19346327 fallacies in mathematics

8/8/2019 19346327 Fallacies in Mathematics

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FALLACIES I,NMATHEMATICS

BY

E. A. MAXWELL, PH.D.Fellow of Queens' College. Cambridge

CAMBRIDGEAT THE UNIVERSITY PRESS

1963


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P U B L I S H E D BYT H E S Y N D I C S O F T H E C A M B R I D G E U N I V E R S I T Y P R E S S

Bentley House, 200 Euston Road, London, !'\.W. 1American Branch: 32 East ,17th Street, !'\ew York 22, :\.Y.

West African Offke: P.O. Box :1:1, Ihadan, l\'igeriuC A M B R I D G E U N I V E R S I T Y P R E S S

1959

First printedReprinted

19591961

F i r < ~ t paperback edition 1963

First printed in Great Britain at the Univprsity Pres , CambridgeReprinted by offset-litho by John Dicken.\' &- Co .. Ltd., Northampton


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CONTENTSPreface page 7

I The Mistake, the Howler and the Fallacy 9I I Four Geometrical Fallacies Enunciated 13

III Digression on Elementary Geometry 20IV The' Isosceles Triangle' Fallacy Analysed 24V The other Geometrical Fallacies Analysed 33

VI Some Fallacies in Algebra andTrigonometry 37VII Fallacies in Differentiation 55

Vln Fallacies in Integration 65IX Fallacy by the Circular Points at Infinity 75X Some' Limit' Fallacies 81

XI Some Miscellaneous Howlers 88


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PREFACEThe aim of this book is to instruct through entertain-ment. The general theory is that a wrong idea may oftenbe exposed more convincingly by following it to itsabsurd conclusion than by merely denouncing the errorand starting again. Thus a number of by-ways appearwhich, it is hoped, may amuse the professional and helpto tempt back to the subject those who thought theywere losing interest.

The standard of knowledge expected is quite elementary; anyone who has studied a little deductivegeometry, algebra, trigonometry and calculus for a fewyears should be able to follow most of the expositionwith no trouble.Several of the fallacies are well known, though I haveusually included these only when I felt that there wassomething fresh to add. There is not (I hope) much of therather outworn type x = 0,

x(x- l ) = 0,x-I = 0,

x = 1,1 = 0.

I have also tried to avoid a bright style; the reader shouldenjoy these things in his own way.My original idea was to give references to the sourcesof the fallacies, but I felt, on reflection, that this was togive them more weight than they could carry. I should,however, thank the editor of the Mathematical Gazettefor his ready permission to use many examples whichfirst appeared there.


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8 PREFACEI t was with pleasure that I received the approval of the

Council of the Mathematical Association to arrange forthe Association to receive one half of the royalties fromthe sale of this book. I welcome the opportunity to recordmy gratitude for much that I have learned and for manyfriendships that I have made through the Association.

I must express my thanks to members (past andpresent) of the staff of the Cambridge University Press,who combined their skill and care with an encouragementwhich, in technical jargon, became real and positive. Iam also indebted for valuable advice from those who readthe manuscript on behalf of the Press, and to my son,who helped me to keep the proofs from becomingunnecessarily fallacious.

CAMBRIDGE28 April 1958

E. A. MAXWELL


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CHAPTER I

THE MISTAKE, THE HOWLERAND THE FALLACY

9

All mathematicians are wrong at times. In most casesthe error is simply a MISTAKE, of little significance and,one hopes, of even less consequence. Its cause may bea momentary a b ~ r r a t i o n , a slip in writing, or the misreading of earlier work. For present purposes it has nointerest and will be ignored.

The HOWLER in mathematics is not easy to describe,but the term may be used to denote an error which leadsinnocently to a correct result. By contrast, the FALLACYleads by guile to a wrong but plausible conclusion.One or two simple examples will illustrate the use ofthese terms. The first is a howler of a kind that manypeople, here and abroad, might think the Britishsystem of units only too well deserves:

To make out a bill :i lb. butter @ 28. lOd. per lb.2t lb. lard @ lod. per lb.3 lb. sugar @ 3id. per lb.6 boxes matches @ 7d. per dozen.4 packets soap-flakes @ 2td. per packet.

(Who but an examiner would include four packets ofsoap-flakes in such an orded)The solution is

8td. + 28. Id. +9d. +3td. + lOd. = 48. 8id.One boy, however, avoided the detailed calculations

and simply added all the prices on the right:28. IOd. + lOd. +3id. +7d. + 21d. = 48. 8id.


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10 FALLACIES IN M A T H E M A T I C SThe innocence of the pupil and the startling accuracy

of the answer raise the calculation to the status ofhowler.The emphasis of this book is on the fallacy, though

a selection of howlers (leading, sometimes, to curiousgeneralisations) will be given in the last chapter. Twosynthetic howlers may, however, be added here beforewe proceed to the real business:

(i) To prove the formulaa2 -b 2 = (a-b)(a+b).Consider the quotient

'Cancel' an a and a b:a * - b ~ ifI,-H'

and then 'cancel' the minus ' into' the minus:+

a ~ T b ~ { , f - fg '

The result is a+b.( . ) T . l f 26 d 1611 0 8 ~ m p y 65 an 64'The correct answers are obtained by cancelling:

and

26 265 516 164 = 4'

Consider next a typical fallacy:To prove that every triangle is isosceles.Let ABC (Fig. 1) be a given triangle. I t is required to

prove that AB is necessarily equal to AG.


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M I S T A K E , H O W L E R AN D F A L L A C Y 11I f he internal bisector of the angle A meets BC in D,

then, by the angle-bisector theorem,DB DCAB = AC'

Now LADB = LACD+ LCAD=C+tA,

A

B Fig. 1so that, by the sine rule applied to the triangle ADB,

DB sinBAD

Further,

so thatHence

AB sinADBsintA

= - - : - - ~ - " - - - - : ; - - - - : 7 sin(C+lA)'LADC = LABD + LBAD

= B+!A,DC sintAAC = sin (B+tA)"

sintA sin tAsin(C+tA) = sin (B+tA)"Moreover sin tA is not zero, since the angle A is not zero,and so sin(C+tA) = sin(B+!A),or C+!A = B+1A,or C = B.The triangle is therefore isosceles.


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12 FALLACIES IN M A T H E M A T I C SThe analysis of this fallacy may be used to illustrate

standard features which we shall meet often.To begin with, of course, the actual error must be

detected. (The false step here follows the linesin(C+tA) = sin(B+!A).

Equality of sine need not mean equality of angle.) Butthis, in a good fallacy, is only a part of the interest, andthe lesser part at that.

The second stage is to effect, as it were, a reconciliationstatement in which (i) the correct deduction is substitutedfor the false and, when possible, (ii) the discrepancybetween the wrong and the correct theorems is accountedfor in full.

Thus the stepsin(C+!A) = sin(B+!A)

leads not only to the stated conclusionC+tA = B + ~ A ,

but also to the alternative

orC+!A = IS00-(B+!A),

A +B+C = 180.The necessity for the angles B, C to be equal is negativedby the fact that the sum of the angles A, B, C is always1800 The error is therefore found and the correct versionsubstituted.

In the work that follows we shall usually begin witha straightforward statement of the fallacious argument,following it with an exposure in which the error is tracedto its most elementary source. I t will be found that thisprocess may lead to an analysis of unexpected depth,particularly in the first geometrical fallacy given in thenext chapter.


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CHAPTER IIFOUR GEOMETRICAL FALLACIES

ENUNCIATED

13

There are four well-known fallacies whose statement requires only the familiar theorems of elementary (schoolboy) geometry. They are propounded without commentin this chapter, the analysis, which becomes very fundamental, being postponed. The reader will doubtless wishto detect the errors for himself before proceeding to theirexamination in the next chapter.

1. THE FALLACY OF THE ISOSCELES TRIANGLE.To prove that every triangleis isosceles.

GIVEN: A triangle ABO(Fig. 2).REQUIRED: To prove that,necessaril y ,

AB = AO.CONSTRUCTION: Let theinternal bisector of the angle

A

B cFig. 2

A meet the perpendicular bisector of BO at O. DrawOD, OQ, OR perpendicular to BO, OA, AB respectively.PROOF: Since DO = DODB=DO

LODB = LODO60DB == 60DO:. on = 00.

(SAS)* Notation such as RAS is uscd as an abbreviation for 'having two

sides and the included angle equal'. Tho symbol :0 denotes congruence.


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14Also

FALLACIES IN MATHEMATICSAO= AO,

LRAO = LQAOLARO = LAQOf,ARO == f,AQO (ASA)

andAR=AQOR = OQ.

Hence, in triangles OBR, OOQ,

Finally,

LORB = LOQO = right angle,OB = 00 (proved)OR = OQ (proved)

!;'ORB == !;'OQO (rt. L, H, S)RB = QO.AB = AR+RB

= AQ+QO=AO.

(proved)Q.E.D.

2. THE FALLACY OF THE RIGHT ANGLE.To prove that every angle is a right angle.

GIVEN: A square ABOD and a line BE drawn outwardsfrom the square so that LABE has a given value, as-sumed obtuse (Fig. 3). (If the value is acute, take LABEto be its supplement.)n EQUIRED: To prove that

LABE = a right angle.CONS'J:RUCTION: Let P, Qbe the middle points ofOD,AB. Take BE equal in length to a side of the square, andlet the perpendicular bisector ofDE meet PQ, producedif necessary, at O.


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FOUR G E O M E T R I C A L FALLACIES 15PROOF: By symmetry, PQ is the perpendicular bisectorof CD and of AB.

Consider the triangles ORD, ORE:OR=OR

LORD = LORE (construction)RD = RE (construction)

f:.ORD == f:.ORE. (SAS)In particular, OD = OE.

E

Fig. 3

Consider the triangles OQA, OQB:OQ=OQ

LOQA = LOQB (perpendicular bisector)QA = QB (construction)!:'OQA == f:.OQB. (SAS)

In particular, OA = OBand (for later reference)

LOAB= LOBA.


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16 FALLACIES IN MATHEMATICSConsider the triangles OAD, uBE:

OA=OB (proved)AD = BE (BE = side of square)OD = OE (proved)

l:!.OAD == l:!.OBE. (SSS)In particular,

LOAD = LOBE.Now the point 0 on PQ may be either (i) between P, Q;

(ii) at Q; (iii) beyond Q (as in Fig. 3).In case (i)

In case (ii)

In case (iii)

LABE = LOBE+ LOBA= LOAD+ LOAB= right angle.

LABE == LOBELOADLBAD

= right angle.LABE = LOBE - LOBA= LOAD- LOAB

= right angle.Hence, in all three cases,

LABE = one right angle.(3 . THE TRAPEZIUM FALLACY.

