whoever said nothing is impossible? - cfl...in the history of mathematics there have been periodical...

GENERAL ARTICLE

25RESONANCE March 1999

Whoever said Nothing is Impossible?Three Problems from Antiquity

Shashidhar JagadeeshanShashidhar JagadeeshanShashidhar JagadeeshanShashidhar JagadeeshanShashidhar Jagadeeshan

Shashidhar Jagadeeshan Shashidhar Jagadeeshan Shashidhar Jagadeeshan Shashidhar Jagadeeshan Shashidhar Jagadeeshan

received his PhD inreceived his PhD inreceived his PhD inreceived his PhD inreceived his PhD in

mathematics frommathematics frommathematics frommathematics frommathematics from

Syracuse University, NY,Syracuse University, NY,Syracuse University, NY,Syracuse University, NY,Syracuse University, NY,

in 1994. His researchin 1994. His researchin 1994. His researchin 1994. His researchin 1994. His research

interests are in the areainterests are in the areainterests are in the areainterests are in the areainterests are in the area

of ‘representation theoryof ‘representation theoryof ‘representation theoryof ‘representation theoryof ‘representation theory

of finite dimensionalof finite dimensionalof finite dimensionalof finite dimensionalof finite dimensional

algebras’. Following hisalgebras’. Following hisalgebras’. Following hisalgebras’. Following hisalgebras’. Following his

interest in the questionsinterest in the questionsinterest in the questionsinterest in the questionsinterest in the questions

raised by J Krishnamurtiraised by J Krishnamurtiraised by J Krishnamurtiraised by J Krishnamurtiraised by J Krishnamurti

on life and education, heon life and education, heon life and education, heon life and education, heon life and education, he

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Learning, a small schoolLearning, a small schoolLearning, a small schoolLearning, a small schoolLearning, a small school

in Bangalore, in 1995. Hein Bangalore, in 1995. Hein Bangalore, in 1995. Hein Bangalore, in 1995. Hein Bangalore, in 1995. He

is interested inis interested inis interested inis interested inis interested in

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mathematics curriculummathematics curriculummathematics curriculummathematics curriculummathematics curriculum

to help studentsto help studentsto help studentsto help studentsto help students

appreciate the inherentappreciate the inherentappreciate the inherentappreciate the inherentappreciate the inherent

beauty and rigour inbeauty and rigour inbeauty and rigour inbeauty and rigour inbeauty and rigour in

mathematics.mathematics.mathematics.mathematics.mathematics.

1. Prologue

In the history of mathematics there have been periodical dis-coveries which have changed the very way we think aboutmathematics. For example, the discovery of non-Euclideangeometries in the early nineteenth century made us questionthe link between mathematics and reality. Godel’s work,in this century, not only introduced the notion of undecid-ability, but also raised very serious questions about the ax-iomatic foundations of mathematics. In a similar vein thediscovery in the nineteenth century that certain problems inmathematics are impossible to solve, revealed a totally newaspect to mathematics. Yes! that’s right, there are prob-lems in mathematics that are impossible to solve. It’s not amatter of being clever (you might be a Gauss or a Ramanu-jan) or how hard you work, but given certain constraintsthese problems are impossible to solve. What’s more, wecan prove that this is so! The present article will focus onproving the following three problems of antiquity:

1. The duplication of a cube, or the problem of construct-ing a cube having twice the volume of a given cube.

2. The trisection of an angle, or the problem of dividinga given arbitrary angle into three parts.

3. Squaring a circle, or the problem of constructing asquare whose area is equal to that of a given circle.

are impossible to solve; under the constraint that we useonly a straight edge and compass. The epilogue will discusssome other impossibilities in mathematics. Before I definethe problems more clearly and give an outline of their proofs,I would like to give a brief historical introduction to them.

All three problems discussed here have their origins in Greekgeometry around the 5th century BC. These problems have

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26 RESONANCE March 1999

Box 1.

The following is the translation (taken from, [5]) of an ancient document supposedly written by

Eratosthenes to King Ptolemy III about the year 240 B.C. :

To King Ptolemy, Eratosthenes sends greetings. It is said that one of the ancient tragic poets

represented Minos as preparing a tomb for Glaucos and as declaring, when he learnt that it was

hundred feet each way: “Small indeed is the tomb thou hast chosen for a royal burial. Let it be double

[in volume]. And thou shalt not miss that fair form if thou quickly doublest each side of the tomb.”

