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Re: Archimedes (and Ahmes) square root of 5, 6 and 7
Posted:
Jan 14, 2013 11:01 PM


Archimedes' square root of 3 problem cited limits
(1271/780)^2 > 3 > 265/153
A. The higher limit 1251/780 was calculated in three steps
1. (1 + 2/3)^2 = error 1
2. 2/9 x (3/10) = 1/15, (1 + 2/3 + 1/15)= (1 + 11/15)
3. 1/15 x (15/52) = 1/52, (1 + 11/15 + 1/52)=
(1 + 573/780) = 1271/780
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)
3. (1 + 111/153)changed to (1 + 112/153) ... good guesses



