MathematicsinAncientChina(69P)

7/30/2019 MathematicsinAncientChina(69P)

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MathematicsinAncientChina

Chapter7


2/69

Timeline

Li ZhiQin Jiushao

Yang HuiZhu Shijie

Liu HuiZu Chongzhi

Gnomon,Nine Chapters

CHINA Yuan / MingSongTangWarring StatesWarring

StatesZhouShang Han

1500 CE1000 CE1000 BCE1500 BCE2000 BCE2500 BCE3000 BCE 500 BCE 0 CE 500 CE

MycenaeanMinoan GREECEChristianRomanHellenisticClassical

Arc hai cDark

500 CE0 CE500 BCE3000 BCE 2500 BCE 2000 BCE 1500 BCE 1000 BCE

MESOPOTAM IA

EGYPTInt

Int

1000 BCE1500 BCE2000 BCE2500 BCE3000 BCE

New KingdomMiddle KingdomIntOld KingdomArc hai c

Assyr iaOld BabylonAkkad iaSumaria


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EarlyTimeline

ShangDynasty:ExcavationsnearHuangRiver,datingto1600BC,showedoraclebones tortoiseshells

withinscriptionsusedfordivination.Thisisthesourceofwhatwe

knowaboutearlyChinesenumbersystems.


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EarlyTimeline


5/69

HanDynasty(206BC220AD)

SystemofEducationespeciallyforcivil

servants,i.e.scribes.

Twoimportantbooks*:

ZhouBiSuan Jing(ArithmeticalClassicoftheGnomonandtheCircularPathsofHeaven)

Jiu ZhangSuan Shu (NineChaptersontheMathematicalArt)

*unlessofcoursewereoffbyamillenniumorso.


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NineChapters

Thissecondbook,NineChapters,becamecentraltomathematicalworkinChinafor

centuries. Itisbyfarthemostimportant

mathematicalworkofancientChina. Later

scholarswrotecommentariesonitinthe

samewaythatcommentarieswerewrittenon

TheElements.


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Chaptersinuh,theNineChapters

1. Fieldmeasurements,areas,fractions

2. Percentagesandproportions3. Distributionsandproportions;arithmeticand

geometricprogressions

4. LandMeasure;squareandcuberoots

5. Volumesofshapesusefulforbuilders.

6. Fairdistribution(taxes,grain,conscripts)

7. Excessanddeficitproblems

8. Matrixsolutionstosimultaneousequations9. GouGu: ;astronomy,

surveying


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LinearEquations

Therearethreeclassesofgrain,ofwhich

threebundlesofthefirstclass,twoofthe

second,andoneofthethird,make39

measures. Twoofthefirst,threeofthe

second,andoneofthethirdmake34

measures.Andoneofthefirst,twoofthe

second

and

three

of

the

third

make

26

measures. Howmanymeasuresofgrainare

containedinonebundleofeachclass?


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LinearEquations

Solution: Arrangethe3,2,and1bundlesof

the3classesandthe39measuresoftheir

grainsattheright. Arrangeotherconditions

atthemiddleandtheleft:

1 2 3

2 3 2

3 1 1

26 34 39


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LinearEquations

Withthefirstclassontherightmultiply

currentlythemiddlecolumnanddirectlyleave

out.(Thatis,multiplythemiddlecolumnby

3,andthensubtractsomemultipleofthe

rightcolumn,toget0).

1 0 3

2 5 23 1 1

26 24 39


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LinearEquations

Dothesamewiththeleftcolumn:

0 0 3

4 5 2

8 1 1

39 24 39


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LinearEquations

Thenwithwhatremainsofthesecondclass

inthemiddlecolumn,directlyleaveout. In

otherwords,repeattheprocedurewiththe

middlecolumnandleftcolumn:

0 0 3

4 5 2

8 1 1

39 24 39

0 0 3

0 5 2

36 1 1

99 24 39


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LinearEquations

ThiswasequivalenttoadownwardGaussian

reduction. Theauthorthendescribedhowto

backsubstitutetogetthecorrectanswer.


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MethodofDoubleFalsePosition

Or,ExcessandDeficit.

Atubofcapacity10dou containsacertainquantityofhuskedrice. Grains(unhuskedrice)areaddedtofillupthetub. Whenthe

grainsarehusked,itisfoundthatthetubcontains7dou ofhuskedricealtogether. Findtheoriginalamountofhuskedrice. Assume1

dou ofunhusked riceyields6sheng ofhuskedrice,with1dou =10sheng.