But he was wrong. For when the sides are doubled, the surface [area] becomes four times as great

and the volume eight times. It became a subject of enquiry among geometers in what manner one

might double the given volume without changing shape. And this problem was called the duplication

of the cube, for given a cube they sought to double it ... .

become famous because they are impossible to solve withinthe constraints imposed. The resolution of these problemshad to wait for nearly 2200 years, for the invention of ab-stract algebra. In the meantime many great mathemati-cians, amateurs and mathematical cranks have had a goat them without much success. To this day mathematicsdepartments around the world receive letters from people(unfamiliar with the fact that these problems are impossibleto solve), claiming to have solved them! Like many otherlong-standing unsolved problems, these too have inspiredthe discovery and creation of a lot of mathematics such asthat of the conic sections, many cubic and quartic curvesand later on algebraic numbers and group theory.

Problem 1 might well have originated in the words of anobscure ancient Greek poet representing the mythical KingMinos as being dissatisfied with the size of the tomb erectedto his son Glaucos (see Box 1). Minos had ordered the tombbe doubled in size. The poet proceeds to say that Minosclaimed this could be accomplished by doubling the lengthof each side. Obviously neither Minos nor the poet passedhigh school algebra! Attempts to correct the mathematicsof the poet led geometers to study the problem of doublingthe cube. There is another story related to this problem,where it is said, that the Delians were instructed by theiroracle to double the size of Apollo’s cubical altar, so as toget rid of a certain pestilence. This is why this problem isalso referred to as the Delian Problem.

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The origins of Problem 2 are most likely in the Greeks’ at-tempt to construct regular polygons. In the constructionof a regular nine-sided polygon, one needs to trisect an an-gle of sixty degrees. In fact this problem has attracted alarge number of so called ‘angle trisectors’, who to this dayattempt to trisect an angle.

The history of Problem 3 is tied up with that of findingthe area of a circle, and information about this is foundin the Rhind (or Ahmes) Papyrus (one of the best knownmathematical manuscripts, purchased in 1858 in Egypt byA Henry Rhind and later acquired by the British Museum).Here it states that the area of a circle is equal to that of asquare whose side is the diameter diminished by one ninth.This is of course based on the crude approximation of π tobe 3.1604 . . .. Since these problems fascinated many mathe-maticians and amateurs, the problem solvers themselves ac-quired names! For example, in Greek times ‘circle squarers’were referred to by the term ‘tetragonidzein’, which meansto occupy oneself with the quadrature.

As you can see these problems come with a very rich history,and have many fascinating anecdotes connected with them,so interested readers can pursue them in standard books onthe history of mathematics [1]. The problems were finallysettled in the nineteenth century. P Wantzel showed in 1837that Problems 1 and 2 are impossible to solve and in 1882,Problem 3 was shown to be impossible to solve by F Linde-mann.

2. Ruler and Compass Constructions

Anyone who has done high school geometry is familiar withconstructions using only a ruler and compass. You may wellremember constructions like bisecting an angle, drawing theperpendicular bisector of a given line segment and so on.Apart from some of these easy and obvious constructionsit’s amazing what one can do with these tools. For example,one can construct a square having the same area as that ofa lune, i.e., a plane figure bounded by two circular arcs! Foran excellent account and proof of this fact, I would urge youto read [2]. It is not clear why the Greeks imposed these

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constraints on themselves. Perhaps it was due to the no-tion of Plato that the only ‘perfect’ figures were the straightline and the circle, or it was an intellectual game, with veryprecise rules. It must be noted here that the Greeks them-selves did not always restrict themselves to these tools anddid not hesitate to use other instruments. We will discussthis briefly in the epilogue.

So what then are the precise rules involved in constructionsusing a ruler and a compass? Remember by ruler we meana ‘straight edge’; one cannot use it to make measurementsor transfer lengths, only draw straight lines. If you noticeall constructions using ruler and compass start with a givenset S = {q1, q2, . . . , qn} of points and we use the ruler andcompass in the following two ways:

Operation 1(ruler) Through any two points in S drawa straight line.

Operation 2 (compass) Draw a circle, whose centre isa point of S, and whose radius is equal to the distancebetween some pair of points in S.

We note here that the Greeks preferred a more restrictedcompass than the one we allow. Their compass was a ‘col-lapsing’ one which when lifted from the paper collapsed, notallowing one to maintain a fixed radius. Their compass al-lowed one to place the compass point at one point of S andthe pencil tip at another point of S and draw a circle or anarc. It can be shown that every construction that can beperformed by a noncollapsing compass such as ours can beperformed by a collapsing one ( see Exercise 5.11, pp.59–60of [3]).

