NPTEL COURSE ON

MATHEMATICS IN INDIA:FROM VEDIC PERIOD TO MODERN TIMES

Lecture 20

Lılavatı of Bhaskaracarya I

M. S. SriramUniversity of Madras, Chennai.

Outline

◮ Introduction, Importance of Lılavatı◮ Arithmetical operations: Inversion method, rule of

supposition◮ Solution of quadratic equations◮ Mixtures◮ Combinations, progressions.◮ Plane figures: Application of Right Triangles.

Introduction

Bhasakaracarya: Perhaps the most well known name among theancient Indian astronomer-mathematicians. Designated asBhaskara-II to differentiate him from his earlier namesake, wholived in the seventh century CE (Bhaskara-I).

According to his own statement in his Goladhyaya, was born insaka 1036 or 1114 CE. He also adds that he came fromVijjalavıda near the Sahyadri mountain. → Bijapur? S.B. Dixit →Bhaskara’s original home was Pat.an in (Khandes) → Inscriptionlists his grandfather, father, son and grandson: Manoratha,Mahesvara, Laks.mıdhara, Cangadeva respectively.

Bhaskara’s Lılavatı :A standard work of Indian Mathematics: Patıor Patıgan. ita: elementary mathematics covering arithmetic,algebra, geometry and mensuration. Still used as a textbook inSanskrit institutions in India. Composed around 1150 CE.

Commentaries and other works

Many commentaries: Paramesvara (about 1430), Gan. es.a

(Buddhivilasinı) (1545), Suryadasa, Munısvara (about 1635), andRamakr. s.n. a (1687). According to R. C. Gupta, a well knownhistorian of Indian mathematics, the best traditional commentaryis Kriyakramakarı (C. 1534) of Sankara Variyar andMahis.amangala Narayan. a (who compiled this after the demise ofSankara). Number of commentaries and versions in regionallanguages, Kannada, Telugu, etc., Three Persian translationsare known: Abul-Fayd Faydi (1587 CE). My lectures based onColebrook’s translation (1817) reprinted along with the originalSanskrit text and commentaries with notes by H. C. Banerji.Others works of Bhaskara: Bıjagan. ita (Algebra), one of the mostimportant treatises on Indian algebra. Siddhantasiroman. i:comprising Grahagan. ita and Goladhyaya also very popular.Other work Karan. akutuhala (epoch 1183 CE).

Operations of arithmatic

Lılavatı : Rules and Examples in about 270 Verses.Chapter 1 on weights and measures, money denominations.Numeration upto 1014.Eight operations of arithmetic.

i) Addition and subtractionii) Multiplication

a) a1 . . . an × B. Multiply an by A, an−1 by A etc. and add (takinginto account place values).

b) A(b + c) = Ab + Ac.

c) A × B = Ax × Bx

, where x is a factor of B.

d) ab = a(b ± c ∓ c) = a(b ± c)∓ ac.

iii Division: Remove common factors.

Squaring , Square roots etc.

Verses 18-19. Squaring

i)

(a1 + · · · an)2 = a2

n + 2an(a1 + · · ·+ an−1)

+a2n−1 + 2an−1(a1 + · · · + an−2)

+ · · ·+a2

2 + 2a2a1

+a21.

ii) (a + b)2 = a2 + 2ab + b2.

iii) a2 = (a + b)(a − b) + b2.

Verses 21-22: Square root and example.Verses 23-26: Cube and example.Verse 27-28: Cube root: Standard Indian method.

Multipulation with fractions

Fractions:ab,

cd

=adbd

,cbdb

for addition, subtraction etc.,

Verse 32-33:

ab× c

d× · · · =

ac...bd ....

ab+

cd× a

b+

ef

(ab+

cd× a

b

)

+gh

[ab+

cd× a

b+

ef( )

]

. . .

=a(d + c)(e + f )(g + h) . . .

bdfh . . .

