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Re: Archimedes (and Ahmes) square root of 5, 6 and 7
Posted:
Jan 16, 2013 6:58 PM


Archimedes' square root of 3 problem cited limits
(1271/780)^2 > 3 > (265/153)^2
A. The higher limit 1251/780 was calculated in three steps
1. (1 + 2/3)^2 error 1 = 2/9
2. 2/9 x (3/10) = 1/15, (1 + 2/3 + 1/15)= (1 + 11/15). error 2 = 1/225
3. 1/225 x (15/52) = 1/780, (1 + 11/15 1/780) = 1251/780
B. the lower limit 265/153 modified step 2, used
2. 1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51)
such that (1 + 111/153)changed to (1 + 112/153) = 265/153
Q.E.D.



