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Re: Archimedes square root of 3
Posted:
Feb 4, 2013 6:29 PM


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