While doing ruler and compass constructions you might haverealized that all constructions involve the intersections of aline with another line, a line with a circle or a circle withanother circle. How do we formalize these ideas? We startby making two definitions.

Definition 2.1. The points of intersection of any two dis-tinct lines or circles using operations 1 or 2, are said to be

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Figure 1. Bisecting a line

segment using ruler and

compass.

Figure 2. Doubling the cube.

p1

p3

p1

p5 p2

p4

p1 d p2

pi pjd

constructible in one step from S. A point p is said to be con-structible from S if there is a finite sequence p1, p2, . . . , pn =p of points such that for each i = 1, . . . , n the point pi isconstructible in one step from the set S ∪{p1, p2, . . . , pi−1}.

Definition 2.2. Let γ be a real number with absolutevalue |γ|. Then γ is said to be constructible if we havetwo constructible points pi,pj whose distance apart is |γ|units, the starting set S for the points pi, pj being the setof points{p0, p1} whose distance apart is 1 unit.

Let us now use a simple construction as an example to il-lustrate our formal set-up. Suppose we are given two pointsp1, p2. Let S = {p1, p2}. (Figure 1)

1. Draw the line p1p2 (operation 1);2. Draw the circle centre p1 of radius p1p2 (operation 2);3. Draw the circle centre p2 of radius p1p2 (operation 2);4. Let p3 and p4 be the points of intersection of these circles;5. Draw the line p3p4 (operation 1);6. Let p5 be the intersection of the lines p1p2 and p3p4.

Then the sequence of points p3, p4, p5 defines a constructionof the midpoint of the line p1p2.

We are now ready to see what all this formal stuff looks likewhen we consider the three problems of antiquity.

Doubling the Cube. We are given a cube and asked toconstruct a cube with double its volume. Given two pointsp1 and p2 whose distance apart is equal to one side of theoriginal cube, our task is then to construct two points pi andpj whose distance apart is exactly 3

√2 times the distance

between p1 and p2(Figure 2).

Trisecting an Angle. We are given points p1, p2 and p3which determine the angle and are asked to construct a pointpi using only operations 1 and 2, such that angle p1p2pi isexactly one third that of the angle p1p2p3. (Figure 3)

Squaring the Circle. Given two points p1 and p2, whosedistance apart is the radius of the circle, construct two pointspi and pj whose distance apart is

√π times the distance

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Figure 3. Trisecting an

angle.

Figure 4. Squaring the

circle.

between p1 and p2.(Figure 4)

3. How Does Algebra Come into the Picture?

We will now show how we convert these problems into thelanguage of algebra and can also sympathize with why theGreeks had so much trouble with them.

As mentioned earlier, not only was the word algebra ‘Greek’to the Greeks, they also did not realise that these problemswere impossible to solve. To convert these problems intoalgebra, the reader will now have to be patient as we buildup the required background. For reasons of brevity we can’tgive a comprehensive background and would recommend anystandard text in algebra say [4]. Moreover, we will not givecomplete proofs, but will only give a sketch. Many of thedefinitions and statements of theorems are taken directlyfrom the books referred below. The algebra that is usedis based on the work of many great algebraists, like Abel,Galois, Lagrange and more recently Artin. The materialis also the beginnings of the beautiful area of mathematicsknown as Galois theory.

Our aim is to characterize constructible numbers algebraically.In other words convert the geometric idea of a constructiblenumber as defined in Definition 2.2 to that of an algebraicone and understand it in the language of algebra. Actually,looked at correctly algebra comes in immediately!

Theorem 3.1. The set of constructible numbers is a subfieldof R. Furthermore every rational number is constructibleand if γ is a positive constructible number then so is

√γ.

Proof: The first part follows from the well known fact thatif α and β are constructible, then so are α+β, αβ and α/β.If we are given a unit length, then we can construct all therationals. Once a given length γ > 0 is constructed, then itis fairly easy to construct (using ruler and compass)

√γ.

2

By repeated applications of the above theorem, for example,

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one can prove that the number

4√

13 +4

3

√√

6 +

√1 + 2

√7

is constructible.

The theorem tells us that the class of constructible numbersis large. In fact our aim is to find the ones that are notconstructible. More specifically, if we are able to double thecube then 3

√2 has to be constructible, and if we can square

a circle√π would be constructible. Before we proceed we

need to understand the notion of a number being algebraicover a field. The set of constructible numbers is a subset ofthe set of algebraic numbers, so information about the latterwill be useful to understand the former.

