My Math Forum > Math Archimedes’ Approximation to SQRT3

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 April 12th, 2016, 07:54 PM #1 Newbie   Joined: Apr 2016 From: uk Posts: 3 Thanks: 0 Archimedes’ Approximation to SQRT3 Archimedes states that 1351/780 > SQRT3 > 265/153, but doesn’t say how he derived this. It struck me that Archimedes may have used geometry to get there, and I came up with a possible solution at Archimedes and the Square Root of Three, A Geometrical Approach. This suggests a general procedure for refining square roots, but I’m puzzled about how I might derive the starting approximation SQRT3=26/15 without knowing the general procedure. Because if I know the general procedure, I can eventually work out, by number crunching, that 1351/780 > SQRT3 > 265/153 simply by starting from SQRT3=7/4, derived from the Babylonian method of approximating square roots. Please, does anyone know of an ancient method of deriving SQRT3 that generates 26/15 or 45/26?
 April 13th, 2016, 03:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,484 Thanks: 2041 I don't know how Archimedes did it. Below is a more recent method, but possibly still fairly old. Consider the sequences (7, 26, 97, 362, 1351, ...) and (4, 15, 56, 209, 780, ...). To generate further terms, multiply the last term by 4, then subtract the previous term. The ratios of corresponding terms approximate √3, and they include 26/15 and 1351/780. Note that 45/26 = 97/26 - 2.

### archimedes square 3 1351

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