Date: Feb 4, 2013 6:29 PM
Author: Milo Gardner
Subject: Re: Archimedes square root of 3

corrected by:

E.J. Dijkerhuis "Archimedes" (1987) page 435 discussed

(27^1/2)/3 in terms of a continuing fraction series

cited 3^1/2, 26/15, 265/153, 1351/780

followed by

"... what a more satisfactory solution of the question could possibly look like like?"

this method solved (27^1/2)/3 in two steps

1. [(5 + 1/5)^]2/3 = (27 + 1/25)/3

2. [(5 + 1/5 + 1/25) x 52/5)]/3 = (1/260/3

such that

(5 + 1/5 + 1/25 - 1/260)/3 = 1351/780

cited in ...

http://planetmath.org/encyclopedia/EgyptianAndGreekSquareRoot.html