The Shulbasutras of Baudhayana, Apastamba, Katyayana and Manava: With Text, English Translation and Commentary c8.t2: Chapter 2: Transformation of Geometrical Figures

The Śulbasūtras of Baudhāyana, Āpastamba, Kātyāyana and Mānava: With Text, English Translation and Commentary c8.t2

Language: English

Chapter 2: Transformation of Geometrical Figures

c8.t2.2

This chapter deals with the method of combination or difference of two separate squares into a square and the transformation of a square into a rectangle, an isosceles trapezium or a circle and vice versa.

c8.t2.3

CONSTRUCTION OF A SQUARE BEING SUM OF, DIFFERENCE BETWEEN, TWO SQUARES

c8.t2.4

**2.1-2.2.** These two rules of Baudhāyana give the methods of construction of a square as the sum and difference of two different squares.

c8.t2.5

Here three technical terms, *hrasīyasaḥ*, *varṣīyasaḥ* and *vṛddhram* are used. According to Kapardisvāmī,a *hrasīyasa* means the side of the smaller square, *varṣīyasa* the side of the larger square and *vṛddhram* the rectangular portion (*dīrghacaturaśram*).

c8.t2.6

*Method of combination* (*samāsa*).

c8.t2.7

For the combination of a smaller square *EBGF* with another square *ABCD*, this rule of Baudhāyana suggests that the rectangular portion *ABGH* is cut off by the side of the smaller square whose side is equal to *BG*. Then *AG* of this cut-off portion will be the side of the combined square (Fig. 12).

c8.t2.8

Fig. 12.

c8.t2.9

Fig. 13.

c8.t2.10

Evidently, $AG² = AB² + BG² = sum of two squares.$ The same method is also given by Āpastamba (*Āśl*. 2.4) and Kātyāyana (*Kśl*. 2.13).

c8.t2.11

PROOF : Dattab has suggested the following proof of this proposition (Fig. 13). $sq. ABCD + sq. ECGF$ $= tr. ABO + tr. AOI + tr. OFG + tr. OFJ + sq. IJED$ $= tr. ADH + tr. AOI + tr. HEF + tr. OFJ + sq. IJED$ $= sq. AOFH$ $or, AB² + CG² = AO²$

c8.t2.12

a *Āśl*. Mysore 73, 39.

c8.t2.13

b Datta (2), 77.

c8.t2.14

*Method of difference (nirhāra).*

c8.t2.15

To construct a square equal to the difference between a smaller square *EBGF* and other square *ABCD*, the rule *Bśl.* 2.2 suggests that the rectangular portion *ABGH* is cut off by the side *BG* of the smaller square. Then the side *GH* of the cut off portion is allowed to fall on *AB*, and *P* is the point where it falls. Here *GH* = *GP*. Then *BP* is the side of a square which is equal to the difference of the squares *ABCD* and *EBGF* (Fig. 14).

c8.t2.16

Fig. 14.

c8.t2.17

Fig. 15.

c8.t2.18

Evidently, BP² & = GP² - BG² / & = GH² - BG² / & = AB² - BG² / & = difference of two squares ABCD and EBGF.

c8.t2.19

The method is also given by Āpastamba (*Āśl.* 2.5) and Kātyāyana (*Kśl.* 3.1).

c8.t2.20

Proof : The following proof based on the knowledge of the *śulbakāras* is due to Dattaa (Fig. 15).

c8.t2.21

Now, sq. *PGFH* = 4 tr. *PGI* + sq. *IJKL* & = 2 tr. PGI + 2 tr. PGI + sq. IJKL / & = rect. PBGI + rect. PBGI + sq. IJKL / & = (rect. PBGI + sq. IJKL) + rect. PBGI / & = (rect. PBGI + sq. IJKL) + sq. MBGJ + rect. PMJI / & = (rect. PBGI + sq. IJKL + rect. PMJI) + sq. MBGJ / & = (rect. PBGI + sq. IJKL + rect. JGEK) + sq. MBGJ / & = sq. PBEL + sq. MBGJ / or, sq PBEL & = sq. PGFH - sq. MBGJ / \therefore BP² & = PG² - BG² / or BP² & = AB² - BG²

c8.t2.22

Transformation of a Square into a Rectangle

c8.t2.23

**2.3-2.4.** Baudhāyana has given two methods for transformation of a square into a rectangle.

c8.t2.24

According to the first method, a square is transformed into a rectangle, such that the diagonal of the square equals the longer side of the rectangle. The method is also given by Kātyāyana (*Kśl.* 3.4).