Definition 3.1. A number α ∈ C is said to be algebraicover a field F contained in C if there exists a non-zero poly-nomial f(x) ∈ F [X] such that α is a zero of f(x). If no suchpolynomial exists then α is said to be transcendental overF .

In other words, if α is algebraic over F there exists a poly-nomial f(X) = a0 + a1X + . . . + anX

n whose coefficientsa0, a1, . . . an are all in F with at least one of these coefficientsbeing non-zero and f(α) = 0. For example

√2 is algebraic

over Q, since it satisfies the polynomial X2 − 2 = 0.

Recall that a polynomial is monic if its leading coefficient is1.

Definition 3.2. Let α be algebraic over F ⊆ C. Thenthere exists a unique monic polynomial f(X) over F of leastdegree such that f(α) = 0 and it is called the irreduciblepolynomial of α over F , and is denoted irr (α, F ).

For example, X2−2 is the irreducible polynomial of√

2 overQ. We denote the degree of irr(α, F ) by deg(α, F ).

Definition 3.3. Let F be a field. A non constant polyno-mial f(X) ∈ F [X] is said to be irreducible over F if it hasno factors in F [X].

The following theorem connects the notion of the irreducible

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polynomial of a number and that of an irreducible polyno-mial over a field. Moreover given α ∈ C algebraic over F , itgives us a way of computing irr(α, F ).

Theorem 3.2. Let F be a subfield of C and let α ∈ C bealgebraic over F . The following conditions on a polynomialf(X) ∈ F [X] are equivalent:(i) f(X) = irr(α, F ).(ii) f(α) = 0 and f(X) is monic irreducible over F .

In general it is not easy to find out if a given polynomialis irreducible or not. One powerful criterion is Eisenstein’scriterion (see Box 2 for details).

Every time we have a field F embedded in another fieldK, i.e.,F ⊆ K, we obtain a vector space K over F . Thedimension of K over F is denoted by [K : F ]. The field Kis called the extension field of F . The next theorem tells uswhen a given α ∈ K is algebraic over F .

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Theorem 3.3. Let F be a subfield of a field K with doubt[K : F ] = n where n is a positive finite integer. Then everynumber α ∈ K is algebraic over F and deg(α, F ) ≤ n.

One can extend a given field F by ‘attaching’ a single ele-ment α (not in F , but in K) to F .

Definition 3.4. Let F be a subfield of C and let α be alge-braic over F with deg(α, F ) = n. The extension of F by α isthe set F (α) ⊆ C where F (α) = {b0+b1α+ . . .+bn−1α

n−1 :b0, b1, . . . , bn−1 ∈ F}.

Theorem 3.4. Let F be a subfield of C and let α ∈ C bealgebraic over F with deg(α, F ) = n. Then(i) F (α) is a finite dimensional vector space over F withbasis {1, α, α2, . . . , αn−1}. The dimension n of F (α) over Fas a vector space is denoted by [F (α) : F ].(ii) F (α) is a field.

For example the set Q (√

2) = {a+ b√

2 : a, b ∈Q} is a fieldextension of Q of degree 2. Actually for our purpose we willbe mainly interested in extensions of fields of degree 2, theso called quadratic extensions.

Now whenever we perform a construction using ruler andcompass it is not enough to have an extension of the rationalsby a single element; in fact we have to iterate the processto obtain a sequence of extensions. For example to find asuitable extension of Q where

4√

13 +4

3

√√

6 +

√1 + 2

√7

lives, we need the following sequence of field extensions :

F0 = Q, F1 = F0(√

7), F2 = F1(√

1 + 2√

7), F3 = F2(√

6),

F4 = F3(

√6 +

√1 + 2

√7), F5 = F4(

√13), F6 = F5(

√√13).

We now discuss briefly what happens if we have a tower ofdistinct subfields of C, E ⊆ F ⊆ K. We notice that thereare three vector spaces that are involved here. F over E, Kover F and K over E.

It is natural to ask if, say, the dimensions of the vector

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Figure 5. Tower of subfields

of C.

spaces F over E and K over F are finite, what is the di-mension of K over E ? The following theorem gives theanswer.

