Ganita and Kuttaka: the calculation of Indian mathematics in the classical period ( )

Transcription

1 ESU-8 Oslo, July 2018 Ganita and Kuttaka: the calculation of Indian mathematics in the classical period ( ) Iolanda Guevara 1 & Carles Puig-Pla 2 (1) Dept. d Ensenyament de la Generalitat de Catalunya; Universitat Autònoma de Barcelona iguevara@xtec.cat (2) Dept. de Matemàtiques, Universitat Politècnica de Barcelona carles.puig@upc.edu

2 In the classroom In relation to the students it is necessary to: - Locate the country - Identify the time - Show the context in which it was developed - State the problem to solve. 2

3 Ganita and Kuttaka In this presentation we will focus on some of the issues related to Aryabhata and Brahamagupta calculation implemented in the classroom ( ). The Reversed procedure from Aryabhata was presented during a Preservice teacher course: Interuniversity Master in Teacher Training -Universitat de Barcelona, Universitat Autònoma de Barcelona, Universitat Politècnica de Catalunya, Universitat Pompeu Fabra- (UB, UAB, UPC, UPF). The Kuttaka method from Aryabhata and Brahamagupta was offered in an elective course of History of mathematics for students of the Degree in Mathematics -Universitat Politècnica de Catalunya- (UPC). 3

4 In the classroom: The Reversed procedure In a Preservice teacher course: Interuniversity Master in Teacher Training (UB-UAB-UPC-UPF) To locate the country and the historical time. To identify the context in which it was developed. To raise the problem that was wanted to solve. A short explanation of the method from the Aryabhatiya of Aryabhata with commentary of Bhaskara I. Students are asked to solve an activity. 4

5 Mathematics in Ancient India Ancient India = Indian subcontinent India, Nepal, Pakistan, Bangladesh and Sri Lanka 5

6 Mathematics in Siddhanta texts In the 4th and 5th centuries: A remarkable mathematical activity in India related to astronomy Siddhanta texts: treatises on mathematical astronomy Texts with instructions to calculate positions of celestial bodies and solve questions related to the calendar, geography or astrology. They included whole chapters devoted exclusively to mathematical calculus (ganita). 6

7 Predictive astronomy The computation of times, locations and appearances of celestial phenomena future or past as seen from any given terrestrial location. Computational astronomy Siddhanta texts Explanations on computational procedures in terms of the geometry of the spherical models (presented in a separate section called gola (sphere). Instruction in general mathematical knowledge Basic arithmetic operations, calculation of interest on loans, rules for finding areas, volumes, sum of series... (ganita). 7

8 Aryabhatiya & Brahma-sphuta-siddhanta Two Indian astronomers and mathematicians, Aryabhata ( ) and Brahamagupta ( ), wrote their Siddhanta texts: Aryabhatiya and Brahma-sphuta-siddhanta, which were references for later astronomers and mathematicians. In Indian mathematics, Aryabhatiya played, in some way, the role of the Elements of Euclid in Greek mathematics. 8

9 Siddhanta s structure The Siddhanta texts include some chapters dedicated exclusively to the calculation in the proper mathematical sense. For example: In Aryabhatiya (4 chapters) Chapter 2 (on ganita or calculation) In Brahma-sphuta-siddhanta (24 chapters) [...] Chapter 12. Calculation with numbers (ganita) [...] Chapter 18. Calculation with unknown quantities (kuttaka) [...] 9

10 Aryabhatiya (499) The earliest completely preserved Siddhanta Written by Aryabhata (born in 476). Chapter 2 is devoted to mathematics (ganita). 33 verses in Sanskrit metric named arya. Bhaskara I (629) wrote Aryabhatiyabhasya, a comment in Sanskrit prose about the 33 verses of chapter 2 of the Aryabhatiya. 10

11 Aryabhatiya. Chapter 2 Salutation (1) Defense of positional notation (2) Geometric and arithmetic procedures, calculation of areas and volumes, and square and cubic roots Area of the circle (two approximations of π) π = 10 & π = / (=3,1416) (3-10) Computation of half-chords (sines), relations between shadows (11-17) 11

12 Aryabhatiya. Chapter 2 Sum of natural numbers, squares and cubes (19) Calculations of interests produced by a capital (25) Methods for resolution of: first-degree equations, quadratic equations, first-degree indeterminate equations (26-33) 12