c8.t2.25

a Datta (2), 79.

c8.t2.26

The square ABCD is divided by its diagonal AC (Fig. 16). The portion ADC is again divided into two equal halves by GD and each is transferred to occupy the position AEB and BFC. Then AEFC is the required rectangle. For, $sq. ABCD = tr. ABC + tr. AGD + tr. GCD$ $= tr. ABC + tr. AEB + tr. BFC$ $= rect. AEFC.$

c8.t2.27

The method is limited in scope, for it only turns a square into a rectangle, the longer side of which is equal to the diagonal of the square.

c8.t2.28

The second method concerns the transformation of a square into a rectangle of which one side is given. The same rule is also given by Āpastamba (*Āsl.* 3.1). Both Baudhāyana and Āpastamba gave no clear exposition of the second half of this *sūtra*. To substantiate this rule, Thibaut,a considered as an instance a square of side 5 units and changed it into a rectangle of 3 units by 8⅓ units. Likewise, Bürkb transformed a square of side 6 units into a rectangle of 4 units by 9 units as follows.

c8.t2.29

Fig. 16.

c8.t2.30

Fig. 17.

c8.t2.31

The sq. ABCD is broken into a rect. ABEF making its side BE (=4 units) equal to the desired shorter side of the rectangle, and rectangle OECP (where EC=2 units), together with a square FOPD. The rectangle OECP is transferred to the other side, and GAFH is its new position. Next the smaller square FOPD (2 units × 2 units) is changed into a rectangle (of 1 unit by 4 units) and IGHJ becomes its new position (Fig. 17). Hence BI (6 + 2 + 1 = 9 units) is the length of the new rectangle. Similarly, if we change a square of 7 units into a rectangle of 5 units by 49/5 (= 7 + 2 + ⅘) units, we have to construct a rectangle of ⅘ unit by 5 units from a square of 2 units by 2 units. This is actually no solution to the problem since the transformation of square FOPD to a rectangle IGHJ is again a problem of fundamental nature.

c8.t2.32

The commentators Dvārakānātha Yajvā and Sundararāja have described a general method as follows : *yāvadicchaṃ pārśvamānyau prācyau vardhayitvā uttarapūrvāṃ karṇarajjumāyacchet sā dīrgha caturaśramadhyasthāyāṃ samacaturaśra tiryaṅmānyāṃ yatra nipatati tata uttaraṃ hitvā dakṣiṇāṃsaṃ tiryaṅmānīṃ kuryāt taddīrghacaturaśraṃ bhavati/* This means: Having increased upto the desired length the two sides (*pārśvamānī*)

c8.t2.33

a Thibaut (1), 246.

c8.t2.34

b Bürk, 56, 334.

c8.t2.35

c Thibaut (1), 247.

c8.t2.36

towards east, the diagonal-cord is stretched towards north-east corner. The (diagonal) line cuts the breadth (*tiryaṅmānī*) of the square lying inside the rectangle; the northern portion is cut off (by drawing a line through this point parallel to *prācī*); the southern side becomes the breadth (*tiryaṅmānī*) of the (desired) rectangle.

c8.t2.37

In Fig. 18, the sides *BA* and *CD* of the square *ABCD* are increased to *E* and *F* respectively, so that each of the sides *BE* and *CF* becomes equal to the given length *PQ*. The diagonal cord *CE* cuts the side *AD* at *O*. Then the northern portion *EBHG* is cut off by drawing a line *HG* passing through *O* parallel to the *prācī* line *BA*. Now *GHCF* is the required rectangle.

