What is so mysterious about Archimedes' approximation of $\sqrt 3$?

Question

In his famous estimation of $\pi$ by inscribed and circumscribed polygons, Archimedes uses several rational approximations of irrational values; a typical example is that he states, without explanation, $$\sqrt 3 > \frac{265}{153}.$$

The sudden appearance of $\frac{265}{153}$ has puzzled many people through the ages. The page Archimedes and the Square Root of 3 lists several such. This example, attributed to W.W. Rouse Ball, is typical:

It would seem...that [Archimedes] had some (at present unknown) method of extracting the square root of numbers approximately.

To my mind, there is a very easy answer to this seeming puzzle. If you want rational approximations to $\sqrt 3$, the very simplest thing you can possibly do is as follows: You want to find integers $a$ and $b$ whose ratio is close to $\sqrt 3$, or equivalently you want to find $a^2$ and $b^2$ whose ratio is close to $3$. So you should tabulate $n^2$ and $3n^2$ and look for integers in the left column that are approximately equal to integers in the right column:

$$\begin{array}{rr} n^2 & 3n^2 \\ \hline 1 & \color{darkgreen}{3} \\ \color{darkgreen}{4} & 12 \\ 9 & \color{darkred}{27} \\ 16 & \color{darkblue}{48} \\ \color{darkred}{25} & 75 \\ 36 & 108 \\ \color{darkblue}{49} & 147 \\ 64 & 192 \\ 81 & 243 \\ 100 & 300 \\ 121 & \color{purple}{363} \\ \vdots & \vdots \\ \color{purple}{361} & 1183 \\ \vdots & \vdots \\ \end{array} $$

The pairs of colored entries give the approximations $\color{green}{\frac21}, \color{maroon}{\frac53}, \color{darkblue}{\frac74}, $ and $\color{purple}{\frac{19}{11}}$ respectively. If you carry the table to 265 entries, you find that $265^2 = 70225$ and $3\cdot 153^2 = 70227$ and there is your $\sqrt 3 \approx \frac{265}{153}$. You don't have to be as clever as Archimedes to think of this. No magical technique was required, no sophisticated theory was required. The calculation is tedious but straightforward and I judge that it could have been carried to 265 entries in at most a few hours, even using whatever awful technology was available at the time.

Archimedes might, of course, have used a better method. (He also produces the approximation $\frac{1351}{780}$, for which the foregoing is not obviously practical. He might have deployed the so-called “Babylonian method”. Or, having tabulated the first few approximations from the table above, he might have noticed the simple recurrence that governs them; this would not be hard for a bright high-school student. Or he might have had a some other method.) But it seems to me that since there is a straightforward and simple method by which Archimedes could have produced the required approximations, there is no puzzle to solve.

However, Rouse Ball and a number of other eminent authors disagree with me, and consider Archimedes' calculation of $\sqrt 3\approx \frac{265}{153}$ to be mysterious. (The page I linked earlier cites: T. Heath, E.T. Bell, C.B. Boyer, M. Kline, P. Beckmann, Sondheimer and Rogerson, and goes on to describe a complicated and elaborate method that Archimedes could have used.)

If it was only popular mathematics authors who thought this calculation required mysterious powers on the part of Archimedes, I would write it off as the authors overlooking the obvious, because the authors of popular books on mathematics can be quite obtuse. But Rouse Ball and Morris Kline are not dummies and I do not quite believe that they could have overlooked the method I described above, which seems not merely obvious but literally the first thing anyone would try.

My question is:

Is there some reason I don't appreciate that Archimedes could not have or certainly did not use the method I described? Or did all those other guys overlook the obvious?

Answer 1

According to the paper of Davies cited by MJD, Archimedes actually gives a double inequality $$\frac{265}{153}<\sqrt{3}<\frac{1351}{780}.$$

As both of these fractions are not just random approximations, but are the 9-th and 12-th convergents for the continued fraction expansion of $\sqrt{3}$, there is no doubt that Archimedes used the continued fraction expansion or some equivalent algorithm, for example Pell's equation. (Otherwise it cannot be explained why Archimedes has chosen these two fractions of infinitely many other fractions aproximating $\sqrt{3}$.

Continued fraction algorithm was known to Euclid in some form (book 10). Davis's arguments against it are totally unconvincing. And his conjecture about Archimedes computation by the "averaging" algorithm that he proposes is not plausible, because it does not explain the appearance of exactly these fractions.