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OurMethod,Maybe

Letxbeamountofhuskedrice,ybeamountof

unhusked rice. Then and

. So ,andsubstituting

wehave

. Simplifying,we

get

,and

,or2dou,5sheng.


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MethodofDoubleFalsePosition

Iftheoriginalamountis2dou,ashortageof2

sheng occurs.Iftheoriginalamountif3dou,

thereisanexcessof2sheng. Crossmultiply2

dou bythesurplus2sheng,andthen3dou by

thedeficiencyof2sheng,andaddthetwo

productstogive10dou. Dividethissumby

thesumofthesurplusanddeficiencyto

obtaintheanswer2dou and5sheng.


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DoubleFalsePosition

Whydoesthiswork?

Wewanttosolve

Ingeneral,well

examine

a

method

for

solving

.


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DoubleFalsePosition

Sosupposewewanttosolve .

Welldoitbymakingtwoguesses and ,

withtherespectiveerrors

,and

.

Thensubtractingtheseequationsgives

. Next,multiplying

equation1by andequation2by weget:


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DoubleFalsePosition

,and

.

Subtractingtheseequationsgives:

. Finally,dividingthisequationby givesus:

.

Finally,

if

is

a

surplus

and

is

adeficit,wecansay

.


20/69

GouGu inZhouBi


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LiuHuis ProofofGouGu


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SongDynasty(900 1279)

TwoBooksbyZhuShijie hadtopicssuchas:

Pascalstriangle(350yearsbeforePascal)

Solutionofsimultaneousequationsusingmatrix

methods

Celestialelementmethodofsolvingequations

ofhigherdegree. (Hornersmethod)

European

algebra

wouldnt

catch

up

to

this

leveluntilthe1700s.


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Numeration

NumeralsontheOracleStones:


24/69

Numeration


25/69

Numeration

Hindu

Arabic

0 1 2 3 4 5 6 7 8 9 10 100 1000

Chinese

Financial


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CountingRods

Countingrodsallowedforanumberofvery

quickcalculations,includingthebasicfour

arithmeticoperations,andextractionofroots.

Someexamples:


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MultiplicationwithCountingRods

1. Setupthetwofactors,inthiscase68and47,suchthattheonesdigitofthebottomisalignedwiththetensdigitofthe

top.

Leave

room

in

middleforcalculations.

2. Multiplytensdigitoftopbytensdigitofbottom,

placeinmiddlewithonesoverthebottomstensdigit.


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3. Multiplyonesdigiton

topbytensdigitonbottom,andaddto

middle.

4. Sinceyouveusedthe

tensdigitontop,eraseit.


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5. Movebottomdigitsto

the

right

one

space.

6. Multiplytensdigitonbottombyonesdigitoftop;placeonesdigitofanswerabovetensonbottom,and

7. Combine.


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8. Erasethetensdigiton

bottom

because

youre

donewithit.

9. Finally,multiplythetwounitsdigits,and

addthemtothemiddle.

10. Combine,anderase

theunitsdigitsontopandbottom.


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DivisionwithCountingRods1. Placethedividend,

407,inthemiddlerowandthedivisor,9,in

thebottomrow.Leave

space

for

the

top

row.

2. 7doesntgointo4,so

shiftthe7totheright


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DivisionwithCountingRods3. Ninegoesforty4

timeswitharemainderof4;write

thequotientinthetop

row,

the

remainder

in

themiddle.

4. Shiftthe9totheright

one

digit.


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DivisionwithCountingRods5. Ninegoesinto47five

times,witharemainderof2. Put5

ontop,remainderin

themiddle.

6. Theansweris,45,

remainder2.


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Fractions

FromNineChapters:

Ifthedenominatorandnumeratorcanbe

halved,halvethem. Ifnot,laydownthe

denominatorandnumerator,subtractthe

smallernumberfromthegreater. Repeatthe

processtoobtainthegreatestcommondivisor

(teng). Simplify

the

original

fraction

by

dividingbothnumbersbytheteng.


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Fractions

Additionandsubtractionweredoneaswedo

thembutwithoutnecessarilyfindingleastcommondenominators thecommon

denominatorisjusttheproductofthetwo

denominators. Thefractionissimplifiedafter

addingorsubtracting.


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Fractions

Multiplicationwasdoneaswedoit.

Divisionwasdonebyfirstgettingcommon

denominators,theninvertingandmultiplying

sothatthecommondenominatorscancel.

Thenthefractionwassimplified.


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Negativenumbers? Red andblackrods,orrodslaiddiagonallyover

others.