c8.t2.38

Fig. 18. Square into a rectangle. Fig. 19. Square into a rectangle of given side.

c8.t2.39

This is a general and perfectly satisfactory method. Both Thibaut and Bürk did not consider this method as that of Baudhāyana on the ground that Baudhāyana himself mentioned this method as *anyaśca prakāraḥ*, meaning 'another method'. Baudhāyana's method was to cut off from a given square a rectangle of side smaller than that of the square while Dvārakānātha's method started from the construction of a rectangle of side greater than that of the square. From our discussion it is clear that in the methods suggested by both Baudhāyana and Dvārakānātha, the final result of constructing a rectangle equivalent to a square is the same but their methods of attaining it are different. For this difference, Sundararāja gave the same line of argument as that of Dvārakānātha in transforming a square into a rectangle with the remark, *ayamatra prakāraḥ*a meaning, 'this is the method taught here'. To keep a symmetry with the original *sūtra* of Baudhāyana, Dattab put the method of Dvārakānātha in the following form.

c8.t2.40

From the square *ABCD*, the portion *AGHD* is cut off, such that AG = DH = PQ, the side of the required rectangle. The diagonal *AH* is produced to meet *BC* (produced) at *E*. The rectangle *ABEF* is completed. Then *AGKF* is the equivalent rectangle (Fig. 19).

c8.t2.41

For, tr. ABE = tr. AFE, tr. AGH = tr. ADH and tr. HCE = tr. HKE. Hence rectangle GC = rectangle DK.

c8.t2.42

Now sq. ABCD &= rect. AH + rect. GC / &= rect. AH + rect DK / &= rect. AK.

c8.t2.43

a *Āsl. Mysore* 49.

c8.t2.44

b *Datta* (2), 90.

c8.t2.45

TRANSFORMATION OF A RECTANGLE INTO A SQUARE

c8.t2.46

**2.5.** This is a most general method of transforming a rectangle into a square given by Baudhāyana. The same method is also taught by Āpastamba (*Āśl.* 2.7) and Kātyāyana (*Kśl.* 3.2). Baudhāyana’s method runs as follows.

c8.t2.47

Let *ABCD* be the given rectangle (Fig. 20). The portion *ABFE* is cut off such that AE = AB = the breadth of the rectangle. The remaining portion *EFCD* is cut off into two equal` halves. One half *GHCD* is placed on the other side and its new position becomes *BKLF*. A small square *FLMH* is fitted at the corner.

c8.t2.48

Fig. 20. A rectangle into a square.

c8.t2.49

Fig. 21. A square into an isosceles trapezium.

c8.t2.50

Now, rect. ABCD = sq. AKMG - sq. *FLMH*, which shows that the rectangle *ABCD* is expressed as the difference of two squares. Since the method of *nirhāra* has already been taught before by Baudhāyana (*Bśl.* 2.2), a square equal to the difference of the two squares mentioned above is found by allowing the side *KM* to fall at *P* over *BH*. Then the square on *BP* will be equal to the difference of two squares, which is equal to the area of the given rectangle.

c8.t2.51

For, BP² & = PK² - BK² / & = MK² - FL² / & = sq. ABFE + rect. EFHG + rect. FBKL / & = sq. ABFE + rect. EFHG + rect. DGHC / & = rect. ABCD.

c8.t2.52

In the case of a rectangle of very great length, Kātyāyana’s (*Kśl.* 3.3) advice is to cut it again and again by its breadth, combine the pieces by the *samāsa* method (*Bśl.* 2.1) and finally to achieve the result by applying the *nirhāra* method (*Bśl.* 2.2). This is clearly no improvement upon the method given by Baudhāyana.

c8.t2.53

TRANSFORMATION OF A SQUARE OR RECTANGLE INTO AN ISOSCELES TRAPEZIUM

c8.t2.54

**2.6.** By this method a square as well as a rectangle are changed into a trapezium of given side (smaller than the side of the square).

