetition rule.
Fibonacci's distinction 7 rule obviously was extended to include 2/53, solving the problem, meaning 2/53 was converted by:
2/53  1/30 = (53 +5 + 2)/1590 = 1/30 + 1/318 + 1/795
exactly following Ahmes 2/n table method.
Summary: Sigler offered no proof that an algorithm was used by Fibonacci. The best way to explain Fibonacci's seventh distinction is to clarify a Medieval context for solving 28/97 and 30/97 examples, thereby drawing a numerical parallel to Ahmes' method. Fibonacci fairly concluded that his seventh conversion method works for all examples, a fact that is attested during by Ahmes by applying a modified 2/n table rule reported in RMP 31 and RMP 36.
Best Regards,
Milo Gardner

