Lakatos

5/28/2018 Lakatos

1/20

Proofs and Refutations

What follows is the first part (minus the introduction) of Imre Lakatos influentialessay Proofs and Refutations. Its written as a dialogue between fictional students and

teacher, as they discover and prove (and disprove?) Eulers V

E+ F= 2 formula,much like we did in class.

One of Lakatos goals in writing this dialogue was to argue that mathematics isa dynamic process and that proofs and discoveries are not final, immutable, bullet-proof kernels of truth. Mathematics proceeds through a dialogue. Although I haveemphasized this in class all quarter, in 1963 it was a revolutionary perspective. Insome ways it still is: college and even high school mathematics is often taught inthe style of definition theorem proof with no room for questions or discovery.Students never get to taste real mathematics a messy process of conjecture, discovery,proofs and refutations1.

Though the characters in Lakatos dialog are made up and the account is fictional,they often play the roles of historical mathematicians. The history of Eulers formulais traced in the footnotes, which you should

The full dialogue is available as a book called Proofs and Refutations (which alsoincludes more chapters of Lakatos philosophy), and online on JSTOR:

part II: http://www.jstor.org/pss/685430part III: http://www.jstor.org/pss/685242

part IV: http://www.jstor.org/pss/685636(also easy to find if you google scholar search for it)

The articles were originally published in the British Journal for the Philosophy ofScience, 1963-64.

1this was a real problem in philosophy of mathematics, especially pre-Lakatos: most philosophers

have not done research-level math and so have some pretty inaccurate ideas about what it means to

do mathematics

1

5/28/2018 Lakatos

2/20

PROOFS AND REFUTATIONS (I)The dialogue ormshouldreflect he dialecticof the story; it ismeantto containa sort of rationallyeconstructedr' distilled'history.Therealhistory illchimen in theootnotes, ost fwhich re obetaken,

therefore,s an organic art of theessay.I A Problem nd a ConjectureThedialogueakesplacenan maginarylassroom. Theclassgetsinterestedn a PROBLEM:s therea relation etween henumber fverticesV, thenumberof edgesE andthe number f facesF of poly-hedra-particularlyof regular olyhedra-analogouso the trivialrelation etween he number f vertices ndedgesofpolygons,amely,thatthereareasmany edgesas vertices:V=E? This latterrelationenablesus to classify olygonsccordingo the numberof edges(or

vertices): triangles,quadrangles,pentagons,etc. An analogousrelationwouldhelpto classify olyhedra.Aftermuch rialanderrortheynotice hat or all regular olyhedraV-- E+ F= 2.1 Somebodyguesseshat this may applyfor any1Firstnoticed by Euler [1750]. His original problem was the classificationofpolyhedra,the difficultyof which was pointedout in the editorialsummary: ' Whilein plane geometry polygons(figuraerectilineae)ould be classifiedvery easily accord-ing to the number of their sides, which of course is always equal to the numberof their angles, in stereometry the classificationof polyhedra (corpora edrisplanisinclusa) represents a much more difficult problem, since the number of facesalone is insufficient for this purpose.' The key to Euler's result was just theinvention of the concepts of vertexand edge: it was he who first pointed out thatbesides the number of faces the number of points and lines on the surface of thepolyhedron determines ts (topological) character. It is interestingthat on the onehand he was eagerto stress he novelty of his conceptual ramework,andthathe hadto invent the term ' acies' (edge) instead of the old ' latus' (side), since latuswas apolygonal concept while he wanted a polyhedral one, on the other hand he stillretained the term ' angulussolidus' (solid angle) for his point-like vertices. It hasbeenrecently generallyaccepted hatthe priorityof theresultgoes to Descartes. Theground for this claim is a manuscriptof Descartes [ca. I639] copied by Leibniz inParis from the original in 1675-6, and rediscovered and published by Foucher deCareil in i86o. The priority should not be granted to Descartes without a minorqualification. It is true that Descartes statesthat the number of plane angles equals2, + 2a - 4 where by 0 he means the numberof facesandby a the numberof solidangles. It is alsotruethathe statesthat there are twice as many planeanglesasedges(latera). The trivial conjunctionof these two statementsof course yields the Eulerformula. But Descartes did not see the point of doing so, since he still thought intermsof angles(planeandsolid)andfaces,anddidnot make a consciousrevolutionarychange to the conceptsof o-dimensionalvertices, I-dimensionaledges and 2-dimen-sional facesasa necessaryandsufficientbasis or thefull topologicalcharacterisationfpolyhedra. 7

5/28/2018 Lakatos

3/20

I. LAKATOSpolyhedron whatsoever. Others try to falsify this conjecture,try to test it in many differentways-it holds good. The resultscorroboratehe conjecture,and suggestthat it could be proved. It isat this point-after the stagesproblem nd conjecture-thatwe enterthe classroom.1 The teacher sjust going to offeraproof

2. A ProofTEACHER: In our last lesson we arrivedat a conjectureconcerningpolyhedra,namely,that for all polyhedraV-- E-+ F= 2, where V isthe number of vertices,E the numberof edges and F the number offaces. We testedit by variousmethods. But we haven'tyet provedit. Hasanybody found a proof?PUPIL SIGMA: 'I for one have to admit that I have not yet beenable to devise a strict proof of this theorem. . . . As however thetruth of it has been establishedn so many cases, here can be no doubtthat it holds good for any solid. Thus the propositionseems to besatisfactorilydemonstrated.'2 But if you have a proof, please dopresent t.TEACHER: In fact I haveone. It consistsof the following thought-experiment. Step1: Let usimaginethe polyhedronto be hollow, witha surfacemadeof thin rubber. If we cut out one of the faces,we canstretch he remainingsurface lat on the blackboard,without tearing t.