c8.t2.55

The square *ABCD* is required to be transformed into an isosceles trapezium *AGCE*, whose shorter side *AE* is equal to the given length *PQ* (Fig. 21). The rectangular portion *EFCD* is divided into two equal halves and the half *ECD* is shifted to its other side, such the *AGB* is its new position. Hence *AGCE* is the required isosceles trapezium.

c8.t2.56

For, sq. *ABCD* = rect. *ABFE* + tr. *EFC* + tr. *ECD*
= rect. *ABFE* + tr. *EFC* + tr. *AGB*
= trap. *AGCE*

c8.t2.57

This method of transformation was known earlier in the *Śatapatha Brāhmaṇa* (*Śat. Br.* 10.2.1.4).

c8.t2.58

TRANSFORMATION OF A RECTANGLE OR SQUARE INTO A TRIANGLE OR RHOMBUS

c8.t2.59

**2.7-2.8.** Baudhāyana has given a method of constructing a triangle or a rhombus whose area is equal to that of a given rectangle. For the construction of a triangle as described in *sūtra* (*Bśl.* 2.7), a square is to be constructed whose area will be twice that of the given rectangle. A rectangle twice the area is first constructed and then transformed into a square by the method described in *Bśl.* 2.5. Let *ABCD* be the transformed square and *E* the middle point of *AD*. Join *EB*, *EC*. Then *EBC* is the required triangle whose area is equal to that of the given rectangle (Fig. 22).

c8.t2.60

Fig. 22. A rectangle into a triangle.

c8.t2.61

Fig. 23. A rectangle into a rhombus.

c8.t2.62

For, tr. *EBC* = ½ sq. *ABCD* = given rectangle.

c8.t2.63

For the construction of a rhombus as in *sūtra* (*Bśl.* 2.8), let the rectangle *ABCD* be so constructed that its area is double that of the given rectangle. Let *E*, *F*, *G*, *H* be the middle points of *AB*, *BC*, *CD* and *DA* respectively. Join *EF*, *FG*, *GH* and *HE* to produce the required rhombus *EFGH* (Fig. 23).

c8.t2.64

For, rhombus *EFHG*
= tr. *EFH* + tr. *GFH*
= ½ (rect. *ABFH* + rect. *CDHF*)
= ½ rect. *ABCD*

c8.t2.65

This is given by both Āpastamba (*Āśl.* 12.8) and Kātyāyana (*Kśl.* 4.4).

c8.t2.66

TRANSFORMATION OF A SQUARE INTO A CIRCLE.

c8.t2.67

**2.9.** The following method of transforming a square into a circle is given by Baudhāyana. The same method has also been taught by Āpastamba (*Āśl.* 3.2), Kātyāyana (*Kśl.* 3.11) and Mānava (*Mśl.* 1.8).

c8.t2.68

Let *ABCD* be the given square and *O* its centre. The half diagonal *OA* is drawn over the east-west line *OE*, such that OA = OE. Then a circle with radius *OF* equal to *OG* plus ⅓ of *GE* i.e. *GF*, is drawn to give the required circle (Fig. 24).

c8.t2.69

Fig. 24.

c8.t2.70

Here, radius [t] = OF & = OG + GF / & = OG + ⅓ GE / & = OG + ⅓ (OA - OG).

c8.t2.71

Let 2a be the side of the square *ABCD*. $OF = a + ⅓ (a √2 - a)$ or r = a [1 + ⅓ (√2 - 1)], where OF = r or r = a/3 (2 + √2)

c8.t2.72

As per *Bśl* 2.12 (*vide infra*), √2 is given by, $√2 & = 1 + ⅓ + 1/3.4 - 1/3.4.34 / & = 577/408 = 1.4142156$

c8.t2.73

Baudhāyana’s more refined value of π is given by (*Bśl.* 4.15), $π & = 4 (1 - ⅛ + 1/8.29 - 1/8.29.6 + 1/8.29.6.8) / & = 3.0885.$

c8.t2.74

Using the above value of √ 2 and π, the area of the transformed circle = π r² = 3.9989 a², which is in close agreement with the area of the given square, 4a².

