# 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," Convergence (May 2010), DOI:10.4169/loci003488