The facesand edgeswill be deformed,the edgesmay become curved,but V,E andF will not alter,so thatif andonly if V--E + F =- 2 forthe original polyhedron,then V-- E+ F - I for this flatnetwork-rememberthat we have removed one face. (Fig. I shows the flatnetworkforthe caseof a cube.) Step2: Now we triangulateourmap-it does indeed look like a geographicalmap. We draw (possiblycurvilinear)diagonalsin those (possiblycurvilinear)polygons which1Euler ested heconjecture uitethoroughlyor consequences.He checkedtfor prisms,pyramids nd so on. He could have added hatthe propositionhatthereareonlyfiveregular odies s also a consequencef theconjecture.Another

suspectedonsequences thehitherto orroboratedropositionhat our coloursaresufficiento coloura map.Thephaseof conjecturingndtestingn thecaseof V-- E+ F- 2 is discussednP6lya([1954],Vol. I, the firstfive sections f the thirdchapter, p. 35-41). P6lyastopped ere,anddoesnotdealwiththephase fproving-though f course epointsouttheneed ora heuristicf' problemso prove' ([1945], . 144). OurdiscussionstartswhereP61lyatops.2 Euler[1750], . 119 andp. 124). Butlater[1751] eproposed proof.8

5/28/2018 Lakatos

4/20

PROOFS AND REFUTATIONS (I)are not already (possibly curvilinear)triangles. By drawing eachdiagonalwe increasebothE andF by one, so thatthe total V-- E+ Fwill not be altered(Fig. 2). Step 3: From the triangulatednetworkwe now remove the trianglesone by one. To remove a triangleweeitherremove an edge-upon which one face and one edge disappear(Fig.3a),or we remove two edgesanda vertex-upon which one face,two edgesand one vertexdisappearFig.3b). Thus if V-- E+ F=

FIG. I FIG. 2

FIG.3a FIG.3bbeforea triangle s removed,it remains o afterthe triangle s removed.At the end of this procedure we get a single triangle. For thisV- E-- F = I holds true. Thus we have proved our conjecture.PUPIL DELTA: You should now callit a theorem. Thereisnothingconjecturalaboutit any more.2PUPIL LPHA: I wonder. I see that this experimentcan be per-formedfor a cube or for a tetrahedron,but how am I to know that itcan be performedfor any polyhedron? For instance,are you sure,Sir,thatanypolyhedron,fterhavingaface removed,anbestretchedflat ontheblackboard? am dubious aboutyour firststep.

1 Thisproof-ideatems romCauchy 1811].2 Delta'sview that thisproofhasestablishedhe 'theorem' beyonddoubtwasshared y manymathematiciansn thenineteenthentury, .g. Crelle[1826-27],II,pp.668-671,Matthiessen1863], . 449,Jonquieres189oa] nd[189ob].To quoteacharacteristicpassage: AfterCauchy's roof, t becameabsolutelyndubitablehatthe elegant elationV-+ F= E+ 2 applies o all sortsof polyhedra,ust as Eulerstatedn 1752. In 1811allindecisionhouldhavedisappeared.'onquieres189oa],pp. 111-112.9

5/28/2018 Lakatos

5/20

I. LAKATOSPUPIL BETA: Are you sure that in triangulatingthe mapone willalways get a newface for any new edge? I am dubious about yoursecondstep.PUPIL GAMMA: Are you sure that thereareonly two alternatives-thedisappearancef oneedgeor elseof two edgesanda vertex-when onedrops hetriangles neby one? Are you even sure that one is left witha single rianglet the endof thisprocess? am dubiousaboutyourthirdstep.1TEACHER: Of courseI am not sure.ALPHA: But then we areworse off thanbefore Instead f one

conjecture e now haveatleast hree And thisyou calla 'proof'TEACHER: I admit that the traditional ame 'proof' for thisthought-experimentay rightlybe considered bit misleading.Ido not think hat t establisheshe truthof theconjecture.DELTA: Whatdoes t do then? Whatdoyouthinkamathematicalproofproves?TEACHER: This is a subtlequestionwhich we shall try to answerlater. Till then I proposeto retain the time-honouredtechnicalterm'proof' for a thought-experiment-orquasi-experiment-which suggestsa decompositionf theoriginal onjecturentosubconjecturesrlemmas,husembeddingt in a possibly quite distant body of knowledge. Our'proof', for instance,has embedded the original conjecture-aboutcrystals,or, say, solids-in the theory of rubber sheets. DescartesorEuler, the fathersof the original conjecture,certainlydid not evendreamof this.2

1The class s a rather dvanced ne. To Cauchy,Poinsot,and to manyotherexcellentmathematiciansf thenineteenthenturyhesequestionsidnotoccur.2Thought-experimentdeiknymi)as the mostancientpatternof mathematicalproof. It prevailedn pre-Euclideanreekmathematics(cf.A. Szab6 1958]).That conjecturesor theorems)precedeproofsin the heuristicorderwas acommonplaceor ancientmathematicians. his followed fromthe heuristicpre-cedenceof 'analysis'over 'synthesis'. (Foran excellentdiscussionee Robinson[1936].) Accordingo Proclos,. . . it is . . . necessaryo knowbeforehand hatis sought' (Heath 1925], I, p. 129). 'They saidthata theorem s that whichisproposedwith a view to the demonstrationf the very thing proposed-saysPappus ibid. , p. io). TheGreeks idnot thinkmuchof propositions hichtheyhappenedo hit upon n thedeductive irectionwithouthavingpreviously uessedthem. They called hemporisms,orollaries,ncidental esultsspringingromtheproofof a theoremor the solutionof a problem, esultsnot directly oughtbutappearing,s it were,by chance,withoutanyadditionalabour, ndconstituting,sProclus ays,a sortof windfallermaion)r bonus kerdos)ibid.I, p. 278). We readin the editorialummaryo Euler[1753]hatarithmeticalheoremswere discovered

IO

5/28/2018 Lakatos

6/20

PROOFS AND REFUTATIONS (I)3. CriticismftheProof yCounterexampleshich reLocal utnotGlobal

TEACHER: his decomposition f the conjecture uggestedbythe proof opensnew vistas or testing. The decompositioneploystheconjecturen a wider ront,so thatour criticism asmoretargets.We now haveat least hreeopportunitiesorcounterexamplesnsteadof oneGAMMA: already xpressedmy dislikeof yourthird emma(viz.that n removing rianglesrom thenetworkwhichresulted rom thestretchingand subsequentriangulation,we have only two possi-bilities: eitherwe removean edge or we removetwo edgesand avertex). I suspecthat otherpatternsmayemergewhenremovingatriangle.

