Suan Shu Shu Bamboo Texts to the Nine Chapters on the Art of Mathematics

Fairbank Center for Chinese Studies Harvard University

The Evolution of Mathematics in Ancient China. From the newly discovered 數Shu and 算數書 Suan shu shu Bamboo Texts to the Nine Chapters on the Art of Mathematics

Joseph W. Dauben Department of History Herbert H. Lehman College and Ph.D. Program in History The Graduate Center City University of New York

Fairbank Center for Chinese Studies Harvard University

The Evolution of Mathematics in Ancient China. From the newly discovered 數Shu and 算數書 Suan shu shu Bamboo Texts to the Nine Chapters on the Art of Mathematics

1934: 楚帛書 Chu boshu The Chu Silk Manuscript State of Chu, Warring States period

Excavated: 1934/1942 Zidanku, Changsha

radiocarbon date: 305±30 BCE

Purchased: 1965 Arthur M. Sackler

數術 “Numerals and Arts/Skills”

If […] and the length of the lunar months becomes too long or too short, then they will not fit the proper degree and spring, summer, autumn, and winter will [not] be […] regular; the sun, moon, and planets will erratically overstep their paths. When (the months) are too long, too short, contrary, or chaotic, (the growth of) the grasses and trees have no regularity. This is [called] yao, “demonic” (influences or omens). When heaven and earth create calamities, the Heaven’s Cudgel (Tianpou) star creates (sweeping) destruction, sending (the destruction) down through all four regions (of the earth). Mountains collapse, springs gush forth geysers. This is called “contravention.” If you contravene the years (and) the months, then upon entering the seventh or eighth day of the month there will be fog, frost, and clouds of dust, and you will not be able to function according (to heaven’s plan) [Li and Cook 1999: 174] 1972–1974: 馬王堆 Mawangdui , Changsha, Hunan (186-168 BCE)

Excavated: 1972 to 1974 Three Western Han dynasty tombs:

Li Cang, chancellor to the prince of Changsha State and the Marquis of Dai; Li Cang’ wife, the Marquise of Dai; and their son.

Tomb No. 1 Tomb No. 2

Tomb No.3 Wooden Tablet with Burial Date Length 30cm Unearthed from Tomb No.3

“On the first Wuchen day of Yisi, the second month of the twelfth year, the chamberlain in charge of funeral services, holding a memorial with funeral articles, presents a list of the delivery for inspection.” i.e. the twelfth year in the reign of Emperor Wendi of Han (168 BCE)

Wooden Outer Coffin from Tomb No.1 Length: 672cm Width: 488cm Height: 280cm

Coffin with Painted Design on Black Lacquer Coating Length 256cm width 118cm height 114cm Unearthed from Tomb No.1

Second Inner Coffin with Painted Design on Vermilion Lacquer Coating Length 230cm width 92cm height 89cm

Innermost Coffin Decorated with Brocade Length:202cm Width:69cm Height:63cm

Xin Zhui Marquise of Dai Scene of Banquet on the T-shaped Paintings on Silk

Book of Changes (Zhou Yi ) Length 30cm width 21.5cm Unearthed from Tomb No.3

“Harmony of Yin and Yang” (He Yin Yang)

Drawing of Daoyin, a Physical Exercise Chart Length 100cm width 50cm (Tomb No. 3)

44 Daoyin positions are illustrated, such as breathing exercises, limb stretching and gymnastics, and each position is labeled with its particular term and function. The exercise of Five Fowls was created by Hua Tuo, a famous doctor in the Eastern Han Dynasty and was based on Dao-yin.

“Ten Questions” (十問 Shi Wen)

Tomb No.3

Prescriptions for 52 Diseases (Wu Shi Er Bing Fang)

Guard of Honor Length 219cm width 99cm Unearthed from Tomb No.3 A painting on silk was found on the western wall of the coffin chamber. At the top left: the tomb owner

Divination by Five Stars (Wu Xing Zhan)

天文氣象雜占 Divination by Astrological and Meteorological Phenomena

Excavated: 1973 Mawangdui Western Han (202 BCE – 9 CE) 29 comets (彗星 hui xing) recorded over a period of about 300 years. ca. 223 BCE

1975: 睡虎地秦简 Shuihudi Qin jian Qin Bamboo Slips from Shuihudi

Excavated: 1975 Tomb #11 at 城關睡虎地

Shuihudi, Yunmeng County, Hubei Province

The tomb belonged to a Qin administrator. From Shuihudi:

Bamboo strips written with documents of law; Painted lacquer spoon with wooden body in the shape of phoenix; kitchen utensil (bottom, height 13.3 cm); Flattened lacquer pot with colored painting of ox and horses: wine vessel or water vessel (in background), height 22.8 cm.

