squre root and quadritic.pdf

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http://www.math.cornell.edu/~dwh/papers/geomsolu/geomsolu.html 1/16

GeometricSolutionsofQuadraticandCubicEquations

by

DavidW.Henderson1

DepartmentofMathematics,CornellUniversity

Ithaca,NY,148537901,USA1

Iamreadytoleadyou,thereader,onapaththroughpartoftheforestofmathematicsapaththathasdelightedmemanytimesandsurprisedme.EverytimeIwalkalongitIseesomethingIhadnotseenbefore.Wewillbringwithusthequestion:Whataresquareroots?Wewillfindwhatisoneoftheoldestwrittenmathematicalproofs,stillverymuchalive,rightalongsidesomenewresultsneverbeforepublished.

Thesewillbecombinedtosolvequadraticequationsby"completingthesquare"arealsquare.TheseinturnleadtoconicsectionsandcuberootsandculminatinginthebeautifulgeneralmethodfromOmaral'Khayyam,thePersiangeometer,philosopher,poet,whichcanbeusedtofindalltherealrootsofcubicequations.AlongthewayweshallclearlyseesomeoftheancestralformsofourmodernCartesiancoordinatesandanalyticgeometry.Iwillpointourseveralinaccuraciesandmisconceptionsthathavecreptintothemodernhistoricalaccountsofthesematters.ButIurgeyoutonotlookatthisonlyforitshistoricalinterestbutratherlookforthemeaningithasinourcurrentdayunderstandingofmathematics.Thispathisnotthroughadeadmuseumorpetrifiedforest,thispathpassesthroughideaswhichareverymuchaliveandwhichhavesomethingtosaytoourmoderntechnological,increasinglynumerical,world.

1.TheBeginningofthePath

FormethepathstartedineighthgradewhenIaskedmyteacher"Whatisthesquareroot?"IknewthatthesquarerootofNwasanumberwhosesquarewasequaltoNbutwherecanIfindit?(Hiddeninthatquestionis"HowdoIknowitalwaysexists?")Iknewwhatthesquarerootsof4and9werenoproblemthere.

Ievenknewthatwasthelengthofthediagonalofaunitsquare,butwhatofor?AtfirsttheteachershowedmeaSquareRootTable(atableofnumericalsquareroots),butIsoondiscoveredthatifItookthenumberlistedinthetableasandsquareditIgot1.999396not2.(Moderndaypocketcalculatorsgiverisetothesameproblem.)SoIpersistedaskingmyquestionWhatisthesquareroot?ThentheteacheransweredbygivingmeTHEANSWERtheSquareRootAlgorithm.DoyouremembertheSquareRootAlgorithmthatprocedure,similartolongdivision,bywhichitispossibletocalculatethesquareroot?Orperhapsmorerecentlyyouweretaughtthe"DivideandAverage"Methodwhichgoeslikethis:IfA1isanapproximationofNthentheaverageofA1andN/A1isanevenbetterapproximationwhichwecouldcallA2.AndthenthenextapproximationA3istheaverageofA2andN/A2.InequationformthisbecomesAn+1=(1/2)(An+(N/An)).Forexample,ifA1=1.5isanapproximationof2,thenA2=1.417,A3=1.414216andsofortharebetterandbetterapproximation.Butwait!Mostofthetimethesealgorithmsdonotcalculatethesquareroottheyonlycalculateapproximationstothesquareroot.ThealgorithmshaveanadvantageoverthetablesbecauseI

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could,atleastintheory,calculateapproximationsascloseasIwished.HowevertheyarestillonlyapproximationsandmyquestionstillremainedWhatisthissquarerootwhichthesealgorithmsapproximate?

Myeighthgradeteacherthengaveup,butlaterincollegeIfoundoutthatmodernmathematicsanswers:"Wemakeanassumption(TheCompletenessAxiom)whichimpliesthatthesequenceofapproximationsfromtheSquareRootAlgorithmmustconvergetosomerealnumber."And,whenIcontinuedtoaskmyquestion,IfoundthatinmodernmathematicsthesquarerootisacertainequivalenceclassofCauchysequencesofrationalnumbersoracertainDedekindcut.Ithenletgoofmyquestionandforgotitintheturmoilofgraduateschool,writingmythesisandbeginningmymathematicalcareer.

Later,IstartedteachingageometrycourseformathematicsmajorsandoneofthetopicswasDissectionTheorywhichleads(amongotherthings)totheresultthateverypolygonalregionintheplanecanbecutup(dissected)intoafinitenumberofpieceswhichcanthenberearrangedtoformasquare.Inthiscasewesaythatthepolygonalregionisequivalentbydissectiontoasquare.Apreliminarysteptothegeneralresultisthe:

Theorem1.Everyrectangleisequivalentbydissectiontoasquare.

