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

Author(s):
Friedrich Katscher (Vienna Technological University)

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 $$2\left (\frac{p}{q}\right)=\frac{2p}{q}$$. 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 $$2pq=2\times 161\times72=23184.$$ Therefore, the third approximation is $$\frac{51841}{23184}$$, or, in decimals, 2.2360679779158\dots\). Its square is $$5.000000001860\dots$$. The excess is $$1.860...\times10^{-9}$$. 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 $$1.730...\times10^{-19}$$ and of the fifth approximation $$1.497...\times10^{-39}$$, really excellent results obtained so easily.

Friedrich Katscher (Vienna Technological University), "Extracting Square Roots Made Easy: A Little Known Medieval Method - A Simpler Way in Many Cases," Loci (May 2010), DOI:10.4169/loci003488