TEACHER:Suspicion is not criticism.GAMMA:Then is a counterexampleriticism?TEACHER:Certainly. Conjectures ignore dislike and suspicion, butthey cannotignore counterexamples.THETA(aside):Conjecturesareobviouslyvery different rom thosewho represent hem.GAMMA:proposea trivialcounterexample. Take the triangularnetwork which resultsfrom performingthe first two operationson acube (Fig.2). Now if I remove a trianglefrom the insideof thisnet-work, as one might take a piece out of ajigsaw puzzle,I remove onetrianglewithoutremovingasingleedgeor vertex. So thethird emmalong beforetheirtruthhasbeen confirmedby rigid demonstrations. Both theEditor ndEuleruse orthisprocess f discoveryhemodern erm inductioninsteadof the ancient analysis' ibid.). The heuristicprecedencef the resultover theargument, f the theoremover the proof,hasdeeproots n mathematicalolklore.Letusquote omevariationsn a familiarheme: Chrysippuss said o havewrittento Cleanthes' Justsendmethetheorems,henI shall indtheproofs'(cf.DiogenesLaertiusca.2oo],VII. 179). Gausss said to havecomplained: I have hadmyresults or a long time; but I do not yet knowhow I am to arriveat them' (cf.Arber1954], .47),andRiemann: If onlyI had hetheorems ThenI shouldindthe proofseasily enough.' (Cf. H6lder[1924],p. 487.) P61lyatresses: 'Youhaveto guessa mathematicalheorembeforeyou prove t ' ([1954],Vol. I, p. vi).The term'quasi-experimentis fromthe above-mentionedditorialsummaryoEuler 1753]. Accordingo theEditor: 'As we mustrefer he numberso thepureintellect lone,we canhardlyunderstandow observationsndquasi-experimentsanbe of use in investigatinghe natureof the numbers. Yet, in fact,asI shallshowherewithvery goodreasons,hepropertiesf the numbers nowntodayhavebeenmostlydiscoveredyobservation . .' (P61lya'sranslation;emistakenlyttributesthequotationo Euler n his[1954], , p. 3).

II

5/28/2018 Lakatos

7/20

I. LAKATOSis false-and not only in the case of the cube, but for all polyhedraexcept the tetrahedron,n the flat network of which all the trianglesare boundarytriangles. Your proof thus proves the Euler theoremfor the tetrahedron. But we alreadyknewthat V- E+ F-= 2 forthe tetrahedron, o why prove it?TEACHER:You are right. But notice that the cube which is acounterexample o the third lemma is not also a counterexample othe main conjecture,since for the cube V-- E+ F= 2. You haveshown the poverty of the argument-the proof-but not the falsityofour conjecture.ALPHA: Will you scrap your proof then?TEACHER:No. Criticism is not necessarily destruction. I shallimprove my proof so that it will standup to the criticism.

GAMMA: HOW?TEACHER:Before showing how, let me introduce the followingterminology. I shall call a ' localcounterexample'n example whichrefutes a lemma (without necessarilyrefuting the main conjecture),and I shallcall a 'globalcounterexample'n examplewhich refutesthemain conjectureitself. Thus your counterexample s local but notglobal. A local, but not global, counterexamples a criticismof theproof, but not of the conjecture.GAMMA:o, the conjecturemay be true,but your proof does notprove it.

TEACHER:But I caneasilyelaborate, mproveheproof,by replacingthe falselemmaby a slightlymodifiedone, which your counter-examplewill not refute. I no longercontend hat theremovalf anytriangleollowsoneofthetwopatterns entioned,utmerely hatateachstageof theremoving perationheremovalof anyboundaryriangleollowsone of thesepatterns. Coming back to my thought-experiment,allthatIhave o dois to insert singleword nmythirdstep, o wit, that' fromthetriangulatedetworkwe now remove heboundaryrianglesonebyone '. Youwillagree hat t onlyneeded trifling bservationto puttheproofright.1GAMMA:donot thinkyourobservation assotrifling; n fact twasquite ngenious. To make hisclearI shallshowthat t is false.Taketheflatnetworkof thecubeagainandremoveeightof the ten

1 Lhuilier,whencorrectingn a similarway a proofof Euler, ays hathe madeonlya 'triflingobservation'[1812-13],. 179). Eulerhimself,however,gavetheproofup,sincehenoticed hetroublebut couldnotmake hat trifling bservation.12

5/28/2018 Lakatos

8/20

PROOFS AND REFUTATIONS (I)triangles n the order given in Fig. 4. At the removal of the eighthtriangle,which is certainlyby then a boundarytriangle,we removedtwo edgesand no vertex-this changesV- E + F by I. And we areleft with the two disconnectedtriangles9 and Io.FKII

7\ ' I 17 2\ aL?i8~

FIG.4TEACHER: Well, I might save face by saying that I meant by a

boundary trianglea trianglewhose removal does not disconnectthenetwork. But intellectualhonesty prevents me from making sur-reptitiouschanges n my position by sentencesstartingwith 'I meantS. . ' so I admit that now I must replacehe second version of thetriangle-removingoperationwith a third version: thatwe remove thetrianglesone by one in such a way that V-- E + F does not alter.KAPPA: I generouslyagree that the lemma corresponding o thisoperation s true: namely, that if we remove the trianglesone by onein such a way that V- E - F does not alter, then V- E + F doesnot alter.TEACHER: No. The lemma is that the trianglesn ournetwork anbeso numberedhat nremovinghemn therightorderV- E+ F will notalter ill we reach he lasttriangle.KAPPA: But how shouldone constructthis right order, if it existsat all?1 Your original thought-experiment gave the instructions:remove the triangles n any order. Your modified thought-experi-ment gave the instruction: remove boundary triangles n any order.Now you say we should follow a definiteorder,but you do not saywhich andwhetherthat orderexistsat all. Thus the thought-experi-ment breaksdown. You improved the proof-analysis, .e. the list of

lemmas; but the thought-experimentwhich you called 'the proof'hasdisappeared.RHO: Only the thirdstephasdisappeared.