(proved)

(proved)

(proved)

To prove that, i f ABCD is a quadrilateral in whichAB = CD, then AD is necessarily parallel to BC.GIVEN: A quadrilateral ABCD (Fig. 4) in whichAB = CD.REQUIRED: To prove that AD is parallel to BC.


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FOUR GEOMETRICAL FALLACIES 17CONSTRUCTION: Draw the perpendicular bisectors ofAD, BG. I f hey are parallel, the theorem is proved. I fnot, let them meet in 0, which may be outside (Fig. 4 (i))or inside (Fig. 4 (ii)) the quadrilateral.

tNN B N ('(i) (ii)Fig. 4PROOF: The perpendicular bisectorofa line is the locus ofpoints equally distant from its ends. Hence

OA =ODOB = OG.

Compare the triangles OAB, ODG.OA =ODOB=OGAB=DG

60AB== 60DGLOAB= LODG.

Also, by isosceles triangles,LOAD = LODA.

(proved)(proved)

(given)(SSS)

(OA = OD)Hence, by subtraction (Fig. 4.(i)) or addition (Fig. 4 (ii)),

LBAD = LGDA.Similarlyand

LOBA = LOGD (60AB == 60DG)LOBG = LOGB. (OB = OC)

Hence, by addition,LABG = LDOB.

2 MF


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18 FALLACIES IN MATHEMATICSThus

LBAD+ LABO = LODA+ LDOB.But the sum of these four angles is four right angles, sincethey are the angles of the quadrilateral ABOD. Hence

LBAD + LABO = 2 right angles.Also these are interior angles for the lines AD, BO withtransversal AB.

Thus AD is parallel to BO.(4. THE FALLACY OF THE EMPTY CIRCLE.

To prove that every point inside a circle lies on its cir-cumference.GIVEN: A circle ofcentre 0 and radius r, and an arbitrarypoint P inside it (Fig. 5).

o p Q

Fig. 5REQUIRED: To prove thatP lies on the circumference.CONSTRUCTION: Let Q be the point on OP producedbeyond P such that OP.OQ = r2,and let the perpendicular bisector ofPQ cut the circle atU, V. Denote by R the middle point of PQ.


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FOUR GEOMETRICAL FALLACIES 19PROOF:

OP = OR-RPOQ = OR+RQ= OR +RP (RQ = RP, construction)

OP.OQ = (OR-RP) (OR+RP)= OR2_Rp2= (OU2-RU2) - (PU2-RU2) (Pythagoras)=OU2_PU2=OP.OQ-PU2 (OP.OQ = r2 = OU2)

PU=o:. P is at U, on the circumference.

2-2


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20

CHAPTER I I I

DIGRESSION ON ELEMENTARYGEOMETRY

Elementary geometry is commonly taught in the schoolsfor two main purposes: to instil a knowledge of thegeometrical figures (triangle, rectangle, circle) met incommon experience, and also to develop their propertiesby logical argument proceeding step by step from themost primitive conceptions. The supreme exponent ofthe subject is Euclid, whose authority remained almostunchallenged until very recent times.

I t is probable that Euclid's own system of geometry isnot now used in many schools, but children studyinggeometry become familiar with a number of thestandard theorems and with the proofs of several ofthem. The general idea of a geometrical proof, if not ofthe details, will thus be familiar to anyone likely to readthis book. Wegiveinillustrationa typical example, whichwe shall, in fact, find it necessary to criticise later. Theproof will first be stated in standard form, and thenature of the geometrical arguments leading towards itwill then be discussed.

To prove that the exterior angle of a triangle is greaterthan the interior opposite angles.GIVEN: A triangle ABC whose side BC is producedbeyond C to P (Fig. 6).REQUIRED: To prove that the anglePCA is greater thanthe angle BAC.


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E L E M E N T A R Y GEOMETRY 21CONSTR UCTIOX: Let 0 be the middle point of AC, andproduce BO beyond 0 to D so that OD = BO.PROOF: Since OA =OC (construction)

that is,

OB = OD (construction)LAOB = LCOD (vertically opposite)t:.AOB == t:.COD (SAS)LBAO = LDCO,LBAC = LDCA.

A D

A:;(B (' PFig. 6But LDCA is part of LPCA, so thatLDCA is less than LPCA.

Hence also LBAC is less than LPCA,or LPCA is greater than LBAC.

The essence of this proof (rarely used nowadays) is thatit is enunciated as a necessary consequence of an earliertheorem, namely that two triangles are congruent whichhave two sides and the included angle equal. Thistheorem, in its turn, has been proved explicitly fromyet earlier work (as in Euclid's treatment) or perhapsenunciated explicitly as one of the foundation stones onwhich the fabric of geometrical argument is to be built.

In this sort of way a structure is obtained involvingpoint, line, angle, congruence, parallelism and so on,each new item being either defined explicitly as it appears


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22 F A L L A C I E S IN MATHEMATICSor derived inductively from first principles or from pre-ceding theory.

Since Euclid's time, his system of geometry has beensubjected to very acute scrutiny, and a number ofseriousgaps have been revealed. In particular (and this is veryrelevant for the next chapter) it appears that he un-wittingly made a number of assumptions from the ap-pearance of his diagrams without realising that hislogical foundations were thereby rendered suspect. Thepreceding illustration shows this very clearly.

pA

DcB Fig. 7

The crux of the argument there is that the angleLDCA is part of the angle LPCA, and this observationis usually followed in standard texts by the unassailableremark,

'but the whole is greater than its part',so that the result follows. There is, however, no step in theargument which proves from earlier results which is thewhole and which the part. The only reason for selectingLPCA as the whole is that it 'looks like it ' ; but whetherit would continue to do so for a triangle of atomic orastronomic dimensions is a very different matter. Theresult may, indeed, be untrue for a triangle drawn on asphere (compare Fig. 7), so something is involved whichis basically different for sphere and plane. The immediate


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ELEMENTARY G E O M E T R Y 23point, however, is that this step of the argument is'diagram', not logic.

Discussion of all that is involved would carry us farbeyond the present aim. There is a very full treatment ofthe subject in Foundations of Euclidean Geometry byH. G. Forder (Cambridge University Press, 1927). Weconclude this chapter by giving explicit mention to twobasic geometrical properties not usually considered inelementary geometry, namely those covered by thewords between and outside (or inside). These are the twocharacteristics of figures in space which remain undefinedin the usual treatments, and the logical gaps left by theiromission are precisely those which are often, uninten-tionally, taken over from a diagram. The work of the nextchapter will reveal the damage that can be done byignoring them.


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24C H A P T E R IV

THE ' ISOSCELES TRIANGLE' FALLACYANALYSED

I t is customary to dispose of this fallacy by drawing anaccurate diagram; but analysis by argument is fruitful,and we shall let the discussion lead us whither it will.

(i) Note first that the internal bisector of the angle Aand the perpendicular bisector of BO both pass throughthe middle point of the arc BO opposite to A of the cir-cumcircle of the triangle ABO (Fig. 8). In the earlierdiagram (Fig. 2, p. 13),0 was placed inside (in the normalintuitive sense of the word) the triangle; it is now seento be outside, though, as we pointed out in ch. III, thisfact cannot be established by any of the results of ele-mentary geometry, which does not define the term.

()Fig. 8

oFig. 9

The removal of 0 to the outside of the triangle, how-ever, does not of itself expose any fatal error, and wemust proceed further. The points D, Q, R (Fig. 9) are nowknown to be the feet of the perpendiculars on the sides of


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TH E ' I S O S C E L E S T R I A N G L E ' FALLACY 25a triangle from a point 0 on the circumcircle. Hence, bythe Simson-line property,

D, Q, R are collinear.We therefore appear to have resolved the difficulty,

since, for collinearity to be possible, one of the pointsQ, R must lie upon a side of the triangle and the other ona side produced. Thus (in the figure as drawn)

butAB = AR+RB,AG = AQ-QG.

The equality of AB and AG is thus negatived.Once again, however, the axioms and theorems of

elementary geometry are not sufficient to establishabsolute proof. I t is not possible to prove by them alonethat a straight line cannot cut (internally) all three sidesof a triangle. The final step requires a supplementaryaxiom, enunciated (like the other axioms) from experience and not from preceding logic:

PASCH'S AX IOM . A straight line cutting one side ofa triangle necessarily cuts one, and only one, of the othertwo sides, except in the case when it passes through theopposite vertex.

The line through D therefore meets either AB or AG,but not both, and the discussion reaches its close.

The crux of this discussion is that the attempt to arguesolely from the usual axioms and theorems has beenfound inadequate. I t is necessary to supplement themeither by reference to a diagram, or by Pasch's axiom.But reference to a diagram is unsatisfactory; it is hardenough in any case to be sure that possible alternativeshave been excluded, and, even more seriously, thereremains the criticism that something outside the logical


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26 F A L L A C I E S IN MATHEMATICSstructure has been called in to bolster it up. This, in adeliberately self-contained system like Euclidean geometry, cannot be allowed.

(ii) The introduction of trigonometry seems to givesome sort of an answer to the problem, and is of interest.Since LOOA = O+lA,

application of the' a/sin A = 2R' formula to the triangleOOA gives

OA = 2Rsin(O+lA) ,where R is the radius of the circumcircle of the triangleABO. Thus

Hence

and

AQ = AR = OAcoslA= 2Rsin (0 + lA) cos lA= R {sin (0 + A) + sin O}= R{sinB+ sinO}= t{2RsinB+ 2RsinO}= l(b+c).

AB-AR =c-l(b+c)= l(c-b),

AO-AQ = b-!(b+c)=i{b-c),

so that one ofthese quantities is positive and the other isnegative; that is, one of the points Q, R lies on a side ofthe triangle and the other lies on a side produced-allwithout Pasch's axiom or any equivalent discussion.

The objection to this treatment lies concealed in thestep 2sin(0+!A)cos!A = sin(O+A)+ sinO,


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TIlE 'I S 0 s eE L E S T R I A N G L E ' FA L LAC Y 27since the proof of this formula involves ultimately theearlier formula 'sin (A +B) = sin A cosB +cos A sin B' ,and, through it, an appeal to a diagram to settle sense(positive or negative). A proof somewhat more excitingand illuminating than the standard one can be given bytransferring the argument to Ptolemy's theorem, as in thenext section.

(iii) Let PQ be a diameter of a circle of radius a andcentre 0 (Fig. 10). Take points M, N on opposite sides ofPQ so that (assuming the angles to be acute)

LQPM = A, LQPN = B.

-+--- - f - - ->1Q

Fig. 10Then LMON = 2(A +B).