c8.t2.75

If we take π = 3 (*Bśl.* 4.15), area of the circle becomes 3.885 a², which falls far short of the area of the given square. Āpastamba made an additional remark on the method of circling a square as *sānityā maṇḍalaṁ yāvaddhīyate tāvadāgantu*, which makes also the interpretation equally difficult as to whether the method is exact or inexact one. The commentator Kapardisvāmī has broken up *sānityā* as *sā* and *anityā* concluding that the method is an inexact one. The passage has been interpreted by Karavindasvāmī as follows: ‘The circle is exactly as large as the square, for as much the circle falls short, so much comes in.’a Thibaut, Bürk and Datta have referred to the same difficulty as to the real sense in which these words were used by Āpastamba.b

c8.t2.76

However, Dvārakānātha Yajvāc, commentator of *Baudhāyana śulba* has proposed the following correction to the formula of Baudhāyana, which gives better result: $r = [a + a/3 (√2 - 1)] (1 - 1/118)$

c8.t2.77

a Āśl. Mysore, 50.

c8.t2.78

b Datta (2), 142-43.

c8.t2.79

c Thibaut (2), 10, 21.

c8.t2.80

The problem of quadrature has also been discussed by Drenckhahn,a Chakrabarty,b and Gurjar.c

c8.t2.81

TRANSFORMATION OF A CIRCLE TO A SQUARE

c8.t2.82

**2.10-2.11.** Baudhāyana describes two methods of finding quadrature of a circle.

c8.t2.83

*First Method.*

c8.t2.84

If 2a be the side of a square and d the diameter of the circle, then

c8.t2.85

$2a = 7d/8 + [ d/8 - 28d/8.29 + ( d/8.29.6 - d/8.29.6.8 ) ]$

c8.t2.86

or, 2a = d - d/8 + d/8.29 - d/8.29 ( ⅙ - 1/6.8 )

c8.t2.87

*Second Method.*

c8.t2.88

$2a = d - 2/15 d$

c8.t2.89

This result is also given by both Āpastamba (*Āśl.* 3.3) and Kātyāyana (*Kśl.* 3.12)

c8.t2.90

*Rationale.*

c8.t2.91

(A) The rationale of the result obtained from the first method is given by Thibaut, Cantor and Müller as follows :

c8.t2.92

(i) Thibautd has suggested that the result was possibly obtained from the previous result of circling a square, r = a/3 (2 + √2) by inversion.

c8.t2.93

$For, 2a = 3/(2 + √2) d$

c8.t2.94

$= 1224/1393 d √2 = 577/408$

c8.t2.95

$= d ( ⅞ + 1/8.29 - 1/8.29.6 + 1/8.29.6.8 )$

c8.t2.96

since, 1) 1/8th of 1393 = 174⅛.

c8.t2.97

$. 2) & 7/8th of 1393 = 1218 7/8 (less by 5 1/8 from 1224) / 3) & 1/8.29 th of 1393 = 6 (approx) / 4) & 1/8.29.6 th of 1393 = 1 / 5) & 1/8.29.6.8 of 1393 = 1/8 } & (i.e. 6 - 1 + 1/8 = 5 1/8 / & surplus by 5 1/8 from 1224)$

c8.t2.98

c Gurjar (2), 11-16.

c8.t2.99

More or less the same method is given by Cantor.e (ii) Müller's derivationa is as follows : $2a = 3/(2 + √2) d = (3 √2)/(2 √2 + 2) d = 3/2 · (√2)/(1 + √2) d$ $= ( 3/2 · (17 - 1/34)/(29 - 1/34) ) d = ( (51 - 3/34)/(58 - 2/34) ) d √2 = 17/12 - 1/12.34$ $= ( 1 - (7 + 1/34)/(58 - 2/34) ) d$ $= d - ⅛ · ( (56 + 8/34)/(58 - 2/34) ) d = d - ⅛ [ 1 - (2 - 10/34)/(58 - 2/34) ] d$ $= d - ⅛ [ 1 - 1/29 ( 1 - (10/34 - 2/34.29)/(2 - 2/34.29) ) ] d$ $= d - ⅛ d [ 1 - 1/29 1 - ⅙ ( 1 - (4 + 5/29)/(34 - 1/29) ) ]$ $= d - ⅛ d + 1/8.29 d [ 1 - ⅙ 1 - ⅛ ( 1 - (2 - 41/29)/(34 - 1/29) ) ]$

c8.t2.100

a Drenckhahn, 1-13.