Theorem 3.5. Consider a tower of subfields of C, E ⊆F ⊆ K. If the vector spaces F over E and K over F havefinite dimensions, then so does K over E and [K : E] =[K : F ][F : E].(Figure 5)

4. The Proofs of the Impossibilities at Last !

We are now almost ready to prove the impossibilities of thethree constructions – we just need the following theorem thatdoes the trick! The background material from the previoussection was necessary and hopefully sufficient to understandthis theorem. We have noticed from the example above thathaving a sequence of quadratic extensions produces a con-structible number. The question then is : ‘is there any otherway to produce constructible numbers ?’. The answer is ‘no’ !

Theorem 4.1.The following are equivalent :

(i) The number γ is constructible.

(ii) There exists a sequence of positive real numbers γ1, γ2, . . . ,γn such thatγ1 ∈ F1 where F1 = Q,γ2 ∈ F2 where F2 = F1(

√γ1),

.

.

.γn ∈ Fn where Fn = Fn−1(

√γn−1),

and, finally,γ ∈ Fn+1 where Fn+1 = Fn(

√γn).

Proof: (ii) implies (i) follows from Theorem 3.1 and induc-tion on n. The proof that (i) implies (ii) rests on the factthat to produce a new constructible point one needs the in-tersection of either a line with a line, or a line with a circleor that of a circle with a circle. In each of the above cases atworst one may need extra square roots to describe the newpoint constructed, in other words a quadratic extension. Weillustrate with an example below. 2

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Figure 6.

Example 4.2. Start with two points P0 = (0, 0) and P1 =(1, 0) at a distance 1 unit apart. Draw circles with centres atP0 and P1 with radius P0P1 then the points of intersection P2

and P3 can be computed to be (1/2,√

3/2) and (1/2,−√

3/2)respectively involving nothing more than

√3. (Figure 6).

Corollary 4.3. If γ is constructible, then γ is algebraic anddeg(γ,Q)= 2s for some integer s ≥ 0.

The proof follows from the above Theorems 4.1 and 3.5.

Corollary 4.4. If a real number γ satisfies an irreduciblepolynomial over Q of degree n, and if n is not a power of 2,then γ is not constructible.

This is an immediate consequence of the above corollary.

We are now really ready to prove that the three great prob-lems of antiquity are impossible to solve!

Theorem 4.5. The cube cannot be duplicated using ruler-and-compass constructions.

Proof: As discussed earlier, if we could duplicate the cubethen 3

√2 would be constructible. However 3

√2 satisfies the

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monic polynomial X3− 2 = 0 over Q, which by Eisenstein’scriterion is irreducible over Q. But the degree of X3−2 = 0is 3 and by Corollary 4.4, 3

√2 is not constructible. Hence it

is impossible to duplicate the cube using ruler-and-compassconstructions.

We now show that the angle π/3 cannot be trisected bya ruler-andcompass construction. Thus there is no ruler-and-compass construction which will provide a trisection ofevery angle. This leaves the possibility of being able totrisect ‘most’ angles, with the method failing only for a finitenumber of angles. However it turns out that once we haveestablished the fact that π/3 is not constructible, it is quiteeasy to deduce that there are infinitely many angles thatcannot be trisected ( see [5], p.106).

Theorem 4.6. The angle π/3 cannot be trisected usingruler-and-compass constructions.

Proof: If the angle π/3 could be trisected, then the angle π/9can be constructed. Using a right angled triangle it is easyto see that the length cos(π/9) and hence β = 2 cos(π/9) canbe constructed. Recalling the fact that cos(π/3) = 1/2 andthe trignometric identity cos(3θ) = 4 cos3(θ) − 3 cos(θ), wesee that β satisfies the monic polynomial β3 − 3β − 1 = 0.Now the polynomial f(X) = X3 − 3X − 1 is irreducibleover Q, since the polynomial f(X + 1) = X3 + 3X2 − 3 isirreducible over Q, by Eisenstein’s criterion. But once againthe degree of f(X) = X3−3X−1 is 3 and by Corollary 4.4,β is not constructible. Hence π/3 cannot be trisected. 2

Theorem 4.7. The circle cannot be squared using ruler-and-compass constructions.

Proof: As discussed earlier if we could square the circle,then

√π would be constructible and hence by Corollary 4.3

algebraic. This implies that π is algebraic over Q, but thereis a famous Theorem of Lindemann (see [3], Chapter 6, fora proof) which asserts that π is not algebraic over Q. Henceit is impossible to square a circle. 2

Neat, isn’t it?

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Suggested Reading Suggested Reading Suggested Reading Suggested Reading Suggested Reading

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York,1964.York,1964.York,1964.York,1964.York,1964.

