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Sat, 25 Mar 2006
Achimedes and the square root of 3
Throughout this proof, Archimedes uses several rational approximations to various square roots. Nowhere does he say how he got those approximations—they are simply stated without any explanation—so how he came up with some of these is anybody's guess.It's a bit strange that Dr. Lindsey seems to find this mysterious, because I think there's only one way to do it, and it's really easy to find, so long as you ask the question "how would Archimedes go about calculating rational approximations to √3", rather than "where the heck did 265/153 come from?" It's like one of those pencil mazes they print in the Sunday kids' section of the newspaper: it looks complicated, but if you work it in the right direction, it's trivial. Suppose you are a mathematician and you do not have a pocket calculator. You are sure to need some rational approximations to √3 somewhere along the line. So you should invest some time and effort into calculating some that you can store in the cupboard for when you need them. How can you do that? You want to find pairs of integers a and b with a/b ≈ √3. Or, equivalently, you want a and b with a^{2} ≈ 3b^{2}. But such pairs are easy to find: Simply make a list of perfect squares 1 4 9 16 25 36 49..., and their triples 3 12 27 48 75 108 147..., and look for numbers in one list that are close to numbers in the other list. 2^{2} is close to 3·1^{2}, so √3 ≈ 2/1. 7^{2} is close to 3·4^{2}, so √3 ≈ 7/4. 19^{2} is close to 3·11^{2}, so √3 ≈ 19/11. 97^{2} is close to 3·56^{2}, so √3 ≈ 97/56. Even without the benefits of Hindu-Arabic numerals, this is not a very difficult or time-consuming calculation. You can carry out the tabulation to a couple of hundred entries in a few hours, and if you do you will find that 265^{2} = 70225, and 3·153^{2} is 70227, so that √3 ≈ 265/153. Once you understand this, it's clear why Archimedes did not explain himself. By saying that √3 was approximately 265/153, had had exhausted the topic. By saying so, you are asserting no more and no less than that 3·153^{2} ≈ 265^{2}; if the reader is puzzled, all they have to do is spend a minute carrying out the multiplication to see that you are right. The only interesting point that remains is how you found those two integers in the first place, but that's not part of Archimedes' topic, and it's pretty obvious anyway. [ Addendum 20090122: Dr. Lindsey was far from the only person to have been puzzled by this. More here. ] In my article about the peculiarity of π, I briefly mentioned continued fractions, saying that if you truncate the continued fraction representation of a number, you get a rational number that is, in a certain sense, one of the best possible rational approximations to the original number. I'll eventually explain this in detail; in the meantime, I just want to point out that 265/153 is one of these best-possible approximations; the mathematics jargon is that 265/153 is one of the "convergents" of √3. The approximation of √n by rationals leads one naturally to the so-called "Pell's equation", which asks for integer solutions to ax^{2} - by^{2} = ±1; these turn out to be closely related to the convergents of √(a/b). So even if you know nothing about continued fractions or convergents, you can find good approximations to surds. Here's a method that I learned long ago from Patrick X. Gallagher of Columbia University. For concreteness, let's suppose we want an approximation to √3. We start by finding a solution of Pell's equation. As noted above, we can do this just by tabulating the squares. Deeper theory (involving the continued fractions again) guarantees that there is a solution. Pick one; let's say we have settled on 7 and 4, for which 7^{2} ≈ 3·4^{2}. Then write √3 = √(48/16) = √(49/16·48/49) = 7/4·√(48/49). 48/49 is close to 1, and basic algebra tells us that √(1-ε) ≈ 1 - ε/2 when ε is small. So √3 ≈ 7/4 · (1 - 1/98). 7/4 is 1.75, but since we are multiplying by (1 - 1/98), the true approximation is about 1% less than this, or 1.7325. Which is very close—off by only about one part in 4000. Considering the very small amount of work we put in, this is pretty darn good. For a better approximation, choose a larger solution to Pell's equation. More generally, Gallagher's method for approximating √n is: Find integers a and b for which a^{2} ±1 = nb^{2}; such integers are guaranteed to exist unless n is a perfect square. Then write √n = √(nb^{2} / b^{2}) = √((a^{2} ± 1) / b^{2}) = √(a^{2}/b^{2} · (a^{2} ± 1)/a^{2}) = a / b · √((a^{2} ± 1) / a^{2}) = a/b · √(1 ± 1/a^{2}) ≈ a/b · (1 ± 1 / 2a^{2}). Who was Pell? Pell was nobody in particular, and "Pell's equation" is a complete misnomer. The problem was (in Europe) first studied and solved by Lord William Brouncker, who, among other things, was the founder and the first president of the Royal Society. The name "Pell's equation" was attached to the problem by Leonhard Euler, who got Pell and Brouncker confused—Pell wrote up and published an account of the work of Brouncker and John Wallis on the problem.
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