# Extracting Square Roots Made Easy: A Little Known Medieval Method - A Simpler Way in Many Cases

Author(s):
Friedrich Katscher (Vienna University of Technology)

In the case of $$\sqrt{5}=\sqrt{2^2+1}$$ with $$a=2$$ and $$r=1,$$ $$\:\frac{2a}{r}=\frac{4}{1} = 4$$ is an integer. Therefore, its reciprocal, $$\frac{r}{2a}=\frac{1}{4},$$ is a unit fraction; that is, a fraction with numerator $$1.$$ For all square roots for which this is true, and only for these, there is a simplification of al-Hassar's method.

First we see that $\left(a\pm \frac{1}{n}\right)^2= a^2\pm\frac{2a}{n}+\frac{1}{n^2}.$ This means that, for both the plus and the minus cases, the excess is $$\frac{1}{n^2},$$ again a unit fraction. This is true for all further approximations. In all of these cases the first approximation is $$a\pm \frac{1}{n}=\frac{an\pm 1}{n}.$$ We call this fraction $$\frac{p}{q}.$$

To get the next approximation according to al-Hassar's rule, the excess $$\frac{1}{n^2}=\frac{1}{q^2}$$ has to be divided by double the preceding approximation, namely $$2\times{\frac{p}{q}}=\frac{2p}{q}.$$ We have $\frac{1}{q^2}\div \frac{2p}{q}=\frac{1}{q^2}\times\frac{q}{2p}=\frac{1}{2pq}.$ This has to be subtracted from $$\frac{p}{q}.$$ But $\frac{p}{q}-\frac{1}{2pq}= {\frac{2p^2-1}{2pq}}.$ And with this simple formula – much easier to use than al-Hassar's – we obtain an increasingly accurate series of approximations. (This formula is not new. It was given in 1766 by the Italian-French mathematician Joseph-Louis Lagrange (1736-1813); see Oeuvres de Lagrange, vol. 1, p. 695.)

Let us apply the formula ${\frac{2p^2-1}{2pq}}$ to get the third approximation of $$\sqrt{5}$$ from al-Hassar's second approximation $$\frac{p}{q}=\frac{161}{72}.$$ We have $2p^2-1= 2\times 161^2-1=2\times 25921-1=51841$ and $2pq=2\times 161\times 72=23184.$ Therefore, the third approximation is $$\frac{51841}{23184},$$ or, in decimals, $$2.2360679779158\dots .$$ Its square is $$5.000000001860\dots .$$ The excess is $$1.860...\times 10^{-9}.$$ This is equal to $$\left(\frac{1}{23184}\right)^2.$$

When we calculate the next approximations with our formula, we find that the excess of the fourth approximation is $$1.730...\times 10^{-19}$$ and of the fifth approximation $$1.497...\times 10^{-39},$$ really excellent results obtained so easily.