1983: 張家山漢簡 Zhangjiashan Han jian (Han Bamboo Slips from Zhangjiashan)

202–186 BCE

Excavated: 1983 Tomb No. 247, Zhangjiashan Jiangling County, Hubei Province

算數書 Suan Shu Shu

Wen Wu (2000) 算數書 Suan Shu Shu (A Book on Numbers and Computations)

Peng Hao, ed. Beijing: Science Press, 2001.

The first of the bamboo slips shown here on the right, with the title of the book clearly shown in the uppermost part of the right most slip: Suan Shu Shu.

Suan Shu Shu, Peng Hao, ed. Beijing: Science Press, 2001. Title of the book: 算數書 Suan Shu Shu (A Book on Numbers and Computations)

筭 suan

Suan Shu Shu, Peng Hao, ed. Beijing: Science Press, 2001. 1993: 郭店楚簡 Guodian chujian Guodian Chu Bambo Slips

Excavated: 1993 Guodian, Jingmen, Hubei Warring States period (mid-4th –early 3rd-century BCE). Tu Weiming, Yenching Institute, Harvard University Guo Yi discussing the Tao Te Ching (1993) 1996: 走馬樓簡牘 Zoumalou jiandu Zoumalou Bamboo Documents

Census count

Wooden tablets unearthed from Ancient Well No. J22

Zoumalou, Changhsha, Hunan Wu Kingdom (222-280) of the Three Kingdoms (220-280) Excavated: 1996

2002: 里耶古城秦代简牍 Liye gucheng Qindai jiandu Liye Ancient City Qin Dynasty Bamboo Documents

Multiplication table Liye, Longshan County, Hunan

“Four times eight is thirty-two; five times eight is forty…”

Excavated: April, 2002 Qin Dynasty (221 BCE-206 BCE) Length 22cm width 4.5cm Unearthed at Liye city site, Longshan in 2002

2007: 嶽麓書院藏秦簡 Yuelu shuyuan cang qinjian Qin Slips Collected by the Yuelu Academy

嶽麓書院

Founded in 976 in the 9th year of the Song Dynasty during the reign of Emperor 開寶 Kaibao

. Zhu Xi and . Zhang Shi .

Emperor Song Taizu 宋太祖

Figure 1, p. 77 Figure 2, p. 78 Figure 3, p. 79

Figures 11 and 12, p. 83

Figure 13, p. 83 2008: 清华简 Qinghua jian Tsing-Hua Bamboo Slips

Booking a Place in History: Rare ancient Chinese bamboo books dating back more than 2,000 years come home By ZAN JIFANG

NO.48 NOV.27, 2008 Tsinghua Warring States Bamboo Strips Professor 李学勤 Li Xueqin

战国竹简重回故土发布时间： 2008-10-30 13:21 来源：文摘报近日，清华大学宣布，2100枚战国时期的竹简入藏清华，它是由校友赵伟国从境外拍卖会买到后捐赠给清华的。经过专家组鉴定，这批“清华简”属于战国中晚期，距今大约2300~2400年左右，出土于古代的楚国境内。 “清华简”最长的有46厘米，最短的不到10厘米。“10 厘米就相当于我们现在的‘口袋书’了，这在以前从未发现过。” 战国竹简重回故土发布时间： 2008-10-30 13:21 来源：文摘报近日，清华大学宣布，2100枚战国时期的竹简入藏清华，它是由校友赵伟国从境外拍卖会买到后捐赠给清华的。经过专家组鉴定，这批“清华简”属于战国中晚期，距今大约2300~2400年左右，出土于古代的楚国境内。 “清华简”最长的有46厘米，最短的不到10厘米。“10 厘米就相当于我们现在的‘口袋书’了，这在以前从未发现过。”

Conservation of the 2388 Warring States Bamboo Strips from Chu at the Tsinghua University Center for Unearthed Documents Research and Protection

The strips include invaluable texts such as “Bao Xun”-- the last words of King Wen of the Zhou Dynasty to his son, and some other important contents, including a multiplication table.

Tsinghua University Center for Unearthed Documents Research and Protection

Peking University Qin Bamboo Slips

June 14, 2012 Sackler Museum of Art and Archaeology, Peking University Guo Han Feng Shuchun Wei Lisheng 北大秦简中的数学文献

韩巍（北京大学中国古代史研究中心）

北大秦简中的数学文献 Mathematical Documents among the Qin Bamboo Strips at Peking University

韩巍（北京大学中国古代史研究中心） Han Wei (Peking University Research Center for Ancient Chinese History) 日書 Ri shu Comparable illustration in black ink only found in the “Day Book” from Shuihudi tomb 11. slips 150r1-154r1 [KALINOWSKI 2008] 湖南大學嶽麓書院

嶽麓書院藏秦簡《數》書釋文·簡注計算田地產量或租稅的算題嶽麓書院藏秦簡整理小组（执筆：肖燦） 1.

