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

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.