Extracting Square Roots Made Easy: A Little Known Medieval Method - Al-Hassar’s Description of the Method

Imagine you live in a time with no computers. If you want to approximate the square root of a non-square number \(N,\) there is a logical way to do it: Look for the square number \(a^2\) nearest to \( N.\) If \(N\) is greater than \(a^2,\) then \(N=a^2+r\) (where we use \( r\) for remainder); if \(N\) is smaller than \(a^2,\) then (using \( r\) again) \(N=a^2-r.\) Now think of the formula \(\left(a \pm b\right)^2=a^2\pm 2ab+b^2,\) and assume that \(b\) is so small that you may neglect its square \(b^2.\) This means \(r\approx 2ab\) and, from this, \(b\approx \frac{r}{2a}.\)

Thus the first approximation of the square root of \(N\) is \[\sqrt{N}=\sqrt{a^2 \pm r}\approx a\pm \frac{r}{2a}.\] Take, for example, \(N=3=2^2-1. \) That is \(a=2,\: r=1,\) and \[\sqrt{3}\approx 2-\frac{1}{4}=1\frac{3}{4}=1.75,\] a very good approximation because \[\left(1\frac{3}{4}\right)^2=\left(\frac{7}{4}\right)^2=\frac{49}{16}=3\frac{1}{16},\] with an excess of only \(\frac{1}{16}\) or \( 0.0625.\)

This method (but not, of course, the symbolic formula) was used in Greek antiquity, especially by the first century A.D. mathematician Heron of Alexandria. It was more accurate the nearer \(N\) was to \(a^2\); that is, the smaller \(r\) was. The square of \(a \pm \frac{r}{2a}\) is \(a^2 \pm r + \left(\frac{r}{2a}\right)^2.\) This means that, for both positive and negative \(r,\) there is always a surplus or excess of \(\left(\frac{r}{2a}\right)^2.\)

Of course, some ancient Greek mathematicians were able to calculate better approximations. Archimedes (287-212 B.C.), for example, used \(\sqrt{3}\approx 1\frac{571}{780}=\frac{1351}{780},\) or, in decimal notation, \(1.7320512\dots.\) The real value of \(\sqrt{3}\) is \(1.7320508\dots.\) (Because decimal fractions began spreading only at the beginning of the 17th century, square roots of non-square numbers were represented only in the form of common fractions prior to that time.)

But the Greeks had no general method to improve the accuracy of the above approximations. We find the earliest known description of such a procedure in the five pages of the “Chapter on the extraction of irrational roots by approximation” of al-Hassar’s book (beginning on the reverse of leaf 125 of the Gotha copy). It is typical that al-Hassar does not give general instructions, rules, or formulas, but only numerical examples that show how to proceed. And he does not derive his method or give a general proof, but shows only that his method gives the right answers in his examples.

Al-Hassar knew the Greek procedure for both the positive and the negative case, and he calculated the example \(\sqrt{5}=\sqrt{2^2+1}\approx 2\frac{1}{4}.\) His “Chapter on the extraction of irrational roots by approximation” begins with the words (English translation of Suter's German translation):

When it is said: which is the square root of \(5,\) so take the next square number to \(5,\) this is equal to \(4,\) subtract it from \(5,\) the remainder is \(1,\) divide this by \(4,\) that gives \(\frac{1}{4},\) and add this to the root of \(4,\) which is equal to \(2.\) That gives \(2\frac{1}{4}.\) And this is the approximate root of \(5.\) Namely, when you multiply \(2\frac{1}{4} \: \left[=\frac{9}{4} \right]\) by itself you get \(\left[\frac{81}{16}=\right]\: 5\frac{1}{16}.\) The deviation is an excess of \(\frac{1}{16}.\)

(Later al-Hassar obtained also \(\sqrt{10}=\sqrt{3^2+1}\approx 3 \frac{1}{6}\) with the square \(10\frac{1}{36}\) – an excess of \(\frac{1}{36}.\)) And now comes the important instruction:

But if you want a closer approximation, then double \(2\frac{1}{4},\) that gives \(4\frac{1}{2}.\) Divide \(\frac{1}{16}\) by \(4\frac{1}{2}.\) This gives \(\frac{1}{72}\) (an eighth of a ninth). Subtract this from \(2\frac{1}{4}\:\left[=2\frac{18}{72}\right].\) This leaves a remainder of \(2\frac{17}{72}\: \left[=\frac{161}{72}\right].\) When you multiply this by itself you get \(\left[\frac{25921}{5184}=\right]\: 5\frac{1}{5184},\) and this is nearer [to \(5\)] than \(5\frac{1}{16}.\)

This procedure, expressed in modern notation, is \[ \sqrt{a^2+r}\approx\left(a+\frac{r}{2a}\right)-\frac{\left(\frac{r}{2a}\right)^2}{2\left(a+\frac{r}{2a}\right)}.\]

Al-Hassar continues:

If you want a closer approximation yet, double \(2\frac{17}{72},\) divide \(\left(\frac{1}{72}\right)^2\: \left[=\frac{1}{5184}\right]\) by the result, and subtract what you get from \(2\frac{17}{72}.\) So the result will be an even closer approximation than the first and second roots. You may continue like this as far as you want.

In other words, apply this method several times. Of course, in the process the numerator and the denominator of the fractions get longer and longer, and the multiplications become more and more unwieldy. Today's calculator or computer spares us the multidigit calculations that had to be done by hand in the past.

A warning about calculating the first approximation: In both the positive case \(N=a^2+r\) and the negative case \(N=a^2-r,\) the \(r\) must not exceed \(a.\) Therefore, \(\sqrt{14}\) must not be calculated by \(\sqrt{3^2+5},\) but only by \(\sqrt{4^2-2}.\) With respect to the square root of \(20,\) al-Hassar remarked that you may take either \(16\) (positive \(r\)) or \(25\) (negative \(r\)) as the nearest square, that is, \(\sqrt{4^2+4}\) or \(\sqrt{5^2-5}.\) In both cases you get the same first approximation, namely \(4+\frac{1}{2}=5-\frac{1}{2}=4\frac{1}{2}.\) This is true for all numbers of the form \( a^2+a=(a+1)^2-(a+1)=a(a+1)\); that is, for \(2, 6, 12, 20, 30, 42, 56, \) and so on. The first approximation is always \(a+\frac{1}{2}.\)