說明 2. 计算土地面积的算題《数》书經整理得到：一，有編號的簡共二百三十余枚。谷物换算殘片二十余片，無編號。 3. 二，有完整算题75例。（现存题设条件和问题或答案能依据简文列出算法式的算题就视为完整算题）。衰分算題又有单独成題的“术”17例。 4. 又有记录有谷物体积重量比率、兑换比率的簡32枚，未計入算題數目。贏不足算題湖南大學又有記錄量制的簡3枚，未計入算題數目。嶽麓書院 5.

凡例 6. 少廣算题一釋文中異體字、假借字一般隨文注出，外加（）號。二簡文原有錯字，一般在釋文中隨注正字，外加計算體積的算題〈〉號。原有脫字或衍文，釋文不加更動， 7. 竹簡殘斷，不能復原。在注釋中說明。三簡文原有殘缺字，可據殘筆或文例補足的， 8. 勾股算题外加〔〕號。不能辨識的殘缺字，用 □ 表示，每字一格。其他符號：營軍之術（此算題暫時未歸類） ……字跡模糊，字數不能確定。 9. 10. 量制

湖南大學嶽麓書院

嶽麓書院藏秦簡《數》書釋文·簡注計算田地產量或租稅的算題嶽麓書院藏秦簡整理小组（执筆：肖燦） 1.

湖南大學嶽麓書院

嶽麓書院藏秦簡《數》書釋文·簡注計算田地產量或租稅的算題嶽麓書院藏秦簡整理小组（执筆：肖燦） 1.

凡例 6. 少廣算题一釋文中異體字、假借字一般隨文注出，外加（）號。二簡文原有錯字，一般在釋文中隨注正字，外加計算體積的算題〈〉號。原有脫字或衍文，釋文不加更動， 7. 竹簡殘斷，不能復原。在注釋中說明。三簡文原有殘缺字，可據殘筆或文例補足的， 8. Gou-Gu Problems 外加〔〕號。不能辨識的殘缺字，用 □ 表示，每字一格。其他符號：營軍之術（此算題暫時未歸類） ……字跡模糊，字數不能確定。 9. 10. 量制

□有圓材薶（埋）地，不智（知）小大，斲之，入材一寸而得平一尺，問材周大幾可（何）。即曰，半平得五寸，令相乘也，以深 [0304] 一寸为法，如法得一寸，有（又）以深益之，即材徑也。 [0457]

Suppose there is a circular [piece of] wood buried in the ground, whose size is unknown, but cutting to a depth of 1 cun gives a chord of 1 chi, it is asked how great is the circumference of the [circular piece of] wood? [The method] says: half the chord is 5 cun, multiply it by itself and using the depth of 1 cun as the divisor, dividing gives the result in cun, again adding the depth [of the cut] gives the diameter of the wood.

Suppose there is a circular [piece of] wood buried in the ground, whose size is unknown, but cutting to a depth of 1 cun gives a chord of 1 chi, it is asked how great is the circumference of the [circular piece of] wood? [The method] says: halving the chord gives 5 cun, multiplying it by itself and using the depth of 1 cun as the divisor, dividing gives the result in cun, again adding the depth [of the cut] gives the diameter of the wood.

[JZSS 9.09]: 今有圆材埋在壁中，不知大小。以锯锯之，深一寸，锯道长一尺。问：径几何? 荅曰：材径二尺六寸。术曰：半锯道自乘，此术以锯道一尺为句，材径为弦，锯深一寸为股弦差之一半，锯道长是半也。

[JZSS 9.09]: 今有圆材埋在壁中，不知大小。以锯锯之，深一寸，锯道长一尺。问：径几何? 荅曰：材径二尺六寸。术曰：半锯道自乘，此术以锯道一尺为句，材径为弦，锯深一寸为股弦差之一半，锯道长是半也。 [JZSS 9.09]: 今有圆材埋在壁中，不知大小。以锯锯之，深一寸，锯道长一尺。问：径几何? 荅曰：材径二尺六寸。术曰：半锯道自乘，如深寸而一，以深寸增之，即材径。

Suppose there is a circular [piece of] wood imbedded in a wall, whose dimensions are unknown. If a saw cuts to a depth of 1 cun, the length of the cut is 1 chi long. The question: What is the diameter [of the piece of wood]? The answer says: The diameter of the timber is 2 chi 6 cun. The method says: Multiply half of the length of the cut by itself, divide by the depth of 1 cun; adding the depth of 1 cun gives the diameter of the timber. 此术以锯道一尺为句，材径为弦，锯深一寸为股弦差之一半，锯道长是半也。