13 Aryabhata in Aryabhatiya (Chapter 2, 28 ) From the Aryabhatiya of Aryabhata with commentary of Bhaskara I, it is known what Aryabhata said about how to teach the Reversed procedure (Chapter 2, 28 ): In a reversed [operation], multipliers become divisors and divisors, multipliers, and an additive [quantity], is a subtractive [quantity], a subtractive [quantity] an additive [quantity]. 13

14 The activity The Reversed procedure of Aryabhata In Aryabhatiya algebraic equations appear (in rhetorical language) solved by the Reversed procedure. This method starts from the final result and performs the reversed operations in the opposite direction as given in the statement. 14

15 The activity The procedure can be illustrated through the following problem: A number is multiplied by 3, the product is added to its three quarters, the sum is divided by 7, the ratio is subtracted from its third part, the difference is multiplied by itself, the square is reduced to 52, from the difference is extracted from the square root, which is added 8, that sum is divided by 10 and the result is finally 2. What is this number? 15

16 The activity Indicate each of the inverse operations from the ellipsis in order to give the solution following the guideline of the beginning of the resolution. 16

17 The activity Guidelines to solve the exercise In order to perform the reversed operation in one of the steps, it may be useful to bear in mind that expressions of the type "remove (or subtract) the third part from an amount" or "add up to three quarters" can be thought of as that the amount has multiplied by a certain fraction. 17

18 The activity Resolution The Reversed procedure is as follows: the final result is 2 and the last operation before reaching 2 is to divide by 10, so we do the "reversed operation" to multiply 2 by 10: 2 x 10 = 20 The previous operation (the penultimate one) consisted of adding 8, therefore what we will do will be... 18

19 Students productions The activity was carried out by 26 students They formed freely 7 groups that had to be about 3 or 4 students 1 group of 2 students 1 group of 3 students 4 groups of 4 students 1 group of 5 students 19

20 Students productions The students had in the statement the pattern to follow (at the beginning of the resolution). They were asked to write a rhetorical explanation about the previous operation to carry out the reversed operation. Although all the groups arrived at the correct solution, only one of them followed "exactly" this pattern. 20

21 Students productions 21

22 Students productions All students indicated correctly the operations to be performed in each of the eight steps but "without" or "with few" rhetorical explanations 22

23 Students productions Two groups introduced modern mathematical formalism by writing equations (using the "x" for the unknown) 23

24 Final Remarks (students) Preservice teacher course: Interuniversity Master in Teacher Training (UB-UAB-UPC-UPF). About Reversed procedure In general, we can say that students easily understood the Reversed procedure although not all of them were able to solve the problem "in the manner of Aryabhata" and without using the modern formalism of algebra. 24

25 In the classroom: Kuttaka method In an Elective course of History of mathematics of the Degree in Mathematics (Universitat Politècnica de Catalunya) To locate the country and the historical time. To identify the context in which it was developed. To raise the problem that was wanted to solve (the planets conjunction). A short explanation of the method of Brahmagupta in Brahma-Sphuta-Siddhanta (chapter 18, 3-6) Students are asked: To solve a problem To assess the method To answer a questionnaire about the history of math. 25

26 The time of Brahmagupta (Harsha Empire) Brahmagupta lived between the Gupta Empire ( ) and the first incursions of Islam into the territory of ancient India (711). Harsha or Harshavardhana (ca ) - Was proclaimed Maharaja (great king) in adopted Buddhism as a religion. - established the court in Kannauj (cosmopolitan center). - held diplomatic relations with the Tang Dynasty 26 Brahmagupta ( )

27 The Harsha Empire Extension of the Harsha Empire Kannauj and its area of influence 27

28 Brahmagupta 628: He wrote the Brahma-sphuta-siddantha (Corrected Teatrise of Brahma). He was 30 years old. In Ujjain? 1300 km from Pataliputra, in the north-east of India (part of the Harsha Empire). Ujjain: the most prominent center of Hindu mathematics and astronomy. It had the best astronomical observatory in India. In Bhillamala? Capital of the Gurjara-Pratihara dynasty. Bhillamala Pataliputra 28

29 The work of Brahmagupta In some way, Brahma-sphuṭa-siddhanta (628) is a response to Aryabhatiya (499) by Aryabhata. He criticized, for example, that in a kalpa (the period of time between the creation and recreation of a world) there were 1008 mahayugas (time period) instead of Khanda-khadyaka, written at the age of 67, is a karana or manual of mathematical astronomy with simplified calculations. 29