1 Cauchyhoughthat heinstructionlo findateach tagea trianglewhichcanberemovedeitherby removing wo edgesanda vertexor one edgecan be triviallycarried ut foranypolyhedron[18II],p. 79). This s of course onnectedwithhisinabilityo imagine polyhedronhat s nothomeomorphic iththesphere.13

5/28/2018 Lakatos

9/20

I. LAKATOSKAPPA:Moreover,didyou improvehe lemma? Yourfirsttwosimpleversions t least ookedtriviallyruebeforetheywererefuted;yourlengthy,patchedupversiondoesnot evenlookplausible. Can

you reallybelieve hat t willescape efutation?TEACHER:Plausible'or even 'triviallytrue' propositions reusuallyoonrefuted:sophisticated,mplausibleonjectures,aturedncriticism,mighthit on thetruth.OMEGA:And what happens f even your 'sophisticated on-jectures'are falsifiedand if this time you cannotreplace hem byunfalsifiednes? Or, if you do notsucceedn improving he argu-ment urtherbylocalpatching? You havesucceededngettingoveralocalcounterexamplehich wasnot globalby replacinghe refutedlemma. Whatif you do not succeed ext time?TEACHER: ood question-it will be put on the agenda or to-morrow.

4. CriticismftheConjectureyGlobalCounterexamplesALPHA:Ihaveacounterexamplehichwillfalsify our irst emma-but this will alsobe a counterexampleo the mainconjecture,.e.thiswill be a globalcounterexampleswell.TEACHER:ndeed Interesting.Letussee.ALPHA:Imagine solidboundedby a pairof nested ubes-a pairof cubes,oneof which s inside,but doesnot touch he other(Fig.5).

FIG.5Thishollow cubefalsifies ourfirst emma,becauseon removingafacefromtheinnercube, hepolyhedronwillnot bestretchablen toaplane. Norwill ithelp o removeaface rom he outercube nstead.Besides, or eachcubeV- E+ F= 2, so thatfor the hollowcubeV- E+ F= 4.

TEACHER: Good show. Let us call it Counterexample1.1 Nowwhat?1 This Counterexamplewasfirstnoticedby Lhuilier([1812-13], . 194). ButGergonne,heEditor, dded(p.186) hathehimself oticed his ongbeforeLhuilier's

14

5/28/2018 Lakatos

10/20

PROOFS AND REFUTATIONS (I)(a) Rejectionf theconjecture.The methodf surrenderGAMMA:Sir, your composurebafflesme. A single counter-example efutes conjectureseffectively s ten. Theconjecturend

its proofhavecompletelymisfired. Handsup You haveto sur-render. Scraphefalseconjecture,orgetabout t andtrya radicallynew approach.TEACHER:agreewithyou thattheconjectureasreceived severecriticism y Alpha's ounterexample.But it is untrue hat theproofhas'completelymisfired'. If, for the timebeing, you agree o myearlierproposalo use the word 'proof' for a 'thought-experimentwhich leads to decompositionf the originalconjecturento sub-conjectures, instead f using t in the senseof a ' guaranteef certaintruth', you need not draw this conclusion. My proof certainlyprovedEuler'sconjecturen thefirstsense,but not necessarilyn thesecond. You are nterestednlyin proofswhich'prove' whattheyhave setout to prove. I aminterestedn proofsevenif theydo notaccomplishheir ntendedask. Columbus id not reach ndiabuthediscoveredomething uite nteresting.ALPHA:So accordingo your philosophy-whilea local counter-example ifit isnotglobalat thesame ime) s a criticism f theproof,butnot of theconjecture-aglobalcounterexamples acriticismf theconjecture,ut not necessarilyf theproof. You agree o surrenderas regards he conjecture,but you defend the proof. But if theconjectures false,what on earthdoes theproofprove?GAMMA:Youranalogywith Columbus reaks own. Acceptinga globalcounterexample ustmean otalsurrender.

(b)Rejectionf thecounterexample.hemethodof monster-barringDELTA:But why acceptthe counterexample?We provedourconjecture-nowit is a theorem. I admit that it clasheswith thisso-called counterexample'. One of them has to give way. Butwhy should he theoremgive way, whenit has beenproved? It isthe 'criticism' thatshouldretreat. It is fakecriticism. Thispairofpaper. Not so Cauchy,who publishedhis proof just a yearbefore. And thiscounterexampleasto be rediscoveredtwentyyears aterby Hessel([1832], . 16).Both Lhuilier ndHesselwere led to theirdiscovery y mineralogicalollectionsnwhichtheynoticed ome doublecrystals,where heinnercrystals not translucent,butthe outer s. Lhuilieracknowledgeshe stimulus f thecrystal ollection f hisfriend ProfessorPictet([i812-13],p. 188). Hessel refers o lead sulphidecubesenclosedn translucentalcium luoride rystals[1832],p. 16).

15

5/28/2018 Lakatos

11/20

I. LAKATOSnestedcubesis not a polyhedronat all. It is a monster, pathologicalcase,not a counterexample.GAMMA:Why not? A polyhedrons a solidwhosesurfaceonsistsofpolygonal aces. And my counterexample is a solid bounded bypolygonal faces.