[Liu Hui Comments]: This [method] uses the length of the cut of 1 chi as the gou, and the diameter of the timber as the xian; the depth of the cut of 1 cun is one-half the difference between the gu and xian, and the length of the cut should also be halved. [ZJSS 9.06]: 今有池方一丈，葭生其中央，出水一尺。引葭赴岸，适与岸齐。问：水深、葭长各几何? 荅曰：水深一丈二尺，葭长一丈三尺。

[ZJSS 9.06]:

Suppose there is a [square] pond with a side of 1 zhang, in the middle of which a reed grows, extending 1 chi above the water. If the reed is pulled to the edge [of the pond], then it just reaches the edge. The question is: How much are both the depth of the water and the length of the reed? The answer says: The depth of the water is 1 zhang 2 chi. The length of the reed is 1 zhang 3 chi.

[ZJSS 9.06]: 术曰：半池方自乘，此以池方半之，得五尺为句，水深为股，葭长为弦。以句、弦见股，故令句自乘，先见矩幂也。以出水一尺自乘，减之，出水者，股弦差。减此差幂于矩幂则除之。余，倍出水除之，即得水深。差为矩幂之广，水深是股。令此幂得出水一尺为长，故为矩而得葭长也。加出水数，得葭长。臣淳风等谨按：此葭本出水一尺，既见水深，故加出水尺数而得葭长也。

The method says: Multiply half the side of the pond by itself. [Liu Hui Comments]: This [general method] uses half the side of the pond, which gives 5 chi as the gou, the depth of the water serves as the gu, and the length of the reed is the xian. Using the gou and xian to find the gu, therefore let the gou be multiplied by itself and first find the area of the gnomon.

The “Hypotenuse Diagram” from a copy of the 周髀算經 Zhou bi suan jing, Asian Studies Library, University of British Columbia

Ancient Chinese Mathematics The In-Out Complementary Method

The 弦 Xian or Hypotenuse Diagram

Liu Hui’s Explanation of the Proof of the Gou-Gu Theorem

Frank J. Swetz and T.I. Kao

Was Pythagoras Chinese?

University Park: Pennsylvania State University Press, 1977.

Harpedonaptai: the rope stretchers

Tomb of Menna, Sheikh abd el Qurna, New York: Metropolitan Museum of Art, 30.4.44.

九章算術 Jiu zhang suan shu

The problem of the reed in the middle of a pond. The Gou-Gu theorem may be reformulated as c2–b2 = a2 (b) (c-b) (c-b) Here, the entire area is c2, the yellow area represents b2, the two red areas are (b) (c–b)(b) and the blue- green area is (c–b)2.

The area of the gnomon is a2, i.e. 2(c–b)(b)+ (c–b)2. (b) (c-b) (c-b) We now follow the directions of the method to find the depth of the (b) pond, i.e. multiply half the side of the pond by itself, this is the a2.

Since a2 = 2(c–b)(b)+ (c–b)2, the next step in the method subtracts the length of the reed above water, (c–b), multiplied by itself, i.e. a2 –(c–b)2, which equals 2(c–b)(b). (b) (c-b) (c-b) Next the method divides the remainder by twice the (b) length of the reed above water, 2(c–b):

[a2 –(c–b)2] / 2(c–b), which equals b.

(b) (c-b) (c-b)

Since [a2 –(c–b)2] / 2(c–b)= b, the method instructs as a final (b) step to add the length of the reed above water, (c–b), and therefore: {[a2 –(c–b)2] / 2(c–b)} + (c–b) = b + (c–b) = c.

Clearly, the method does give the length of the reed, c, and the procedure is given in general terms, along with the specific numerical details of the given problem. (b) (c-b) (c-b)

We now translate this (b) same method into dealing with the analogous case of the log buried in the ground or imbedded in a wall. The only difference between the reed-in-the-pond problem and the log-in-the-wall is that the dimensions of the lengths in the latter are exactly half those of the former. But since this is the only difference, the method proceeds exactly as in the case of the reed-in-the-pond problem. ½(b) ½(c-b) ½(c-b)

To find the diameter of the ½(b)½ circle c, the method begins as before by squaring the gou, a, which is half the length of the cut i.e. a/2:

Since (a2)/4 = 2(¼ )(c–b)(b)+ ¼ (c–b)2, the next step in the method divides by the depth of the cut, ½ (c–b):, and after simplifying: a2 / 2(c–b), which equals b + ½ (c–b). ½(b) ½(c-b) ½(c-b)

The final step, to find the ½(b)½ diameter of the circle c, proceeds as the method instructs, by adding the depth of the cut:

[a2 /2(c–b)] + ½ (c–b) = b + ½ (c–b) + ½ (c–b) = c, the diameter obtained in complete generality. 湖南大學嶽麓書院

嶽麓書院藏秦簡《數》書釋文·簡注計算田地產量或租稅的算題嶽麓書院藏秦簡整理小组（执筆：肖燦） 1.