Division of fractions.abcd=

adbc

Operations with zeroVerses 44-45 :ya;ea;gea KMa [ea;pa;sa;mMa va;ga;Ra;d;Ea KMa Ka;Ba;a;Æa:$a;ta;ea .=+a; a;ZaH ÁKa;h:=.H .~ya;a;t,a Ka;gua;NaH KMa Ka;gua;Na; a;(ãÉa;ntya;(ãÉa Zea;Sa;a;va;Da;Ea Á Á 44 Á ÁZUa;nyea gua;Na;ke .$a;a;tea KMa h;a:=+(ãÉea;t,a :pua;na;~ta;d;a .=+a; a;ZaH ÁA;a;va;kx +:ta O;;va :℄ea;ya;~ta;TEa;va Kea;na;ea;�a;na;ta;(ãÉa yua;taH Á Á 45 Á Á

“In addition, zero makes the sum equal to the additive.In involution and (evolution) the result is zero. A definitequantity , divided by zero, is the submultiple of nought.The product of zero is nought: but it must be retainedas a multiple of zero, if any operations impend. Zerohaving become a multiplier, should nought afterwardsbecome a divisor, the definite quantity must beunderstood to be unchanged. So likewise any quantity,to which zero is added, or from which it is subtracted,(is unaltered).

Operations with zero

a±0 = a,a0= Kha-hara: a fraction with zero for its denominator.

( In Bıjagan. ita. Gan. esa: Infinity).

a × 0 = 0

But it must retained as a multiple of zero, if further operations,impend. “Zero having become a multiplier, should noughtafterwards become a divisor, the definite quantity must beunderstood to be unchanged”.

a × 00= a? (But

00

is indeterminate )

.

Inversion method

Inversion method to find the solution of a problem in Verse 47-48 :Ce +.dM gua;NMa gua;NMa Ce +.dM va;g a mUa;lM :pa;dM kx +:�a;ta;m,a Á�+.NMa .~vMa .~va;mxa;NMa ku +:ya;Ra;d, dx ;Zyea .=+a; a;Za;pra;Æa;sa:;dÄâ ;yea Á Á 47 Á ÁA;Ta .~va;Ma;Za;a;�a;Da;k+:ea;nea tua l+.va;a;Q.�a;ea;nea h:=+ea h:=.H ÁAM ;Za;~tva;a;va;kx +:ta;~ta:�a ;a;va;l+.ea;mea Zea;Sa;mua;�+:va;t,a Á Á 48 Á Á“To investigate a quantity, one being given, make the divisor amultiplicator; and the multiplier, a divisor; the square, a root; andthe root, a square; turn the negative into positive and the positiveinto negative. If a quantity is to be increased or diminished by itsown proportional part, let the (lower) denominator, being theincreased or diminished by its numerator, becomes the (corrected)denominator, and the numerator remain uncharged*; and thenproceed with the other operations of inversion, as before directed.”

(* If we have 1 +ab=

a + bb

, in the inversion, it will be ba+b .)

Inversion method: ExampleExample in Verse 49 :ya;�///�a;~:�a.éÈåîÅéé:a;�///�a;~:�a;Æa;Ba:=+�///�a;nva;taH .~va;. a:=+NEa;BRa;�+:~ta;taH .sa;�a;Æa;BaH.~va:�yMa;Zea;na ;a;va;va;Æa:$Ra;taH .~va;gua; a;Na;ta;ea h� ;a;na;ea ;a;dõ ;pa:úãÁ*.a;a;Za;ta;a Áta;n}å.Ua;le ;Y;�;yua;tea &+.tea .. a d;Za;Æa;Ba:$a;Ra;tMa dõ ;yMa b.rUa;a;h tMa.=+a; a;ZMa :vea;�//////�a;tsa ;a;h .. a:úãÁ*.a;l+.a;Æa;[a ;a;va;ma;l+.Mava;a;mea ;a;va;l+.ea;ma;a;kÒ +:ya;a;m,a Á Á 49 Á Á

“Pretty girl, with tremulous eyes, if thou know thecorrect method of inversion, tell me the number, whichmultiplied by three and added to three quarters of theproduct, and divided by seven, and reduced bysubtraction of half a third part of the quotient and thenmultiplied into itself, and having fiftytwo subtracted fromthe product and the square root of the remainderextracted, and eight added and the sum divided by tenyields two (2).”

Inversion method: Example

110

8 +

√√√√√√√√

−52 +

(1 − 1

3)

︸ ︷︷ ︸23

17(1 − 3

4)

︸ ︷︷ ︸74

×3 × x

2

= 2

Answer:

x =√

(2 × 10 − 8)2 + 52 × 32× 4

7× 1

3= 28.