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Morris and Kennneth R.Morris and Kennneth R.Morris and Kennneth R.Morris and Kennneth R.Morris and Kennneth R.

Pearson, Pearson, Pearson, Pearson, Pearson, Abstract Alge-Abstract Alge-Abstract Alge-Abstract Alge-Abstract Alge-bra and Famous Impossi-bra and Famous Impossi-bra and Famous Impossi-bra and Famous Impossi-bra and Famous Impossi-bilitiesbilitiesbilitiesbilitiesbilities, Springer-Verlag,, Springer-Verlag,, Springer-Verlag,, Springer-Verlag,, Springer-Verlag,

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5.Epilogue

In the process of writing this article, I have discovered thatI have only touched the tip of the iceberg, both in terms ofgeometric constructions and in terms of mathematical im-possibilities.

As far as geometric construction goes, it has been knownsince the time of the Greeks that if one relaxed the con-dition on the use of ruler and compass alone, then thesethree problems could be solved. For example, Problem 1was solved by the Greeks using a ruler in the form of a rightangle (see p.146 of [6], for details) and using the spiral ofArchimedes one can square a circle and trisect an angle (seepp. 95–96, of [1] for details). There are even variations withregard to the traditional ruler and compass as tools. If weagree that a line can be drawn if we are given two points onit, then it can be shown that all the ruler and compass con-structions can be constructed using a compass alone (firstdiscovered by Georg Mohr in 1672, and later discovered byand attributed to Lorenzo Mascheroni). It is also knownthat all ruler and compass constructions can be performedby using a two-edged ruler, whose edges are either parallelor meet at a point.

One of the reasons Gauss is revered so much is that at the ageof 19, he demonstrated how to construct a regular polygonof 17 sides, using straight edge and compass! (See box itemby Renuka Ravindran in Resonance, Vol.2, No.6, p.62) Thisafter nearly 2000 years of stagnation as far as constructionof regular polygons was concerned. Till then the Greekshad constructed regular polygons with 3 sides, 5 sides and ingeneral regular polygons with 2m sides, for m ≥ 2. They alsoknew that if they could construct polygons with r sides ands sides, then they could construct one with side rs. Gaussshowed further, that a regular polygon of n sides can beconstructed whenever n is prime and is of the form n = 2k+1for some integer k ≥ 0. These primes are actually calledFermat primes. You can check that 17 is one such. Sofrom our discussion we see that we can construct a regularpolygon of n sides provided it is of the form n = 2i+2 orn = 2ip1p2 . . . pj , where i ≥ 0 and pi, 1 ≤ i ≤ j, are Fermat

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primes. The question that naturally comes up is, what if n isnot of this form. Well! Pierre Wintzel in 1837, proved thatif n is not of this form, then it is impossible to construct aregular polygon with n sides.

The next famous impossibility comes from the area of solvingpolynomial equations with real coefficients. We all know

the quadratic formula x = −b±√b2−4ac2a , where a 6= 0, is the

solution to the equation ax2 + bx + c = 0. If you noticethe solution involves only the coefficients of the equationand square roots. Such a solution is referred to by the termsolutions by radicals. That is, the solutions are obtainedusing field operations and extraction of roots alone. Onceagain it is natural to ask: is there a solution by radicalsfor a polynomial of arbitrary degree? The answer is yes ifthe degree is 3 or 4. But for degree 5, Niels Abel showedin 1824 that it is impossible to solve an arbitrary fifthdegree equation by radicals alone. Just as in the case of thetrisection of an angle, one should note that there may besome fifth degree equation, which does have a solution byradicals, but there is no general formula which takes careof all fifth degree equations. Actually the punchline to thisstory was given by Evariste Galois, who showed that in factfor an integer n ≥ 5 it is impossible to solve an arbitraryequation of degree n by radicals. The tools used by Abeland Galois, form what is today known as Galois theory.

The last example we choose is from the area of calculus.We all know that no matter how hard we try we are unableto find

∫ex

2dx in ‘closed’ form, i.e., in terms of functions

usually defined in introductory calculus courses. Joseph Li-ouville proved that it is impossible to find

∫ex

2dx in closed

form. In [7] a more modern proof, based on the ideas of Ga-lois theory has been provided.

So next time you come across the phrase ‘nothing is impos-sible’ don’t believe it!

whoever said nothing is impossible? - cfl...in the history of mathematics there have been periodical...

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