chinese math

Chinese Mathematics 200 BC to 1400 ADJOHN HISCOCKSSchool of Education, University of Exeter

1. Introduction

IT is unfortunate that to many people the greatachievement of Chinese mathematicians appears tobe the study of magic squares and the use of theabacus. In fact neither of these topics featured pre-dominantly in the period under discussion and muchmore impressive mathematics was considered.

Amongst the achievements of Chinese mathe-maticians in the period 200 BC to 1400 ADwere their approach to evaluating n, their work onfinding volumes, square root and cube root evalua-tion and their ability to solve more or less any poly-nomial equation.

2. Estimating n

The Chinese approach to estimating limitsbetween which n could lie was not dissimilar to thattaken by Archimedes. In one of the so called tenclassic mathematical books, "The Sea-Island Arith-metic" by Liu Huf written about 260 AD, thefollowing method was adopted.

Fig. 1

Suppose a In sided regular polygon is inscribed ina unit circle and AB is one of its edges as shown inFig. 1. If O is the centre of the circle and by use ofPythagoras theorem it can be shown that

AC will be the edge of a An sided regular polygon.Volume 1 No. 2,1982

Thus at this stage Liu Hui had an iterative typeformula for finding the length of edge for any regularpolygon with an even number of edges.

The next step was to connect the area of a Ansided polygon with an expression involving thelength of edge to a 2« sided polygon. Referring againto Fig. 1,

area OACB =AB.OC

and since OC = 1,

area OACB =AB

The area of a An sided regular polygon will beIn x area OA CB or

ABIn x = n AB.

2

Thus the iterative procedure in the first stage couldnow be used to find areas of other regular polygons.

The final step was to consider the configuration inFig. 2.

Fig. 2

ABQP is a rectangle.Area of ABQP = 2 x A ABC.

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All such triangles formed on the sides of a 2«sided regular polygon will collectively represent thedifference in area between a An sided regularpolygon and a 2n sided regular polygon.

Using the notationA4n = Area of 4n sided regular polygon, andA w = Area of In sided regular polygon, we have:

AM < Area of circle < Am + all rectanglesA A f i l A liA A)< Area of circle <A

< Area of circle <A< Area of circle < l

- Ay,)

— A

By considering 96 and 192 sided regularpolygons, Liu Hui came to the conclusion that n hada value between 3.141024 and 3.142704 (moderndecimal equivalent). He suggested that anapproximation to n be thus taken as either 3.14 or

3. Volumes

In the classic, Chiu-Chang Suan-Shu ("Arith-metic in Nine Sections") dated from between200 BC to as late as 0 AD, section 5 deals withvolumes and n is here taken to be 3. A formula forthe volume of a sphere is given as equivalent to $ r3.The way this formula was arrived at is of note sinceit involved a line of thinking not found in WesternEurope until 1500 when Cavalieri considered similarideas.

The Chinese realised the principle that if twosolids have all their corresponding cross-sectionsequal in area (see Fig. 3) then their volumes will alsobe equal. Likewise, if their corresponding cross-sectional areas are in a fixed ratio, then theirvolumes will be in this ratio.

Fig. 3

Consider a circle inscribed in a square. If n = 3then the area of the circle: area of the square = 3 : 4 .

Using this result led to the idea that if a cube ofside 2r has each one of its horizontal cross-sections46

Fig. 4

replaced by an inscribed circle the resulting shape isa cylinder of height 2r and hence the volumes beingin the ratio 3 :4 implies that the volume of thecylinder is 6H.

Fig. 5

So far so good, the error in this formula arisesfrom taking n as 3.

The Chinese then made a jump in their thinking,suggesting that the volume of a sphere: volume ofcircumscribing cylinder was 3 :4, which in turnimplied that the volume of a sphere was \ r3. Theerror now was no longer due to approximating n to3 (which of course suggests an answer of 4r3). It wasto be another Chinese mathematician Tsu Chung-Chih (429-500 AD) who realised the error andcorrected i t

Tsu's approach to the problem was as follows.Consider a hemisphere made as it were from a pileof circular discs (Fig. 6).