IpresentedtotheclassthefollowingproofwhichIfoundslightlymodifiedinastandardgeometrytextbook,Eves(1963):

"Lets=abbethesideofthesquareequivalenttotherectanglewithsidesaandb.Placethesquare,AEFH,ontherectangle,ABCD,asshownin[thefigure].

DrawEDtocutBCinRandHFinK.LetBCcutHFinG.FromthesimilartrianglesKDHandEDAwehaveHK/AE=HD/AD,or

HK=(AE)(HD)/AD=s(as)/a=ss/a=sb.

Therefore,...wehaveEFKRCD,EBRKHD."

(IncasethatABCDissolongandskinnythatKendsupbetweenGandFwecan,bycuttingABCDinhalfandstackingthehalves,reducetheprooftotheabovecase.)

IwassatisfiedwiththeproofuntilinthesecondyearofthecoursewhenIstartedsensingstudentuneasinesswiththeproof.AsIlistenedtotheiruneasinesstherestartedtocomeupthequestionWhatisab?Howdoyoufindit?Oh,yes,Irememberthatusedtobemyquestion!

ThestudentsandIalsonoticedthatthefactsusedaboutsimilartrianglesintheaboveproofareusuallyprovedusingthetheoryofareasoftrianglesandthusthatthisproofcouldnotbeusedaspartofa

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concretetheoryofareasofpolygons,whichwasourpurposeinstudyingDissectionTheoryinthefirstplace.

Thatstartedmeoffonanexplorationwhichcontinuedonandoffovermanyyears.SomeofwhatIfoundIwillnowshowyou(butinadifferentorderfromtheorderIfirstsawthen).

2.WhatisaSquareRoot?

WhilereadinganarticleaboutsomethingelseIranacrossanitemthatsaidthattheproblemofchangingarectangleintoasquareappearedintheSulbasutrambyBaudhayana(seePrakash(1968))."Sulbasutram"means"Rulesofthecord"andisanancient(atleast600BC)bookwritteninSanskritasahandbookforpeoplewhowerebuildingaltarsandtemples.Mostofthebookgivesdetailedinstructionsontempleconstructionanddesign,butthefirstchapterisageometrytextbookwhichcontainsgeometricstatementscalled"Sutra".Sutra54is:(Here"oblong"means"rectangle".)

"Ifyouwishtoturnanoblongintoasquare,takethetiryanmani,i.e.theshortersideoftheoblongforthesideofsquare.Dividetheremainder(thatpartoftheoblongwhichremainsafterthesquarehasbeencutoff)intotwopartsandinverting(theirplaces)jointhosetwopartstotwosidesofthesquare.(Wegetthusalargesquareoutofonecornerofwhichasmallsquareiscutoutasitwere.)Filltheemptyplace(inthecorner)byaddingapiece(asmallsquare).Ithasbeentaughthowtodeductit(theaddedpiece).

"Byaddingthesmallsquareinthecornerwegetalargesquarewhichisequaltotheoblongplusthesmallsquare,thereforewemustdeductthesmallsquarefromthelargesquare(seeSutra51)andthenwehaveasremainderasquarewhichisequaltotheoblong."

HereisadiagramforSutra54:

Soourrectanglehasbeenchangedintoalargesquarefromwhichasmallsquarehasbeenremoved(ordeducted).NowSutra51:

"IfyouwishtodeductonesquarefromanotherSquare,cutoffapiecefromthelargersquarebymakingamarkonthegroundwiththesideofthesmallersquarewhichyouwishtodeduct(theprocessisthesameasthatdescribedinSutra50anoblongiscutoff,thesidesofwhichareequaltothesidesofthetwogivensquares)drawoneofthesides(THECORDREPRESENTINGoneOFTHElongerSIDESoftheoblong)acrosstheoblongsothatittouchestheothersidewhereittouches(theotherside),bythislinewhichhasbeencutoffthesmallsquareisdeductedfromthelargeone(i.e.thecutofflineisthesideofasquaretheareaofwhichisequaltothedifferenceofthetwosquares.)"

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Thislastassertionfollowsfromsutra50:

"Ifyouwishtocombinetwosquaresofdifferentsizeintoone,scratchupwiththesideofthesmallersquareapiececutofffromthelargerone(i.e.cutoffapiecefromthelargersquarebyscratchingupthegroundormakingamarkuponthegroundatadistancefromoneendofasideofthelargesquare,whichisequaltothelengthofthesideofthesmallersquareandbyrepeatingthisprocessontheoppositesideofthelargersquareandjoiningthetwomarksonthegroundbyalineorcord,anoblongiscutoff,ofwhichthetwolongersidesareequaltothesideofthelargesquare).Thediagonalofthiscutoffpieceisthesideofthecombinedsquares(ofthesquarewhichcombinesthetwosquares)."

Doessutra50soundfamiliar?ItshoulditisaclearstatementofwhatwecallthePythagoreanTheorem,writtenbeforePythagoraswasborn!