城 cheng [city walls]

[S] 城之深四尺，廣三丈三尺，袤二丈五尺，積尺三千三百。术曰：以廣乘袤有（又）乘深即成。唯築城止與此等。 [1747] (4)(33)(25) = 積 = 3,300 尺

[S] 救（求）城之述（術）曰：〔并上〕下厚而半之，以袤乘之，即成尺。 [0767]

a ½(a –a ) 1 2 1

除 chu [tunnel/ tomb entrance]

[JZSS 5.17] 今有羡除，下广六尺，上广一丈，深三尺；末广八尺，无深；袤七尺。问：积几何? 荅曰：八十四尺。术曰：并三广，以深乘之，又以袤乘之，六而一。

方亭 fang ting [square pavilion/ truncated pyramid]

[JZSS 5.10] 假令方亭，上方一尺，下方三尺，高一尺。其用棋也，中央立方一，四面堑堵四，四角阳马四。上下方相乘为三尺，以高乘之，约积三尺，是为得中央立方一，四面堑堵各一。下方自乘为九，以高乘之，得积九尺，是为中央立方一，四面堑堵各二，四角阳马各三也。上方自乘，以高乘之，得积一尺，又为中央立方一。凡三品棋皆一而为三。故三而一，得积尺。用棋之数：立方三，堑堵、阳马各十二，凡二十七，棋十三。更差次之，而成方亭者三，验矣。 From 芻童 chu tong to 鄆都 yun du

From the 芻童 chu tong the central parallelepiped and two 塹堵 qian du are removed:

鄆都 yun du

This leaves the 鄆都 yun du, which is comprised of two 塹堵 qian du and four 陽馬 yang ma:

圜亭 yuan ting [conical pavilion/ frustrum of a cone]

[JZSS 5.11] 术曰：上、下周相乘，又各自乘，并之，以高乘之，三十六而一。此术周三径一之义，合以三除上下周，各为上下径，以相乘；又各自乘，并，以高乘之，三而一，为方亭之积。假令三约上下周，俱不尽，还通之，即各为上下径。令上下径相乘，又各自乘，并，以高乘之，为三方亭之积分。此合分母三相乘得九，为法，除之。又三而一，得方亭之积。从方亭求圆亭之积，亦犹方幂中求圆幂。乃令圆率三乘之，方率四而一，得圆亭之积。前求方亭之积，乃以三而一，今求圆亭之积，亦合三乘之。二母既同，故相准折。惟以方幂四乘分母九，得三十六，而连除之。算數書 Suan Shu Shu (A Book on Numbers and Computations)

Peng Hao, ed. Beijing: Science Press, 2001. [53] 方田 Fang tian: Square Fields

(Given) a field of 1 mu, how many (square) bu are there? (The answer) says: 15 15/31 (square) bu. The method says: a square 15 bu (on each side) is deficient by 15 (square) bu; a square of 16 bu (on each side) is in excess by 16 (square) bu. (The method) says: combine the excess and deficiency as the divisor; (taking) the deficiency numerator multiplied by the excess denominator and the excess numerator times the deficiency denominator, combine them as the dividend. Repeat this, as in the “method of finding the width.” [53] 方田 Fang tian: Square Fields deficiency 不足子赢子 excess numerator (bu zu zi) 15 16 (ying zi) numerator

deficiency 不足母 15 16 赢母 excess denominator (bu zu mu) (ying mu) denominator . 九章算術 JIU ZHANG SUAN SHU [4.12]

. SUPPOSE THERE IS A [SQUARE] AREA OF 55,225 BU. . THE QUESTION: WHAT IS THE [LENGTH OF THE] SIDE? . THE ANSWER: 235 BU. In the Nine Chapters, the method of finding square roots has been highly developed into an algorithm with its own technical designation as the 開方術 Kai fang shu or “Method of Finding the Square Root” (lit. “Method of Opening the Square” [Qian Baocong 1963, vol. 1, p. 150]. This latter method allows determination of the square root by an iterative procedure based upon successive approximations by completing squares.

Oldest surviving diagram showing the derivation of the square root extraction method, from the Yongle dadian.