30 Brahma-sphuṭa-siddhanta Chapters 1-10: Basic topics of astronomy: Average lengths of planets; true lengths of the planets; the problems of daytime rotation; lunar eclipses; solar eclipses; rising and setting of the sun; phases of the Moon; the shadow of the moon; conjunctions of the planets and conjunctions of the planets with the fixed stars. Chapter 11: Criticism of Aryabhatiya Chapter 12: Arithmetic or ganita Chapters 13-17: Other topics related to astronomy Chapter 18: Calculation with unknowns Chapters 19-20: Other topics related to astronomy Chapter 21: Construction of Sines Chapters 22-24: Other topics related to astronomy 30

31 Chapter 18: kuttaka Calculation with unknowns, first and second degree. Kuttaka or pulverizer" method of calculation to treat equations or systems of indeterminate equations A master [acarya] among those who know treatises [is characterized] by knowing the pulverizer, zero, negative [and] positive [quantities], unknowns, elimination of the middle [term, that is, solution of quadratics], single-color [equations, or equations in one unknown], and products of unknowns, as well as square nature [problems, that is, second-degree indeterminate equations]. 31

32 Chapter 18: kuttaka Arithmetic of the positive and negative numbers & zero [The sum] of two positives is positive, of two negatives negative; of a positive and a negative [the sum] is their difference; if they are equal it is zero. The sum of a negative and zero is negative, [that] of a positive and zero positive, [and that] of two zeros is zero [...]. The product of a negative and a positive is negative, of two negatives positive, and of positives positive; the product of zero and a negative, of zero and a positive, or of two zeros is zero. [BSS, Ch. 18, 30 and 33, respectively] 32

33 Chapter 18: kuttaka Verses 43 through 59, refer to techniques and examples to solve equations with an unknown, both first and second degree. In the case of second degree, when a multiple "b" of the unknown added to a multiple "a" of the square of the unknown is equal to a number c The calculation algorithm is focused on removing the middle term 33

34 Chapter 18: kuttaka Pulverizer (kuttaka): successive steps to transform an equation with two unknowns into a simpler one. Given an equation ax + c = by where a and b do not have common divisors, by means of a change of variable, the initial equation is transformed into another equivalent, until reaching an equation that has one of the coefficients equal to 1. From here the solutions are reconstructed until they reach the initial equation. For which part of astronomy were these calculations necessary? 34

35 The problem to solve A Planet A

36 The problem to solve A 22,5 d Planet A

37 The problem to solve A 45 d Planet A

38 The problem to solve A 77,5 d Planet A

39 The problem to solve A 90 d Planet A

40 The problem to solve A 90 d 22,5 d 77,5 d 45 d Planet A

41 The problem to solve A 90 d Planet A

42 The problem to solve A 90 d B Planet A Planet B

43 The problem to solve A 90 d B 8,25 d Planet A Planet B

44 The problem to solve A 90 d B 16,50 d Planet A Planet B

45 The problem to solve A 90 d B 24,75 d Planet A Planet B

46 The problem to solve A 90 d B 33 d Planet A Planet B

47 The problem to solve A 90 d B 33 d 8,25 d 24,75 d 16,5 d Planet A Planet B

48 The problem to solve A 90 d B 33 d Planet A Planet B

49 The problem to solve A 90 d B 33 d A & B Planet A Planet B Planets A & B

50 The problem to solve A 90 d B 33 d A & B 22,5 d Planet A Planet B Planets A & B

51 The problem to solve A 90 d B 33 d A & B 22,5 d 19 d Planet A Planet B Planets A & B

52 The problem to solve A 90 d B 33 d A & B 8,25 d 19 d Planet A Planet B Planets A & B

53 The problem to solve A 90 d B 33 d A & B 22,5 d 8,25 d 19 d 28 d Planet A Planet B Planets A & B

54 The problem to solve A 90 d B 33 d A & B 22,5 d 8,25 d 19 d 28 d How many days (N) have passed since the last conjunction?