A.Seidenberg(1961)inanarticleentitledTheRitualOriginofGeometrygivesadetaileddiscussionofthesignificanceoftheSulbasutram.Hearguesthatitwaswrittenbefore600BC(Pythagoraslivedabout500BCandEuclidabout300BC)Hegivesevidencetosupporthisclaimthatitcontainscodificationofknowledgegoing"farbackof1700BC"andthatthisknowledgewasthecommonsourceofIndian,Egyptian,BabylonianandGreekmathematics.Combinedtogethersutras50,51and54describeaconstructionofasquarewiththesameareaasagivenrectangle(oblong)andaproof(basedonthePythagoreanTheorem)thatthisconstructioniscorrect.YoucanfindstatedinmanybooksandarticlesthattheancientHindus,ingeneral,andtheSulbasutraminparticular,didnothaveproofsordemonstrationsortheyaredismissedasbeing"rare".Isuggestyoudecideforyourself.

BaudhayanaavoidstheCompletenessAxiombygivinganexplicitconstructionofthesideofthesquare.

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Theconstructioncanbesummarizedinthediagram:

ThisisthesameasEuclid'sconstructioninPropositionII14(seeHeath(1956),page409).ButEuclid'sproofismuchmorecomplicated.

NotethatneitherBaudhayananorEuclidgiveaproofofTheorem1becausetheuseofthePythagoreanTheoremobscuresthedissection.However,theydogiveaconcreteconstructionandaproofthattheconstructionworks.Inaddition,ifsupplementedwithadissectionproofofthePythagoreanTheoremsuchasinTheorem3,below,bothBaudhayana'sandEuclidmethodsprove(withoutusingcompleteness):

Theorem2:ForeveryrectangleRtherearesquaresS1andS2suchthatR+S2isequivalentbydissectiontoS1+S2andthusRandS1havethesamearea.

Theorem3:(DissectionversionofPythagoreanTheorem).Inanyrighttriangle,theunionofthesquaresonthetwosidesisequivalentbydissectiontothesquareonthehypotenuse.

ProofofTheorem3knowntotheancientChinese:

Aboutadozendifferent(andcorrect)proofsofTheorems1havebeenfoundbythestudentsinmygeometrycourse.Oneparticularlyclearonefollows:(AsfarasIknowthisproofhasneverbeforebeenpublished.)

LetABGHbetherectangleandextendthelineABtoCsothatBCBH.DrawthesemicircleSonACandletDbetheintersectionofSwiththeextensionofBH.

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ThenADCisarightangleandtheanglesarecongruentasindicatedinthediagram.ConstructthesquareDBEF.ThenDBCIGAandDBCDFJbothbyAngleSideAngle.EasilyAEJIHD.ThustherectangleABHGisequivalentbydissectiontothesquareDBEF.

Noticethatthisproofavoidsassumingthatthesquarerootexists(andthusavoidstheCompletenessAxiom)andavoidsusinganyfactsaboutsimilartriangles.Theproofexplicitlyconstructsthesquareandshowsinanelementarywaythatitsareaisthesameastheareaoftherectangle.Thereisnoneedfortheareaorthesidesoftherectangletobeexpressedinnumbers.Alsogivenarealnumber,b,thesquarerootofbcanbeconstructedbyusingarectanglewithsidesband1.

So,finally,IhaveananswertomyquestionWhatisasquareroot?Isay"ananswer"becauseeveryyearIseemoreorseeitfromadifferentpointofview.

3.QuadraticEquations

Findingsquarerootsisthesimplestcaseofsolvingquadraticequations.Ifyoulookinsomehistoryofmathematicsbooks,youwillfindthatquadraticequationswereextensivelysolvedbytheBabylonians(numerically)andbytheGreeks(geometrically).However,theearliestknowngeneraldiscussionofquadraticequationstookplacebetween800and1100ADintheMuslimEmpire.BestknownareMohammedIbnMusaal'Khowarizmi(wholivedinBaghdad)andOmaral'Khayyam(wholivedinPersia,nowIran,andismostlyknownintheWestforhispoetryTheRubaiyat.BothwrotebooksentitledAljabrW'almugabalah,al'Khowarizmiabout820ADandal'Khayyamabout1100AD.Fromal'Khowarizmiwegetourword"algorithm"andfromthetitleoftheirbooksourword"algebra".AnEnglishtranslationofbothbooksisavailableinmanylibraries,ifyoucanfigureoutwhosenameitiscataloguedunder(seeReferences,Karpinski(1915)andKasir(1931)).

