Author(s):
Robert J. Wisner (New Mexico State University)
The classical Greek ladder for \(\sqrt{3}\) to sixplace 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 sixplace accuracy that was presented at the outset of this paper. The sevenplace denominator of \(1000000\) has been beaten by the threeplace \(780\) – quite an improvement.
Robert J. Wisner (New Mexico State University), "A Disquisition on the Square Root of Three  The Classical Greek Ladder," Convergence (December 2010), DOI:10.4169/loci003514