Rule of false position

Verse 50:o+.�e +Za;k+:a;l+.a;pa;va;a;d;�:=+a; a;ZaH [ua;NNa;ea &+.ta;eMa.Y;ZEa .=+a;h;ta;ea yua;ta;ea va;a ÁI+.�;a;h;tMa dx ;�;ma;nea;na Ba;�M .=+a; a;Za;BRa;vea;t,a :pra;ea;�+:Æa;ma;t�a;a;�;k+:mRa Á Á 50 Á Á“Any number assumed at pleasure is treated as specified inthe particular question, being multiplied and divided, raisedor diminished by fractions; then the given quantity, beingmultiplied by the assumed and divided by that (which hasbeen found) yields th number sought. This is called theprocess of supposition.”

Let some operations be done on the number x , represent it as A. Let ityield A(x) = y . Take any x ′ and do the same operations. Let it yield y ′,A(x ′) = y ′. Then

x = y × x ′

(y ′)

Rule of false positionExample in Verse 51.:pa:úãÁ*.a.Èåî ÁÁ*+H .~va;a:�a;Ba;a;ga;ea;na;ea d;Za;Ba;�;H .sa;ma;�///�a;nva;taH Á.=+a; a;Za:�yMa;Za;a;DRa;pa;a;dE H .~ya;a;t,a k+:ea .=+a; a;Za;dõùÅ;aRU ;na;sa;�a;�a;taH Á Á 51 Á Á

” What is that number, which multiplied by 5 and having the thirdpart of the product subtracted and the remainder divided by ten,and one-third, a half and a quarter of the original quantity added,gives two less than seventy?”

Solution: Multiplier 5. Subtractive 13 of itself. Divisor 10. Additive

13,

12,

14

, of

the quantity.

Given 68 = y . Take x ′ = 3; 5 × 3 = 15, (1 −

13)15 = 10;

110

× 10 = 1;

1 + 3(13+

12+

14) =

174

; This is y ′. So, y ′ =174

.

∴, x = y ×

x ′

y ′= 68 ×

3174

= 48.

One gets x = 48, whatever value one chooses for x ′. y ′ will be

correspondingly different.

Sankraman. a

Rule of concurrence or ‘Sankraman. a’.

Verse 55:ya;ea;ga;ea.Y;nta:=e +Na;ea;na;yua;ta;ea.Y;�a;DRa;ta;~ta;Ea.=+a;Z�a;a .smxa;ta;Ea .sa;ñÍö�ÅÅ*:" +.ma;Na;a;K.ya;mea;ta;t,a Á Á 55 Á Á“The sum with the difference added or subtracted,being halved, gives the two quantities. This is termedConcurrence.”.

If x + y = k and x − y = l where k and l are given quantities.

x =12(k + l) and y =

12(k − l)

Verse 57. If x − y and x2 − y2 are given, find x + y =x2 − y2

x − y.

Then x and y are found by Sankraman. a.

Rational Squares

Verse 59 and 60: Examples of generation of rational squares:Consider the pairs

a)

x1 =1

2n(8n2 − 1), x2 =

12{ 1

2n(8n2 − 1)}2 + 1,

b)

x1 =1

2n+ n, x2 = 1,

c)x1 = 8n4 + 1, x2 = 8n3.

For all these pairs, x21 + x2

2 − 1 are squares.

Solution of Quadratic equations

Next, Bhaskara gives the solution of a quadratic equation .Verse 62-63: Suppose we have

x ± a√

x = b

Then√

x = ∓a2+√

b + (a/2)2 and x =

(

∓a2+

√

b + (a/2)2

)2

.

(At least here, only one root is being talked about)

Similarly, given: x +cd

x ± a√

x = b, we obtain

x ± a√

x

1 ± cd

=b

1 ± cd

.

Then proceed as before to find√

x and then x .

