Mathematics in Ancient China. Chapter 7

Transcription

1 Mathematics in Ancient China Chapter 7

2 Timeline Archaic Old Kingdom Int Middle Kingdom Int New Kingdom EGYPT 3000 BCE 2500 BCE 2000 BCE 1500 BCE 1000 BCE Sumaria Akkadia Int Old Babylon Assyria MESOPOTAM IA 3000 BCE 2500 BCE 2000 BCE 1500 BCE 1000 BCE 500 BCE 0 CE 500 CE Minoan Mycenaean Dark Classical Archaic Hellenistic Roman Christian GREECE 3000 BCE 2500 BCE 2000 BCE 1500 BCE 1000 BCE 500 BCE 0 CE 500 CE 1000 CE 1500 CE CHINA Shang Zhou Warring States Han Warring States Tang Song Yuan / Ming Gnomon, Nine Chapters Liu Hui Zu Chongzhi Li Zhi Qin Jiushao Yang Hui Zhu Shijie

3 Shang Dynasty: Excavations near Huang River, dating to 1600 BC, showed oracle bones tortoise shells with inscriptions used for divination. This is the source of what we know about early Chinese number systems. Early Timeline

4 Early Timeline

5 Han Dynasty ( 206 BC 220 AD) System of Education especially for civil servants, i.e. scribes. Two important books*: Zhou Bi Suan Jing (Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) Jiu Zhang Suan Shu (Nine Chapters on the Mathematical Art) *unless of course we re off by a millennium or so.

6 Nine Chapters This second book, Nine Chapters, became central to mathematical work in China for centuries. It is by far the most important mathematical work of ancient China. Later scholars wrote commentaries on it in the same way that commentaries were written on The Elements.

7 Chapters in uh, the Nine Chapters 1. Field measurements, areas, fractions 2. Percentages and proportions 3. Distributions and proportions; arithmetic and geometric progressions 4. Land Measure; square and cube roots 5. Volumes of shapes useful for builders. 6. Fair distribution (taxes, grain, conscripts) 7. Excess and deficit problems 8. Matrix solutions to simultaneous equations 9. Gou Gu: ; astronomy, surveying

8 Linear Equations There are three classes of grain, of which three bundles of the first class, two of the second, and one of the third, make 39 measures. Two of the first, three of the second, and one of the third make 34 measures. And one of the first, two of the second and three of the third make 26 measures. How many measures of grain are contained in one bundle of each class?

9 Linear Equations Solution: Arrange the 3, 2, and 1 bundles of the 3 classes and the 39 measures of their grains at the right. Arrange other conditions at the middle and the left:

10 Linear Equations With the first class on the right multiply currently the middle column and directly leave out. (That is, multiply the middle column by 3, and then subtract some multiple of the right column, to get 0)

11 Linear Equations Do the same with the left column:

12 Linear Equations Then with what remains of the second class in the middle column, directly leave out. In other words, repeat the procedure with the middle column and left column:

13 Linear Equations This was equivalent to a downward Gaussian reduction. The author then described how to back substitute to get the correct answer.

14 Method of Double False Position Or, Excess and Deficit. A tub of capacity 10 dou contains a certain quantity of husked rice. Grains (unhusked rice) are added to fill up the tub. When the grains are husked, it is found that the tub contains 7 dou of husked rice altogether. Find the original amount of husked rice. Assume 1 dou of unhusked rice yields 6 sheng of husked rice, with 1 dou = 10 sheng.

15 Our Method, Maybe Let x be amount of husked rice, y be amount of unhusked rice. Then and we have get. So, and substituting, and. Simplifying, we, or 2 dou, 5 sheng.

16 Method of Double False Position If the original amount is 2 dou, a shortage of 2 sheng occurs. If the original amount if 3 dou, there is an excess of 2 sheng. Cross multiply 2 dou by the surplus 2 sheng, and then 3 dou by the deficiency of 2 sheng, and add the two products to give 10 dou. Divide this sum by the sum of the surplus and deficiency to obtain the answer 2 dou and 5 sheng.

17 Double False Position Why does this work? We want to solve In general, we ll examine a method for solving.

18 Double False Position So suppose we want to solve. We ll do it by making two guesses and, with the respective errors, and. Then subtracting these equations gives. Next, multiplying equation 1 by and equation 2 by we get:

19 Double False Position, and. Subtracting these equations gives:. Finally, dividing this equation by gives us: a deficit, we can say. Finally, if is a surplus and is.