Fig. 6

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Replace each disc by its circumscribing square andwe obtain a rather odd looking "pyramid" (it is not apyramid, its slant edges, for example, are quadrantsof a circle).

Now area of circle : area of corresponding square= nr1 : 4r*= 7i : 4

orVolume of hemisphere _ TU _

Volume of new shape 4

Tsu's next problem was to connect the volume ofi this new "pyramid" type of shape with a shapewhose volume was known. He considered | of the"pyramid" inscribed in a cube (see Fig. 7).

Fig. 7

If the original sphere had a radius of r then byconsidering one cross-section of the \ "pyramid" atheight h above the base, it is not difficult to showthat

x2 + h2 = r2 or r2-x2 = h2.But r2 — x2 is the cross-sectional area of the solid

resulting from removing the | "pyramid" from the

cube (see Fig. 8). Thus the volume of this latter solidis equivalent to the volume of a solid with cross-sectional area = h2.

Fig. 9

Such a solid is the real pyramid shown in Fig. 9.Volume of difference solid = J r2

or volume of | "pyramid" is

i.e., Volume of "pyramid" is 5 r3.

But Volume of hemisphere _ TT

Volume of "pyramid" 4

thus volume of hemisphere

= JL x fr» = ?jir3

4and of course the volume of a sphere will beJnH.

Fig. 8

Volume 1 No. 2.1982

4. Square roots

The frequent use of Pythagoras' theorem inproblems led the Chinese to devise means of findingsquare roots. Perhaps it should be said that theChinese did not seem to be deterred either byirrational numbers or the equivalent of decimalfractional answers.

To find square roots the Chinese used a methodthat most older readers will recognise since it wastaught extensively in British schools as an algorithmup to say the 1960*8.

In 263 AD Liu Hui justified the methodgeometrically. His argument was equivalent tosaying: suppose we wish to find the square root of223 729 and suppose the answer is the three digitnumber 100a + 106 + c, then the square of thislatter number would be as in Fig. 10.

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10b

lOOa

lO2ac

lO3ab

1 0 V

lObc

l o V

103ab

it

4

lOOa

The actual numerical working with its algebraicequivalence is shown as:

•lObc Fig. 10

10b c

100a = 400

200a + 106 =870

200a + 206 + c = 943

22 37 29

16 00 00 = (100a)2

6 37 29 = A r 2-(100a)2

6 09 00 = (200a + 106) x 106

28 29 = A ' 2 - (100a) 2 - l (200a+ 106)106]

28 29 = (200a + 206 + c) x c

The hundreds digit a = 4 is found by inspectionand its square removed. The tens digit 6 = 7 is foundby considering what value 6 must have if .800 + 106is to divide into 63 729. The rest of trie workingfollows similarly.

Cube roots were also found using a similarmethod.

5. Solving polynomial equations

The Chinese approach to equation solving isinteresting in that they evolved a method that is notdissimilar to what we would now call Homer'sMethod. First, however, let us consider theirgeometrical approach to quadratics of the formx(a—x)=c, dating from about 400 AD. The left handside of this equation could be thought of as the areaof a rectangle with dimensions as shown (Fig. 11).

Area c

Fig. 11

By placing four of these rectangles together alarge square is formed with a square hole in its

centre, as shown in Fig. 12. The large square has anedge length of a and hence an area of a2. Subtractingthe four rectangles of area c from the large squaregives the area of the small square as a2 — 4c, hencethe edge of this small square could be found. But theedge of the small square is a — 2x, thus a simpleequation could then be solved to find x.

I tk

iT

48

Fig. 12

By the 7th Century AD the Chinese were able tosolve cubic equations but their methods are notclear. However, in 1023 AD Chia Hsien was toclarify the method and once again we see a methodof solution that seeks to find the answer digit by digitas in the Chinese square root and cube rootmethods.