Quadratic equations: Example

An example in Verse 67.:pa;a;TRaH k+:NRa;va;Da;a;ya ma;a;gRa;Na;ga;NMa kÒu +.;dÄâ ;ea .=+Nea .sa;nd;Deata;~ya;a;DeRa;na ;�a;na;va;a;yRa ta;. C+=+ga;NMa mUa;lE +.(ãÉa;tua;Æa;BRa;hR ;ya;a;n,a Á ÁZa;�yMa :Sa;a:ñ ÅÅå*.+.=+Tea;Sua;Æa;Ba;�///�a;~:�a;Æa;Ba:=+a;pa C+�Ma ;Dva:$Ma k+:a;mRua;kM;�a;. a;. Ce +.d;a;~ya ; a;Za:=.H Za:=e +Na k+:�a;ta .tea ya;a;na:$Rua;naH .sa;nd;Dea Á Á 76 Á Á“The Son of Pr.tha, irritated in fight, shot a quiver ofarrows to slay Karn.a. With half his arrows he parriedthose of his antagonist; with four times the square rootof the quiver-full, he killed his horses; with six arrows,he slew Salya; with three he demolished the umbrella,standard and bow; and with one, he cut off the head ofthe foe. How many were the arrows which Arjuna letfly?”

Solution of Quadratic equations

Let x denote the number of arrows.

12+ 4

√x + 6 + 3 + 1 = x

12

x − 4√

x − 10 = 0

∴ x − 8√

x − 20 = 0√

x = 10; x = 100.

(Here√

x = −2 is also a solution. But in the problem√

x has tobe +ve.)

Rule of 3, 5, 7, 9, ...

Investigation of mixtures

Rule in Verse 90.A;Ta :pra;a;NEa;gRua; a;Na;ta;aH .~va;k+:a;l+.aH :pra;t�a;a;ta;k+:a;l;éÈåî Åéé:a;P+.l+.ea:;dÄâx ;ta;a;~tea Á.~va;ya;ea;ga;Ba;�+:a;(ãÉa ;a;va;Æa;ma;(ra;�a;na.Èåî ÁÁ*+;aH :pra;yua;�+:Ka;Nq+.a;�a;na :pxa;Ta;g,a Ba;va;�////�a;nta Á Á“The arguments taken into their respective times onedivided by the fruit taken into the elapsed times; theseveral quotients, divided by their sum, and multipliedby the mixed quantity, and the parts as severally lent.”

Mixtures

Let x , y , z be the portions lent at r1, r2, r3 percent per month andlet I = common interest in t1, t2, t3 months respectively.

Let x + y + z = a, a given quantity. Then,

x × r1 × t1100

=y × r2 × t2

100=

z × r3 × t3100

= I.

∴ x : y : z ::100 × 1r1 × t1

:100 × 1r2 × t2

:100 × 1r3 × t3

.

x =100 × 1r1 × t1

× a100 × 1r1 × t1

+100 × 1r2 × t2

+100 × 1r3 × t3

Here: Argument 100, Time: 1, fruit: r, elapsed time : t1.

Similar expressions for y and z.

Mixtures: Example

Example in Verse 91:ya;t,a :pa:úãÁ*.a;k+:a:�a;k+:. a;tua;Sk+:P+.le +.na d:�MaKa;NqE +.�///�a;~:�a;Æa;Ba;gRa;Na;k ;�a;na;Sk+:Za;tMa :Sa;qU +.na;m,a Áma;a;sea;Sua .sa;�a;d;Za;pa:úãÁ*.a;sua tua;�ya;ma;a;�MaKa;Nq+�a;yea.Y;a;pa ;a;h :P+.lM va;d Ka;Nq+.sMa;K.ya;a;m,a Á Á 91 Á Á“The sum of six less than a hundred nis.kas being lent inthree proportion at interest of 5, 3, and 4 percent, anequal interest was obtained on three portions in 7, 10and 5 months respectively. Tell mathematician, theamount of each portion.”

SolutionsHere a = 94, r1 = 5, r2 = 3, r3 = 4; t1 = 7, t2 = 10, t3 = 5..

Now r1t1 = 35, r2t2 = 30, r3t3 = 20..

∴10035

+10030

+10020

=(6 + 7 + 10.5)

210× 100 =

23521

.

∴ x(portion 1) =10035 × 94

23521

=10035

× 21235

× 94 = 24,

y(portion 2) =10030 × 94

23521

=10030

× 21235

× 94 = 28,

z(portion 3) =10020 × 94

23521

=10020

× 21235

× 94 = 42.

Common interest = I =24 × 5 × 7

100=

840100

= 825.