Now PM = 2acosA, QM = 2asinA,PN = 2acosB, QN = 2asinB,MN = 2{tMN} = 2{asin (A +B)}

= 2a sin (A +B).But, by the theorem of Ptolemy,

PQ.MN = QM.PN +PM.QN,so that4a 2 sin(A+B) = 4a 2 sinAcosB+4a2 cosAsinB,


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28 FALLACIES IN M A T H E M A T I C Sor sin (A +B) = sin A cosB + cosA sinB.This seems to establish the basic theorem authorisingthe step

2 s i n ( C + ! - A ) c o s ~ A = sin(C+A)+ sinCin the preceding discussion.

We now give further attention to the theorem ofPtolemy. I t can be restated in the form:

I f A, B, C, D are four concyclic points and i f the threeproducts BC.AD, CA.BD, AB.CDare formed, then the sum of two of them is equal to the third.

The difficulty confronting us is to decide which is 'thethird'. The usual method is to say that ' the third' is theproduct of the two diagonals of the cyclic quadrilateralwhose vertices are A, B, C, D, but this is precisely theappeal to diagram that we seek to avoid. There are other-wise three distinct possibilities:

CA.BD+AB.CD = BC.AD,AB.CD+BC.AD = CA.BD,BC.AD+CA.BD = AB.CD,

and, for any given configuration, the criterion is thediagram. Without it we cannot tell which of the alter-natives to select.

A proof of Ptolemy's theorem itself may be developedby a trigonometrical argument which leads directly tothis dilemma. Let A, B, C, D be any four points on acircle of centre 0 and radius a (Fig. 11). Imagine a radiusvector to rotate about 0 from the initial position OD inthe counterclockwise sense, passing through A after anangle a, B after an angle p, and C after an angle y, wherea, fl, yare unequaL and all between 0 and 21T. I t is not


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TH E ' I S O S C E L E S T R I A N G L E ' FALLACY 29at present known in what order the points lie round thecircle. Then, whether the angles a, (3, yare acute, obtuseor reflex, it is true that

Also

AD = 2asin ta,BD = 2asini(3,CD = 2asin ty.

BC = 2asintl(3-yl,CA = 2asintly-al,AB = 2asintla-(3I.

B A

- - - - - tDo

cFig. 11

where, for example, 1(3 - yl means the numerical val'ueof ((3 -y) .

Now it is easy to eRtablish by direct expansion theidentity

sin la sin t((3 - y) +sin t(3sin t(y - a)+ sin h sin t(a - (3) = 0,

and the point to be made is that this simple identity coversthe three possible cases of the theorem of Ptolemy for fourconcylic points A, B, C, D:

Of the three unequal angles a, (3, y, one must beintermediate in value between the other two; supposethat this is a. Then either

(i) (3 > a > y,so that1(3-yl = (3-y, Iy-al = a-y , la-(31 = fJ-aand the identity is

sin ta sin! IfJ - YI- sin t(3sin tly - al- sinhsintla-(31 = 0,

or AD.Be' = BD.CA +CD.AB,


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30 FALLA CIES IN MATHEMATICSwhich is Ptolemy's theorem when AD, BO are thediagonals; or

(ii) I ' > a > fJ,so that

IfJ-yl = y-fJ, Iy-al = y-.x, la-fJl = a-fJ,leading to the same relation

AD.BO = BD.OA+OD.AB.When fJ is the intermediate angle the relation isBD.OA = OD.AB+AD.BO

and when I ' is the intermediate angle the relation isOD.AB = AD.BC+BD.CA.

The three 'Ptolemy' possibilities are thus obtained,the conceptofdiagonalsfor the quadrangle being replacedby that of between-ness for the angles a, fJ, y.

The point can be further emphasised by a moreadvanced argument from analytical geometry. Let therectangular Cartesian coordinates of A, B, 0, D be(xv Yl)' (x2, Y2)' (x3, Ya), (x 4 , Yt), respectively, and denoteby lij the distance between the points (Xi' Yi)' (Xi' Yj)'Consider the product of two determinants:

Xi+Yt -2X1 -2Y1 1x ~ + y ~ -2X2 -2712 1x ~ + y ~ -2x J --2713 1xi+yi -2X4 -2714 1

1 1 1 1Xl X2 X3 X4

X711 712 713 Y4

xi+yi x ~ + y ~ x ~ + y ~ xi+yi


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THE ' I SOSCELES TRIANGLE ' FALLACY 31On multiplication according to the 'matrix' rule theelement in the ith row andjth column is

(X7 +Y7) - 2XiXj - 2YiYj + (x, +Y'),or (Xi - X )2 + (Yi - Yj)2,or l ~ j ' Hence the product is

0 li2 lia li41 ~ 1 0 1 ~ 3 1 ~ 4 l51 l52 0 154III 112 l ~ a 0

This determinant may be evaluated by direct computation. Alternatively, write

lia = 152 =J; 151 = lia = g; li2 = l ~ l = h;li4 = 111 = I; 1 ~ 4 = 112 = m; 1 ~ 4 = 1 ~ 3 = n,

and replace zeros temporarily by the letters a, b, c.Then the determinant is

a h g 1h b J mg J c n1 m n 0

which, equated to zero, may be recognised as giving thetangential equation of the conicax2 +by2+ ez2+ 2Jyz +2gzx+2hxy = 0,namely

Al2+Bm2+Cn2+2Fmn+2Gnl+2Hlm = 0,where A = be-j2, F = gh-aJ,or (with a, b, c returned to zero)

-j2l2- g2m2-h2n2+2ghmn+2hfnl+2fglm = o.


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32 F A L L A C I E S IN MATHEMATI CSFurther, the value of the determinant is indeed zero ifA,B, 0, D are taken to be concyclic, the condition for thisbeing xi+ Xl Yl 1

x ~ + y ~ x2 Yz 1X ~ + y ~ Xs Y3 1xi+yi x4 Y4 1

= O.

(This is found by substituting the coordinates ofA, B, 0, D in the standard equation

X2+y2+2gx+2fY+c = 0for a circle, and then eliminating determinantally theratios 1:2g:2f:c.)

The condition for the four points to be concyclic is thusf2l2 + g2m2+h2n2- 2ghmn - 2hJnl- 2fglm = 0,

or .J(fl) .J(gm) .J(hn} = O.(This is a well-known formula for the tangential equationofa conic through the vertices of the triangle of reference.)

In terms of lii' the condition is .J ( l ~ s l i 4 ) .J ( l 5 l l ~ 4 ) .J(li2154) = 0,

or l2Sl14 Isll24 l12134 = o.This gives the three distinct possibilities

l31l24 + 12 l34 = l23 l14'112134 + 23 l14 = lSl lU.l2Sl14 + 31l24 = l12 l31'

remembering that the distances are positive so that thethree alternative signs cannot be all alike.

We have therefore recovered the three relations givenon p. 28, and this is all that can be done without furtherinformation about relative positions round the circle.


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CHAPTER VTHE OTHER GEOMETRICAL

FALLACIES ANALYSED

33

THE FALLACY OF THE RIGHT ANGLE. Hereagain a proof of the point at which fallacy intrudes leadsto considerations of some depth. We seek as a prelim-inary step an alternative interpretation for the positionof the point 0 (p. 14).

We have seen that PO, RO are the perpendicularbisectors of OD, DE, and so 0 i8 the circumcentre of thetriangle ODE. Hence 0 lies also on the perpendicularbisector of 0E.

But, by construction, BO = BE, so that the perpendi-cular bisector of OE also passes through B. Moreover,and this cannot be proved by the usual axioms andtheorems alone, this line is the internal bisector of theangle LOBE. Thus OB passes down the internal bisectorof the angle LOBE, so that the angle LOBEi8reflex, beingequal to the sum of two right angles plus! LOBE. Thetriangle OBE is therefore so placed that OE is on the sideofB remote from the rest of the figure. The step

LABE = LOBE LOBAis therefore illegitimate.

Note that the exposure of the fallacy has involvedarguments not supported by the usual axioms andtheorems. I t has been necessary to consider the internalbisector ofan angle; and ideas ofin8ide and out8ide appearin Euclid by reference to diagrams and not by reason only,as we have already suggested in ch. III.

3 MI'


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34 FALLACIES IN MATHEMATICSTHE TRAPEZIUM FALLACY. The argument here is un-expectedly close to that of the preceding Fallacy. I t is,of course, perfectly possible for the result to be true; weare interested only in the cases where it is not. In sucha case, draw the line through B parallel to AD (Fig. 12).Let the circle with centre D and radius AB = CD meetthis line in F and in G (where G, not shown in the dia-gram, is the point such that ABGD is a parallelogram),with DF not parallel to AB.

A

B ' - - = : : : : : : : : ~ t = : : = P = = ~ \ F 11 cFig. 12

Then by symmetry (or by easy argument) OM is alsothe perpendicular bisector ofBF; and ON is the perpendi-cular bisector of BC. Hence 0 is the circum-centre of thetriangle BCF, so that, as in the preceding Fallacy, OD liesalong the internal bisector of the angle CDF, and theexposure follows as before.THE FALLACY OF THE EMPTY CIRCLE. The fallacyis most convincingly exposed by algebra. Let

OP=p,so that OQ = r2jp.


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T H E O T H E R G E O M E T R I C A L F A L L A C I E S 35Now (r--p)2> 0,

since the left-hand side is a perfect square. Hencer2 - 2rp+p2 > 0,or p2+r2> 2rp,

or (dividing by the positive number 2p)i,(p+r2jp) > r.

Hence OR> r,that is, the point R lies outside the circle, and so thepoints U, V do not exist and no calculations can be madeinvolving them.

The result is, however, startling when approached withthe technique of coordinate geometry. Take OP to be thex-axis and the line through 0 perpendicular to OP to bethe y-axis. Then Pis the point (p, 0), Q is the point (r2/p, 0)and R is the point {t(r2jp +p), o}.

The circle is X2+y2 = r2,and this cuts the line through R perpendicular to OP,that is, the line x = l(r2jp+p),where y2 = r2_!(r2jp+p)2= _ !(r2/p _ p )2.

In a real geometry no such value of y exists and theargument stops. We therefore continue it in complexCartesian geometry. Then

y = ~ i ( r 2 i p - p),where i = .J( - 1). Thus the complex perpendicular bisector of PQ cuts the complex circle in two points, ofwhich a typical one is

U{i(r2/p +p), !i(r2jp- pH.Now apply the formula

d2= (X1-X2)2+ (YI-Y2)2


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36 FALLACIES IN MATHEMATICSto calculate the distance PU. Thus

PU2 = {t(r2jp +p)_p}2+ {!i(r2jp-p)-0}2= 1(r2jp-p)2+1i2(r2jp-p)2= 1(r2/p-p)2(1+i2)= 0,

so that the distance between P and U is zero.(This is less odd, however, than it sounds, for it de

pends to a large extent on the transference of the word'distance' to this context.)The circle of centre P and passing through U is thus

given by the equation(X_p)2+y2 = 0,

and this meets the given circleX2+y2 = r2

on their radical axis

oror

{(x- p)2 +y2} _ {X2 +y2_ r2} = 0,_2px+p2+r2 = 0,

x = !(r2/p+p),which is precisely the line UV already considered.