Lam Lay Yong and Ang Tian Se, Fleeting Footsteps. Tracing the Conception of Arithmetic and Algebra in Ancient China, Singapore: World Scientific, 1992, p. 76. 55,225

甲 55,225

甲

2002=40,000 55,225–40,000=15,225 15,225

甲 15,225

甲

(30)(200)=6,000 2(6,000)=12,000 15,225

甲

乙

12,000+302=12,900 15,225–12,900=2325 2325

甲

乙 2325

甲

乙

(5)(230)=1150 2(1150)=2300 2325

甲

乙

(2300)+52=2325

甲

乙

√55,225 = 235

算數書 Suan Shu Shu (A Book on Numbers and Computations)

Peng Hao, ed. Beijing: Science Press, 2001. [Problem 46] 負炭 Fu Tan : Transporting Charcoal [WW 2000, p. 82; PH 2001, p. 92; ZJS 2001, p. 264]

Carrying charcoal from a mountain, in 1 day it is possible to carry 7 dou of charcoal to a wagon; the next day, 1 dan of the gathered charcoal taken to the wagon is transported to a government post (官 guan). Now wishing to go to the government post, carrying charcoal from ( the mountain) and transporting charcoal the long distance to the government post, the question is how much charcoal is delivered in (1) day? (The answer) says: in one day 4 dou 2/[17] sheng of charcoal (is delivered). The method says: taking the 7 dou multiplied by 10 (days) gives 7 dan; and (it takes) 7 days as well to transport it to the post; i.e. take the 10 days and the 7 days and combine them as the divisor; (take 7 dan as the dividend); dividing gives the answer in dou. 通訊 Tongxun 2001

《算數書》校勘 Suan shu shu jiaokan (Collation of the Suan shu shu), HPM tongxun (HPM Newsletter), 3(11)(2000), pp. 2-20.

“The meaning of this sentence [problem] is not clear…”

[Problem 47] 盧唐 Lu Tang: Bamboo Ladles/Utensils [WW 2000, p. 82; PH 2001, p. 94; ZJS 2001, p. 265]

The norm says: in 1 day 60 stalks of bamboo are cut down; in 1 day (it is possible to make) 15 lutang, one stalk of bamboo equals 3 lutang. If 1 person is told to cut down bamboo himself to make lutang, how many can be made in 1 day? (The answer) says: 13 and 3/4 lutang. The method says: take 60 as the divisor, and take 55 times 15 as the dividend.

(55×15)/60 = 13 3/4 lutang 郭世榮《算數書》勘误《Suan shu shu》 Guo Shirong kanwu (Corrections for the Suan shu shu), 内蒙古师范大学学报(自然科学汉文版 Neimenggu Shifan Daxue Xuebao (Ziran kexue hanwenban) (Journal of Inner Mongolia Normal University, Natural Science Edition), 30(3)(2001), pp. 276- 285.

(60×15)/65 = 13 11/13 lutang/day (55×15)/60 = 13 3/4 lutang/day

郭書春算數書校勘 Suan shu shu Guo Shuchun jiaokan (Collation of the Suan shu shu), 中国科技史料 Zhongguo kexu shiliao (China Historical Materials of Science and Technology), 22(3)(2001), pp. 202-219.

(60×15)/65 = 13 11/13 lutang/day

彭浩 Peng Hao

張家山漢簡《算數書》注釋 Zhangjiashan hanjian Suan shu shu zhushi (Commentaries on the Book on Calculating with Numbers on Bamboo Strips, unearthed from Zhangjiashan), Beijing: Kexue chubanshe, 2001.

(60×15)/(60+15) = 12 lutang/day

[Problem 47] 盧唐 Lu Tang: Bamboo Ladles/Utensils [WW 2000, p. 82; PH 2001, p. 94; ZJS 2001, p. 265]

To cut one stalk of bamboo takes 1/60 of a day. To make one lutang takes 1/15 of a day. One stalk of bamboo suffices for 3 lutang, which takes: 1/60 + 3(1/15) = 13/60 of a day. 4 stalks of bamboo will result in 12 lutang in 52/60 of a day. This leaves 8/60 of the day. To cut one more bamboo 7/60 of the day. It takes 4/60 of leaves a day to make one lutang, resulting in 13 complete lutang, with 3/60 of the day left. Since it takes 4/60 of a day to make one lutang, this leaves only enough time to complete 3/4 of a lutang, for a total of 13 3/4 lutang.

[Problem 47] 盧唐 Lu Tang: Bamboo Ladles/Utensils [WW 2000, p. 82; PH 2001, p. 94; ZJS 2001, p. 265]

Thus, the answer as given is correct.

[Problem 47] 盧唐 Lu Tang: Bamboo Ladles/Utensils [WW 2000, p. 82; PH 2001, p. 94; ZJS 2001, p. 265]

Thus the answer, 13 3/4 lutang is correct, but now, what about the method?