TEACHER: Let us call this definition Def. 1.1DELTA: Your definition is incorrect. A polyhedron must be asurface:t has faces,edges,vertices, it can be deformed,stretchedouton a blackboard,and has nothing to do with the conceptof' solid'.A polyhedrons a surfaceonsistingf a systemof polygons.TEACHER:CallthisDef.2.2DELTA:So reallyyou showedustwopolyhedra-two urfaces,necompletelynside he other. A womanwitha child n herwombisnot a counterexampleo the thesis hathumanbeingshave one head.ALPHA: So My counterexamplehas bred a new concept ofpolyhedron. Or do you dare to assert that by polyhedron youalwaysmeanta surface?TEACHER: or the moment let us accept Delta's Def. 2. Can yourefuteour conjecturenow if by polyhedronwe mean a surface?ALPHA: Certainly. Take two tetrahedrawhich have an edge incommon (Fig. 6a). Or, take two tetrahedrawhich have a vertex incommon (Fig. 6b). Both these twins are connected,both constituteone singlesurface. And,you maycheckthatforboth V-- E+ F= 3

TEACHER:Counterexamples a and 2b.31 Definition occurs irst n the eighteenthentury;e.g.: 'One givesthenamepolyhedralolid, rsimplypolyhedron,o anysolidboundedby planes r plane aces'(Legendre1794],p. i6o). A similardefinitions given by Euler([175o]). Euclid,whiledefining ube,octahedron,yramid, rism,does not define hegeneralermpolyhedron,utoccasionallyses t (e.g.BookXII,SecondProblem,Prop.17).2 We findDefinition implicitlyn oneofJonquieres'apers ead o the FrenchAcademyagainsthosewho meant o refuteEuler'sheorem. Thesepapers reathesaurusf monsterbarringechniques.He thunders gainstLhuilier'smonstrouspairofnested ubes:' Suchasystemsnotreallyapolyhedron,utapairof distinctpolyhedra, achindependentf the other. . . A polyhedron, t leastfrom the

classicalointof view, deserveshenameonlyif, beforeall else,a pointcanmovecontinuouslyver ts entire urface;here his snot thecase. . . This irstexceptionof Lhuilier an hereforee discarded'[189ob],. 170). Thisdefinition--as pposedto Definition -goes downverywellwithanalyticalopologistswho arenotinter-estedatallin thetheoryof polyhedrassuchbutas a handmaidenorthetheoryofsurfaces.3Counterexamplesa and2bwere missedby Lhuilier nd irstdiscoveredonlybyHessel([1832], p. 13). 16

5/28/2018 Lakatos

12/20

PROOFS AND REFUTATIONS (I)DELTA: admireyour perverted magination,but of courseI didnot mean thatanysystemof polygonsisapolyhedron. By polyhedronI meant a systemofpolygonsarrangedn sucha way that(1) exactly wo

polygonsmeetateveryedgeand(2) it ispossibleoget romtheinsideof anypolygon o the insideof anyotherpolygonbya routewhichnever rossesanyedgeata vertex. Your first twins will be excludedby the first criterionin my definition,your secondtwins by the secondcriterion.

FIG. 6a FIG.6bTEACHER: Def. 3.1ALPHA: I admire your perverted ingenuity in inventing onedefinitionafter another as barricadesagainstthe falsificationof yourpet ideas. Why don't you just define a polyhedron as a system ofpolygons for which the equation V- E + F= 2 holds, and thisPerfect Definition . .

KAPPA: Def P.2ALPHA: . . . would settle the dispute for ever? There would beno need to investigate he subjectany further.DELTA:But there isn't a theoremin the world which couldch't efalsifiedby monsters.1 Definition3 first turnsup to keep out twintetrahedran M6bius ([1865], p. 32).We find his cumbersome definition reproducedin some modern textbooks in theusualauthoritarian take it or leave it' way; the story of its monsterbarringback-

ground-that would at leastexplainit-is not told (e.g. Hilbert-CohnVossen [1956],p. 290).2DefinitionP accordingto which Euleriannesswould be a definitionalcharacter-istic of polyhedra was in fact suggested by R. Baltzer: ' Ordinarypolyhedra areoccasionally (following Hessel) called Eulerian polyhedra. It would be moreappropriate o find a specialname for non-genuine (uneigentliche)olyhedra' ([I86o],Vol. II, p. 207). The reference o Hessel is unfair: Hessel usedthe term 'Eulerian'simply as an abbreviation or polyhedrafor which Euler'srelationholds in contra-distinctionto the non-Eulerianones ([1832],p. 19). For Def. P see also the Schlifliquotationin footnote pp. 18-I9.

B 17

5/28/2018 Lakatos

13/20

I. LAKATOSTEACHER:am sorryto interruptyou. As we have seen, refuta-tion by counterexamplesdepends on the meaning of the terms inquestion. If a counterexamples to be an objectivecriticism,we have

to agree on the meaning of our terms. We may achieve such anagreementby definingthe term where communicationbroke down.I, for one, didn'tdefine'polyhedron '. I assumedfamiliaritywith theconcept, i.e. the ability to distinguisha thing which is a polyhedronfrom a thing which is not a polyhedron-what some logicians callknowing the extension of the conceptof polyhedron. It turnedoutthat the extension of the conceptwasn'tat all obvious: definitionsrefrequentlyproposedand arguedaboutwhen counterexamplesmerge. Isuggestthatwe now considerthe rival definitionstogether,and leaveuntil later the discussion of the differences in the results whichwill follow from choosing differentdefinitions. Can anybody offersomethingwhich even the most restrictivedefinitionwould allow asarealcounterexample?KAPPA:IncludingDef. P?TEACHER:ExcludingDef. P.GAMMA: can. Look at this Counterexample: a star-polyhedron-I shall call it urchin(Fig. 7). This consists of 12 star-pentagons(Fig. 8). It has 12 vertices, 30 edges, and 12 pentagonalfaces-you

