# Continued fractions and the square root of 3.

I’m a Big Fan of both @standupmaths and @sparksmaths, two mathematicians who fight the good fight.

I was interested to see Ben tackling the square root of 3 using the ‘long division’ method. It’s a method I’ve tried hard to love. It’s a method I just can’t bring myself to do or recommend. @colinthemathmo has an explainer here, probably as nice an explainer as the method permits.

I have two alternative methods for finding the square root of three.

The first, Ben alludes to in the video: use the binomial expansion. My instinct is to find a number $3k^2$ (where $k$ is an integer) that’s fairly close to an even power of 10 - for example, $108 = 3(6^2)$.

Now you can apply the binomial expansion: $6\sqrt{3} = \left( 10^2 +8\right)^{1/2} = 10 + \frac{1}{2}\left(\frac{1}{10}\right)(8) + \frac{-1}{8}\left(\frac{1}{1000}\right)(64) + \dots$. This will converge pretty quickly, at least one decimal digit per term, but the divisions quickly become cumbersome and if you’re going to use fractions continually, you may as well…

… use continued fractions. Or a matrix version thereof.

Like Ben, I’m going to skip the derivation and jump straight to the method: the key matrix is $\mathbf {M}_n = \mattwotwo{1}{1}{2}{3}^n$ - for a large value of $n$.

Applying this matrix to $\mattwotwo{0}{1}{1}{1}$ gives a two-by-two matrix in which the bottom-right element divided by the bottom-left one is a good approximation to $\sqrt{3}$. Perhaps more simply, this is just the sum of the bottom row of $\bb M$ divided by its bottom right element.

And we can be clever about calculating $\mathbf M_n$: repeatedly squaring the matrix gives us answers that approach $\sqrt{3}$ quite rapidly - although the multiplications can get big quite quickly.

The approximation from $\bb M_1$ is $\frac{5}{3}$, which is not a bad starting point.

From $\bb M_2$, it’s $\frac{19}{11} \approx 1.727$, which is not at all far from the true answer.

Squaring $\bb M_2$ gives $\bb M_4$, and the approximation from there is $\frac{265}{153}$ - the numbers are already getting big, but we’re within $2.5\times 10^{-5}$ of the true answer.

Squaring $\bb M_4$ gives $\bb M_8$, where numbers are in the ten-thousands; the estimate is now $\frac{51409}{29681}$, which is around $6\times10^{-10}$ away from $\sqrt{3}$.

The operations here are not *trivial* ((hush, sensei)), but they’re more tedious than difficult. Multiplying and adding large numbers is typically easier than long division – and in fact, multiplying repeatedly by $\bb M_1$ is not all that hard, even with big numbers (I believe each extra matrix gives roughly one decimal place of accuracy).

You’re left with one big division to do at the end, which *is* a bit more difficult than the other sums – but it’s just one division!

Now, Sir Isaac may not have had continued fractions available to him - but the maths involved here is certainly achievable on quill and parchment. I’m definitely not saying this is the method Ben should have used in the video (I mean, the whole point was to do it an absurd and 17th-century way), but figured it was a nice method to share.