20 Gou Gu in Zhou Bi

21 Liu Hui s Proof of Gou Gu

22 Song Dynasty ( ) Two Books by Zhu Shijie had topics such as: Pascal s triangle (350 years before Pascal) Solution of simultaneous equations using matrix methods Celestial element method of solving equations of higher degree. (Horner s method) European algebra wouldn t catch up to this level until the 1700 s.

23 Numeration Numerals on the Oracle Stones:

24 Numeration

25 Numeration Hindu Arabic Chinese 〇一二三四五六七八九十百千 Financial 零壹贰叁肆伍陆柒捌玖拾佰仟

26 Counting Rod System

27 Counting Rods Counting rods allowed for a number of very quick calculations, including the basic four arithmetic operations, and extraction of roots. Some examples:

28 Multiplication with Counting Rods 1. Set up the two factors, in this case 68 and 47, such that the ones digit of the bottom is aligned with the tens digit of the top. Leave room in middle for calculations. 2. Multiply tens digit of top by tens digit of bottom, place in middle with ones over the bottom s tens digit. 𝍡 𝍬 𝍬 𝍬 𝍥 𝍦 𝍥 𝍦 𝍰 𝍰

29 Multiplication with Counting Rods 3. Multiply one s digit on top by tens digit on bottom, and add to middle. 𝍥 𝍡 𝍬 𝍬 𝍡 𝍬 𝍦 𝍰 4. Since you ve used the tens digit on top, erase it. 𝍡 𝍰 𝍡 𝍬 𝍦 𝍰

30 Multiplication with Counting Rods 5. Move bottom digits to the right one space. 6. Multiply tens digit on bottom by ones digit of top; place ones digit of answer above tens on bottom, and 7. Combine. 𝍡 𝍰 𝍡 𝍣 𝍡 𝍰𝍫 𝍡𝍡 𝍣 𝍢 𝍩 𝍣 𝍣 𝍰 𝍯 𝍰 𝍯 𝍰 𝍯

31 Multiplication with Counting Rods 8. Erase the tens digit on bottom because you re done with it. 9. Finally, multiply the two units digits, and add them to the middle. 10. Combine, and erase the units digits on top and bottom. 𝍰 𝍢 𝍩 𝍣 𝍯 𝍰 𝍢 𝍩 𝍣𝍤 𝍮 𝍯 𝍢 𝍩 𝍨 𝍮

32 Division with Counting Rods 1. Place the dividend, 407, in the middle row and the divisor, 9, in the bottom row. Leave space for the top row. 𝍬 𝍱 𝍯 2. 7 doesn t go into 4, so shift the 7 to the right 𝍬 𝍨 𝍯

33 Division with Counting Rods 3. Nine goes forty 4 times with a remainder of 4; write the quotient in the top row, the remainder in the middle. 4. Shift the 9 to the right one digit. 𝍣 𝍣 𝍨 𝍣 𝍣 𝍯 𝍯 𝍱

34 Division with Counting Rods 5. Nine goes into 47 five times, with a remainder of 2. Put 5 on top, remainder in the middle. 6. The answer is, 45, remainder 2. 𝍣 𝍣 𝍭 𝍪 𝍱 𝍭 𝍪

35 Fractions From Nine Chapters: If the denominator and numerator can be halved, halve them. If not, lay down the denominator and numerator, subtract the smaller number from the greater. Repeat the process to obtain the greatest common divisor (teng). Simplify the original fraction by dividing both numbers by the teng.

36 Fractions Addition and subtraction were done as we do them but without necessarily finding least common denominators the common denominator is just the product of the two denominators. The fraction is simplified after adding or subtracting.

37 Fractions Multiplication was done as we do it. Division was done by first getting common denominators, then inverting and multiplying so that the common denominators cancel. Then the fraction was simplified.

38 Negative numbers? Red and black rods, or rods laid diagonally over others. For subtractions with the same signs, take away one from the other; with different signs, add one to the other; positive taken from nothing makes negative, negative from nothing makes positive. For addition with different signs subtract one from the other; with the same signs add one to the other; positive and nothing makes positive; negative and nothing makes negative.

39 Approximations of π Liu Hui, 260 AD: (by inscribing hexagon in circle, using the Pythagorean Theorem to approximate successively polygons of sides 12, 24,.,96). Zu Chongzhi, 480 AD: between and (by similar method, but moving past 96 to oh, say 24,576).