(55×15)/60 = 13 3/4 lutang [Problem 48] 羽矢 Yu Shi: Feathering Arrows [WW 2000, p. 82; PH 2001, p. 95; ZJS 2001, p. 265]

The norm: one person in one day makes 30 arrows or feathers 20 arrows. If one now wishes to have one person both make arrows and feather them, in 1 day how many can be made? The answer: 12. The method: combine the arrows and the feathering as divisor; taking the arrows and the feathering, mutually multiply them together as the dividend. [Problem 48] 羽矢 Yu Shi: Feathering Arrows [WW 2000, p. 82; PH 2001, p. 95; ZJS 2001, p. 265]

In one day 1 person can make 30 arrows, or feather 20 arrows. If one person does both, how many arrows can be made in one day?

1/30 + 1/20 = (20+30)/(20×30) = 50/600 = 1/12 of a day, so that 12 arrows can be made and feathered in one day.

This is exactly what the method says: to total number of feathered arrows will be:

(20×30)/(20+30) = 600/50 = 12 feathered arrows/day. [Problem 47] 盧唐 Lu Tang: Bamboo Ladles/Utensils [WW 2000, p. 82; PH 2001, p. 94; ZJS 2001, p. 265]

Why doesn’t (60×15)/(60+15) = 12 lutang work?

Correct method: (55×15)/60 = 13 3/4 lutang [Problem 46] 負炭 Fu Tan : Transporting Charcoal [WW 2000, p. 82; PH 2001, p. 92; ZJS 2001, p. 264]

In 1 day 1 person can collect 7 dou of charcoal The next day it is possible to deliver 1 dan (10 dou) of charcoal Thus it takes 1/7 of a day to collect 1 dou of charcoal and 1/10 of a day to deliver 1 dou: 1/7 + 1/10 = (10+7)/70, i.e. it will take 17/70 of the day to gather and deliver 1 dou of charcoal; therefore, in one day 70/17 dou can be gathered and delivered, i.e. = 4 2/17 dou/day.

Alternatively: If 7 dou can be collected in one day, in 10 days, 70 dou or 7 dan can be collected; to deliver 7 dan will take 7 days, so altogether, 70 dou collected and delivered in 17 days amounts to:

70 dou/17 days = 4 2/17 dou/day.

Standard Measure (Bronze), Xin Dynasty (9-23 CE).

Trésors du Musée national du Palais, Taipei, a special number of Conniassance des Arts, 127 (1998), Catalogue of the exhibition at the Grand Palais, p. 75. 九章算術 Nine Chapters on the Art of Mathematics

. 方田 Fangtian - Rectangular fields. Areas of fields of various shapes; manipulation of fractions. . 粟米 Sumi - Millet and rice. Exchange of commodities at different rates . 衰分 Cuifen - Proportional distribution. Distribution of commodities and money at proportional rates. . 少廣 Shaoguang – Small width. Division by mixed numbers; extraction of square and cube roots; areas and volumes of circles and spheres. . 商功 Shanggong – Construction Consultations. Volumes of solids of various shapes. 九章算術 Nine Chapters on the Art of Mathematics

. 均輸 Junshu - Equitable taxation. More advanced problems on proportion. . 盈不足 Yingbuzu - Excess and deficit. Linear problems solved using the principle known later in the West as the rule of false position. . 方程 Fangcheng - Rectangular arrays. Systems of linear equations, solved by a principle similar to Gaussian elimination. . 勾股 Gougu - Base and altitude. Problems involving the principle known in the West as the Pythagorean theorem

A woodblock print illustrating the double distance method used to determine the height of an island at sea.

Tu Shu Ji Cheng (1726).

Archimedes and Liu Hui on Circles and Spheres

A page of the Archimedes codex, reconstructed.

Johan Ludvig Heiberg (1854–1928) “Archimedes,” by Giuseppe Nogari (1699-1766) Pushkin Museum, Moscow

The highlighted portion indicates the appearance of the original Archimedes codex on parchment as created in minuscule sometime in the 10th century, CE, probably in Constantinople.

Creation of the Euchologion, completed on April 14, 1229, by Ioannes Myronas, probably in Jerusalem. Note the original Archimedes codex has been scraped clean, and the text is divided on the verso/recto sides of the previously single-sheet of the original parchment text. sc

Constantine Tischendorf Jerusalem, 1846 Constantine Tischendorf Jerusalem, 1846 Cambridge University Library, 1876

Constantine Tischendorf Jerusalem, 1846

Cambridge University Library, 1876

Archimedes: Sphere and Cylinder, identified by Nigel Wilson, 1971 σχήμά…αύτό, οϊον εϊρεται περιεχόμενον… Archimedes, Proposition 35, On the Sphere and Cylinder I

Cambridge University Library, MS Add, 1879.23

Tohru Sato, “A Reconstruction of The Method Proposition 17, and the Development of Archimedes’ Thought on Quadrature—Why did Archimedes not notice the internal connection in the problems dealt with in many of his works?” Historia Scientiarum, 31 (1986), pp. 61-86; p. 74.

