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Mathematics in Ancient China

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Mathematics in Ancient China Mathematics in Ancient China Chapter 7 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 Classical Minoan Mycenaean Dark Archaic Christian Hellenistic Roman GREECE 3000 BCE 2500 BCE 2000 BCE 1500 BCE 1000 BCE 500 BCE 0 CE 500 CE 1000 CE 1500 CE Warring CHINA Shang Zhou Han Warring States Tang Yuan / Ming States Song Gnomon, Liu Hui Nine Chapters Zu Chongzhi Li Zhi Qin Jiushao Yang Hui Zhu Shijie Early Timeline • 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 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. 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. 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 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? 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:” 123 232 311 26 34 39 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). 103 252 311 26 24 39 Linear Equations • Do the same with the left column: 003 452 811 39 24 39 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: 003 003 452 052 811 36 1 1 39 24 39 99 24 39 Linear Equations • This was equivalent to a downward Gaussian reduction. The author then described how to “back substitute” to get the correct answer. 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. Our Method, Maybe Let x be amount of husked rice, y be amount of unhusked rice. Then and . So , and substituting we have . Simplifying, we get , and , or 2 dou, 5 sheng. 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. Double False Position ∙∙ • Why does this work? • We want to solve In general, we’ll examine a method for solving . 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: Double False Position , and . Subtracting these equations gives: . Finally, dividing this equation by gives us: . Finally, if is a surplus and is a deficit, we can say . Gou Gu in Zhou Bi Liu Hui’s Proof of Gou Gu Song Dynasty (900 – 1279) • 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. Numeration • Numerals on the Oracle Stones: Numeration Numeration Hindu ‐ 0 1 2 3 4 5 6 7 8 9 10 100 1000 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 Counting Rod System Counting Rods • Counting rods allowed for a number of very quick calculations, including the basic four arithmetic operations, and extraction of roots. • Some examples: 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. Multiplication with Counting Rods 3. Multiply one’s digit on top by tens digit on bottom, and add to middle. 4. Since you’ve used the tens digit on top, erase it. 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. 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. 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. 2. 7 doesn’t go into 4, so shift the 7 to the right 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. 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. 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. 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. 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. 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.” Approximations of π • Liu Hui, 260 AD: 3.1416 (by inscribing hexagon in circle, using the Pythagorean Theorem to approximate successively polygons of sides 12, 24, ….,96). • Zu Chongzhi, 480 AD: between 3.1415926 and 3.1415927 (by similar method, but moving past 96 to oh, say 24,576). 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. Fan Fa: 11‐7 ‐321 • Starting with a guess of 1 ‐6 ‐9 1, we do synthetic 1 ‐6 ‐912 division to get 11‐6 ‐9 remainder 12. Then 1 ‐5 ignoring the remainder, 1 ‐5 ‐14 we do another synthetic 11‐5 division on the 1 quotient, and repeat 1 ‐4 until we get down to a 11 constant. 1 Fan Fa: 11‐7 ‐321 • The remainders on the 1 ‐6 ‐9 first and second lines 1 ‐6 ‐9 12 are divided and 11‐6 ‐9 multiplied by ‐1 to 1 ‐5 obtain the next 1 ‐5 ‐14 “adjustment” to the 11‐5 guess for a root. 1 1 ‐4 • Also, 11 1 GUESS: 0.857143 Fan Fa: 11‐7 ‐321 • The remainders give 1 ‐6 ‐9 you the polynomial for 1 ‐6 ‐9 12 the next round of 11‐6 ‐9 synthetic division.
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