Thus there are two points on the circle whose distancefrom P in complex Cartesian geometry is zero. They are,however, quite distinct from P.

Corollary: given any circle and any point in its plane,there exist in complex Cartesian geometry two points on thecircle whose distance from that point is zero, where distanceis defined by the formula

d2 = (Xl - X2)2+ (Yl - Y2)2.Note that this corollary, as stated, is not a fallacy, butis inherent in the definition of distance for complexgeometry.


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C H A P T E R VI

SOME FALLACIES IN ALGEBRAAND TRIGONOMETRY

37

We gather together here some fallacies based on a disregard of the more elementary rules of algebraic andtrigonometric manipulation. As the reader may prefernot to have the exposure immediately beside the fallacyitself, we reserve the comment (here and in later fallacies)till the end of the chapter.

1. THE FALLACIES1. THE FALLACY THAT 4 = O.Since cos 2 X = 1 - sin2 x,

it follows thatl+cosx= 1+(I - s in2 x)!,

or, squaring each side,

or

(1 +COSX)2 = {I + (1 - sin2 X)!}2.In particular, when x = 'TT,

(1-1)2 = {I +(1 - O)!}2,0= (1 + 1)2

= 4.n . THE FALLACY THAT + 1 = -1 .Since 1 = Jl

= ~ { ( - 1 ) ( - 1 ) } = ~ ( - 1 ) ~ ( - 1 ) ,


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38 FALLACIES IN MATHEMATICSit follows, writing .J( - 1) == i, that

1 = i . i

= -1 .

The two fallacies which follow are somewhat similarat the critical point. I t is perhaps worth while to includeboth for the sake of the geometry with which they areclothed.n . THE FALLACY THAT ALL LENGTHS AREEQUAL.GIVEN: A, C, D, B are four points in order on a straightline.REQUIRED: To prove that AC = BD, necessarily.CONSTRUCTION (Fig. 13): Let 0 be one ofthe points ofintersection of the perpendicular bisector ofCD with thecircle of centre A and radius .J(AC. AB), so that

A

A02 = AC.AB.o

a

Fig. 13PROOF: Since A02 = AC .AB, the line AOis the tangentat 0 to the circle OCB,

LAOC = LCBO= (), say.


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SinceALGEBRA AND TRIGONOMETRY 39

00= ODLOOD = LODO

= a, say,LOAO = a-fJ =LBOD

=rp, say.Write OA = a, OB = b; AO = u, BD = v, OD = w;00 = OD = k.From [:'OAO, a2 = k2 +u2+ 2kucosa.From [:'OBD,

b2 = k2+v2+2kv cos a.Hence (a2- k2 -u2)V = 2kuvcosa

so thatoror

= (b2 - k2 _ V 2 ) u,(k2+v2 ) U - (k2+u2 ) V = b2u - a2v,(u-v) (k2 -uv) = b2u -a 2v,

b2u -a 2vu -v = -k2 -uv .Now the triangles OAO, BOD are similar, so that

a2 [:'OAO AO ub2 [:'BOD BD v

Hence

u -v = 0,or u=v .4. THE FALLACY THAT EVERY TRIANGLE IS

ISOSCELES.GIVEN: A triangle ABO.REQUIRED: To prove that AB = AO, necessarily.


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40 FALLACIES IN MATHEMA TICSCONSTRUCTION: Draw the bisector of the angle A tomeet BC at D, and produce AD to P so that

AD.DP = BD. DC(Fig. 14).PROOF: For convenience of notation, write AB = c,AC = b, BD = u, CD = v, AD = x, PD = y, PB = r,PC = q, LPDB = fJ. The definition of P gives therelation xy = UV.

A

B ~ - - - - - - - - , - ~ = - - - - - - - - 7 C

PFig. 14

Consider the triangles ADC, BDP:LADC = LBDP

AD BDDC Dp (contruction)

The triangles are similar (2 sides about equal anglesproportional) . AD DC AC

= =BD DP BP'x v b

or - .U Y r


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ALGEBRA AND TRIGONOMETRY 41Similarly the triangles ADB, CDP are similar, and

x u e= -.v y qIn particular, v u b ey'y = ,'g'

or uvqr = bey2.Now the length of the bisector AD is given by the

standard theoremAB.AC = AD2+BD.DC: . be = x2+uv.

Hence uvqr = x2y2 + uvy2= U2V2+uvy2

qr = UV+y2.Also, from the triangle PBD,r2 = U2+y2_2uycosO,

and, from the triangle PCD,q2 = v2+ y2+ 2vy cos O.

Multiply these equations by v, u, respectively, and add:vr2 +uq2 = uv(U+V)+y2(U+V)

= (u + v) {uv + y2)= (u+v)qr,

by the previous result.Let us now rearrange this relation to give the ratio ujv:

u(q2-qr) = v(qr-r2),u qr-r2 r(q-r)V= q2_ qr = q(q-r)r

r= - .q


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42 F A L L A C I E S IN MA TH EMA TICSThat is, DB PBDC = PC'so that PD bisects the angle BPC.Finally, compare the triangles APB, APC:

In particular,

AP = AP,LBAP = LCAPLAPB = LAPC6.APB == 6. APC.AB=AC.

5. THE F A L L A C Y THAT + 1 = -1 .To solve the equation cot () + tan 3(J = O.

cot (J + tan ((J + 2(J) = 0. t (J tan (J + tan 2(J 0" c o + (J (J=1 - tan tan2cot (J - tan 2(J + tan (J + tan 2(J = 0

: . cotO+tanO = 0tan2 (J+ 1 = 0: . tan 20= -1: . tan(J = i

(construction)(proved)

(ASA)

(i = .J( -1)).But there are genuine values* for 0, in the form

For example,

o= l1T + !n1T (n integer).tan 30 = -co t 0= tan (O+i1T): . 30 = O+i1T+n1T: . 0 = I1T+in1T.

tan i1T = - cot I1T


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ALGEBRA AND TRIGONOMETRY 43Thus, with, say, 0=117,

tan2 117 = -1:.+1=-1.

6. THE FALLACY THAT EVERY ANGLE IS AMULTIPLE OF Tw o RIGHT ANGLES .

Let 0 be an angle (complex) satisfying the relationtanO=i.

Then, if A is any angle,tan(A+O) = tanA+tanO1-tanA tanO

tan A +i=-------1-i tanA=t

= tanO.Thus tan (A + 0) = tan 0,80 that A +0 = n17+O,or A = n17for any angle A.

7. THE FALLACY THAT 17 = O.Since e2ni = cos 217 + i sin 217

= 1,it follows that, for any value of x,

Raise each side to the power i:(eiX)i = (ei (x+2nl)i,

or


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44 FALLACIES IN MATHEMATICSMultiply each side by eX+271, which cannot be zero for anyvalue of x. Then e271 = 1,so that 21T = o.([ 8. THE FALLACY THAT THE SUM OF THESQUARES ON Two SIDES OF A TRIANGLE ISNEVER LESS THAN THE SQUARE ON THETHIRD.GIVEN: A triangle ABO so named that

a> b.REQUIRED: To prove that a2 +b2 > c2 , necessarily.PROOF: Since a> b,it follows that a cos 0 > bcos O.But, by standard formula,

a = bcosO+ccosBb = acosO+ccosA

bcosO = a-ccosB,acosO = b-ccosA.

Hence b-ccosA > a-ccosBccosB-ccosA > a-b.Multiply each side by 2ab and use the cosine formula.

b(a2+c2-b2)-a(b2+c2-a2) > 2ab(a-b): . a3 -b 3 - a 2b+ab2> c2(a-b),

(a-b)(a2+b2) > c2(a-b),

division by a - b being legitimate since a - b > O.ILL U STRAT ION. Consider the particular case

a = 4, b = 3, c = 6.


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ALGEBRA AND T R I G O N O M E T R Y 45This triangle exists, since

b+c > a, C+ a > b, a + b > c.But a2+b2 = 25, c2 = 36.Hence, by the theorem,

25 > 36.II. THE COMMENTARY

FALLACIES 1 AND 2.These are both based on the ambiguity of sign whicharises whenever square roots are taken. An equationhas two solutions, x = +a, x = -a ,and care must always be taken to select the appropriateone. I t is not necessarily true that both solutions arerelevant in the problem giving rise to the equation, andit is always essential to check independently.

This phenomenon is well known in the case of thesolution of equations involving surds. Such equationscan, indeed, be stated to give an elementary form offa.1lacy:

To prove that 1 = 3.Consider the equation

.J(5-x) = 1 +.Jx.Square each side: 5 -x = 1+2.Jx+x,or 4-2x = 2 ~ x , or 2 -x = .Jx.Square each side: 4 - 4x+x2 = x,or x2 -5x+4 = 0,


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46orso that

FALLACIES IN MATHEMATICS(x -4 ) (x - l ) = 0,

x = 4 or 1.In particular, the value x = 4 gives, on substitution in thegiven equation, ~ ( 5 - 4 ) = 1 + ~ 4 , or 1 = 1 +2

= 3.The fact is that the whole of the argument has moved

in one direction only, and there are several points whereit cannot be reversed. The step5 -x = 1+ 2 ~ X + X 2

is a consequence not only of the given equation but alsoof the distinct equation

- ~ ( 5 - x ) = 1 + ~ x . We have in effect solved four equations, ~ ( 5 - x ) = 1 ~ x ,

and each of the two prospective solutions must bechecked against the equation actually proposed.

The following synthetic example of false square rootsaffords another exceedingly simple illustration:By direct computation of each side,

9-24 = 25-40.Add 16 to each side:9-24+ 16 = 25-40+ 16

(3-4)2 = (5-4)2.Take square roots:Hence

3 -4 = 5-4 .3 = 5.

With these remarks in mind, the reader should haveno difficulty in disposing of Fallacy 1. The second one is,however, worthy of further comment.


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A L G E B R A AN D TRIGONOMETRY 47The steps 1 = ,.jl

=,.j{( -I) ( - I ) }are correct. The following step,.j{( -I) ( - I ) } =,.j( -I), . j( -I)may be regarded as correct, but it needs care in inter-pretation. The next step

,.j( -I) , . j ( -I) = i.iis definitely wrong, for each of the square roots ,.j( - I)and,.j( - 1) has, in the first instance, two possible values,and the ambiguities cannot be resolved without reference tothe rest of the problem. I f the symbol i is used for, say,the first square root, then it is in the nature of the pro blemthat the second square root is necessarily - i.