~tf~=5?-----z-~ ;-=trtrrrrcr

M

E

A DFIGS. and 8. Kepler(Fig. 7) shadedeach face in a differentway to showwhich trianglesbelong to the samepentagonal ace.

may checkit if you like by counting. Thusthe Descartes-Eulerhesisis not trueat all, sincefor thispolyhedronV - E + F - - 6.11 The

'urchin.'was first discussedby Kepler in his cosmological theory ([1619],Lib. II, XIX and XXVI, on p. 52 and p. 60 and Lib. V, Cap. I, p. 182, Cap. III,p. 187 and Cap.IX, XLVII). The name' urchin' is Kepler's ' cuinomenEchinofeci).Fig. 7 is copied from his book (p. 52) which containsalso anotherpictureon p. 182.

18

5/28/2018 Lakatos

14/20

PROOFS AND REFUTATIONS (I)DELTA:Why do you think thatyour 'urchin' is a polyhedron?GAMMA:Do you not see? This is a polyhedron, whose faces arethe twelve star-pentagons. It satisfiesyour last definition: it is 'a

system of polygons arrangedin such a way that (i) exactly twopolygons meet at every edge, and (2) it is possibleto get from everypolygon to every otherpolygon without ever crossinga vertex of thepolyhedron'.DELTA: But then you do not even know what a polygon is Astar-pentagon s certainlynot a polygon A polygonis a systemofedgesarrangedn sucha waythat(1) exactly woedgesmeetat everyvertex,and(2) theedgeshavenopoints n commonexcept he vertices.TEACHER: Let us call thisDef 4.GAMMA: I don't seewhy you includethe secondclause. The

rightdefinition f thepolygonshouldcontain hefirstclause nly.TEACHER: Def. 4'.GAMMA: The second clause has nothing to do with the essence ofa polygon. Look: if I lift an edge a little, the star-pentagons alreadya polygon evenin your sense. You imaginea polygon to be drawninchalk on the blackboard,but you should imagine it as a woodenstructure:then it is clearthat what you thinkto be a point in commonis not really one point, but two differentpoints lying one above theother. You aremisledby your embeddingthe polygon in a plane-you shouldlet its limbs stretchout in space xPoinsot ndependentlyediscoveredt, and t washe who pointedout thattheEulerformuladidnot apply o it ([1809], . 48). Thenow standarderm smallstellatedpolyhedronis Cayley's[1859], . 125). Schlliflidmittedtar-polyhedran general,butneverthelessejected ursmall tellated odecahedronsa monster. Accordingto him 'this is not a genuinepolyhedron,or it does not satisfy he conditionV- E+ F= ' ([1852],34).1The disputewhetherpolygonshouldbe defined o as to includestar-polygonsor not(Def.4orDef.4') s averyoldone. Theargumentut orwardn ourdialogue-that star-polygonsanbe embedded s ordinary olygons n a spaceof higherdimensions--ismodem opological rgument,utone canputforwardmanyothers.ThusPoinsotdefending isstar-polyhedrarguedorthe admissionf star-polygonswithargumentsaken romanalyticaleometry:'... allthesedistinctionsbetween"ordinary and"star"-polygons)remoreapparenthan eal,and heycompletelydisappearn the analyticalreatment,n which the various peciesof polygonsarequite nseparable.To theedgeof a regular olygontherecorrespondsnequationwithrealroots,whichsimultaneouslyields heedgesof all theregular olygonsofthe sameorder. Thus t is not possible o obtain he edgesof a regularnscribedheptagon,withoutat thesame ime inding dgesofheptagonsf thesecond nd hirdspecies. Conversely, iventheedgeof a regular eptagon, nemaydeterminehe

19

5/28/2018 Lakatos

15/20

I. LAKATOSDELTA: Would you mind telling me what is the area of astar-pentagon? Or would you say that some polygons have noarea?GAMMA: Was it not you yourselfwho said that a polyhedronhasnothing to do with the idea of solidity? Why now suggest that theidea of polygon should be linked with the idea of area? We agreedthat a polyhedron is a closed surface with edges and vertices-thenwhy not agreethat a polygon is simply a closedcurvewith vertices?But if you stick to your ideaI am willing to definethe area of a star-polygon.'TEACHER: Let us leave this disputefor a moment, and proceedasbefore. Considerthe last two definitionstogether-Def. 4 and Def.4'. Can anyone give a counterexample o our conjecturethat will

comply with bothdefinitionsof polygons?ALPHA: Here is one. Considera picture-frameike this (Fig. 9).This is a polyhedron accordingto any of the definitionshithertopro-posed. Nonethelessyou will find, on countingthe vertices,edgesandfaces,that V-- E+- F 0.radiusof a circle in which it can be inscribed,but in so doing, one will find threedifferent circlescorresponding o the three speciesof heptagon which may be con-structedon the given edge; similarlyfor other polygons. Thus we arejustified ingiving the name " polygon " to thesenew starredfigures'([1809], p. 26). Schrbderuses the Hankelianargument: 'The extension to rational fractions of the powerconcept originallyassociatedonly with the integershasbeenvery fruitful n Algebra;this suggeststhat we try to do the samething in geometry whenever the opportunitypresents tself ...' ([1862],p. 56). Then he shows that we may find a geometricalinterpretation or the concept of p/q-sided polygons in the star-polygons.1 Gamma'sclaimthat he can definethe area or star-polygonss not a bluff. Someof those who defendedthe wider conceptof polygon solved the problem by puttingforwarda wider conceptof the area of polygon. Thereis an especiallyobviouswayto do this in the case of regularstar-polygons. We may take the area of a polygonas the sum of the areasof the isoscelestriangleswhichjoin the centreof the inscribedor circumscribed ircle to the sides. In this case, of course,some ' portions' of thestar-polygonwill count more than once. In the caseof irregularpolygons where wehave not got any one distinguishedpoint, we may still take anypoint as origin andtreatnegativelyorientedtrianglesashaving negativeareas(Meister 1769-70],p. 179).It turnsout-and this can certainlybe expected from an ' area -that the areathusdefined will not depend on the choice of the origin (Mibius [1827], p. 218). Ofcoursethereis liable to be a disputewith those who think that one is not justified incallingthe numberyielded by this calculationan' area ; though the defendersof theMeister-M6biusdefinitioncalled t ' therightdefinition' which ' aloneis scientificallyjustified' (R. Haussner'snotes [1906], pp. 114-115). Essentialismhas been a per-manent feature of definitionalquarrels.