55 The problem to solve Planet A Planet B Planets A & B A 90 d B 33 d A & B 28 d 19 d 22,5 d 8,25 d If x is the integer number of revolutions performed by A from the conjunction If y is the integer number of revolutions performed by B from the conjunction N = x N = y

56 The problem to solve How many days (N) have passed since the last conjunction? If x is the integer number of revolutions performed by A from the conjunction If y is the integer number of revolutions performed by B from the conjunction N = x = y 90x - 33y = x - 33y = 9 N = x N = y Linear equation with two unknowns (with infinite integer solutions) We look for the positive integers such as 90x - 33y = 9 56

57 Brahmagupta in Brahma-Sphuta-Siddhanta (chapter 18, 3-6) 1. Divide by the divisor having the greatest remainder [agra] by the divisor having the least remainder [Indication: Euclid's algorithm]. 2. Once mutually divided, the last remainder will be multiplied by an arbitrary number [integer] such that if the product that we obtain is added [if the number of quotients in the process is odd] or we take it [if it is pair] the difference of remainders [the additive], which results is divisible by the penultimate remainder. 57

58 Brahmagupta in Brahma-Sphuta-Siddhanta (chapter 18, 3-6) 3. Place the quotients of the mutual divisions one below the other in columns, until you reach the optional divisor and then the quotient you have obtained. 4. Multiplies the penultimate by the previous one and the one that follows is added to him. Repeat the process [the result is saved in the next column and occupy the penultimate position and then the penultimate number of the previous column]. 5. Divide the last number obtained [agranta] by the divisor having the least remainder. Then multiply the remainder by the divisor having the greatest remainder and add the largest remainder. The result will be the remainder of the product of the divisors [to obtain a smaller solution]. 58

59 The activity Follow the steps indicated by Brahmagupta to solve a problem of the Aryabhatiya by Aryabhata with comments from Bhaskara I (chapter. 2, 33): [A quantity when divided] by twelve has a remainder which is five, and furthermore, it is seen by me [having] a remainder which is seven, when divided by thirty-one. What should one such quantity be? 12y + 5 = N 31x + 7 = N 31x + 2 = 12 y 59

60 A student s solution of the activity 60

61 A student s solution of the activity [A quantity when divided] by twelve has a remainder which is five, and furthermore, it is seen by me [having] a remainder which is seven, when divided by thirty-one. What should one such quantity be? 61

62 A student s solution of the activity 1. Divide by the divisor having the greatest remainder [agra] by the divisor having the least remainder [Indication: Euclid's algorithm] 62

63 A student s solution of the activity 2. Once mutually divided, the last remainder will be multiplied by an arbitrary number (integer) such that if the product that we obtain is added (if the number of quotients in the process is odd) or we take it (if it is pair) the difference of remainders (the additive), which results is divisible by the penultimate remainder. 63

64 A student s solution of the activity 3. Place the quotients of the mutual divisions one below the other in columns, until you reach the optional divisor and then the quotient you have obtained. 64

65 A student s solution of the activity 4. Multiplies the penultimate by the previous one and the one that follows is added to him. Repeat the process [the result is saved in the next column and occupy the penultimate position and then the penultimate number of the previous column]. 65

66 5. Divide the last number obtained (agranta) by the divisor having the least remainder. Then multiply the remainder by the divisor having the greatest remainder and add the largest remainder. The result will be the remainder of the product of the divisors [to obtain a smaller solution]. 66

67 Another student s solution 67

68 The same student using Bézout s identity Let a and b be integers wit h greatest common divisor d. Then, there exist integers x and y such that ax + by = d. 68

69 Questionnaire for students 1.- Did you have any knowledge about Mathematics in ancient India? If yes, how did you get acquainted with it? Explain briefly what you knew about it 69

70 Q1: Students answers Student A: positional notation and decimal base of numbers. Student B: research project on High School about Indian Math (only early civilizations). She remembered that the actual system of numbers came from Indian numbers. 70

71 Questionnaire for students 2.- Before studying the Kuttaka method, did you know any other way to solve this kind of equations (linear with two unknowns)? If yes, where did you learn this other method? Which one? Advantages and disadvantages of both methods? 71

72 Q2: Students answers All of them knew an other method to solve this kind of equation. They learn it in Foundations of Mathematics (1 st year). Only two of them (students A & B) knew the name: Bézout Identity and only one of them (student B) said the full name: Étienne Bézout. In general they prefered Bézout s method. Students A & B like Bézout Identity to solve this equations because they think is a proof but they thought both methods were similar. 72

73 Questionnaire for students 3.- The Kuttaka method is a calculation procedure to solve indeterminate equations. Can you cite applications or contexts (current or historical) in which indeterminate equations are used? 4.- Throughout the Degree in Mathematics or in High School, have you studied the solutions of a linear equation with two variables o unknowns? 73