This may be made clearer by examination of theexponential forms of the square roots. I t is well knownthat, if n is any integer,

e(l+2n)7Ti = cos (t + 2n) TT + sin (i+2n) TT= cos iTT + sin tTT=

and that, if m is any integer, then, similarly,e ( - ~ + 2 m ) 7 T i = - i .

If, in particular, we use the identification (with n = 0)i = e!7Ti,

then the argument is1 = ,.j( -1)-J( - I )

= el7Ti x (1),and it is at once evident that the disputed square root ise-!7Ti, or - i, in agreement with the formula

1 = eO= ehie-hi .


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48 FALLACIES IN MATHEMATICSFALLACIES 3 AND 4.

The resolution of these fallacies depends on the illegitimate step usually referred to as division by zero. I t isusually true that, if am = an,then m=n;but this need not be so if a is zero. For example,

0.5 = 0.3,butThe division ofeach side of the equation by a is proper if,and only if, a is not zero.

There is an alternative way of regarding this processwhich is perhaps more significant. We may call it themethod of the false factor. The relation

am=anmay be expressed in factorised form

a(m-n) = 0,from which it follows that one at least of a, m - n is zero.If, therefore, a is not zero, then m - n must be.

The point is that the reader is presented with twofactors and then subjected to pressure to accept thewrong one.

There is a crude type of fallacy based on division by zero,of which the following example is typical:

Let x = 2.Multiply each side by x - I :

x2 - x = 2x-2.Subtract x from each side:

x2 -2x = x-2 .Divide each side by x - 2 :

x = 1.


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A L G E B R A A N D TRIGONOMETRY 49But it is given that x = 2. Hence

2 = 1.The forced insertion of the factor x - I is, however, veryartificial. In the given fallacies the alternative factor arises inits own right out ofthe geometry and must be accounted for.Note that the extraction of square roots referred to

above is itself a particular case of false factors; for theequationmay be written (x-a) (x+a) = 0,and here again the wrong factor is insinuated into theargument.

Having these ideas in mind we turn to Fallacy 3. Wereached correctly the step

(u-v) (k 2 -uv) = b2u -a 2v,and later the equally correct step

b2u -a 2v = O.Hence it is true that

(u-v) (k2-uv) = O.The reader was then presented with the relation

u -v = 0,but the alternative

also required consideration. Now the triangles OAC,BOD are similar, so that

Ie uor v k'

4 MF


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50 FALLACIES IN MATHEMATICSThe relationis therefore true always, so that the deduction

u -v = 0cannot be made.The reader interested in geometry will notice that the

relation k 2 = uvis an extension of a well-known result of elementarygeometry. I f L AOB is a right angle, then 0 and Dcoincide at the foot of the altitude from 0 and

002 = OD2 = OA .OB.The figure under consideration is, so to speak, a 'widening out' of the more familiar one.

The algebra in Fallacy 4- is surprisingly similar to thatofnumber 3, but the underlying geometry has an interestof its own. The step

U(q2 - qr) = v(qr - r2)is correct, and leads to the factorised form(uq-vr) (q-r) = 0,from which attention was drawn to the factor uq - vr.The possibility q - r = 0 must, however, be consideredtoo. Now the relation

xy = uvleads to the result that the quadrilateral ABPO is cyclic,so that, since LBAP = LOAP, we have

BP = OP.Hence, necessarily,q -r = 0,

and the deduction uq - vr = 0 is thus inadmissible.


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ALGEBRA AND T R I G O N O M E T R Y 51We may now examine the basic geometry behind this

algebraic f a ~ a d e . The primary and standard theorem isthat, if the internal bisector of the angle BAO of a trianglemeets the base in D, then

AB.AO = AD2+BD.DO.The problem (which we applied to the triangle PBO) is todetermine whether the converse is also true, that theexistence of a point]) for which the above relation holdsnecessarily implies that AD bisects the angle A. Theanswer is that in general it does, but the argumentbreaks down if the triangle ABO is isosceles withAB = AO. In the latter case the relation holds for allpositions of D on BO. By ignoring the fact that thetriangle P BO is isosceles, we obtained P D as the anglebisector, leading at once to the fallacy.F A L L A C I E S 5 AND 6.

These two fallacies are similar in essential point, butit is interesting to remark that the first of them arose asan attempt to solve the given equation by a perfectlyreasonable method. One suspects that the trap is onethat might deceive many experienced mathematicians.

Suppose that any number k, real or complex, is given.Then, with proper definition of complex circular func-tions, it is always possible to solve the equation

tan x = k,sinx cosx 1k 1 = .;-(-+--=k-c-) ,since thenThere is, however, one case of exception, namely thatwhich arises when lc2+ 1 = 0,or k = i.


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52 FALLACIES IN M A T H E M A T I C SThus no (complex) angle exists whose tangent has eitherof the values i.

Fallacy 6 therefore collapses completely, since theangle 0 is non-existent, and Fallacy 5 must have ano m i s s i o ~ o at some stage, since the step

is excluded.The omission is one that might easily pass detection

were it not for the fallacy to which it leads. There is infact a solution 0 = !7T, so that 2() = !7T; and tan 20 has novalue (or, as is sometimes said, is infinite). The phenomenon may be exhibited more easily by taking theequation in i ts ' inverse' form

cotO = - tan 30tan 0 = - cot 30

: . tanO+cot30 = o.This givest 0 cot 0 cot 20 - Ian + = 0cot 0 + cot 20

I + tan 0 cot 2() + cot 0 cot 2() - I = 0,so that cot 20 (tan 0+ cot 0) = o.Since tan 0 + cot () ::p 0, the solution is

cot 20 = 0,ororThe zero value of cot 20 corresponds to the 'infinite'value of tan 20.


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A L G E B R A AND TRIGONOMETRY 53FAL L ACY 7.

This fallacy reinforces the lesson of Fallacies 1, 2, thatunless n is an integer (positive or negative), then thepoweris a many-valued function; the particular value musttherefore be defined explicitly when this becomes necessary. Thus the stepis correct, but examination is required when raisingeither side to power i. Since

e ix = ei (X+2prr) (p integral),the functionhas the infinite set of values

e- x . e-2]JTr (p 2 1 1 2 )... , - ,- , , , , ....The function (ei (X+2 rr)ihas similarly the infinite set of values

e-(x+2rr) .e- 2IJrr (q = ... , -2 , - 1,0,1,2, ... ).When the values ofp, q are chosen correctly the paradoxdisappears. The most obvious choice is p = 0, q = - I,giving, impeccably,

FAL L ACY 8.Everybody who devotes any time at all to the subject

knows that he must exercise the greatest care whenmultiplying the two sides of an inequality by some givennumber. For example, it is true that

[) > 4,


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54 F A L L A C I E S IN l I IATHEMATICSbut not necessarily true that

5a> 4a.I f a = 3, the relation is 15 > 12,which is correct; if a = - 3, the relation is

- 1 5 > -12 ,which is not correct. It is necessary to make sure that themultiplier is positive.Most inequality fallacies are bascd, rather tediollsly, onthis principle, but one example ought to be included inthe collection. The present illustration has at any ratethe virtue of emphasising, what is well known otherwise,that a2+b2 is greater than c2 when the angle C is acuteand less than c2 whcn C is obtuse.


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55

CHAPTER VII

FALLACIES IN DIFFERENTIATIONThe following fallacies illustrate one or two points ofdanger in the use of the differential calculus. As before,comments are given in the later part of this chapter.

I. THE FALLACIES1. THE FALLACY THAT T H E R E IS NO POINT

ON THE C I R C U M F E R E N C E OF A CIRCLE N E A R E S TTO A POINT INSIDE IT .

Choose the given inside point as the origin 0 ofCartesian coordinates and the x-axis as the line joiningo to the centre C. Let the coordinates of C then be(a, 0) and the radius b, so that the equation ofthe circle is

(x-a)2+y2 = b2,orThe distance from 0 of the point P(x, y) of the circle is r,where r2 = X2+y2,so that, since P is on the circle,

r2 = 2ax-a2+b2The distance OP is least, or greatest, at points for

which = O. But, differentiating the relation for r2, wehave dr

r -=a .dxNow r is not zero or 'infinity', and this equation can onlybe satisfied when :: = 0 if a = 0, that is, if 0 is at the


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56 FALLACIES IN MATHEMATICScentre of the circle-in which case all points of the cir-cumference, being at constant distance from 0, are alikenearest to it and furthest from it.

If, as is true in general, a is not zero, then there is nopoint on the circumference whose distance from 0 iseither a maximum or a minimum.If 2. THE F A L L A C Y OF THE RADIUS OF A CIRCLE.

To prove that the radius of a circle is indeterminate.The working is based on the standard formula

drp=r-dpfor the radius of curvature ofa plane curve, where r is thedistance of the point P of the curve from an origin 0 andp is the length of the perpendicular from 0 on the tangentat P (Fig. 15).

--------p

o

Fig. 15For a circle, take the origin at the centre, and let the

radius (for anyone determination of it) be denoted by k.Then r = p = k for all positions of P. Hence p, rareconnected by the relation

kmrn = pm+nfor any values of m and n.


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DIFFERENTIATION

Take logarithms:mlogk+nlogr = (m+n)logp.

Differentiate with respect to p:n dr m+n' dp = pso that dr m+n r2r- - - - -dp - n .p

=m+n knp = {I + (min)} k.

57

Since this is true for any values of m, n, it follows thatp is indeterminate.

3. THE FALLACY THAT EVERY TRIANGLE ISISOSCELES.

Let ABC be an arbitrary triangle. We prove that,necessarily, AB = AG.

Denote the lengths of the sides by the usualsymbols a, b, c. I fD is the middle point ofBG (Fig. 16),A

B ~ / ~ - - - ~ ~ - - - - ~

Fig. 16


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58 F A L L A C I E S IN MAT HE MAT ICSthen, by the sine rule applied to the triangles ADB,ADO, BD AB

sinBAD = sinADB'OD AO= - , - - ~ ~ sinOAD sinADO'

Write LBAD = ct, LOAD = fJ, and divide corresponding sides of these equations, remembering thatBD = OD; thus sinfJ c

sina = 1)"Now subject the base BDO to a small displacement

into the position B'D'O' by letting the points B, D, 0'slide', as it were, down AB, AD, AO. The angles a, fJ areunchanged in this displacement. Adjust the variation,as is legitimate, so that

8c == BB' = k, 8b == 00' = k,where the displacement k is small, being the same atboth ends of the base. (The line B'O' is not in generalparallel to BO for such a displacement.)

