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

Author(s):
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].