Inthesebooksyoufindgeometricandnumericalsolutionstoquadraticequationsandgeometricproofsofthesesolutions.Butthefirstthingthatyounoticeisthatthereisnotonegeneralquadraticequationasweareusedtoit:2ax+bx+c=0.Rather,becausetheuseofnegativecoefficientsandnegativerootswasavoided,theylistsixtypesofquadraticequations(wefollowal'Khayyam'sleadandsettheleadingcoefficientequalto1):

1. x=c,whichneedsnosolution,2. x=bx,whichiseasilysolved,3. x2=c,whichhasrootx=c,4. x2+bx=c,withrootx=(b/2)2+c]b/2,5. x2+c=bx,withrootsx=b/2(b/2)2c],ifc

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6. x2=bx+c,withrootx=b/2+[(b/2)+c].

Herebandcarealwayspositivenumbersorageometriclength(b)andarea(c).Thesetypesaretheonlypossibilitieswithpositivecoefficientsandpositiveroots.(x2+bx+c=0hasnopositiveroots.)

Butwhydidmathematiciansavoidnegativenumbers?Theavoidanceofnegativenumberswaswidespreaduntilafewhundredyearsago.IntheSixteenthCentury,Europeanmathematicianscalledthenegativenumbersthatappearedasrootsofequations,"numerifictici"fictitiousnumbers(seeWitmer(1968),page11).

Togetafeelingforwhy,thinkaboutthemeaningof2x3astwo3'sand3x2asthree2'sandthentrytofindameaningfor3x(2)and2x(+3).Anotheranswerisfoundintherelianceongeometricjustifications,asal'Khayyamwrote(seeAmirMoez(1963),page329):

"Whoeverthinksalgebraisatrickinobtainingunknownshasthoughtitinvain.Noattentionshouldbepaidtothefactthatalgebraandgeometryaredifferentinappearance.Algebras(jabbreandmaqabeleh)aregeometricfactswhichareprovedbypropositionsfiveandsixofBooktwoof[Euclid's]Elements".

Somehistorianshavequotedthispassagebuthaveleftoutallthewordsappearingafter"proved".Inmyopinion,thisomissionchangesthemeaningofthepassage.Euclid'spropositionsthatarementionedbyal'KhayyamarethebasicingredientsofEuclid'sproofofthesquarerootconstructionandformabasisfortheconstructionofconicsectionsseebelow.Geometricjustificationwhentherearenegativecoefficientsisatleastverycumbersomeifnotimpossible.(Ifyoudoubtthistrytomodifysomeofthegeometricjustificationsbelow.)Inanycase,Euclid,uponwhichthesemathematiciansrelied,didnotallownegativequantities.

Forthegeometricjustificationof(III)andthefindingofsquareroots,al'KhayyamreferstoEuclid'sconstructionofthesquarerootinPropositionII14.

For(IV)wehaveasgeometricjustification:

andthus,by"completingthesquare"onx+b/2,wehave(x+b/2)2=c+(b/2)2.NotethesimilaritybetweenthisandBaudhayana'sconstructionofthesquareroot(seeSection2).

For(V),firstassumex

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Thisleadstox=b/2[(b/2)2c].Notethatifc>(b/2)2thenthisgeometricsolutionisimpossible.Whenx>(b/2),usethedrawings:

Forthesolutionof(VI)usethedrawing:

Dotheabovesolutionsfindthenegativeroots?Well,first,theanswerisclearly,No,ifyoumean:Didal'Khowarizmiandal'Khayyam(ortheearlierGreeksandBabylonians)mentionnegativeroots?Butletusnotbetoohasty,supposer(r,positive)isthenegativerootofx2+bx=c.Then(r)2+b(r)=corr2

=br+c.Thusrisapositiverootofx2=bx+c!Theabsolutevalueofthenegativerootofx2+bx=cisthepositiverootofx2=bx+candviceversa.Also,theabsolutevaluesofthenegativerootsofx2+bx+c=0arethepositiverootsofx2+c=bx.So,inthissense,Yes,theabovegeometricsolutionsdofindalltherealrootsofallquadraticequations.Thusitismisleadingtostate,asmosthistoricalaccountsdo,thatthegeometricmethodsfailedthefindnegativeroots.Theusersofthesemethodsdidnotfindnegativerootsbecausetheydidnotconceiveofthem.However,themethodscanbeeasilyanddirectlyusedtofindallthenegativeroots.

4:ConicSectionsandCubeRoots

TheGreeksnoticedthat,ifa/c=c/d=d/b,then(a/c)2=(c/d)(d/b)=(c/b)andthusc3=a2b.Nowsettinga=1,weseethatwecanfindthecuberootofb,ifwecanfindcanddsuchthatc2=dandd2=bc.Ifwethinkofcanddasbeingvariablesandbaconstant,thenweseetheseequationsastheequationsoftwoparabolaswithperpendicularaxesandthesamevertex.TheGreeksalsosawitthiswaybutfirstthey

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hadtodeveloptheconceptofaparabola!

TotheGreeks,andlateral'Khayyam,ifABisalinesegment,thentheparabolawithvertexBandparameterABisthecurvePsuchthat,ifCisonP,thentherectangleBDCE(seethedrawing)hasthepropertythat(BE)2=DBAB.SinceinCartesiancoordinatesthecoordinatesofCare(BE,BD)thislastequationbecomesafamiliarequationforaparabola.