…to read the entire proposition 14, it starts in column 1 of folio 110 recto of the palimpsest; you would then have to turn the codex 90 degrees to read the Archimedes text, but towards the bottom of the Archimedes text, it disappears into the gutter of the palimpsest; the Archimedes text then reappears five folios earlier, on 105 verso, but the first few lines would again be hidden by the gutter of the palimpsest, lines that Heiberg was unable to see. The next continuation of the Archimedes text appears on folio 158.

Netz and Noel, The Archimedes Codex, p. 125.

算數書 Suan Shu Shu (A Book on Arithmetic)

Peng Hao, ed. Beijing: Science Press, 2001.

The first of the bamboo slips shown here on the right, with the title of the book clearly shown in the uppermost part of the right most slip: Suan Shu Shu.

Peng Hao, ed. Beijing: Science Press, 2001.

筭 suan 數 shu 書 shu

Peng Hao, ed. Beijing: Science Press, 2001.

Heiberg photographed the palimpsest, and his photographs survive in the Royal Library of Denmark in Copenhagen.

3 August 2006, XRF Imaging at SLAC 110v-105r Blue-Yellow XRF Pseudocolor (reversed) as from side 105v-110r

From the Archimedes Palimpsest website: http://www.archimedespalimpsest.org/index.html

……Let ABCD be the given circle, and K the area of the triangle described (of height equal to the radius and base equal to the circumference of the circle. Then, if the circle is not equal to K, it must be either greater or less.

Euclid XII.2

Archimedes (287-212 BCE) ……the Greek preference for the method of exhaustion is thus evidence both of their demand for rigour and of their avoidance of infinite processes wherever possible.

Geoffrey Lloyd, Adversaries and Authorities (1996), p. 150. Dai Zhen’s reconstruction of Liu Hui’s method of the rule for finding the area of a segment of a circular field.

Christopher Cullen

“Learning from Liu Hui? A Different Way to Do Mathematics”

Notices of the AMS, 49(7)(August, 2002), pp. 783- 790.

From the side of an inscribed polygon, Liu Hui can compute the area, and for an inscribed regular polygon of 96 sides, he computes the area as 314 64/625. Liu Hui: Lü 率 50 and 157

Liu Hui: Lü 率 50 and 157

Li Chunfeng: Lü 率 7 and 22

Liu Hui: Lü 率 50 and 157

Li Chunfeng: Lü 率 7 and 22

Zu Chongzhi: 密率 Mi Lü 113 and 355

The Method, Proposition 2 …If in a cube a cylinder is inscribed whose bases lie in opposite squares and whose surface touches the other four planes, and if in the same cube a second cylinder is inscribed whose bases lie in two other squares and whose surfaces touch the four other planes, then the body enclosed by the The Method, Proposition 2 surface of the cylinder and comprehended within both cylinders will be equal to 2/3 of the whole cube.

Jiuzhang suanshu

The Nine Chapters

From the Southern Song edition, Shanghai Library. 九章算术 Jiu Zhang Suan Shu 刘徽 Liu Hui (Nine Chapters on (fl. 263 CE) Mathematical Procedures)

Chapter 4:

d = [3√16/9) ]V

From the Southern Song edition, Shanghai Library.

Chapter 4:

A copper cube of diameter 1 cun weighs 16 ounces, while a copper ball of the same diameter weighs 9 ounces; this is the origin of the lü 16 and 9.

From the Southern Song edition, Shanghai Library.

Liu Hui’s “mou he fang gai” 牟合方蓋

Liu Hui’s “mou he fang gai” two “inverted 牟合方蓋 umbrellas”

“…the stacked blocks 棋 (qi) form the volumes, the areas being identical, the volumes cannot differ from one another.”

Archimedes, The Method, Propositions 14 and 15; Proposition 14 establishes the volume of the sphere by examining the proportional relations between cuts of infinitesimal lamina; Proposition 15 then establishes this result rigorously using the method of exhaustion.

Liu Hui reduces the problem of finding the volume of the sphere to that of finding the volume of the two “inverted umbrellas.”

Zu Gengzhi managed to solve this part of the problem by appealing to the similar proportions of the areas for any cut of a plane through the two “inverted umbrellas” applying a line of reasoning equivalent to that employed by Archimedes. Fairbank Center for Chinese Studies Harvard University

The Evolution of Mathematics in Ancient China. From the newly discovered 數Shu and 算數書 Suan shu shu Bamboo Texts to the Nine Chapters on the Art of Mathematics

Joseph W. Dauben Department of History Herbert H. Lehman College and Ph.D. Program in History The Graduate Center City University of New York