Ancient Indian Rope Geometry in the Classroom - Approximating the Square Root of 2

Cynthia J. Huffman (Pittsburg State University) and Scott V. Thuong (Pittsburg State University)

The Śulba-sūtras of Āpastamba and Kātyāyana contain an ingenious approximation of the square root of \(2\) by a rational number, \[\sqrt{2}\approx 1+\frac{1}{3}+\frac{1}{3\cdot 4}-\frac{1}{3\cdot4\cdot34}=1+\frac{1}{3}+\frac{1}{12}-\frac{1}{408}=1.41422\dots.\] Indeed, the relative error in this approximation is less than 0.0003%! The original text in the Śulba-sūtras of Kātyāyana (Section 2.9) may be restated as [Joseph, p. 334]:

Increase the measure by its third and this third by its own fourth, less the thirty-fourth part of that fourth. This is the value with a special quantity in excess.

Note that this passage demonstrates knowledge that the approximation is a slight overestimate of the square root of \(2.\) However, no explanation of this approximation appears in the Śulba-sūtras. We will present the rather plausible explanation found in [Joseph, p. 334-336], originally due to Bibhutibhusan Datta in 1932 [Datta].

Begin with two unit squares, \(SPQR,\) and \(ADCB,\) as shown in the GeoGebra applet in Figure 9 below. The square with area \(2\) will of course have side length equal to the square root of \(2.\) The square \(SPQR\) is divided into two rectangles of dimension \(1\) by \(1/3,\) one square of dimension \(1/3\) by \(1/3,\) and \(8\) rectangles of dimension \(1/{12}\) by \(1/3.\) These rectangles are then arranged around \(ADCB,\) so that they fit into square \(GAEF,\) which has side lengths \[1+\frac{1}{3}+\frac{1}{12} = \frac{17}{12}.\] Notice that these rectangles do not fill up \(GAEF\) completely; that is, the red shaded rectangle in the upper right hand corner is not filled with a piece of \(ADCB.\) Therefore, the area of \(GAEF\) is precisely \(2\) plus the area of the red shaded rectangle, which is \(1/{144},\) being a square of side length \(1/{12}.\)

Figure 9. This applet outlines a preliminary approximation of the square root of \(2\) as the rational number \(17/12.\) Click "Go" to demonstrate that a square with side length \(17/12\) has area slightly greater than \(2.\) The overestimate of the area, shaded in red in the final step, is \(1/144.\) Therefore, \(17/12\) is an overestimate of the square root of \(2.\)

Indeed, square \(GAEF\) is the basis for our approximation, for its area is slightly greater than \(2,\) and, therefore, its side length of \[1+\frac{1}{3}+\frac{1}{12} = \frac{17}{12}\] is slightly greater than the square root of \(2.\) Our goal is to modify \(GAEF,\) so that its area is as close to \(2\) as possible. To this end, imagine cutting off two rectangular strips, of equal thickness, from the left and bottom edges of \(GAEF.\) This forms the “L” shaped region \(GAELNM,\) shown in the GeoGebra applet in Figure 10 below. We would like the area of \(GAELNM\) to be \(1/144,\) for then the area of square \(MNLF\) would be precisely \(2,\) and thus its side lengths would be precisely the square root of \(2.\)

Figure 10. This applet demonstrates the procedure described in the article for calculating just by how much \(17/12\) overestimates the square root of \(2.\) The key idea is to shave off an "L" shaped region of area \(1/144\) from the square with side length \(17/12.\)

Let \(X\) denote the “thickness” of region \(GAELNM.\) In the GeoGebra applet in Figure 11 (zoomed in around the lower left corner of \(GAEF),\) this is the length of segment \(HI.\)

Figure 11. After zooming in to the lower left corner of square, this applet demonstrates that the desired thickness of the "L" shaped region is approximately \(1/408.\) As a result, the square root of \(2\) is approximately \(17/12-1/408.\)

The area of region \(GAELNM,\) in terms of \(X,\) is \[2\left(\frac{17X}{12}\right)-X^2.\]

To see this, observe that \(GAELNM\) is the union of two rectangular strips of dimensions \({17}/{12}\) by \(X,\) with intersection a square of side length \(X.\) We therefore add the areas of the rectangular strips and subtract the area of the overlap \(X^2.\) Setting this equal to \(1/{144},\) ignoring the small term of \(X^2,\) and solving for \(X\) we obtain:

\[2\left(\frac{17X}{12}\right)-X^2\approx 2\left(\frac{17X}{12}\right)=\frac{1}{144},\] so that

\[X = \frac{1}{408} = \frac{1}{3\cdot4\cdot34}=0.0245\]

Therefore, the side length of a square of area \(2\) – that is, the square root of \(2\) – is approximately \[1 + \frac{1}{3} + \frac{1}{3\cdot 4} - \frac{1}{3\cdot4\cdot34}.\]

This is an overestimate of the square root of \(2,\) because we ignored the small, but positive, term \(X^2,\) when solving for \(X.\)

We caution the reader that knowledge of fractions, and operations with fractions, which we have used above, were most likely not known to the ancient Indians. In 2006, a reconstruction of the approximation of \(\sqrt{2}\) using only manipulation of measuring rope was discovered, which the reader may find in the article published that year by Satyanad Kichenassamy [Kichenassamy].