PointsoftheparabolamaybeconstructedbyusingtheconstructionforthesquarerootgiveninSection2.Inparticular,EistheintersectionofthesemicircleonADwiththelineperpendiculartoABatB.(TheconstructioncanalsobedonebyfindingD'suchthatAB=DD',thenthesemicircleonBD'intersectsPatC.)Iencourageyoutotrythisconstructionyourselfitisveryeasytodoifyouuseacompassandgraphpaper.

Nowwecanfindthecuberoot.LetbbeapositivenumberorlengthandletAB=bandconstructCsothatCBisperpendiculartoABandsuchthatCB=1.

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ConstructaparabolawithvertexBandparameterABandconstructanotherparabolawithvertexBandparameterCB.LetEbetheintersectionofthetwoparabolas.DrawtherectangleBGEF.Then(EF)2=BFABand(GE)2=GBCB.But,settingc=GE=BFandd=GB=EF,wehaved2=cbandc2=d.Thusc3=b.Ifyouuseafinegraphpaperitiseasytogetthreedigitaccuracyinthisconstruction.

TheGreeksdidathoroughstudyofconicsectionsandtheirpropertieswhichculminatedinAppolonius'sbookConicswhichappearedin200BC.YoucanreadthisbookinEnglishtranslation,seeHeath(1961).

Tofindrootsofcubicequationsinthenextsectionweshallalsoneedtoknowthe(rectangular)hyperbolawithvertexBandparameterAB.ThisisthecurveH,suchthatifEisonHandACEDisthedeterminedrectangle(seedrawing),then(EC)2=BCAC.

ThepointEcanbeconstructedusingSection2.LetFbethebisectorofAB.ThenthecirclewithcenterFandradiusFCwillintersectatDthelineperpendiculartoABatA.Fromthedrawingitisclearhowthesecirclesalsoconstructtheotherbranchofthehyperbola(withvertexA.)

NoticehowthesedescriptionsandconstructionsoftheparabolaandhyperbolalookverymuchliketheyweredoneinCartesiancoordinates.TheancestralformsofCartesiancoordinatesandanalyticgeometryareevidenthere.Alsotheyareevidentinthesolutionsofcubicequationsinthenextsection.Theideas

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ofCartesiancoordinatesdidnotcometoDescartesoutofnowhere.TheunderlyingconceptsweredevelopinginGreekandMuslimmathematics.OneoftheapparentreasonsthatfulldevelopmentdidnotoccuruntilDescartesisthat,aswehaveseen,negativenumberswerenotaccepted.ThefulluseofnegativenumbersisessentialfortherealizationofCartesiancoordinates.

5:RootsofCubicEquations

InhisAlJabrwa'lmuqabalahOmaral'Khayyamalsogavegeometricsolutiontocubicequations.Weshallseethathismethodsaresufficienttofindgeometricallyallreal(positiveornegative)rootsofcubicequationshoweverinhisfirstchapteral'Khayyamsays:(seeKasir(1931),page49.)

"When,however,theobjectoftheproblemisanabsolutenumber,neitherwe,noranyofthosewhoareconcernedwithalgebra,havebeenabletoprovethisequationperhapsotherswhofollowuswillbeabletofillthegapexceptwhenitcontainsonlythethreefirstdegrees,namely,thenumber,thethingandthesquare."

By"absolutenumber",al'Khayyamisreferringto,whatwecall,algebraicsolutionasopposedtogeometricone.Thisquotationsuggests,contrarytowhatmanyhistoricalaccountssay,thatal'Khayyamexpectedthatalgebraicsolutionswouldbefound.

Al'Khayyamfound19typesofcubicequations(whenexpressedwithonlypositivecoefficients).(SeeKasir(1931),page51).Ofthese19,fivereducetoquadraticequations(e.g.,x3+ax=bxreducestox2+ax=b).Theremaining14typesal'Khayyamsolvesbyusingconicsections.Hismethodsfindallthepositiverootsofeachtypealthoughhefailedtomentionsomeoftherootsinafewcasesand,ofcourse,heignoresthenegativeroots.Insteadofgoingthroughhis14types,Iwillshowhowasimplereductionwillreduceallthetypestoonly3typesinadditiontotypesalreadysolvedsuchas,x3=b.Iwillthengiveal'Khayyam'ssolutionstothesetypes.

Inthecubicy3+py2+gy+r=0(where,here,p,g,r,arepositive,negative,orzero)sety=x(p/3).Tryit!Theresultingequationinxwillhavetheformx3+sx+t=0,(where,here,sandtarepositive,negativeorzero).Ifwerearrangethisequationsoallthecoefficientsarepositivethenwegetfourtypesthathavenotbeenpreviouslysolved:

(I)x3+ax=b,(II)x3+b=ax,(III)x3=ax+b,and(IV)x3+ax+b=0,

whereaandbarepositive,inaddition,totypespreviouslysolved.Now(IV)hasnopositiverootsandtheabsolutevalueofitsnegativerootsarethe(positive)rootsof(I).Also,theabsolutevalueofthenegativerootsof(II)aretherootsof(III)andviceversa.Thus,weneedonlyfindthepositiverootsoftypes(I),(II),and(III).

