A Disquisition on the Square Root of Three - An Archimedean Conundrum

A conundrum is a riddle having only conjectural answers, and a long-standing mathematical conundrum with numerous conjectural answers is centered on how Archimedes, in his Measurement of a Circle, arrived at his famous inequality \[\frac{265}{153}<\sqrt{3}<\frac{1351}{780}.\] In the first pages of [1], Heath offers only what are called "probable steps" that would lead to these inequalities. In [2], Dickson states that the two approximations, " . . . can be explained in connection with \(x^{2}-3y^{2}=-2\), \(x^{2}-3y^{3}=1,\)" but gives no hint that Archimedes proceeded in such a manner.

In [3], Dijksterhuis argues that Archimedes might have used what is called "the Babylonian rule," equivalent to Newton’s Method for the function \(y=x^{2}-3,\) to arrive at \(\frac{1351}{780}\), but that rule doesn't explain the other fraction, except indirectly by means of what seems to be a curious supposition and roundabout calculations. (The Babylonian Rule asserts that in estimating a square root, make a simple but reasonable guess, then average your guess with the radicand divided by your guess to obtain a better estimate, then repeat. In our case, beginning with \(x=1\), and computing the average of \(x\) and \(\frac{3}{x}\) through four iterations yields the sequence of four estimates obtained on page 4.) Heath (again) in [4] seems to mention the inequalities only in a passing manner, and Stein [5] curiously does not mention these famous Archimedean inequalities at all. To see how much attention has been paid to speculation about the above inequalities, visit the website Archimedes and the Square Root of 3.

Recall that Occam's Razor asserts that among competing explanations, the simplest is the most likely to be correct. To that end, notice that the two fractions involved here are contained in rungs nine and twelve of the Greek ladder for \(\sqrt{3}\) on the second page of this paper. That would seem to be the simplest explanation all right, but did Archimedes know of Greek ladders or something equivalent? Did he realize that

if \(\frac{a}{b}\) is an approximation to \(\sqrt{3},\)

then \(\frac{3a+b}{a+b}\) is a better one?

Because this, beginning with \(\frac{1}{1}\) and iterating, yields – exactly – the rungs of the classical Greek ladder for \(\sqrt{3}\). It would seem that knowing of Greek ladders or something equivalent is a reasonable supposition, as is suggested on the second page of [6] in a quote from [2], wherein it is urged that the ancients seemed to know of Greek ladders or something equivalent to them.

On many classroom occasions over the years, I have used Greek ladders to get simple approximations. I have also on occasion described this conundrum about the Archimedean approximations to \(\sqrt{3}\), and students always seem surprised that there can be such current disagreements and open questions among mathematicians about so ancient a topic. This opens the door for more dialog about the open-ended nature of mathematical discovery as well as of our current understanding of the ancient history of mathematics.