Take logarithms of both sides of the relation

so thatcb

sinfJ. ,smalog c - log b = log sin fJ -log sin a.

Differentiate, remembering that a, fJ are constant:

orHenceso that

8c _ 8b = 0c b 'k k--- = o.c b

c = b,AB=AO.


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DIFFERENTIATION 59n . THE FALLACY THAT l/x IS INDEPENDENTOF x.

We prove first a more general result:AXLet x, y, z be given functions of u, v, w. Evaluate au and

axexpress it in terms of x, y, z. Then ou so expressed isindependent of x.

;: s a function ofx, y, z, and will be independent of x if~ ( a x ) = o.ox au

By definition of second-order partial differentiation, theleft-hand side is 82x

3x 21l ,or, reversing the order of differentiation,

(l2xouox'

Now this, by definition again, is

a aX)au ax ;and ax = 1ax 'so that a : ( ~ : ) = o.Hence i(2X) = 0,ax auand so is independent of x.

The particular example quoted above is a corollary forthe simple case:

x = (2u)!, y = (2v)!, z = (2w)!.


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DIFFERE NTIATION 61Since r cannot be zero (unless ( /2 = b2 ), the condition

for a turning value is . lJ 0SInu = ,so that 0=0, () = 7T.This gives the two ends of the diameter through O.FALLACY 2.

This is akin to the preceding fallacy. The relationrnlogk+nlogr = (rn+n)logp

is correct, and the relation in differentials isndr (rn+n)dpr p

This is perfectly true. But since rand p are both constantfor the circle, dr and dp are both zero, so that no furtherdeduction can be drawn.

The real interest of this fallacy, however, lies in a moredetailed examination of the equation

To make ideas precise, take the simplest catse, whenrn = n = 1, so that the curve is

p2 = kr.This we recognise as the 'pedal' or 'p-r' equation ofa parabola whose latus recturn is 4k and whose focus is atthe origin. To verify this statement geometrically, let Pbe an arbitrary point on the parabola of focus Sandvertex A, where SA = k (Fig. 17). I f the tangent at Pmeets the tangent at the vertex at T, it is known that

(i) ST is perpendicular to the tangent,(ii) LAST = LTSP.


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62 FALLACIES IN MAT HEMA 1'1 CRHence the triangles AST, TSP are similar, so that

AS TSST SP'or, in the notation of the text,

k pp r'

so that p2 = kr.

Fig. 17Hence the equation

p2 = krnot only represents (jor p = r = k) the circle of centre theorigin and radius k, but also a parabola of focus the originand latus rectum 4k. The analysis of the text shows thatat a general point of the parabola the radius of curvatureis 2r2/p.

More generally, an equation between r awl p of theform f(r, p) = 0


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DIFFERENTIATION 63gives a certain curve (as in standard theory); but it isalso satisfied by the circles whose radii are the roots of theequation

f(r, r) = o.These are the circles through the apses of the curve, thatis, the points where the tangent is perpendicular to theradius. (The formula p = rdrjdp requires care at suchpoints of the curve, since dr and dp are both zero. Seealso the preceding fallacy.)FALLACY 3.

The explanation of this fallacy is comparativelysimple, for it is a case of pure misconduct. The formula

c sinfJb sinarequires D to be the middle point ofBC; butD' is not the

middle point of B'C', so the differentiation is not legiti-mate-a and fJ would have to vary too.

I t is essential to check that variations do not violatethe conditions on which formulae are based.FALLACY 4.

This is an example of hypnosis; the notationa2x axaxau' auox

lures one into believing that the differentiations can bereversed in order. But ( ? ~ ) ax au

is formed on the understanding that the other coefficientsare a OX) a ax)ay cu ' az au '


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64 FALLACIES IN MATHEMATICS

whereas

impliesThe symbolis short for

a OX)au ax

a OX)ov ax ' ~ ( O X ) .w axa ax)

ax au

ax { ~ : } (y, II: constant) (v, w constant)and the abbreviation to

is unwarranted.


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65

C H A P T E R VII I

FALLACIES IN INTEGRATIONThe fallacies arise chiefly through failure to observecertain well-known elementary precautions. The resultsto which this failure leads may emphasise more vividlythan warnings the need for care.

I. THE F A L L A C I E Sn . THE FALLACY THAT 0 = 1.Consider the integral

1== fdX.Integrate by parts: x

1= f . (l jx)dx= x(ljx) - f ( - IJx2)dx= 1+ fd;= 1+1.Hence 0=1 .n . THE FALLACY THAT 2 = 1.

Let f(x) be any given function. Thenf>(X)dX = f>(X)dX- f>(X)dX.

I f we write x = 2y in the first integral on the right, thenf>(X)dX = 2f>(2Y)dY

= 2f>(2X)dX,on renaming the variable.s MF


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66 FALLACIES IX M A T H E M A T I C SSuppose, in particular, that the function f(x) is such

that f(2x) == H(x)for all values of x. Then

J>(X)dX = 2 J ~ ~ f ( X ) d X - J>(X)dX= 0.

N ow the relationf(2x) == U(x)

is satisfied by the function

Henceso thator

1f(x)==-.xJ2 dX= 0,1 Xlog 2 = 0,

2 = 1.

([ 3. THE F A L L A C Y THAT 1T = 0.To prove that, if (O) is any function of 0, then

Substituteso thatand write

J:f(O) cos 0 dO = 0.sinO = t

cosOdO = dt,f{sin- 1 t} == (J(t).

The limits of integration are 0, 0, since sin 0 = 0 andsin 1T = 0. Hence the integral is

J>(t)dt = O.


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INTEGRATION 67Corollary. The special case when f(O) == cos 0 is of

interest. Then the integral is

I: cos2 ()dO = i I: (1 +cos 20) dO= [iO +! sin 20]:= t1T

Hence 21T = O.4. THE FALLACY THAT 1T = o.Consider the integralII adx1== / 2 22cosA\(1-ax)

= [sin-l (ax)]l2cosA

= sin-la-sin-1 (2acosA).Suppose, in particular, that a = sin A. Then

I = sin-1 (sin A) - sin-1 (sin 2A)

Now letso that

Then

Thus

and so

=A-2A = -A .A = i1T

2cosA = 1.

= o.0= -A= -i1T,

1T = o.5-2


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68 FALLACIES IN MATHEMATICSn . THE FALLACY THAT 17 = 2.j2.

Consider the integral1== f : xf(sinx)dx,

wheref(sin x) is any function ofsin x. In accordance witha standard treatment, make the substitution

x = 17-X',and then drop dashes. Thus

I = f : (17 - x) f{sin (17 - x)} d( - x)= f:(17-X)f(SinX)dx.

Hence 2f: xf(sinx) dx = 17f:f(SinX) dx.Take, in particular,

f(u) == usin-1 (u),so that f(sinx) = sinx.x = xsinx.Then the relation is

2f : x2 sinxdx = 17f:xsinxdx.But f : x2 sin xdx = 172 - 4,

f : x sin xdx = 17.Henceorso that 17=2.j2.


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INTEGRATION 69n . THE FALLACY THAT A CYCLOID HAS ARCHESOF ZERO LENGTH.

To prove that the length of an arch of the curvex = t+sint, y = 1+cost

is zero.An arch is covered by values oft running from 0 to 27T.

I f 8 denotes length ofare, then the required length is, bystandard formula,

f2"d8Nowso that

-d dt.o tdx dy- = 1+cost - dt ' dt -sint,(dS) 2dt = (l+cost)2+(-sint)2

= 1+ 2 cos t + cos2 t + sin2 t= 2 +2cost= 2 (1 + cost)= 4 cos2 it.

Thus the length isf:" 2 cos itdt = [4 sin itJ:"= o.

II. COMMENTARYFALLACY 1.

The error is brought about through the omission of thearbitrary constant for indefinite integrals. The anomalydisappears for definite integration:

I f1== fbdX ,aX


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70 FALLACIES IN M A T H E M A T I C S

[J fbdX= 1 + -a a X= 0+1= I.

(Note that [ 1J: = 0 since the function 1 has the value1 at x b and also at x a.)FALLACY 2.

The integralf>(X)dX

does not exist when f(x) == Ijx. A reconciliation state-ment may be effected by returning to the original inte-gral. I t is true that for any value of 0 greater than zero

Put x = 2y in the first integral on the right; it becomesJl 2dx = Jl dx.M 2x !J X

The relation is thus


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INTEGRATION 71The assumption made in the text is that this integraltends to zero with o-as is indeed plausible, since bothlimits then vanish. But we know that

Jo dx = [IOgXJ8X ~ . t = logo-loglo= log (:0)= log 2.

This is in fact the value ofJ2dX1 x '

so the problem is resolved.

FALLACY 3.The fallacy may be exposed by considering the par-

ticular example in detail. Let

Substituteso that

1== J: cos 2 0dO.sinO=t,

cos ada = dt.Then, as an intermediate step,

1= J os()dtbetween appropriate limits.

Now cos () is given in terms of t by the relationcos() = ..J(1-t2 ) ,


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72 FA L LAC I E S IN MAT II E MAT I e Swhere the positive sign must be taken for the part (0, t7T) ofthe interval of integration and the negative sign for the part(!1T, 7T). Thus

1= +f~ ( 1 - t 2 ) d t - f~ ( 1 - t 2 ) d t , where the first integration corresponds to () l1Tand the second to t7T () 7T, so that the limits for tare0,1 and 1, 0, respectively. Hence

1= f: ~ ( 1 - t 2 ) d t - f: ~ ( 1 - t 2 ) d t = 2f: ,,(1- t2)dt= 17T

F A L L A C Y 4.I t is immediately obvious that the error must lie in the

interpretation of the inverse sines in the stepsin-1 (sin A) - sin-1 (sin 2A) = A - 2A,

but it is less easy to see exactly what is correct. We have toexamine the expression

[ sin-1 (XsinA)]l ,2C08'&

where, to avoid further complications, we suppose thatthe lower limit is not greater than the upper, so that

cosA t,or, for an acute angle, A ;:: i-1T.The essential point is that the inverse sine must varycontinuously as x rises from 2 cos A to 1.


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Draw the curveINTEGRATION

SIn'ltv=--sin A

73

for values of u between 0 and 1T (Fig. 18). Since, by theassumption,A !1T,itfollowsthat2A i1T,asindicatedin the diagram. The values of v for u = A, 2A are 1,2cosA, respectively, where 2cosA :s; 1. The line v = 1cuts the curve at P, where u = A, and at R, whereu = 1T - A; the line v = 2 cos A cuts the curve at Q, whereu = 2A, and at S, where u = 1T - 2A.

v

l/sin A -----------------1 ----------p --------------- R

2 cos A - - - - ~ ' ) ----- ------------------

1/-A 2A uFig. 18

To get the variation in the inverse sine as v rises from2 cos A to I, it is possible to take either the journey fromS to P, giving A _ (1T - 2A) = 3A -1T,or the journey from Q to R, giving

(1T - A) - 2A = 1T - 3A,that choice ofsign being taken here which, by the natureof the integral, gives a positive result: that is, 3A -1T .Thus 1= 3A -1T,which vanishes when A = i1T.