20

5/28/2018 Lakatos

16/20

PROOFS AND REFUTATIONS (I)TEACHER:Counterexample .1BETA: o that'she endof ourconjecture.Itreallys apity,sinceit heldgoodforso manycases. Butit seems hatwe have ustwasted

our time.ALPHA:elta,I amflabbergasted.Yousaynothing? Can'tyoudefine hisnewcounterexampleut of existence? I thought herewasno hypothesisn theworldwhichyoucouldnot save romfalsificationwith a suitableinguistic rick. Areyou givingup now? Do youagreeat last that thereexist non-Eulerianpolyhedra? Incredible

FIG.DELTA:You shouldreally inda moreappropriateame oryournon-Eulerianestsandnotmislead sallby callinghem polyhedra.ButI amgraduallyosing nterestn yourmonsters. I turn n disgustfrom your lamentable'polyhedra', for which Euler'sbeautiful

theorem oesn'thold.2 Ilook fororderandharmonynmathematics,but you only propagate narchy ndchaos.3 Our attitudes reir-reconcilable.1WefredCounterexampletooinLhuilier'slassical[1812-13], onp.I85--Gergonneagainadded hathe knewit. But Grunert id not know it fourteenyears ater([1827]) ndPoinsot orty-five earsater [1858], . 67).2 This s paraphrasedrom a letterof Hermite'swritten o Stieltjes: I turnasidewith a shudder f horror rom thislamentableplagueof functionswhichhavenoderivatives'[18931).3' Researchesealingwith . . . functions iolatingawswhichonehopedwereuniversal, ereregardedlmostasthepropagationf anarchyndchaoswherepast

generations ad soughtorder and harmony'(Saks[1933],Preface). Saksrefershereto thefiercebattlesof monsterbarrerslikeHermite) andof refutationistshatcharacterisedn the last decades f the nineteenth entury and indeed in thebeginningof the twentieth)he developmentf modernreal unctiontheory,' hebranch f mathematicshichdealswithcounterexamples(Munroe1953],Preface).Thesimilarlyiercebattle hatragedaterbetween heopponents ndprotagonistsfmodemmathematicalogic andset-theorywas a directcontinuationf this. Seealso ootnote2 on p. 24 andI on p. 25.21

5/28/2018 Lakatos

17/20

I. LAKATOSALPHA: You are a real old-fashionedTory You blame thewickedness f anarchistsor the spoilingof your 'order' and'har-mony', andyou 'solve' thedifficultiesy verbal ecommendations.TEACHER:etushear he latest escue-definition.ALPHA:You mean helatest inguisticrick, helatestcontractionof the conceptof 'polyhedron' Delta dissolvesreal problems,instead f solving hem.

FIG. O

FIG. IIa FIG. IIbDELTA: I do not contractoncepts. It is you who expand hem.For instance,this picture-frames not a genuinepolyhedronat all.ALPHA:Why?DELTA:akeanarbitraryoint n the' tunnel-the space oundedby the frame. Laya planethrough hispoint. You will find thatany such plane has alwaystwo differentcross-sectionswith the

22

5/28/2018 Lakatos

18/20

PROOFS AND REFUTATIONS (I)picture-frame, making two distinct, completely disconnectedpolygons (Fig. io).

ALPHA: So what?DELTA: In the case of a genuinepolyhedron,hroughany arbitrarypoint in space herewill be at least oneplanewhosecross-sectionith thepolyhedronill consist f onesinglepolygon.In the caseof convexpolyhedra llplaneswill complywith thisrequirement,hereverwetake hepoint. Inthecaseof ordinaryoncavepolyhedraomeplaneswill havemore ntersections,uttherewill alwaysbe some thathaveonly one. (Figs. Iia and iib.) In the case of thispicture-frame, fwe take the point in the tunnel, all the planeswill have two cross-sections. How then can you call this a polyhedron?TEACHER: This looks like anotherdefinition,this time an implicit

one. Call it Def. 5.1ALPHA: A seriesof counterexamples,a matchingseriesof defini-tions, definitionsthat are alleged to contain nothing new, but to bemerelynew revelationsof the richnessof that one old concept, whichseemsto have asmany ' hidden' clausesas there are counterexamples.ForallpolyhedraV- E+ F= 2 seemsunshakable, n old and' eternaltruth. It is strange o think that once upon a time it was a wonderfulguess,full of challengeandexcitement. Now, becauseof your weirdshifts of meaning, it has turned into a poor convention, a despicablepiece of dogma. (He leaves heclassroom.)DELTA: I cannotunderstand ow an ablemanlike Alphacanwastehis talenton mereheckling. He seemsengrossedn the productionofmonstrosities. But monstrositiesnever foster growth, either in theworld of natureor in theworld of thought. Evolutionalwaysfollowsan harmoniousand orderlypattern.GAMMA: Geneticistscan easily refute that. Have you not heardthat mutations producing monstrositiesplay a considerablerole inmacroevolution? They call such monstrous mutants 'hopeful