c8.t2.101

b Chakrabarty (2), 23-28.

c8.t2.102

d Thibaut (1), 254.

c8.t2.103

e Datta (2), 145.

c8.t2.104

Thus, $2a = d - d/8 + d/8.29 - d/8.29 ( ⅙ - 1/6.8 ) - d/8.29.6.8 · (2 - 41/29)/(34 - 1/29)$

c8.t2.105

The last term is neglected, it being very small.

c8.t2.106

However, Dvārakānāthab has suggested a more correct result of the above formula as follows : $2a = [ d - d/8 + d/8.29 + d/8.29 ( ⅙ - 1/6.8 ) ] × ( 1 + ½ · 3/133 )$

c8.t2.107

(B) The rationale of the second method may be obtained as follows :

c8.t2.108

The average of two squares, one circumscribed and the other inscribed, determines the approximate area of the circle.

c8.t2.109

a Müller, 201.

c8.t2.110

b Thibaut (2), 10, 21.

c8.t2.111

∴ Area of the circle = (4r² + 2r²)/2 = 3r²

c8.t2.112

Since the square is taken to be equal in area to the circle, 4a² = 3r² or a = (√3)/2 r

c8.t2.113

The value of √3 may be obtained by the method of successive approximation as follows:

c8.t2.114

(i) √A = √(a² + c) = a + c/(2a + 1),

c8.t2.115

where 2a + 1 is the difference between the squares of c and the next positive integer. Therefore,
√3 = √(1² + 2) = 1 + ⅔ = 5/3

c8.t2.116

(ii) For finding the next approximation e, √A is written as

c8.t2.117

√A = a + c/(2a + 1) + e
Then squaring both sides and cancelling the value of e², since it is very small, the value of e is obtained.
Here √3 = 5/3 + e

c8.t2.118

Squaring and cancelling the value of e² we get
10/3 e + 25/9 = 3
or e = 1/15

c8.t2.119

then √3 = 1 + ⅔ + 1/15 = 26/15

c8.t2.120

Obviously,

c8.t2.121

a = ½ · √3 r = ½ · 26/15 r = 13/15 r
= r - 2/15 r [ r = radius ]
or 2 a = d - 2d/15 [ d = 2r = diameter ]

c8.t2.122

*The value of √2*

c8.t2.123

**2.12 :** The value of √2 given by Baudhāyana is √2 = 1 + ⅓ + 1/3.4 - 1/3.4.34 (approx.)

c8.t2.124

The same *sūtra* is also given by Āpastamba (*Āśl.* 1.6) and Kātyāyana (*Kśl.* 2.9).

c8.t2.125

In decimal fraction, the above value of √2 = 1.4142156. According to modern calculation, √2 = 1.4142135. Thus it is clear that the ancient Indians attained a remarkable degree of accuracy in calculating an approximate value of √2. The *śulbakāras* gave methods, for constructing a square equal to the sum of two equal squares, but gave no method of calculating the value of its diagonal.

c8.t2.126

Thibaut, Rodet, Datta, and others gave possible methods of solution for arriving at the value as follows:—

c8.t2.127

(i) *Thibaut's proof.*a

c8.t2.128

Now, 17² = 2.12² - 1. Thibaut argued, by how much the side 17 must be diminished in order that the square on it may be 2.12² exactly. Since 2 × 17 × 1/34 = 1, he observed, two strips each of 1/34 (approximately) are to be cut off from a square with 17 as side to obtain the square 2.12² (i.e. 12² + 12²).