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Consider the equation

= 4807.227.

Chia Hsien worked as follows (Modern notation).The root lies between 10 and 20. Put JC, = x — 10and the equation reduces to

x,3 + 2 6 V + 221*! = 4197.227.

By trial and error the root to this equation liesbetween 8 and 9. Note at this stage it has beenestablished that the root is thus 18 and somefraction.

Put x2 = x, — 8 and the last equation becomes

x23 + 50X22 + 829x2 = 253.227.

To cater for the decimal part of the answer, thisequation is transformed by putting x3 = 10x2 andbecomes

O.OOIXJ3 + 0.5x32 + 82.9*3 = 253.227.

By trial and error x3 is found to be exactly 3. Thusthe solution is 18.3.

The method can be extended to any polynomialequation and indeed there is evidence of solutionsbeing found to equations of order 15. It is interestingto speculate whether Cardano or Tartaglia, centurieslater in Europe, would have bothered with their workon the cubic had they known this Chinese idea.

It would appear that because the Chinese couldsolve polynomial equations like this, they oftencreated unnecessary complicated equations to solveproblems rather than search for easier solutions. Infact the 7th Century cubic equation used to solve aproblem involving the measurements of a dam was

+2502x2 175493* 2466431

15 60

The problem can be solved, however, by theequation 34. lx = 3.1(JC + 31). The Chinese did notconsider the simpler equation.

The numerical approach as opposed to"algebraic" appears to be a feature of Chinesemathematics. So it is not too surprising to find thatthe Chinese had various forms of interpolation/extrapolation formulae long before Western Europeconsidered such matters.

Lin Chuo in the late 6th Century AD gave aformula for positioning the sun between two timeintervals, t and ( + 1 as follows (again modernnotation):

yx/+i) =

Volume I No. 2,1982

(I)2(A,-A2)

where/is a function of I and

A, =Rt+1) - / ( / ) , A2 =f(t+2) -f(l+1).

The reader might care to compare this with theNewton-Gregory forward difference formula

Sui Yo in the 9th Century AD gave the formula:

In 1280 Kul Sho-Ching used extrapolation ideas tofind the constants for an unknown polynomial. Hisformula was essentially

rt \ /Yr>\ Al x(x— 1)A2

fix) =yxo) + A1* + '

x(x-lX*-2)A3

For example, given the data .A2)=9, J\3)=3l,A4)=83,/(5)=177,/(6)=325,/(7)=539, he drew upa table for the above and their respective A,, A2, A3

values. He used these results to extrapolatebackwards, see the entries below (Table I) which areshown above the dotted line, to obtain/(0) and itsA1, A2, A3 values:

Table I

X

0

1

2

3

4

5

6

7

Ax)7

5

9

31

83

111

325

539

A1

-2

4""- — ̂22

52

94

148

214

A2

6

18

30 ̂

42

54

66

A3

12

12—~

12

12

12

Thus, using his formula:

fix) = 7 + x(-2)+ * ( * ~ 1 ) X 6 + *C*-lXx-2) x 12,2 6

i.e.,J\x) = 2xi-3x2-x49

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6. The Remainder Theorem

Sun-Tsu in the 4th Century AD posed and solvedthe following problem. A number when divided by 3leaves two over, when divided by 5 three are leftover, when divided by 7 two are left over. What isthe number?

