# A Disquisition on the Square Root of Three

## Introduction

The Square Root of Three” is the title of at least one poem already, written by David Feinberg and recited by Kumar in the movie "Harold and Kumar Escape from Guantanamo Bay." Inspired by Feinberg's verses, and with unabashed shamelessness, I submit this “more mathematical” doggerel about the same algebraic irrational number.

 Exactly one-half of $$2\pi – e$$ Is about $$3\%$$ more Than the $$\sqrt{3}.$$

But using the transcendental numbers $$\pi$$ and $$e$$ to approximate $$\sqrt{3}$$ is very far afield of what is of interest herein. Far afield indeed, for the topic here concerns the very ancient concept of Diophantine approximations – that is, approximations of irrational numbers by rational numbers – with $$\sqrt{3}$$ as the center of attention. The adjective Diophantine salutes Diophantus of Alexandria (circa 207–291 AD), whose book was entitled Arithmetica.

The first part of the exposition compares three methods of approximating $$\sqrt{3}$$: Greek ladders, continued fractions, and Newton's Method. The second part addresses $$\sqrt{3}$$ as the center of attention in what has become a long-standing disputation that is associated with Archimedes – and, with this paper, we enthusiastically enter that fracas.

## Approximation by Rational Numbers

That the center of attention here is irrational can be seen by being reminded of this argument: If $$\sqrt{3}$$ were rational, it could be written as the reduced fraction $\sqrt{3}=\frac{a}{b}$ for integers $$a$$ and $$b.$$ Then by squaring both sides, the result is the equation $a^{2}=3b^{2}.$ Now if $$a$$ and $$b$$ are each written as their unique product of primes, then the prime $$3$$ occurs on the left side of this equation an even number of times, while on the right side, $$3$$ occurs an odd number of times. Such a situation violates the Unique Factorization Theorem, so the equation just above is impossible, whence $$\sqrt{3}$$ must be irrational.

Note, however, that, to six decimal places, $\sqrt{3}\approx 1.732051=\frac{1732051}{1000000}.$ This is a reduced fraction since $$1732051$$ is a prime. We will now look at three other ways to arrive at fractional approximations that are "equivalent" in that they also have six-place decimal accuracy: first, by use of the classical Greek ladder for $$\sqrt{3};$$ second, by examining the convergents of a continued fraction for $$\sqrt{3};$$ and, third, by using iterates of Newton's Method.

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.

## 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.

## Newton's Method

Using Newton's Method for approximating $$\sqrt{3}$$ (the relevant function being $$y=x^{2}-3$$, whence $$\frac{dy}{dx}=2x),$$ the first estimate will be chosen as $$a_{0}=$$$$\frac{1}{1}$$ so as to make the comparisons fair. Recall that the method for approximating $$\sqrt{3}$$ consists of following each estimate $$a_{n}$$ by $a_{n+1}=a_{n}-\frac{a_{n}^{2}-3}{2a_{n}}.$ So the calculations begin as follows.

$\frac{1}{1}-\frac{\left(\frac{1}{1}\right)^{2}-3}{2\left(\frac{1}{1}\right)}=\frac{2}{1}=2.000000$

$\frac{2}{1}-\frac{\left(\frac{2}{1}\right)^{2}-3}{2\left(\frac{2}{1}\right)}=\frac{7}{4}=1.750000$

$\frac{7}{4}-\frac{\left(\frac{7}{4}\right)^{2}-3}{2\left(\frac{7}{4}\right)}=\frac{97}{56}\approx 1.732143$

$\frac{97}{56}-\frac{\left(\frac{97}{56}\right)^{2}-3}{2\left(\frac{97}{56}\right)}=\frac{18817}{10864}\approx 1.732051$

Notice that the first three of these steps are the 2nd, 4th, and 8th rungs, respectively, of the Greek ladder on page 3, and you can check that if the Greek ladder were extended to sixteen rungs, the result would be the rung $$\langle 10864\quad 18817 \rangle$$.

Such is the "doubling" pattern that was proved in [6]. In any case, the Greek ladder gives six-place accuracy with but a three-digit denominator $$(780)$$, as do the continued fractions, while Newton's Method requires the five-place denominator $$10864.$$ Moreover, the classical Greek ladder and the continued fraction calculations on pages 3 and 4 use only simple arithmetic, while Newton's Method requires calculating derivatives.

## 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.

## About the Author / References

Robert "Bob" Wisner is Professor Emeritus of Mathematics at New Mexico State University. He was founding editor of SIAM Review, a publication of the Society for Industrial and Applied Mathematics. He was the first full-time Executive Director of the MAA's Committee on the Undergraduate Program in Mathematics (CUPM), 1960-1963. Bob also authored or coauthored numerous K-12 textbooks for Scott, Foresman; was Consulting Editor in Mathematics for Brooks/Cole for over 25 years; coauthored a liberal arts mathematics textbook; and recently coauthored a series of interactive calculus, business calculus, pre-calculus, and AP calculus textbooks, available on CDs from Hardy Calculus.

References

1.       Archimedes, The Works of Archimedes, Edited by T. L. Heath, Dover Publications, Mineola, New York, 2002, pp. lxxx - lxxxi, also pp. 94 and 96; originally published in 1897 by Cambridge University Press, second edition including The Method originally published in 1912.

2.       Leonard Eugene Dickson, History of the Theory of Numbers, vol. II: Diophantine Analysis, Dover Publications, Mineola, New York, 2005, p. 342; originally published in 1919-1923 by Chelsea Publishing, New York.

3.       E. J. Dijksterhuis, Archimedes, Princeton University Press, 1987, pp. 234-236; originally published in 1938.

4.       Thomas L. Heath, Diophantus Of Alexandria: A Study in the History of Greek Algebra, Cambridge University Press, 1910, p. 278; originally published in 1885.

5.       Sherman Stein, Archimedes: What Did He Do Besides Cry Eureka?, Mathematical Association of America, Washington, D.C., 1999.

6.       Robert J. Wisner, “The Classic Greek Ladder and Newton's Method,” Loci: Convergence (August 2009), DOI: 10.4169/loci003330, http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3330