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

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].