Sun-Tsu's solution is equivalent to:

M = tlN+Rl

N =t2R,+R2

Rx = t3R2+R3

a, = t-,1

a4 = r4a3+a2

3x5x7= 105 = 0(Mod5x7= 35 = 2 (Mod3x7= 21 = 0(Mod

21x3= 0(Mod3x5= 15 = 0(Mod

15x2= 0(ModNow consider:

35 = 2 (Mod63 = 0(Mod30 = 0 (Mod

Adding 128 = 2 (ModBut 105 = 0(Mod

Subtracting

3)3)3)3)3)3)

3)3)3)

3)3)

0(Mod0(Modl(Mod3 (Mod0(Mod0(Mod

0(Mod3 (Mod0(Mod

3 (Mod0(Mod

5)5)5)5)5)5)

5)5)5)

5)5)

0(Mod0(Mod0(Mod0(Modl(Mod2 (Mod

0(Mod0(Mod2 (Mod

2 (Mod0(Mod

7)7)*7)7)*7)7)*

7)"7)7)-

7)7)

Rn-2 =

When Rn=\, an is a solution provided n is even. Inthe proof of this, shown below, any multiple of M iswritten as TM:

t^N-tjR, =N-t2(M-a,N)= N+att2N-TM= N(,\+alt2)-TM= a^N-TM

23 = 2 (Mod 3) 3 (Mod 5) 2 (Mod 7) R3=Rl-hR2 = M-a,N-t3(a2N-TM)f From lines marked * above.

byNotice 128 would of course be a solution butsubtracting 105 the smallest solution is obtained.

Nine centuries later Chin Chiu-shao gave anothermethod of solution which is in fact the same as thatgiven by Gauss (1777-1855). The divisors must bemutually prime. Chin's method for a number Ndivided by three numbers m,, m2 and m3 andyielding remainders of Rx, R2 and R3, respectively,was as follows.

First find numbers a,, a2 and a3 such that

= rM-Ar(a,+/3o2)= TM-Na3

= a2N-TM-tA(TM-Na3)

a,m2m3=1 (Mod m,), a ^ m jand 0 ^ , ^ 2 = 1 (Mod m3).Then N=Rl(alm2m3) + /?2(a2wi,m3) + R3(a3mym^).

Using Chin's method on the last problem yields ananswer of 233. There is a vector flavour to the work.

It might be thought that Chin's method wasstraightforward but consider the snag in say findingthe number a, such that a,ffi2m3=l (Mod m,). Forexample if m2/n3=42 and m,=97 it would benecessary to solve the equation 42a,=1 (Mod 97).The solution to this equation is not immediatelyapparent but is in fact 67, namely that 42x67=2814and 2814=1 (Mod 97). Chin evolved a method forsolving all such equations and his reasoningamounted to the following.Consider the equation aN= 1 (Mod M).First find numbers /,, /2, / 3 . . . Rlt R2, R} ... and a,,a2, a 3 . . . such that:

50

17a

= N(a2+lta3)-TM= Na4-TM.

In general if n is even then:

«„=/?„.,-/„/?„., =Nan-TM.

Thus if/?,, = 1 => l=Nan-TM,hence since Nan=\+TM then Na^^l (Mod M) andan is the required solution.

Consider for example the solution to(Mod 63)63=3x17+12 a,=317=1x12+5 a2=a, + l=412=2x5+2 a 3=2a 2+a,=l l5=2x2+1 a4=2a3+a2=26

26 x 17=442, 442=1 (Mod 63)Thus a=26 is the solution.

The next question to arise from this problem is:what happens when n is odd? The Chinese solutionmerely involves an additional line in the working,writing a number, say K, in the form (K— 1)+1.For example consider the very first problem posed,namely 42a=l (Mod 97)


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97=2x42+13 a,=242=3x13+ 3 a2=3a, + l=713=4x 3+ 1 a3=4a2+a,=30

But the remainder of 1 has occurred in an odd row,thus a new row is worked as:

3=2x1 + 1 a4=2a3+a2=67

and thus the solution of a=67 is obtained.