The argument on p. 67 wrongly associated P with Q,leading to the fallacy.


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74 FALLACIES IN M A T H E M A T I C SFALLACY 5.

This is very similar to its predecessor and need not begiven in much detail. The point is that

f : (11 - x) f{sin (11 - x)} dx= f:(11-X).[Sinx.(11-x)] dx= f : (11-x)2 sinxdx.

Since 1= f>2 sin xdx,the correct relation is

f : x 2 sin xdx = f : (11 - X)2 sin xdx.FALLACY 6.

The error here lies in the false extraction of the squareroot: (d8) 2dt = 4cos2 ttThe correct step isfor the interval (0, 11), and

ds- = - 2cos J,tdt wfor the interval (11, 211) where the cosine is negative. Thus

S = f: 2 cos ~ t d t - f:" 2 cos !tdt= [4sintJ: - [ 4 s i n ~ t J : " =4 - ( - 4 )= 8.


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75

CHAPTER IXFALLACY BY THE CIRCULAR POINTS

AT INFINITY1. THE FALLACIES

([ 1. THE FALLACY THAT THE FOUR POINTS OFINTERSECTION OF Two CONICS ARE COLLINEAR.GIVEN: Two conics and a pair of common tangents PUand QV touching one conic at P, Q and the other at U, V.The conics meet in four distinct points A, B, C, D, and thechords of contact PQ, UV meet at R (Fig. 19).

j{

Fig. 19REQUIRED: To prove that the line DR contains each ofthe three points A, B, C.CONSTRUCTION: To prove that, say, A lies on DR,project B, C into the circular points at infinity. The twoconics then become circles, with A, D as common pointsand with PU, QV as common tangents (Fig. 20).PROOF: By symmetry, or by easy independent proof,the lines PQ, UV, AD are all parallel. That is, PQ meets


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76 FALLACIES IN MATHEMATICSU V in a point R at infinity, andDA also passes through R.Thus A lies on DR in the projected figure, and so A lieson DR in the original figure.

In the same way, Band C also lie on DR, and so thefour points A, B, 0, D are collinear.

Fig. 20 Fig. 21

([ 2. THE F A L L A C Y THAT CONCENTRIC CIRCLESINVERT INTO CONCENTRIC C I R C L E S WITH RE-SPECT TO AN A R B I T R A R Y POINT.

Let 81> 8 2 be two given concentric circles and 0 anarbitrary point, the centre ofa circle (Fig. 21). Invert8 1 and 8 2 with respect to L.

The circular points I, J lie on L, and so each invertsinto itself. Moreover, concentric circles touch at I andat J, and the property of touching is not affected byinversion. Hence 8 1, 8 2 invert into cirlces 8 ~ . 8; whichalso touch at I, J. They are therefore concentric circles.


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CIRCULAR POINTS AT INF INITY 77

II. THE COMMENTARYFALLACY l .This example illustrates the danger, often ignored, of'circular points' argument when real and complex geo-metryare confused. The given theorem belongs to com-plex geometry, where two conics with four distinct pointsalways have four common tangents. Projection transfersattention to real geometry, where two circles through twopoints have only two common tangents.

Call the four tangents of the complex geometrya, fl, y, o. When B, 0 are projected into the circular pointsat infinity, two of them, say fl, y, are projected into thetwo real tangents of the circles, and it is from them thatthe collinearity ofD, R, A is established. But when twoother points, say 0, A, are projected into the circularpoints, it is a different pair of tangents that ' appears' inthe real plane, so that the intersection of chords of con-tact is not R but some other point R'; and all that wehave proved is that D, B and this point R' are collinear.Similarly projection ofA, B into the circular points givesthe collinearity of D, 0 with yet another point RI/.

The basis of the proof is therefore unsound.

FALLACY 2.There are two things that must be kept in mind aboutinversion as it is normally understood: first, that the

definitions and proofs are based on real geometry, anddo not necessarily apply to complex; second, that whenextensions are made beyond Euclidean geometry thecorrespondence does not remain (1, 1) for all points ofthe plane.


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78 F A L L A C I E S IN M A T H E M A T I C SSuppose that P, P' are inverse points with respect to

the circle of centre 0 and radius a. Then

oP.or = a2,so that OP' = 5 ~ . I f P is conceived as moving further and further awayfrom 0 along a given radius, then pi moves nearer andnearer to 0 along that radius. In a geometry in which theline at infinity is admitted, the' points' of that line correspond to the directions of approach to O. The' line' itselfis concentrated by inversion into the point O. Since 1, Jlie on the line at infinity, the language 'each inverts intoitself' cannot be accepted without reservation.

But this is not the whole story--otherwise, indeed, weshould reach the conclusion that every circle passedthrough O. The fact that the circular points belong to acomplex geometry must now be considered. In homogeneous coordinates, these are the points

1(1, i, 0), J(I, - i , 0),each lying on every circle

In so far as they can be expresRed at all in non-homogeneous coordinates, they are the points lying 'a tinfinity' along the line

x y1= i '

and so the non-homogeneous coordinates of a point P on01 (where the origin is at 0) can be expressed in the form


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CIRCULAR POINTS AT INF IN ITY 79(1\., il\.). But such a point is at zero distance from 0, sincethat distance d is given by the formula

d2 = (Xl - X 2)2 + (Yl - Y2)2= (I\. - 0)2 + (il\. - 0)2=A2(1+i2)= O.

Thus if we allow any meaning to distance in general, andto infinite distance in particular, we come to the conclusion that the inverse of every point of the line 01 is at 1.

We are, in fact, attempting to combine real and complex geometries, and have reached a sensitive spot.

p

o

Fig. 22There is a more satisfactory interpretation in terms of

pure geometry. Let the tangents at points 1, J on a givenc o n i c ~ meet atO (Fig. 22). From any point Pin the planeanother point P' can be defined as follows:Join OP, and let P' be the point on OP such that P, P'are conj1lgate with respect to that is, such that the polarof either passes through the other.

This defines P' uniquely for given P, and, when 1, Jare the circular points at infinity, the two points areinverse with respect to


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80 F A L L A C I E S IN MATHEMATI CSThe cases of exception are readily seen:(a) I f P lies on IJ, its polar passes through 0 and so

meets OP at O. Thus all points of the line IJ have theirinverses at O.

(b) I f P is on 01, then the polar of P passes throughI and therefore meets OP at I. Thus all points of the line01 (or OJ) have their inverses at I (or J).

We have therefore broken down the statement (p. 76)that I inverts into itself; its inverse is not unique, butconsists of every point on the line 01.


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CHAPTER X

SOME 'LIMIT' FALLACIESI . THE FALLACIES

1. THE BINOMIAL F ALL.A.CY.The binomial theorem states that(a+b)n = an+nan-1b +

81

n(n-1)2! an - 2b2 + ... + nabn- 1 + bn Put n = O. Then

(a+b)O = 1, aO = 1, bO = I;and all terms on the right except the two outside oneshave n as a factor and therefore vanish. Thus

2. THEEQUAL.Let m,

function

1 = 1+0+0+ ... +0+1= 2.

FALLACY THAT ALL NUMBERS ARE

n be two given numbers, and consider themx+nyx+y

Whenever y is 0, its value is

or

mxxm,

and whenever x is 0, its value is

or6

nyyn.

MF


81/94

82 FALLACIES IN MATHEMATICSIllcluded in each of these two cases is the particular

case when x = y = O. Since y = 0, the value of thefunction is m; since x = 0, the value of the function is n.Hence m=n.

(The reader suspicious of the stepmx-=mx

at x = hould note that, in any case,1. mxIm-=mx--*O Xand, similarly, that1. ny )1m - = n.y--*O y

3. THE FURTHER F A L L A C Y THAT ALL N U M B E R SARE E Q U A L .

Let rn, n be any two number::;, and let x denote theirdifference, so that x = rn-n.Then rn{x2-(rn-n)x} = n{x2-(m-n)x}.

I t is, of course, illegitimate to divide this equation byx2- (m-n)x, which is zero, but, alternatively, we mayuse the limiting argument

m . x 2 - (m - n) x- = hm 2 n x--*m-n X - (m-n)x

Now we know that, i f f(x), g(x) are two functions suchthat f(k) = 0, g(k) = 0, but such that g'(k) =j; 0, whereg'(k) is the differential coefficient, then

lim f(x) = j'(k)x--*k g(x) g'(k) .


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SOME ' L IMIT ' FALLACIESHere, J,; = m - nand

so thatand

f(x) = g(x) = x2-(m-n)x ,f'(x) = g'(x) = 2x- (m-n)

f ' (m-n) = g'(m-n) = tn-no

83

I f m = n, thcl'C iH nothing to prove. I f m =1= n, thenm -n =1= 0, so that r/(m-n) =1= O. Hence

. x 2 - (m - n) x 1n - nhm 2 --- = -- = l .x--+m-n x - (m-n)x m -nThus m = 1n 'or m=n .([ 4. SOME INFINITE SERIES FALLACIES .

A number of somewhat synthetic fallacies (some ofthem, though, important in their day) can be obtainedfrom infinite series. The following examples arc rea'Sonably bricf.

(i) TO PROVE THAT 1 = O. WriteS==. 1 - 1+1 - 1+1 - 1+ ....

Then, grouping in pairs,s= (1 -1)+(1-1)+(1-1)+ ...

= 0+0+0+ ...= O.Also, grouping alternatively in pairs,

Hence

s= 1 - (1 -1 ) - ( 1 -1 ) - ( 1 -1 ) - ...= 1-0-0-0- ...=1. 1 = O.

6 ~ 2


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8t FA LLA C IE S IN MA TIIEMA TICS(ii) TO PRO VET HA T - 1 IS PO SITIVE. Write

8 == 1 + 2 + 4 + 8 + 16 + 32 + ....Then 8 is positive. Also, multiplying each side by 2,

28= 2+4+8+16+32+ ...= 8 -1 .

Hence 8 = -1 ,so that - I is positive.

(iii) TO PROVE THAT 0 IS POSITIVE (THAT IS,GREATER THAN ZERO).

Write u = l+!+i+t+ ... ,v = !+!+ i+ t+ .. ..