c8.t2.129

Hence, (17 - 1/34)² = 2.12²

c8.t2.130

or, (17 - 1/34)/12 = √2

c8.t2.131

Again, 17 - 1/34 = 12 + 4 + 1 - 1/34

c8.t2.132

or, 17 - 1/34 = 12 (1 + ⅓ + 1/3.4 - 1/3.4.34)

c8.t2.133

or, (17 - 1/34)/12 = 1 + ⅓ + 1/3.4 - 1/3.4.34

c8.t2.134

or, √2 = 1 + ⅓ + 1/2.4 - 1/3.4.34

c8.t2.135

In Baudhāyana's selection of units of 12 *aṅgulas* (= 1 *pada*) and 34 *tilas* (= 1 *aṅgula*) Thibaut found justification for the choice of the arbitrary relation 17² = 2.12² (approx.) leading to the origin of the formula of √2, as given in the text.

c8.t2.136

(ii) *Rodet's approximation.*b

c8.t2.137

According to Rodet, the approximation adopted by *śulbakāras* may be obtained by successive approximation.

c8.t2.138

a Thibaut (1) 239-41

c8.t2.139

b Rodet, 162-165

c8.t2.140

$√(a² + r) = a + r/(2a + 1) + (r/(2a + 1)(1 - r/(2a + 1)))/(2(1 + r/(2a + 1))) + e$

c8.t2.141

where e is a fourth term approximation. Rodet might have obtained the result as follows : √(a² + r) = a + r/(2a + 1) [two term approximation] where 2a + 1 is the difference of the squares of a and the next positive integer a + 1. For third term approximation, assume

c8.t2.142

√(a² + r) &= a + r/(2a + 1) + e_1 / &= (2a + r + 1)/(2a + 1) + e_1

c8.t2.143

Squaring and neglecting e₁², we get (2(2a + r + 1))/(2a + 1)e_1 &= a² + r - ((2a + r + 1)/(2a + 1))² / &= (r(2a + 1 - r))/((2a + 1)²) / \therefore e_1 &= (r(2a + 1 - r))/(2(2a + 1)(2a + 1 + r)) / &= (r/(2a + 1)(1 - r/(2a + 1)))/(2(1 + r/(2a + 1)))

c8.t2.144

Likewise, the fourth term approximation is obtained. Obviously, following above, we write,

c8.t2.145

$√2 = √(1² + 1) = 1 + ⅓$

c8.t2.146

Let √2 = 1 + ⅓ + e = 4/3 + e

c8.t2.147

Squaring both sides and cancelling e² from both sides, we get $8/3e = 2 - 16/9 = 2/9$

c8.t2.148

$∴ e = 2/9 × ⅜ = 1/12 = 1/3.4$

c8.t2.149

$∴ √2 = 1 + ⅓ + 1/3.4$

c8.t2.150

Let √2 = 1 + ⅓ + 1/3.4 + e

c8.t2.151

$= 17/12 + e$

c8.t2.152

Squaring both sides and cancelling e² from both sides, $17/6 e = 2 - (17/12)² = - 1/144$ $∴ e = - 1/144 × 6/17 = 1/12.34$ $= - 1/3.4.34$ $∴ √2 = 1 + ⅓ + 1/3.4 - 1/3.4.34 (approx.)$

c8.t2.153

The methods described later by Gurjara and Guptab are the same and no improvement over Rodet's method.

c8.t2.154

(iii) *Datta's proof.*c

c8.t2.155

Datta's proof is an improvement over that of Thibaut and maintains the method of construction followed in the *śulba*.

c8.t2.156

The method consists in constructing a square with area equal to the sum of the areas of the two other squares having sides of one unit in length (Fig. 25).

c8.t2.158

Fig. 25.