Davies's argument that the first published theory of Pell's equation appeared in 18-th century is also not convincing. Indeed, there is a well known text which is attributed to Archimedes, called "Archimedes Cattle problem" which shows that the author's familiarity with very complicated Pell's equations:-)

It is absolutely clear that for the author of the Cattle problem, the Pell equation $m^2-3n^2=\pm1$ was a trivial example.

In general, there are indications that Hellenistic Greeks knew much more of number theory than is contained in their work that survived. For example, it was recently discovered that Hipparchus knew the so-called Schroder numbers (rediscovered in 19-th century).

EDIT. I just found a text whose author shares my opinion and elaborates on the question. So I can add a reference:

http://mathpages.com/home/kmath038/kmath038.htm

(I do not know who this author is, but his web pages covering a variety of topics in physics, mathematics and history are so good that I value his opinion.)

EDIT2. In the paper by D. H. Fowler, Ratio in early Greek mathematics (Bull. AMS, 1(1979) 6, 807-846) the author explains that the continued fraction algorithm was called in Greek anthyphairesis, and that it pre-dates Euclid's Elements. He sites Archimedes' approximation of $\sqrt{3}$ as an example of the use of this algorithm.

Answer 2

Archimedes might, of course, have used a better method. (He also produces the approximation $\frac{1351}{780}$, for which the foregoing is not obviously practical...)

I disagree on the foregoing not being practical: it's a trivial computation compared to some which were carried out by hand by later mathematicians. But I think that really is the nub: any explanation for how Archimedes produced the approximation $\frac{265}{153} < \sqrt{3} < \frac{1351}{780}$ should also explain why he didn't produce the approximation $\frac{989}{571} < \sqrt{3} < \frac{1351}{780}$. Here both your approach and the continued fraction approach fall down.

There is also the point that this tabulation method is rather ungeometrical, and so not really in the expected style. I suspect that Archimedes would have preferred not to publish a result than to face the embarrassment of explaining that it was calculated thus when someone asked him for the proof.

Answer 3

There's a very simple geometrical response that provides a rounded solution which might reflect on this problem at http://www.gjbath.com/SQR3.htm. It's so much easier seeing it!

This involves circumscribing an equilateral triangle and taking a diameter from the apex. Then, applying $26/15$ as an approximation to $\sqrt 3$ ($\tan 30^\circ\approx 15/26$) and reiterating in the similar triangle on the base gives $15/(225/26)$. Thence, multiplying out by 26, and using the fact that the diameter is divided by the base in the ratio 3:1, and then combining, results in a further approximation of $$\frac{3\cdot225+26\cdot26}{15\cdot2\cdot26}=\frac{1351}{780}.$$

Thereafter, subtracting the original approximation's numerator and denominator $26/15$ gives $$\frac{1351-26}{780-15}=\frac{1325}{765}=\frac{265}{153}$$.

This is such straightforward geometry and reasoning that you wouldn't expect any comment from a Classical geometer. And it doesn't involve Pell!

Answer 4

Some information can be found here 1

Using the secant method on the parabola $y = x^2$, a new estimate $x_{n+m}$ for $\sqrt{3}$ can be obtained from $x_n$ and $x_m$ as $$x_{n+m} = \frac{3 + x_n x_m}{x_n + x_m}$$ where if we start with $x_1 = \frac{5}{3}$ then $x_{2n} > \sqrt{3}$ and $x_{2n+1} < \sqrt{3}$.
We therefore obtain $$x_1 < x_3 < \sqrt{3} < x_4 < x_2$$ Where $$x_1 = \frac{5}{3}$$ $$x_2 = \frac{3 + x_1 x_1}{x_1 + x_1} = \frac{26}{15}$$ $$x_3 = \frac{3 + x_2 x1}{x_2 + x_1} = \frac{265}{153}$$ $$x_4 = \frac{3 + x_3 x_1}{x_3 + x_1} = \frac{3 + x_2 x_2}{x_2 + x_2} = \frac{1351}{780}$$

Answer 5

We don't know how Archimedes compute square root of 3 not because there is no known method, but because there are several methods that Archimedes could use, and we don't know which one he actually used.

Answer 6

There is a very simple geometric approach to continued fraction that yields the expected approximation:

For the minoring sequence : You alternatively stack a double square above and a square to the right.

For the majoring sequence you start with a quare with a 3:1 rectangle (that creates a 3:4 rectangle) then, as above, You alternatively stack a double square above and a square to the right.