He considers the filling of a cistern from n fountains : Same as inGan. itasarasangraha

Another kind of problem involving mixtures

Let there be a mixture of n items. Let the relative proportion oftype i in the Mixture = βi .

∴ Relative proportion = β1 : β2 : · · · βn.

(Fraction of type i in the mixture = β′

i =βi

∑βi;∑

β′

i = 1).

Let the price of type i be xi per measure (volume or weight).

Let the total amount of the mixture = A (volume or weight).

∴ Total price =∑

(Aβ′

i ) xi =A∑

xiβi∑

βi= Mixed Sum = X ,

Formal solution

As the amount of item i in the mixture is Aβ′

i , and xi is the price permeasure.

∴ A =X∑

βi∑

xiβi.

∴ Amount (Measure) of type i = β′

i A =βiX

∑xiβi

=Portion × Mixed sum

Sum of quotients.

(as βi = β′

i

∑

βi)

.

Price of item of type i =xiβi × X∑

xiβi=

(Price of i) × Portion × Mixed SumSum of quotients

.

.

Mixed quantities: Example

Example in Verse 98.k+:pRUa:=+~ya va:=+~ya ;�a;na;Sk+:yua;ga;le +.nEa;kM :pa;lM :pra;a;pya;tea;vEa;Zya;a;na;nd;na;. a;nd;na;~ya .. a :pa;lM dÒ +mma;a;�;Ba;a;gea;na .. ea;t,a ÁA;�;Ma;Zea;na ta;Ta;a.Y;ga:=+eaH :pa;l+.d;lM ;�a;na;Ske +:Na mea :de ;a;h ta;a;n,aBa;a;gEa;=e +k+:k+:Sa;ea;q+.Za;a;�;k+:Æa;ma;tEa;DRUa;pMa ;�a;. a;k +:a;Sa;Ra;}ya;h;m,a Á Á 98 Á Á“If a pala of best camphor may be had for two nisk ′s(=32 drammas) and a pala of sandalwood for the eighthpart of a dramma and half a pala of alae wood also foreighth of a dramma, good merchant, give me the valuefor one nis.ka (=16 drammas) in the proportions of 1, 16,and 8, for I wish to prepare a perfume.”

Solution

(1) (2) (3)Here proportion: Camphor: Sandalwood: Alaewood = 1 : 16 : 8

Prices: 32 1 18

14

(x1) (x2) (x3)

∑xiβi = 32 × 1 +

18× 16 +

14× 8 = 36.

Total price = Mixed amount = 16 drammas = X =16.

Quotient, q =Mixed sum

Sum of quotients=

X∑xiβi

=1636

=49.

Amounts of item i =βiX∑

xiβi= βi

49, Price = xi

βi X∑xiβi

= xi × Amount.

∴ Amount of Camphor: 1×49, Sandalwood = 16×

49=

649, Alaewood = 8×

49=

329

Prices : Camphor: 32×49=

1289

, Sandal:18×

649

=89, Alaewood: =

14×

329

=89.

Merchants, trading. Alligation: Mixture of gold of different quantities.

Section six on combinations

Verse 130-132.O;;k+:a;dùÅ;ae ;k+:ea:�a:=+a;ñÍö�ÅÅ*:+.a v.ya;~ta;a Ba;a:$ya;aH :pxa;Ta;�//////////////////�a;#~Ta;tEaH Á:pa:=.H :pUa;veRa;Na .sMa;gua;Nya ta;tpa:=+~tea;na .tea;na .. a Á Á 130 Á ÁO;;k+:a;dõ :�ya;a;a;d;Bea;d;aH .~yuaH I+.dM .sa;a;Da;a:=+NMa .smxa;ta;m,a ÁC+.nd; a;(ãÉa:�yua:�a:=e C+.nd;~yua;pa;ya;ea;ga;ea.Y;~ya ta;a;dõ ;d;a;m,a Á Á 131 Á Ámxa;Sa;a;va;h;na;Bea;d;a;d;Ea Ka;Nq+.mea:=+Ea .. a ; a;Za;�pa;ke Á;vEa;dùÅ;a;ke .=+sa;Bea;d� ;a;yea ta;�a;ea;�M ;a;va;~txa;tea;BRa;ya;a;t,a Á Á 132 Á Á“Let the figures of one upwards, differing by one, put in theinverse order, be divided by the same (arithmeticalsequence) in direct order; and let the subsequent bemultiplied by the preceding, and the next following by theforegoing (result). The several results are the changes, ones,twos, threes, etc. This is termed a general rule.”

