A Disquisition on the Square Root of Three

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.

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

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