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A Disquisition on the Square Root of Three - The Classical Greek Ladder

Author(s): 
Robert J. Wisner (New Mexico State University)

The classical Greek ladder for \(\sqrt{3}\) to six-place accuracy begins like this:

\(1\) \(1\) \(\frac{1}{1}=1.000000\)
\(1\) \(2\) \(\frac{2}{1}=2.000000\)
\(3\) \(5\) \(\frac{5}{3}\approx 1.666667\)
\(4\) \(7\) \(\frac{7}{4}=1.750000\)
\(11\) \(19\) \(\frac{19}{11}\approx 1.272727\) where each rung \(\langle a\quad b\rangle\) is
\(15\) \(26\) \(\frac{26}{15}\approx 1.733333\) followed by \(\langle a+b\quad 3a+b\rangle,\)
\(41\) \(71\) \(\frac{71}{41}\approx 1.731707\) written in reduced form,
\(56\) \(97\) \(\frac{97}{56}\approx 1.732143\) with \(\sqrt{3}\) approximated by \(\frac{b}{a}.\)
\(153\) \(265\) \(\frac{265}{153}\approx 1.732026\)
\(209\) \(362\) \(\frac{362}{209}\approx 1.732057\)
\(571\) \(989\) \(\frac{989}{571}\approx 1.732049\)
\(780\) \(1351\) \(\frac{1351}{780}\approx 1.732051\)

While the ladder could begin with any pair of nonnegative integers, not both zero, the rung \(\left\langle 1\quad 1\right\rangle\) was used here because it yields the “classical” Greek ladder. The ladder stops where it did because that's where it yields the six-place accuracy that was presented at the outset of this paper. The seven-place denominator of \(1000000\) has been beaten by the three-place \(780\) – quite an improvement.

Dummy View - NOT TO BE DELETED