Al'Khayyam'ssolutionfortype(I):x3+ax=b.

"Acubeandsidesareequaltoanumber.LetthelineAB[seefigure]bethesideofasquareequaltothegivennumberofroots,[thatis,(AB)2=a,thecoefficient.]ConstructasolidwhosebaseisequaltothesquareonAB,equalinvolumetothegivennumber,[b].Theconstructionhasbeenshownpreviously.LetBCbetheheightofthesolid.[I.e.BC(AB)2=b.]LetBCbeperpendiculartoAB...ConstructaparabolawhosevertexisthepointB...andparameterAB.ThenthepositionoftheconicHBDwillbetangenttoBC.DescribeonBCasemicircle.Itnecessarilyintersectstheconic.LetthepointofintersectionbeDdropfromD,whosepositionisknown,twoperpendicularsDZandDEonBZandBC.Boththepositionandmagnitudeoftheselinesareknown."

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TherootisEB.Al'Khayyam'sproof(usingamorecompactnotation)is:Fromthepropertiesoftheparabola(Section4)andcircle(Section2)wehave

(DZ)2=(EB)2=BZABand(ED)2=(BZ)2=ECEB,

thus

EB(BZ)2=(EB)2EC=BZABEC

andtherefore

ABEC=EBBZand(EB)3=EB(BZAB)=(ABEC)AB=(AB)2EC

So

(EB)3+a(EB)=(AB)2EC+(AB)2(EB)=(AB)2CB=b.

ThusEBisarootofx3+ax=b.Sincex2+axincreasesasxincreases,therecanbeonlythisoneroot.

Al'Khayyam'ssolutionsfortypes(II)and(III):x3+b=axandx3=ax+b.

Al'Khayyamtreatedtheseequationsseparatelybutbyallowingnegativehorizontallengthswecancombinehistwosolutionsintoonesolutionofx3b=ax.LetABbeperpendiculartoBCandasbeforelet(AB)2=aand(AB)2BC=b.PlaceBCtotheleftifthesigninfrontofbisnegative(type(III))andplaceBCtotherightisthesigninfrontofbispositive(type(II)).ConstructaparabolawithvertexBandparameterAB.ConstructbothbranchesofthehyperbolawithverticesBandCandparameterBC.

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Eachintersectionofthehyperbolaandtheparabola(exceptforB)givesarootofthecubic.SupposetheymeetatD.ThendropperpendicularsDEandDZ.TherootisBE(negativeiftotheleftandpositiveiftotheright).Again,ifyouusefinegraphpaperitiseasytogetthreedigitaccuracyhere.Ileaveitforyou,thereader,toprovidetheproofwhichisverysimilartotype(I).

Alittlemorehistory:Mosthistoricalaccountsassertcorrectlythatal'Khayyamdidnotfindthenegativerootsofcubics.However,theyaremisleadinginthattheyallfailtomentionthathismethodsarefullysufficienttofindthenegativerootsaswehaveseenabove.Thisisincontrasttothecommonassertion(see,forexample,Davis&Hersch(1981))thatGirolamoCardano(16thcenturyItalian)wasthefirsttopublishthegeneralsolutionofcubicequationswheninfact,asweshallsee,hehimselfadmittedthathismethodsareinsufficienttofindtherealrootsofmanycubics.

Cardanopublishedhisalgebraicsolutionsinhisbook,ArtisMagnae(TheGreatArt)whichwaspublishedin1545.ForareadableEnglishtranslationandhistoricalsummary,seeWitmer(1968).Cardanousedonlypositivecoefficientsandthusdividedthecubicequationsintothesame13types(excludingx3=candequationsreducibletoquadratics)usedearlierbyal'Khayyam.Cardanoalsousedgeometrytoprovehissolutionsforeachtype.Aswedidabovewecanmakeasubstitutiontoreducethesetothesametypesasabove:

(I)x3+ax=b,(II)x3+b=ax,(III)x3=ax+b,and(IV)x3+ax+b=0.

Ifweallowourselvestheconvenienceofusingnegativenumbersandlengthsthenwecanreducethesetoonetype:x3+ax+b=0,wherenowweallowaandbtobeeithernegativeorpositive.

Themain"trick"thatCardanousedwastoassumethatthereisasolutionofx3+ax+b=0oftheformx=t1/3+u1/3.Pluggingthisintothecubicweget

(t1/3+u1/3)3+a(t1/3+u1/3)+b=0.