Multiplication involves only a few rules. If mul-tiplying by a constant, everything is multiplied in thematrix and the answers recorded in the same cells. Ifmultiplying by a multiple of x, all answers arerecorded one row lower down. (For multiplying byx2, two rows down; x3, three rows down etc.) Formultiplication by multiples of y, y2, y3 . . . answersare recorded one column, two column, threecolumns . . . to the left, respectively. For example

7. Algebraic manipulation

The Chinese used a grid board approach to"algebra," In the book "Precious Mirror of the FourElements"-by Chu Chi-Chieh, written about 1303,the grid board technique is described and up to fourunknowns used. For the purpose of this descriptionwe will limit the ideas to two unknowns and slightlysimplify the Chinese method.

y2 y1 cc

X1

X2

X3

y2 y

l

3

c

- 7

2

cX

X2

The board was essen-tially an nxm matrix withrows and c o l u m nheadings (not shown bythe Chinese) as follows:

To represent an expressionsuch as 2x2+3xy+y—7 theboard would be numbered inthe appropriate cells:

(x2 - 2xy + 3y - 1) x 2

- 2

- 1

{x1—Ixy+3y— 1) x Ax

- 2

- 1

- 2

=2x2-4xy+6y-2

12 - 4

=4x2-8x2y+1 Ixy—Ax

Addition (or subtraction) between two sucharrays entails adding corresponding cells. Forexample (x2+3xy—4)+(2x1+xy+5y)

Volume 1 No. 2,1982

C . - 4

=2>x1+'lxy+5y-A

(x2-2xy+ly— 1) x Ay1

c

- 2

- 1 12

- 8

=Ax1-ixyi+\2y3-4y1

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The binomial multiplication of say(3x—2yX5x+7y) could be done as

- 2

3

- 2

3

x 5x

xly

-10

IS

-14

21 •

1

-14

11

15

(3x-2yX5x+ly)

One wonders whether the Chinese use of countingboards was less inhibiting in evolving methods forsolving equations and manipulating algebraicexpressions than the written recording that took solong to perfect in Western Europe.

8. CommentsBy the end of the 13th Century, the Chinese had

made significant advances in mathematicalknowledge, and had there been communicationbetween the mathematicians of East and West whoknows what a change in the course of history wouldhave been made.

Sad to relate, even the Chinese could not keep upthe quest for more mathematical knowledge, anddynasties such as the Ming did not encouragefurther development The latter, despite its reputa-tion for fine vases, was particularly inhibiting onoriginal thought, demanding even from its writers astandard uniform mode of written communication.By the time the Chinese experienced a newmathematical awakening, the West's influence onEastern thought was so marked, it is difficult todiscern what was original Chinese mathematics andwhat was not

9. BibliographyOne of the main difficulties in the study of the

history of Chinese mathematics is the lack of writtenmaterial on the subject, in particular that written inEnglish! Most of the mathematical histories cur-

52

rently available make little more than passingreference to Chinese mathematics and tend merelyto list topics studied by the Chinese and not detailtheir ideas or methods.

For those who can read Chinese:Li Nien, "A Study of the History of Chinese Mathematics," 5

Vols, 1954.Chin Pao-tsung, "A History of Chinese Mathematics," 1964.

Books In English:D. E. Smith, "History of Mathematics," Volumes 1 and 2,

1958.J. Needham, "Science and Civilization in China," 1959.

(This latter work is the result of a lifetime study of the subjectand contains a fairly comprehensive section on mathematics.)

D. Chu Hole Koo, "A History of Chinese Mathematics Priorto the Sixteenth Century." Dissertation for award of degreeBPhilEd, Exeter University.(This work has been the main source of reference for this article.)

Y. Mikami, "The Development of Mathematics in China,"1913.

Various articles have also been published over the years onaspects of the subject, most notably by the MathematicalGazette.

The author is an honours graduate of ExeterUniversity; and holds the Mathematical Associa-tion Diploma in mathematics teaching. He was amathematics teacher in middle and secondaryschools until 1968, a mathematics lecturer at St. •Luke's College, Exeter until 1978 and is currentlya lecturer in Education at Exeter University.


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