1 Definitionwasputforward y theindefatigable onsterbarrer.deJonquieresto getLhuilier'solyhedron ithatunnelpicture-frame)utof theway: 'Neitheris thispolyhedralomplexa truepolyhedronntheordinaryenseof theword, orifone takesanyplane hroughanarbitrary oint nsideone of thetunnelswhichpassrightthrough hesolid,theresultingross-sectionill be composed f two distinctpolygonscompletelyunconnected ith eachother; thiscanoccur n an ordinarypolyhedronorcertainositions f theintersectinglane,namelyn thecaseof someconcavepolyhedra, utnot for all of them' ([i89ob],pp. 170-171). OnewonderswhetherdeJonqui&resasnoticed hathisDef.5 excludes lsosomeconcavespheroidpolyhedra. 23

5/28/2018 Lakatos

19/20

I. LAKATOSmonsters. It seemsto me that Alpha'scounterexamples,houghmonsters,re'hopefulmonsters.'DELTA: Anyway, Alphahas given up the struggle. No moremonsters ow.GAMMA: Ihaveanew one. ItcomplieswithalltherestrictionsnDefs. I, 2, 3, 4, and5, but V- E+ F= I. ThisCounterexampleis a simple cylinder. It has 3 faces(the top, the bottom and thejacket),2 edges(two circles)and no vertices. It is a polyhedronaccordingo yourdefinition: i) exactly wo polygonsat everyedgeand (2) it is possible o get fromthe insideof any polygonto theinsideofanyotherpolygonbyaroutewhichnevercrossesanyedgeata vertex. Andyouhave o accepthe faces sgenuine olygons, stheycomplywithyourrequirements:I) exactly wo edgesmeetateveryvertexand(2)theedgeshaveno points n commonexcept hevertices.DELTA: Alpha stretchedconcepts,but you tear them Your'edges' arenotedges Anedgehas woverticesTEACHER: Def.6?GAMMA: Butwhy denythestatus f' edge' to edgeswithone orpossibly erovertices? You used o contractconcepts, utnow youmutilate hemso thatscarcely nything emainsDELTA: Butdon'tyouseethefutilityof these o-calledefutations?'Hitherto,when a new polyhedronwas invented, t was for somepracticalnd; today they areinventedexpresslyo put at faultthereasoningsf ourfathers, ndoneneverwill get fromthemanythingmorethan that. Oursubjects turnednto a teratologicalmuseumwheredecentordinary olyhedramaybe happy f they can retainaverysmallcorner.'

1 'We mustnot forget hatwhatappearso-dayas a monsterwill be to-morrowtheoriginof a lineof special daptations. . . I further mphasizedheimportanceof rarebutextremely onsequential utationsaffectingatesof decisive mbryonicprocesses hichmightgiverise o whatonemight ermhopefulmonsters,monsterswhichwouldstartanewevolutionaryine f fitting ntosomeemptyenvironmentalniche'(GoldschmidtI9331, p.544and547). My attention asdrawnothispaperby KarlPopper.2 ParaphrasedromPoincard[90o8],pp. 131-132). Theoriginalull textis this:'Logicsometimesmakesmonsters. Sincehalfa centurywe haveseenarisea crowdof bizarre unctionswhichseem to try to resemble slittle aspossiblehehonestfunctionswhichservesomepurpose. No longercontinuity, r perhapsontinuity,butno derivatives,tc. Naymore, rom he ogicalpointof view,it is these trangefunctionswhicharethe mostgeneral, hoseone meetswithoutseekingno longerappearxceptasparticularases. Thereremainsorthemonlyaverysmallcorner.

24

5/28/2018 Lakatos

20/20

PROOFS AND REFUTATIONS (I)GAMMA:I think that if we want to learn about anything reallydeep, we have to studyit not in its 'normal ', regular,usualform, butin its critical tate, n fever,in passion. Ifyou wantto know the normal

healthy body, study it when it is abnormal,when it is ill. If youwant to know functions, study their singularities. If you want toknow ordinary polyhedra, study their lunatic fringe. This is howone cancarrymathematicalanalysis nto the very heart of the subject.'But even if you were basicallyright, don't you see the futilityof yourad hocmethod? If you want to draw a borderlinebetween counter-examplesand monsters,you cannotdo it in fits andstarts.TEACHER: think we should refuseto acceptDelta's strategyfordealingwith globalcounterexamples, lthoughwe shouldcongratulatehim on his skilful execution of it. We could aptly label his methodthe methodof monsterbarring. sing thismethod one can eliminateanycounterexampleto the original conjectureby a sometimes deft butalwaysad hocredefinitionof the polyhedron,of its defining terms, orof the defining terms of its defining terms. We should some-how treat counterexampleswith more respect, and not stubbornlyexorcise them by dubbing them monsters. Delta's main mistakeisperhapshis dogmatistbias in the interpretation f mathematicalproof:he thinks that a proof necessarilyproveswhat it has set out to prove.My interpretationof proof will allow for a false conjectureto be'proved', i.e. to be decomposed into subconjectures. If the con-jectureis false,I certainlyexpectat least one of the subconjectureso befalse. But the decompositionmight still be interesting I am notperturbedat finding a counterexampleto a 'proved' conjecture;I am even willing to set out to 'prove' a falseconjectureTHETA: I don't follow you.KAPPA: He just follows the New Testament: 'Prove all things;hold fast that which is good' (I Thessalonians 5: 21).

(to be continued)'Heretofore when a new function was invented, it was for some practicalend;to-day they are invented expresslyto put at fault the reasoningsof our fathers,and

one never will get from them anythingmore than that.'If logic were the sole guide of the teacher,it would be necessary o begin withthe most generalfunctions, that is to say with the most bizarre. It is the beginnerthat would have to be set grappling with this teratological museum . . .' (G. B.Halsted'sauthorised ranslation,pp. 435-436). Poincar6discusses he problem withrespectto the situationin the theory of real functions-but that does not make anydifference.1 Paraphrased from Denjoy ([1919], p. 21).

25