74 Q3: Students answers Can you cite applications or contexts (current or historical) in which indeterminate equations are used? Five of them relate this procedure to Astronomy, the context used in the classroom to introduce Kuttaka method. The student who knew Étienne Bézout (student B) said that linear equations with two unknowns are also used to calculate the intersection of varieties. Student C referred to Fermat s theorem and Pythagorean Triples. 74

75 Q4: Students answers Throughout the Degree in Mathematics or in High School, have you studied the solutions of a linear equation with two variables o unknowns? All students said they had studied how to solve this equations with integer solutions in Foundations of Mathematics (1 st year). One of them studied Diophantus s method in High School. It s the student who did a research project on High School about Indian Math. 75

76 Questionnaire for students 5.- Is there any subject in the Degree in Mathematics in which the history of the concepts involved was introduced? 6.- In High School were you introduced to the historical development of any mathematical concepts? 76

77 Q5: Students answers Is there any subject in the Degree in Mathematics in which the history of the concepts involved was introduced Five students said: no Three students said: Student A: yes a little, some times teacher introduced the historical context and the biography of the mathematician who discovered a theorem. Student B: yes in many subjects, the biography of the mathematician who discovered a theorem, but never in context. Student C: yes but it depends on the teacher of the subject. 77

78 Q6: Students answers In High School were you introduced to the historical development of any mathematical concepts? Six students said: no. One of them said: I would like to know something. Two students said: no, only about Pythagoras and his theorem. One of them said: the goal of High School is that students understand the concepts to pass the exam to enter university. 78

79 Questionnaire for students 7.- Give your opinion about whether the History of Mathematics can help to understand better the topics studied in the Degree. 8.- For which mathematical topics would you recommend to introduce the history of mathematics? 79

80 Q7: Students answers Give your opinion about whether the History of Mathematics can help to understand better the topics studied in the Degree. All students agree with the importance of knowing the History of Mathematics. They have chosen this elective course. There are two kind of points of view: a) In order to understand better modern concepts. b) Although it is not useful for a better understanding of modern mathematics, it could be interesting for understanding ancient mathematics. Four students are in position a) and four in b). 80

81 Q8: Students answers For which mathematical topics would you recommend to introduce the history of mathematics? Some students associate history with specific topics: geometry, root of a polynomial, limit, calculus, Some others said: history of math must be taught at the beginning, in order to understand why math developed. And in the same sense: in High School or in the 1s year of the Degree. Some others, in general in all the subjects and in the two ways: a) Related to concepts, to understand why other civilizations use other methods. b) Related to problems, to compare motivations to solve problems and to have a wider view of the problem. 81

82 Final Remarks (students) Elective course of History of mathematics of the Degree in Mathematics (Universitat Politècnica de Catalunya) About Kuttaka method They have related the Kuttaka method with other methods of resolution, although they did not remember the name (Bezóut Identity) studied in the first course of the Degree. They prefer the method which they have learning first that the second one (kuttaka). They have liked to learn the new method because it gives them more information and they can appreciate the advantages of the current method. 82

83 Final Remarks (students) Elective course of History of mathematics for students of the Degree in Mathematics About History of Math All students agreed with the importance of knowing History of Mathematics. They have chosen this elective course. They like History in two ways: a) Related to concepts, in order to understand why other civilizations use other methods. b) Related to problems, in order to compare motivations to solve a problem and to have a wider view of the problem. They would have liked to have done History of Math in the High School and in the first courses of the Degree of Math. 83

84 The calculation of Aryabhata and Brahamagupta Aryabhata's (Aryabhatiya) and Brahamagupta's (Brahmasputa-siddhanta) method to solve indeterminate equations in secondary education or in the university. Establish bridges between current methods of resolution and the ancient Indian methods (Reversed procedure and Kuttaka method). Mathematical competences: connections communication & representation. 84

85 Final Remarks The activities based on the analysis of historical texts connected to the curriculum, contribute to improve the students' integral formation giving them additional knowledge of the social and scientific context of the periods involved. Students achieve a vision of mathematics not as a final product but as a science that has been developed on the basis of trying to answer the questions that mankind has been making throughout times about the world around us. 85

86 बह त बह त धन यव द Moltes gràcies Tusen takk Thank you very much Merci beaucoup 86