Forsubtractions withthesamesigns,takeawayonefromtheother;withdifferentsigns,addonetotheother;positivetakenfromnothing

makesnegative,negativefromnothingmakespositive.

Foraddition withdifferentsignssubtractonefromtheother;withthesamesignsaddonetotheother;positiveandnothingmakespositive;negativeandnothingmakesnegative.


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Approximationsof

LiuHui,260AD: 3.1416(byinscribing

hexagonincircle,usingthePythagoreanTheoremtoapproximatesuccessively

polygonsofsides12,24,.,96).

Zu Chongzhi,480AD:between 3.1415926

and3.1415927(bysimilarmethod,but

movingpast96tooh,say24,576).


40/69

Solving

Polynomials PreciousMirroroftheFourElementsbyShu

Shijie,1303CE. MethodknownasFanFa,todayknownas

HornersMethod, andusingwhatyoumayknowassyntheticdivision.


41/69

Fan

Fa:

Startingwithaguessof

1,wedosyntheticdivisiontoget

remainder12. Then

ignoringtheremainder,

wedoanothersynthetic

divisiononthe

quotient,andrepeat

untilwegetdowntoa

constant.

1 1 7 3 21

1 6 9

1

6

9 12

1 1 6 9

1 5

1 5 14

1 1 5

1

1 4

1 1

1


42/69

Fan

Fa:

Theremaindersonthe

first

and

second

lines

aredividedand

multipliedby 1to

obtainthenext

adjustmenttothe

guessforaroot.

Also,

1 1 7 3 21

1 6 9

1 6 9 12

1 1 6 9

1 5

1 5 14

1 1 5

1

1 4

1 1

1 GUESS: 0.857143


43/69

Fan

Fa:

Theremaindersgive

you

the

polynomial

for

thenextroundof

syntheticdivision.

1 1 7 3 21

1 6 9

1 6 9 12

1 1 6 9

1 5

1 5 14

1 1 5

1

1 4

1 1

1


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Fan

Fa:

Oursuggestednext

guess

was

0.857143.

Wetry0.8,butthe

negativesignonthe

nextguess( 0.06753)

tellsusour0.8wastoo

large. Wegobackto

0.7.

0.8 1 4 14 12

0.8 2.56 13.248

1 3.2 16.56 1.248

0.8 1 3.2 16.56

0.8 1.92

1 2.4 18.48

0.8 1 2.4

0.8

1 1.6

0.8 1

1 GUESS: 0.06753


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Fan

Fa:

Thistime,wegeta

positive

next

guess

of

0.03217. Sowetake

0.03asourguessfor

thenextdigit,andgo

again.

And,again,wetakethe

enddigitsfromeach

resultlinetopopulate

ournexttopline.

0.7 1 4 14 12

0.7 2.31 11.417

1 3.3 16.31 0.583

0.7 1 3.3 16.31

0.7 1.82

1 2.6 18.13

0.7 1 2.6

0.7

1 1.9

0.7 1

1 GUESS: 0.032157


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Fan

Fa:

Runningthealgorithm

again,

we

get

a

next

guessof.002.

Sofarthen,ourapproximaterootis

1.73andisheadingfor

about

1.732

0.03 1 1.9 18.13 0.583

0.03 0.0561 0.54558

1 1.87 18.1861 0.037417

0.03 1 1.87 18.1861

0.03 0.0552

1 1.84 18.2413

0.03 1 1.84

0.03

1 1.81

0.03 1

1 GUESS: 0.002051


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Fan

Fa:

Ournextguessis

positive

and

very

small,

sotheerrorinour

currentapproximation

issmall. Bestguess:

1.732(+0.00005ish?)

2050808

0.002 1 1.81 18.2413 0.037417

0.002 0.00362 0.03649

1 1.808 18.2449 0.000927

0.002 1 1.808 18.2449

0.002 0.00361

1 1.806 18.2485

0.002 1 1.806

0.002

1 1.804

0.002 1

1 GUESS: 5.08E05


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Fan

Fa:

Aw,heck. Justforfun,

lets

go

one

more:

As youcansee,were

gettingveryclosetoour

calculatorprovided

approximation.

0.00005 1 1.804 18.2485 0.000927

0.00005 9E05 0.00091

1 1.80395 18.2486 1.47E05

0.00005 1 1.80395 18.2486

0.00005 9E05

1 1.8039 18.2487

0.00005 1 1.8039

0.00005

1 1.80385

0.00005 1

1 GUESS: 8.08E07


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Fan

Fa and

Horner ThisgeneralmethodwasrediscoveredbyWilliam

Horner(1786 1837)andpublishedinapaperin1830.