40 Solving Polynomials Precious Mirror of the Four Elements by Shu Shi jie, 1303 CE. Method known as Fan Fa, today known as Horner s Method, and using what you may know as synthetic division.

41 Fan Fa: Starting with a guess of 1, we do synthetic division to get remainder 12. Then ignoring the remainder, we do another synthetic division on the quotient, and repeat until we get down to a constant

42 Fan Fa: The remainders on the first and second lines are divided and multiplied by 1 to obtain the next adjustment to the guess for a root. Also, GUESS:

43 Fan Fa: The remainders give you the polynomial for the next round of synthetic division

44 Fan Fa: Our suggested next guess was We try 0.8, but the negative sign on the next guess ( ) tells us our 0.8 was too large. We go back to GUESS:

45 Fan Fa: This time, we get a positive next guess of So we take 0.03 as our guess for the next digit, and go again. And, again, we take the end digits from each result line to populate our next top line GUESS:

46 Fan Fa: Running the algorithm again, we get a next guess of.002. So far then, our approximate root is 1.73 and is heading for about GUESS:

47 Fan Fa: Our next guess is positive and very small, so the error in our current approximation is small. Best guess: ( ish?) GUESS: 5.08E 05

48 Fan Fa: Aw, heck. Just for fun, let s go one more: As you can see, we re getting very close to our calculator provided approximation E E E GUESS: 8.08E 07

49 Fan Fa and Horner This general method was rediscovered by William Horner ( ) and published in a paper in Except it was pretty much identical to a method published in 1820 by Theopholis Holdred, a London watchmaker. Of course, Paolo Ruffini ( ), who we will discuss later in another context, already won a prize for outlining this method in Italy. And, of course, there s Shu Shi jie, more than four centuries earlier.

50 Magic Squares

51 The semi mythical Emperor Yu, (circa 2197 BC) walking along the banks of the Luo River, looked down to see the Divine Turtle. On the back of his shell was a strange design. Lo Shu

52 Lo Shu When the design on the back was translated into numbers, it gave the 3x3 magic square. Saying the 3x3 magic square is appropriate because it is unique up to rotations and reflections.

53 According to legend, the He Tu is said to have appeared to Emperor Yu on the back of (or from the hoof prints of) a Dragon Horse springing out of the Huang (Yellow) River. He Tu

54 He Tu When it was translated into numbers, it gave a cross shaped array. To understand its meaning is to understand the structure of the universe, apparently. Or, at least to understand that, disregarding the central 5, the odds and evens both add to

55 Magic Squares Yang Hui, Continuation of Ancient Mathematical Methods for Elucidating the Strange Properties of Numbers, 1275.

56 Order 3 Arrange 1 9 in three rows slanting downward to the right

57 Order 3 Arrange 1 9 in three rows slanting downward to the right. Exchange the head (1) and the shoe (9)

58 Order 3 Arrange 1 9 in three rows slanting downward to the right. Exchange the head (1) and the shoe (9). Exchange the 7 and

59 Order 3 Arrange 1 9 in three rows slanting downward to the right. Exchange the head (1) and the shoe (9). Exchange the 7 and 3. Lower 9, and raise

60 Order 3 Arrange 1 9 in three rows slanting downward to the right. Exchange the head (1) and the shoe (9). Exchange the 7 and 3. Lower 9, and raise 1. Skootch* in the 3 and 7 *technical term

61 Order 3 The Lo Shu

62 Order 4 Write 1 16 in four rows

63 Order 4 Write 1 16 in four rows. Exchange corners of outer square

64 Order 4 Write 1 16 in four rows. Exchange corners of outer square Exchange the corners of inner square

65 Order 4 Write 1 16 in four rows. Exchange corners of outer square Exchange the corners of inner square

66 Order 4 Write 1 16 in four rows. Exchange corners of outer square Exchange the corners of inner square. Voila! Sum is

67 Order 4 Other magic squares of order 4 are possible for different initial arrangements of the numbers

68 Order 5, 6, 7,. Yang Hui constructed magic squares of orders up through 10, although some were incomplete.

69 A Little About Magic Squares Normal magic squares of order n are n x n arrays containing each number from 1 through They exist for all. The sum of each row, column, and diagonal is the magic number M which for normal magic squares depends only on n.. For the first few n s this is 15, 34, , For n odd, the number in the central cell is