# 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.