A Disquisition on the Square Root of Three - Continued Fractions

A "standard" sequence of continued fractions for approximating \(\sqrt{3}\) follows.

\[1+\frac{2}{2}=2=2.000000\]

\[1+\frac{2}{2+\frac{2}{2}}=\frac{5}{3}\approx 1.666667\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}=\frac{7}{4}=1.750000\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}=\frac{19}{11}\approx 1.727273\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}=\frac{26}{15}\approx 1.733333\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}=\frac{71}{41}\approx 1.731707\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}}=\frac{97}{56}\approx 1.732143\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}}}=\frac{265}{153}\approx 1.732026\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}}}}=\frac{362}{209}\approx 1.732057\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}}}}}=\frac{989}{571}\approx 1.732049\]

\[1+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}{2}}}}}}}}}}}=\frac{1351}{780}\approx 1.732051\]

These approximations give the same fractions as the Greek ladder table. Thus, it seems fair to declare that the continued fraction method of estimating \(\sqrt 3\) ties the Greek ladder method.