Combinations

It serves in prosody, for those versed therein, to find thevariations of metre; in the arts (as in architecture) tocompute the changes upon apertures (of a building);and (in music) the scheme of musical permutations; inmedicine, the combinations of different savours. Forfear of prolixity, this is not (fully) set forth.”

n1,

n · (n − 1)1 · 2

, . . . ,n(n − 1)(n − 2)(n − r) . . . (n − r + 1)

1 · 2 · 3 · · · r =n!

r !(n − r)!

Out of n: 1 at a time: 2 at a time: 3 at a time, ... , r at a time.

Combinations: Example

Example in Verse 134.O;;k+:a;dõ :�ya;a;a;d;mUa;Sa;a;va;h;na;Æa;ma;�a;ta;ma;h;ea b.rUa;a;h mea BUa;Æa;ma;Ba;tRuaH h;}yeRa.=+}yea.Y;�;mUa;Sea .. a;tua:=+a;va:=+�a;. a;tea (ìÉÅ;+[Na;Za;a;l+.a;a;va;Za;a;le ÁO;;k+:a;dõ :�ya;a;a;d;yua;��+.a;a ma;Dua:=+k+:fu ;k+:Sa;a;ya;a;}åÅÉì++.k+:[a;a:=+�a;ta;�E ;H O;;k+:�/////////�a;sma;n,a:Sa;q" +.sEaH .~yua;gRa;Na;k k+:�a;ta va;d v.ya:úêÁÁ*+;nea v.ya;�a;�+:Bea;d;aH Á Á“In a pleasant, spacious and elegant edifice with eight doors,constructed by a skillful architect, as a palace for the lord ofthe land, tell me combinations of doors taken one, two, three,etc. Say, mathematician, how many are the combinations inone composition, with ingredients of six different tastessweet, pungent, astringent, sour, salt and bitter, taking themby ones, twos, threes etc.”

8 7 6 5 4 3 2 1

1 2 3 4 5 6 7 8

Example contd.

No. of ways: 1 at a time = 81 = 8, 2 at a time = 8×7

1×2 = 28, 3 at a time= 8×7×6

1×2×3 = 56, 4 at a time = 8×7×6×51×2×3×4 = 70. 5 at a time

= 8×7×6×5×41×2×3×4×5 = 56, 6 at a time = 8×7×6×5×4×3

1×2×3×4×5××6 = 28, 7 at a time= 8×7×6×5×4×3×2

1×2×3×4×5×6×7 = 8, 8 at a time = 8×7×6×5×4×3×2×11×2×3×4×5×6×7×8 = 8.

Total no. of variations = 28 − 1 = 255.

Combinations of 6 different tastes:

6 5 4 3 2 1

1 2 3 4 5 6

No. of different types of compositions: 6, 15, 20, 15, 6, 1. Total= 26 − 1 = 63.

Chapter 5. Progressions.

Standard results on∑

r ,∑

r2,∑ r(r+1)

2 ,∑

r3, etc., Arithmetic andGeometric Progressions.

Number of verses with a fixed number of syllables

Verse 130-131::pa;a;d;a;[a:=+Æa;ma;ta;ga;. Ce gua;Na;va;gRa;P+.lM .. a;yea ;a;dõ ;gua;Nea Á.sa;ma;vxa:�a;a;na;Ma .sa;*ñÍËÉ ùÁ+;a;a ta;dõ ;ga;eRa va;gRa;va;gRa;(ãÉa Á.~va;~va;pa;d;ea;na;Ea .~ya;a;ta;Ma A;DRa;sa;ma;a;na;Ma .. a ;a;va;Sa;ma;a;Na;a;m,a Á Á 130, 131Á Á“The number of syllables in a verse being taken for theperiod, and the increase twofold, the produce ofmultiplication and squaring (as above directed) will bethe number of (variations) of like verses. Its square andsquare’s square, less their respective roots, will be (thevariations of) alternately similar and of dissimilarverses.”

This rule refers specifically to the example in Verse 132.