Exceptitwasprettymuchidenticaltoamethodpublishedin1820byTheopholis Holdred,a

Londonwatchmaker. Ofcourse,PaoloRuffini (1765 1822),whowe

willdiscusslaterinanothercontext,alreadywonaprizeforoutliningthismethodinItaly.

And,ofcourse,theresShu Shijie,morethanfourcenturiesearlier.


50/69

Magic

Squares


51/69

Lo

Shu Thesemimythical

Emperor

Yu,

(circa

2197

BC)walkingalongthe

banksoftheLuo River,

lookeddowntoseethe

DivineTurtle.Onthebackofhisshellwasa

strangedesign.


52/69

Lo

Shu Whenthedesignonthe

back

was

translated

into

numbers,itgavethe

3x3magicsquare.

Saying

the

3x3

magic

squareisappropriate

becauseitisuniqueup

torotationsand

reflections.


53/69

He

Tu Accordingtolegend,

the

HeTu is

said

to

have

appearedtoEmperor

Yuonthebackof(or

fromthehoofprintsof)

aDragonHorse

springingoutofthe

Huang(Yellow)River.


54/69

He

Tu Whenitwastranslated

intonumbers,itgavea

crossshapedarray.

Tounderstanditsmeaningistounderstand

the

structure

of

the

universe,apparently.

Or,atleasttounderstandthat,disregardingthe

central5,theoddsandevensbothaddto20.

7

2

8 3 5 4 9

1

6


55/69

Magic

Squares YangHui,ContinuationofAncient

MathematicalMethodsforElucidatingtheStrangePropertiesofNumbers,1275.


56/69

Order

3 Arrange19inthree

rows

slanting

downwardtotheright.

1

4 2

7 5 3

8 6

9


57/69

Order

3 Arrange19inthree

rows

slanting

downwardtotheright.

Exchangethehead(1)

and

the

shoe

(9).

9

4 2

7 5 3

8 6

1


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Order

3 Arrange19inthree

rows

slanting

downwardtotheright.

Exchangethehead(1)

and

the

shoe

(9). Exchangethe7and3.

9

4 2

3 5 7

8 6

1


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Order

3 Arrange19inthree

rowsslanting

downwardtotheright.

Exchangethehead(1)

andtheshoe(9).

Exchangethe7and3.

Lower9,andraise1.

4 9 2

3 5 7

8 1 6


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Order

3 Arrange19inthree

rowsslanting

downwardtotheright.

Exchangethehead(1)

andtheshoe(9).

Exchangethe7and3.

Lower9,andraise1.

Skootch*inthe3and7*technicalterm

4 9 2

3 5 7

8 1 6


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Order

3

The

Lo

Shu


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Order

4 Write1 16infour

rows.1 2 3 4

5 6 7 8

9 10 11 1213 14 15 16


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Order

4 Write1 16infour

rows.

Exchangecornersof

outersquare

16 2 3 13

5 6 7 8

9 10 11 124 14 15 1


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Order

4 Write1 16infour

rows.

Exchangecornersof

outersquare

Exchangethecornersofinnersquare.

16 2 3 13

5 11 10 8

9 7 6 124 14 15 1


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Order

4 Write1 16infour

rows.

Exchangecornersof

outersquare


16 2 3 13

5 11 10 8

9 7 6 124 14 15 1


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Order

4 Write1 16infour

rows.

Exchangecornersof

outersquare


Voila!Sumis34.

16 2 3 13

5 11 10 8

9 7 6 124 14 15 1


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Order

4 Othermagicsquaresoforder4arepossible

fordifferentinitialarrangementsofthenumbers1 16.

13 9 5 1

14 10 6 2

15 11 7 3

16 12 8 4

4 9 5 16

14 7 11 2

15 6 10 3

1 12 8 13

Order 5 6 7


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Order5,6,7,.

YangHui constructedmagicsquaresoforders

up

through

10,

although

some

were

incomplete.

A Little About Magic Squares


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ALittleAboutMagicSquares

Normalmagicsquaresofordernarenxnarrays

containingeachnumberfrom1through Theyexistforall .

Thesumofeachrow,column,anddiagonalisthemagicnumber Mwhichfornormalmagic

squares

depends

only

on

n.

. Forthefirst

few

ns this

is

15,

34,

65.

111,175...

Forn odd,thenumberin

thecentralcellis

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MathematicsinAncientChina(69P)

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