Then 2v = I + t +1+ ! + .. .= n+v,so that u -v = o.But, on subtracting corresponding terms,

u -v = ( l - i )+(!- i)+(i- i )+(t- t )+ ... ,where each bracketed term is greater than zero. Thus

u - v is greater than zero,or o s greater than zero.

II. THE CO MMEN TA RYF A L L A C Y 1.

I t is easy to dispose of the fallacy in this argument, forthe binomial theorem for positive integers begins atn = I and the proof does not apply when n = O. I t s lesseasy to see where the extra 1 comes from, and the argument is worth examining in detail.


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SOME ' J , IM IT ' F . \LLACIES 85Since the binomial expallsion, when n is not a positive

integer, involves an infinite series, we take, in accordancewith normal practice, the value a = 1 and assume thatIbl < 1. Consider, then, the series

1 b n(n-l)b2 n(n - l ) (n -2 )b3+n + 2 t + 3 t + ... ,III which the coefficient of bk , where k is a positiveinteger, is n(n - 1) (n - 2) ... (n - k + 1)lc(lc-l)(lc-2) ...1 ".This is a function ofnand k, which we now examine on itsown merits, ignoring the restriction for n to be a positiveinteger. Write

f(n,k) == n(n- l ) (n -2 ) ... (n -k+l ) .k (k - l ) (k -2 ) ... 1

Let n approach the value Te; thuslim f(n, k) = 1.11-+1.:

Now let k approach the value 0; thuslim {limf(n, Te)} == lim (1) = 1.k-+O n-+k k-+O

Alternatively, let n approach the value 0; thufi,lim f(n, I;) = o.n-+O

Now let k approach the value 0; thuslim {limf(n,k)} == lim (0) = O.k-+O n-+O k-+O

Both limiting processes finish with nand k zero, but thedifferent approach gives a result which is 1 in the firstcase and 0 in the second.


85/94

86 FALLACIES IN : \ [ATIIE:\[ATICSThese results may be applied to the particular fallacy.For the first, ifn is identified with the positive integer k,

then a coefficient of value 1 is obtained. The subsequentstep k = s meaninglesfl as picking ont a term of theseries, but the value 1 persists arithmctically, giving(since ,&0 = 1) an extra 1 in the expansion.

In the second case, hm\-ever, the immediate identification of n with zero precludes the possibility of anadditional term, and so leads to the correct result.

This example is very instructive as affording a simpleillustration of the way in which the passage to limitingvalues in two stages may be profoundly affected by theorder in which the calculations are performed.FALLACY 2.

This fallacy, like the preceding one, gives an illustration of the care that must be taken when two limitingprocesses are reycrsed in order of operation.

A slight change of notation allows the use of thelanguage of polar coordinates. Consider the function

mx2+m/u = --- ' - --- X2+y2 .In terms of polar coordinates, with

x = r cos 0, y = r sin 0(so that x2+ y2 = r2), we have

u = mcos20+nsin 2 0.The denouement x = 0, y = arises when r = 0, but the limiting processes towards thatend depend on 0 as 'well as r. The origin x = 0, y = maybe regarded as approached along a line 0 = a, and thelimit is then m cos2 a +n sin 2 a.


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SOME 'L IM IT ' FALLACIES 87But this depends on ex. In particular, it has the value mwhen ex = 0 and the value n when ex = trr, correspondingto the two particular limits given in the text.FALLACY 3.

The fallacy itself is rather elementary, but it does serveto emphasise a point that is sometimes ignored. Thetrouble lies in the expression

. x2 - (m-n)xhm 2 - ,x ~ m - n X - (m-n)xwhrre the limiting process itself is meaningless, since xalways has precisely the value m - n and never any other.The subsequent theory therefore collapses.FALLACY 4.

These fallacies are included chiefly because the collection might appear incomplete without some referenceto infinite series. They are brought about by ignoringstandard rules governing the convergence of series, details of which may be found in appropriate text-books.The series are all non-convergent.

There is some interest in a comparison of (ii) with thecorresponding correct form for a convergent series:

Writeso that

Hence

8 == l+ i+ i+ i+ ... ,28 = 2 + 1 + ! + ! + ...

= 2+8.8 = 2.


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88

C H A P T E R X I

SOME MISCELLANEOUS HOWLERSI t is believed that the howlers with which we concludethis book are all genuine, in the sense that they wereperpetrated innocently in the course of class study or ofexamination. Teachers will be familiar with the typeof mind producing them, and little comment seemsnecessary.

The first two howlers are so astonishing that an investigation is added giving the generalised theory underwhich they become possible. To get the full benefit fromthe others, the reader should trace the mistakes to theirsource, not being content with merely locating thewrong step: for example, the statement

362 = 336in no. I I is the stroke of genius on which the wholesol u tion turns, and arises, presumably, from theargument

362 = 3(6)2 = 3(36) = 336.The author cannot follow the subsequent step

I. To solve the equation

EitherorCorrect.

(x+3)(2-x) = 4.x+3=4 : . x= l ,2 - x = 4 : . x = - 2.

Oommentary. Every quadratic equation can be reduced toa form leading to this method of solution:


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S O M E M I S C E L L A N E O U S H O W L E R S 89The quadratic equation with roots p, q is

(x-p) (x-q) = 0x2_px_qX+pq = 0

I -p+q = 1+q-x-p-pq+px+X+qx-x2

' ... ' either

or

= ( l+q-x) -p( l+q-x)+x( l+q-x)= ( l +q-x ) ( l - P +x)

( l+q-x)( l -p+x) = I -p+q1+q-x = l - p+q

: . x =p,I - p+x= I -p+q

: . x = q.2. To find the largest angle of the triangle with sides 4, 7,9.

Butand

Correct.

sinO == 12857.

1 =sin 90,2857 = sin 16 36': . 12857 = sin 106 36': . 0 = 106 36'.

Commentary. Let OBO be a triangle in which the angle atC is a right angle. Draw the circle with centre B and radiusBO+BO to cut the circle with centre 0 and radius CO +C Bin A. Then ABC is a triangle whose angle C can be foundby the above method.

LetBO = a, 00 = x, OB = y. ThenCA = x+a, AB = y+a

and


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90 F A L L A C I E S IN M A T H E M A T I C SThe method gives

If, then,

, . 0 y+a y -xSIn = - - = 1+-- .x+a x+ay -xsinO = --,x+a

On this basis, cos 0 = - sin 0 = x - y .x+aBut the more usual formula gives

a2+b2 - c 2 a2+(x+a)2-(y+a)2cos 0 = - - - - - ~ - - = ----- -- - - - - - - ~ 2ab 2a(x+a)x2_y2 +a2+ 2ax- 2ay 2ax-2ay x - y--------_ ._------ . - - - -2a(x+a) 2a(x+a) x+a

The two formulae therefore agree.3. To prove that, if

I = f ~ " XU sin xdxI ,othen In +n(n-l)In_2 = n(!7T)n-l.

On 'integration by parts',In = f ~ " x n s i n x d x

= [ n X I ! - l S i n x J ~ 1 T - f ~ " nxn-1cosxdx= n( t7T)n-l- [n(n - 1)xn- 2cos XJ:"

+ f!" n(n - 1)xn-2( - sin x)dx= n(!7T)n-l- 0 - n(n - 1)In_2: . In +n(n-l)In_2 = n(lTT)n-l.


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S 0 ::\[ E ;,1 IS eEL LA!>; E 0 C S 11 0 W L E R is II I4. To prove the converse of the theorem of Apollonius, that,if D is a point in the base BO of a triangle ABO such that

AB2+A02 = 2DA2+DB2+D02,then D i8 the middle point of BO.Let E be the middle point ofBO. Then, by the theorem

of Apollonius,AB2+A02 = 2EA2+2EB2

= 2EA2+EB2+E02.Thus 2DA2+DB2+lJ02 = 2EA2+EB2+E02

D(2A2+B2+02) = E(2A2+B2+C2).But 2A2+1J2+02 i : 0,otherwise A, B, 0 would all be zero. Hence

D=E ,so that D is the middle point of AB.

c

A1-------- - B.7V I

Y---Fig. 235. To prove that, i f OD is a chord of a circle perpendicularto a diameter AB and meeting it in N, then the sum of theareas of the circles on AN, BN, ON, DN as diameters isequal to the area of the whole circle (Fig. 23).


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92 FALLACIES IN M A T H E M A T I C SThe sum of the areas is

!1T(AN2 +BN2 + CN2 + DN2)= t1TN2(A+B+C+D).

But A, B, C, D are all points on the circumference of thebig circle. Hence the sum of the areas is equal to the areaof the whole circle.6. To prove that, i f 0 is a point inside a rectangle ABCD,then OA2+0C2 = OB2 +OD2.I f he equation is divided by 0, its value is unchanged.ThusNowbecause the figure is a rectangle. Thus

A 2 = B2 = C2 = D2,and so A2+C2 = B2+D2,but this is equal to

OA2+0C2 = OB2+0D2.Commentary. The fusion of the branches algebra, geometry, etc., into the single subject mathematics isstrongly urged by many teachers today. The aboveexamples afford interesting illustrations of the process.7. To prove that 8-1 x 4i = 1.

(i) 2-5x 25 = 45- 5 = 4 = 1.(ii) 83 45 43 64 185 x 42 = 82 = 64 = .(iii) (41)5+ (81)5 = 4t+ 8! = 2 + 2 = 1.(iv) 5.. j4+5f/8= 10+10= 1.(v) tlog 4-;.-t log 8 = 5log2-;.-5log2 = 1.


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SOME MISCELLANEO {IS HOWLERS fla8. To solve the equation 2.jx = x - 3.

2xt = x-:-;: . (x l ) t=x-3: . xi= x - 3x i -x: = 3: . x3 = 3

x = 3 x 3 = 9.Correct.23 48 16- 119. To simplify the expression 1 ~ 2 . 4 - 2 n .

Correct.

22 11 - n202- 2n

= 22 11 - n - 20n= 211.

10. To solve the equation

Correct.

( 5 - 3x) (7 - 2x) = (II - 6x) (3 - x).5-3x+7-2x = I I -6x+3-x

12- 5x = 14-7x: . 2x = 2: . x = 1.

I I . To solve the equation

Correct.

x2+(x+4)2 = (x+36)2.x2+x2+ 42 = x 2+ 362

:. X2+X2 + 16 = x2+336X 2 +X 2_ X 2 = :536-16

:. X4 = 320: . x = 80.


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SOME MISCELLANEOUS HOWLERS 95Each ratio is

Hence

a+b+c+db+c+d+a

=1.a = c.

I f a#- c, thena+b#-b+cc+d#-d+a

a+b+c+d #- a+b+c+d,which is not true unless

a+b+c+d = o.

19346327 fallacies in mathematics

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