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Topic: Archimedes (and Ahmes) square root of 5, 6 and 7
Replies: 8   Last Post: Jan 16, 2013 6:58 PM

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 Milo Gardner Posts: 1,105 Registered: 12/3/04
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

Date Subject Author
12/18/12 Milo Gardner
12/19/12 Milo Gardner
12/26/12 Milo Gardner
12/26/12 Milo Gardner
12/27/12 Peter Duveen
12/28/12 Milo Gardner
1/14/13 Milo Gardner
1/15/13 Milo Gardner
1/16/13 Milo Gardner