Example

Example 132..sa;ma;a;na;a;ma;DRa;tua;�ya;a;na;Ma ;a;va;Sa;ma;a;Na;Ma :pxa;Ta;k, :pxa;Ta;k, Ávxa:�a;a;na;Ma va;d mea .sa;*ñÍËÉ ùÁ+;a;Ma A;nua;�u ;pC+.nd;Æa;sa dÒu +ta;m,a Á Á 140 Á Á“Tell me directly the number (of varieties) of like,alternating-like, and dissimilar Verses respectively inthe metre named anus. tup.”

4 caran. as

.

x x x x x x x x 8 Syllablesx x x x x x x x 8x x x x x x x x 8x x x x x x x x 8 Total 32 Syllables.

◮ Variations of like Verses. (All the caran. as are alike.)= 28 = 256.

Example

◮ Alternatively, alike caran. as. To find the number of these, wefind the number of varieties of the syllables in two caran. as

= 216 = 65536. If we place each one of these under itself,we get all the cases included in “alternatively like”. But ofthese, the number of cases of “all like” are also included.Therefore number of cases of “only alternatively like”= 216 − 28 = 65536 − 256 = 65280.

◮ Total no. of variations = 232. Subtracting the case of “alllike” and “alternatively like”, Number of dissimilar Verses= 232 − 216. Note that the “dissimilar” only means itexcludes “all like” and “alternatively like” and does not meanthat all the caran. as are dissimilar. For instance, it includescases in which “first two caran. as are alike, as also last two,etc.”

Plane FiguresChapter 6 is on this topic.

As expected, begins by considering Right Triangles.

Diagonal Upright

side A right triangle

Concentrates on Rational triangles. One type is described inVerse 139.I+.�e ;na Bua:$a;ea.Y;sma;a;d, ;a;dõ ;gua;Nea;na;�a;na.Èåî ÁÁ*+;a;t,a I+.�;~ya kx +:tyEa;k+:a;va;yua;�+:ya;a;�a;m,a Ák+:ea;�a;fH :pxa;Tua;k, .sea;�;gua;Na;a;Bua:$a;ea;na;a k+:Na;eRa Ba;vea;t,a �ya;(ra;Æa;ma;dM tua .$a;a;tya;m,a Á Á

A Rational right triangle

“A side is put. From this multiplied by twice some assumednumber, and divided by one less than the square of the assumednumber, the upright is obtained. This, being set apart, ismultiplied by the arbitrary number, and the side as put issubtracted; the remainder will be the hypotenuse. Such atriangle is termed right-angled.”

a(n2+1)

(n2−1)

2ann2−1

a

It is stated that side = a, upright =2an

n2 − 1yields a right triangle

with hypotenuse =2an

n2 − 1× n − a =

a(n2 + 1)n2 − 1

.

Explanation by Suryadasa in his commentaryConsider two triangles. The first triangle has side = n2

− 1, upright = 2n anddiagonal n2 + 1. Correct as (n2 + 1)2 = (n2

− 1)2 + (2n)2. Let the secondtriangle be similar to the above, with side = a.

n2+ 1 2n

n2− 1

(a)

?2ann2−1

a

(b)Similar right triangles

The upright of the second triangle =a

n2− 1

× 2n.

Now, hypotenuse of the first triangle= upright × n − side = 2n × n − (n2

− 1) = n2 + 1.

Then, hypotenuse of the second triangle

= Upright × n − side =2an

n2− 1

× n − a =a(n2 + 1)

n2− 1

.

References

1.H.T.Colebrooke, Algebra with Arithmetic and Mensuration fromthe Sanskrit of Brahmagupta and Bhaskara, London 1817; Rep.Sharada Publishing House, New Delhi, 2006.

2. Bhaskaracarya’s Lılavatı with Colebrooke’s translation andnotes by H.C.Banerji, Second Edition,The Book Company,Calcutta, 1927; Rep. Asian Educational Series, New Delhi, 1993.

3.Lılavatı of Bhasakaracarya II: Ed. with Bhaskara’s Vasana andBuddhivilasanı of Gan. esa Daivajna by V.G.Apte, 2 Vols., Pune,1937.

4.Lılavatı of Bhasakaracarya , A Treatise of Mathematics of VedicTradition, Tr. by K.S.Patwardhan, S.M.Naimpally and S.L.Singh,Motilal Banarsidass, Delhi, 2001.

Thanks!

Thank You