c8.t2.159

For this one of the two squares having side of unit length is divided into three equal parts by lines drawn parallel to one of its sides. Each of these parts forms a rectangular piece of one unit in length and one-third unit in width. Two of these rectangular parts are then joined length-wise to the two adjacent sides of the other unit square. This leaves a square hole at one of the corners of the enlarged unit square. This square hole will have a side of one-third unit in length. The remaining rectangular piece of the divided unit square is again subdivided into three equal parts each forming a square of side one-third unit in length. One of the squares is fitted into the square hole mentioned above. Each of the remaining two squares is again subdivided into four equal rectangular pieces having length of ⅓ unit and width of 1/3.4 unit. Eight of these small rectangular pieces are placed length-wise side by side on the two adjacent sides of the enlarged square with four on each side. This again leaves a square hole at the corner having a side of length 1/3.4 unit. Now two equal strips have to be deducted from the two adjacent sides of the enlarged square under construction; the width of each of the strips is therefore given by

c8.t2.160

a Gurjar (1), 6-10.

c8.t2.161

b Gupta, 77-79.

c8.t2.162

c Datta (2), 192-94.

c8.t2.163

((1/3.4)²)/(2 × (1 + 1/3 + 1/3.4)) = 1/3.4.34

c8.t2.164

Hence √2, the side of the desired square is given approximately by 1 + 1/3 + 1/3.4 - 1/3.4.34.

c8.t2.165

*Value of √2 in other culture areas.*

c8.t2.166

A small cuneiform tablet (Yale Babylonian collection No. 7289) of the old Babylonian times (c. 1800—1600 B.C.) shows a square with its two diagonals, with three numbers in sexagesimal system inscribed on it. These three numbers are interpreted by Neugebauera as the value of the diagonal, a side and the value of √2 (since d = √2 a). Here √2 is given as 1, 24, 51, 10, which in terms of decimals comes out to be 1.41421,291......, a little more accurate than the Indian value. The Indian value is smaller while the Babylonian value larger than the actual value. Moreover, their first fractional terms are different. The suggestion that the Indian value might have been obtained from a Babylonian source is groundless. As regards Greekb sources, many approximations to the value of √2 are known, but not a value of this order of accuracy.

c8.t2.167

*Irrationality of √2.*

c8.t2.168

Baudhāyana, Āpastamba and Kātyāyana gave the value of √2, as mentioned above, with an additional term *viśeṣa* (approximate). Many scholars expressed doubt whether, by the term *viśeṣa*, the *śulbakāras* recognized the irrationality of √2. According to Karavindasvāmī,c a commentator on the *Āpastambaśulbasūtra*, the root *śiṣ* when prefixed by *vi* denotes in all cases a 'correction in excess'. Dattad has discussed the matter in detail, and the commentator can be relied upon in this interpretation. Looking into the ancient literature of India, we find in the early canonical works of the Jainas many instances of the employment of the term *viśeṣa* in the same connection as we find in the *śulba*. A few instances are given here.

c8.t2.169

a Neugebauer, 34, vide also *Plate 6a*.

c8.t2.170

b Heath (2), 155.

c8.t2.171

c *ĀŚl.*, Mysore 73.

c8.t2.172

d Datta (2), 198-202.

c8.t2.173

(i) The diameter of the circle is 99640 *yojanas*, the circumference is 315089 and a little over (*kiñcid-viśeṣādhika*) (*Sūrya-prajñapati*, *sūtra* 20).

(ii) The diameter is 100000 *yojanas*, circumference is 316227 *yojanas* 3 *gavyutis* 128 *dhanus* 13½ *aṅgulas* and a little over (*kiñchid-viśeṣādhika*) (*Jambūdvīpa-prajñapti*, *sūtra* 3).

c8.t2.174

Hence *viśeṣa* refers to a small quantity, which is either in excess or in deficit, and cannot be accurately determined. *Śulbakāras* gave no proof for it, since it was beyond their tradition.

Linked texts (7)

Āpastambaśulbasūtra — Āpastamba, Sanskrit
Kātyāyanaśulbasūtra — Kātyāyana (Vājasaneyi sūtrakāra), Sanskrit
Baudhāyanaśulbasūtra — Baudhāyana, Sanskrit
Śatapathabrāhmaṇa — Yājñavalkya & Śāṇḍilya (Vedic teacher), Sanskrit
Mānavaśulbasūtra — Sanskrit
Sūriyapannatti — Ardhamāgadhī Prakrit
Jambuddīvapannatti — Ardhamāgadhī Prakrit