Ifyouexpandandsimplifythisyougetto

t+u+b+(3t1/3u1/3+a)(t1/3+u1/3)=0.

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Thusx=t1/3+u1/3isarootif

t+u=bandtu=(a/3)3.

Solving,wefindthattanduaretherootsofthequadraticequationz2+bz(a/3)3=0whichCardanosolvedgeometrically(andyoucanusethequadraticformula)toget

t=b/2+(b/2)a3)3]andu=b/2[(b/2)a3)3].

Thusthecubichasroots

x=t1/3+u1/3={b/2+(b/2)a3)3]}1/3+{b/2[(b/2)a3)3]}1/3.

ThisisCardano'scubicformula.But,astrangethinghappened,Cardanonoticedthatthecubicx3=15x+4hasapositiverealroot4but,forthisequation,a=15andb=4,andifweputthesevaluesintohiscubicformulawegetthattherootsofx3=15x+4are

x={2+}1/3+{2}1/3.

InCardano'stimetherewasnotheoryofcomplexnumbersandsohereasonablyconcludedthathismethodwouldnotworkforthisequationCardanowrites(Witmer(1968),page103):

"Whenthecubeofonethirdthecoefficientofxisgreaterthanthesquareofonehalftheconstantoftheequation...thenthesolutionofthiscanbefoundbythealizaproblemwhichisdiscussedinthebookofgeometricalproblems."

Itisnotclearwhatbookheisreferringtobutthe"alizaproblem"presumablyreferstoal'Hazen,anArab,wholivedaround1000ADandwhoseworkswereknowninEuropeinCardano'stime.Al'Hazenhadusedintersectingconicstosolvespecificcubicequationsandtheproblemofdescribingtheimageseeninasphericalmirrorthislaterproblemisinsomebookscalled"Alhazen'sproblem".

Inaddition,weknowtodaythateachcomplexnumberhasthreecuberootsandsotheformulax={2+}1/3+{2}1/3isambiguous.Infact,somechoicesforthetwocuberootsgiverootsofthecubicandsomedonot.(Experimentwithx3=15x+4.)FacedwithCardano'sFormulaandequationslikex3=15x+4,Cardanoandothermathematiciansofthetimestartedexploringthepossiblemeaningsofthesecomplexnumbersandthusstartedthetheoryofcomplexnumbers.Thisleadstoanotherinterestingpathwhichwemaytakeanotherday.

6:SoWhatDoesThisAllPointTo?

Itpointstodifferentthingsforeachofus.Iconcludethatitisworthwhilepayingattentiontothemeaninginmathematics.Ofteninourhastetogettothemodern,powerful,analytictoolsweignoreandtroduponthemeaningsandimagesthatarethere.Sometimesitishardeventogetaglimpsethatsomemeaningismissing.Onewaytogetthisglimpseandfindmeaningistolistentoandfollowquestionsof"Whatdoesitmean?"thatcomeupinoneselfandinone'sstudents.Wemustlistencreativelybecauseweandourstudentsoftendonotknowhowtoexpresspreciselywhatisbotheringus.

Anotherwaytofindmeaningistoreadthemathematicsofoldandkeepasking"Whydidtheydothat?"or"Whydidn'ttheydothis?"Whydidtheearlyalgebraists(upuntilatleast1600andmuchlaterIthink)

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insistongeometricproofs?Ihavesuggestedsomereasonsabove.Today,wenormallypassovergeometricproofsinfavorofanalyticonesbasedonthe150yearoldnotionofCauchysequencesandtheAxiomofCompleteness.However,formoststudentsand,Ithink,mostmathematicians,ourintuitiveunderstandingoftherealnumbersisbasedonthegeometricrealline.Asanexample,thinkaboutmultiplication:Whatdoesaxbmean?Comparethegeometricimagesofaxbwiththemultiplicationoftwoinfinite,nonrepeating,decimalfractions.Whatis2x?

Thereisanotherreasonforwhyageometricsolutionmaybemoremeaningful:Sometimeswewantageometricresultinsteadofanumericalone.Asanexample,IshalldescribeanexperiencethatIhadwhileafriendandIwerebuildingasmallhouseusingwood.Theroofofthehouseconsistsof12isoscelestriangleswhichtogetherforma12sidedcone(orpyramid).Itwasnecessaryforustodeterminetheanglebetweentwoadjacenttrianglesintheroofsothatwecouldappropriatelycutthelografters.Iimmediatelystartedtocalculatetheangleusing(numerical)trigonometryandalgebra.ButthenIranintoaproblem.ForfindingsquarerootsandvaluesoftrigonometricfunctionsIhadonlyasliderulewiththreeplaceaccuracy.AtonepointinthecalculationIhadtosubtracttwonumbersthatdifferedonlyinthethirdplace(e.g.5.685.65)thusmyresulthadlittleaccuracy.AsIstartedtofigureoutadifferentcomputationalprocedurethatwouldavoidthesubtraction,IsuddenlyrealizedIdidn'twantanumber,Iwantedaphysicalangle.Infact,anumericalanglewouldbeessentiallyuselessimaginetakingtworoughboardsandputtingthematagivennumericalangleapartusingonlyanordinaryprotractor!WhatIneededwasthephysicalangle,fullsize.SoIconstructedtheangleonthefloorofthehouseusingaropeasacompass.NotetherelationshipbetweenthisandBaudhayana'sdescriptionsofusingcords.Thisgeometricsolutionhadthefollowingadvantagesoveranumericalsolution:

Thegeometricsolutionresultedinthedesiredphysicalangle,whilethenumericalsolutionresultedinanumber.Thegeometricsolutionwasquickerthanthenumericalsolution.Thegeometricsolutionwasimmediatelyunderstoodandtrustedbymyfriend(andfollowbuilder),whohadalmostnomathematicaltraining,whilethenumericalsolutionwasbeyondmyfriend'sunderstandingbecauseitinvolvedtrigonometry(suchasthe"LawofCosines").And,sincetheconstructionwasdonefullsize,thesolutionautomaticallyhadthedegreeofaccuracyappropriatefortheapplication.

Iclosewiththewordswrittenin1934bythe"fatherofFormalism",DavidHilbert,fromthePrefacetoGeometryandtheImagination(seeHilbert,CohnVossen(1952),pageiii):

"Inmathematics,asinanyscientificresearch,wefindtwotendenciespresent.Ontheonehand,thetendencytowardabstractionseekstocrystallizethelogicalrelationsinherentinthemazeofmaterialthatisbeingstudied,andtocorrelatethematerialinasystematicandorderlymanner.Ontheotherhand,thetendencytowardintuitiveunderstandingfostersamoreimmediategraspoftheobjectsonestudies,aliverapportwiththem,sotospeak,whichstressestheconcretemeaningoftheirrelations.

"Astogeometry,inparticular,theabstracttendencyhashereledtothemagnificentsystematictheoriesofAlgebraicGeometry,ofRiemannianGeometry,andofTopologythesetheoriesmakeextensiveuseofabstractreasoningandsymboliccalculationinthesenseofalgebra.Notwithstandingthis,itisstillastruetodayasiteverwasthatintuitiveunderstandingplaysamajorroleingeometry.Andsuchconcreteintuitionisofgreatvaluenotonlyfortheresearchworker,butalsoforanyonewhowishestostudyandappreciatetheresultsofresearchingeometry.

"Inthisbook,itisourpurposetogiveapresentationofgeometry,asitstandstoday,initsvisual,intuitiveaspects.Withtheaidofvisualimaginationwecanilluminatethemanifoldfactsandproblemsofgeometry,...

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"Inthismanner,geometrybeingasmanyfacetedasitisandbeingrelatedtothemostdiversebranchesofmathematics,wemayevenobtainasummarizingsurveyofmathematicsasawhole,andavalidideaofthevarietyofitsproblemsandthewealthofideasitcontains."

HilbertisemphasizingthepointwhichIamtryingtomakeinthispaper:Meaningisimportantinmathematicsandgeometryisanimportantsourceofthatmeaning.

References:

AmirMoez,A.R.(1963).APaperofOmarKhayyam,ScriptaMathematica,26,323337.

Davis,P.J.&Hersh,R.(1981).TheMathematicalExperience.Boston:Birkhuser.

Eves,H.(1963).ASurveyofGeometry,Vol.1.Boston:AllynandBacon.

Heath,T.L.(1956).TheThirteenBooksofEuclid'sElements.NewYork:Dover.

Heath,T.L.(1961).AppoloniosofPerga,TreatiseonConicSections.NewYork:Dover.

Hilbert,David,&CohnVossen(1952).GeometryandtheImagination.NewYork:Chelsea.

Karpinski,L.C.,editor(1915).RobertofChester'sLatinTranslationoftheAlgebraofalKhowarizmi.NewYork:Macmillan.(ThisisanEnglishtranslation.)

Kasir,D.S.,editor(1931).TheAlgebraofOmarKhayyam.NewYork:ColumbiaTeachersCollege.

Prakash(1968).BaudhayanaSulbasutram.Bombay.

Seidenberg,A.(1961).TheRitualOriginofGeometry,ArchivefortheHistoryoftheExactSciences,1,488527.

Valens,E.G.(1976).TheNumberofThings:Pythagoras,GeometryandHummingStrings.NewYork:Dutton.

Witmer,T.R.,editor(1968).TheGreatArtortheRulesofAlgebrabyGirolanoCardano.Cambridge:TheMITPress.

11ThispaperwaswrittenwhileIwasavisitingmemberofthefacultyatBirzeitUniversity,aPalestinianuniversityintheIsraelioccupiedWestBank.Iappreciatethehospitalityandsupportgivenmebythestudents,facultyandstaffduringmyvisit.

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