**Square Root of 2, by Baudayana**

L V Nagarajan

Baudhayana is a great mathematician of ancient India estimated to have lived during 800BC. He was an expert mathematician, architect, astronomer and a Hindu high priest. He has proposed several mathematical formulas (Sulva Sutras), some of them with proofs. His statement and proof of the so-called Pythagoras theorem is so simple and elegant.

In ancient times, a Square was held as an important geometrical figure. Every area was expressed in so many squares. There was considerable interest in finding an equivalent square for every area, including circle, rectangle, triangle etc.

Baudhāyana, gives the length of the diagonal of a square in terms of its sides, which is equivalent to a formula for the square root of 2:

**samasya dvikaraṇī. pramāṇaṃ tṛtīyena vardhayet**

** tac caturthenātmacatustriṃśonena saviśeṣaḥ**

*Sama* – Square; *Dvikarani* – Diagonal (dividing the square into two), or Root of Two

*Pramanam* – Unit measure; *tṛtīyena vardhayet* – increased by a third

*Tat caturtena (vardhayet)* – that itself increased by a fourth, atma – itself;

*Caturtrimsah savisesah* – is in excess by 34^{th} part

In English syntax, it will read as below:

**The diagonal of a square of unit measure (is given by) increasing the unit measure by a third and that again by a fourth (of the previous amount). This by itself is in excess by a 34**^{th} part (of the previous amount).

That is,

√2 = 1 + 1/3 + ¼ (1/3) = 17/12

But as per the sulba-sutra above, this in excess by **a 34**^{th} part of the previous amount.

Hence

√2 = 1 + 1/3 + ¼ (1/3) – 1/34[(¼ (1/3)]

= 1 + 1/3 + 1/(3*4) + 1/(3*4*34) = 577/408

The above value is correct to five decimals.

There have been several explanations as to how this formula was evolved. Apparently, initially (even on works of late 6^{th} century AD) an approximate value of 17/12 was used for √2, which is nothing but [1+ 1/3 + 1/(3*4)].

**1.** One theory is they just used actual measurement by ropes to arrive at these fractions – they first tried unit rope length and then 1/2 of the same length. As it was too long they next tried 1/3. It was just short and hence ¼(1/3) was added to it. It was quite close and hence 17/12 was used initially. However it was found slightly longer (*savisesah*) and when measured by rope again it was found longer by 1/34[1/4(1/3)]. This was also found minutely longer (*savisesah*) but accepted as a sufficiently accurate value. – That was a simple explanation.

**2.** Another explanation for the evolution of this formula was based on geometrical construction. – Two equal squares, each with side of one unit were taken. One of the squares was vertically divided into three rectangles. Two pieces of the above was placed along the two adjacent sides of the square to form an approximate square of side 1+1/3, but for missing a small square 1/3×1/3. This was also made up by using a piece from remaining rectangle. Hence we get the first approximate value of (1+1/3) for √2*. *

But a small piece is still remaining of size (1/3 x 2/3). This was made into 4 equal strips of size [1/4(1/3) x 2/3]. Two pieces, end to end, were kept along one side of the above augmented square and the other two pieces on the other side. Now we get a total area which is the sum of the two squares. The above augmented square is of side (1 + 1/3 + 1/(3×4)) = 17/12, which was found good enough initially. But, we still miss a small portion of (1/12 x 1/12), to complete the square. Hence the size has to be reduced by this extra area (*savisesah*). – To find this extra measure, (1/12 x 1/12) should be divided by (17/12 +17/12) and that gives 1/(3x4x34). Of course still we have some minute extra area (*savisesah*). The above construction is explained in the above figures.

Yes, this is really an interesting explanation. But this amazing formula (sutra) evolved in 800 BC deserves a better explanation which I will offer now.

**3.** Way back in 1967, I was in a class room of IIT/Kanpur. The teacher was Professor Dr. V. Rajaraman, the pioneer of computer education in India. He was teaching us the basic algorithms for programming in Fortran, a (then) popular programming language. One of the very early recursive algorithms he taught us was, to find the square root of a number. It goes thus:

Let N be the number for which square root is required

Make first guess of the square root as r0 – Later, we will know, the guess may be as bad as 10 times N; still the method works as smoothly as ever.

The next guess can be made as r1 = ½ (r0 + N/r0)

Keep improving this value using the recursion: r(n+1) = ½ [r(n) + N/r(n)]

Surprisingly the value converges very fast to √N, to the required level accuracy.

As you have seen in my earlier blog of **Evolution of Sine Table by Hindu Maths, **our ancient mathematicians have always preferred recursive steps to solve any problems. Hence, in this case also, Baudayana preferred to use recursive steps, exactly as above. However like others, he preferred to calculate individual step sizes as below:

r(n+1) – r(n) = ½[N/r(n) – r(n)], which is same as the above recursive statement.

To find square root of 2, Boudayana used (1+1/3) = 4/3, as the first guess, r(0).

Hence next step, r(1) – r(0) = ½(3/2 – 4/3) = (3/4 – 2/3) = 1/(3×4)

And hence, r(1) = 1 + 1/3 + 1/(3×4) = 17/12 = 1.4166

Next step, r(2) – r(1) = ½(24/17 – 17/12) = [12/17 – 17/(3x4x2)] = -1/(3x4x34)

And hence, r(2) = 1 + 1/3 + 1/(3×4) – 1/(3x4x34) = 577/408 = 1.414216 (correct up to 5 decimals)

Baudayana would have gone to the next minute step also (as his *savisesah*, indicates), as below

Next step, r(3) – r(2) = ½(816/577 – 577/408) = 408/577 – 577/(3x4x34x2) = -1/(3x4x34x1154)

Hence, r(4) = 1 + 1/3 + 1/(3×4) – 1/(3x4x34) -1/(3x4x34x1157) = 1.41421356237469

The above value is correct up to 13 places.

The recursive algorithm is always the first approach of ancient Indian mathematicians.

Just to satisfy myself that it is not just an isolated case, I tried this logic for finding square root of three also. The step sizes came out to be as below:

Start = 1

First Step = ½ , Root = 1.5

2^{nd} Step = 1/(2×2), Root = 1.75

3^{rd} Step = – 1/(2x2x14), Root = 1.7321429

4^{th} Step = – 1/(2x2x14x194), Root = 1.7320508

The above value is correct up to 7 decimal places.

**Again hats-off to Baudayana!**