s. ramanan distinguished professor (retd.) tata institute...

Negative results are often more interesting!

S. RamananDistinguished Professor (Retd.)

Tata Institute of Fundamental Research

27.11.11

S. Ramanan Distinguished Professor (Retd.) Tata Institute of Fundamental Research ()Negative results are often more interesting! 27.11.11 1 / 34

Caveats

The title is NOT to be taken too seriously.


Caveats


After all any statement can be presented negatively by saying thatits negation is not true!

So you may say it is a non-sequitur.


Caveats


After all any statement can be presented negatively by saying thatits negation is not true!

So you may say it is a non-sequitur.

However, you will know instinctively what I mean by negativeresults. If not, hopefully it will become clear soon.


Man created Integers!

Consider the two statements below.

There is a solution in natural numbers of the equation 2 + x = 5.

There is no solution in natural numbers of the equation 5 + x = 2.






You can yourself decide which one is a negative statementand which one is more interesting!






You can yourself decide which one is a negative statementand which one is more interesting!

The second one led to the creation of negative integers (no wonder!)and we inherited the wonderful (man-made) ring of integers Z.


Euclid was irrational!

In a similar vein, I may consider the following two statements.

There is a solution in integers of the equation 2x = 4.

There is no solution in integers of the equation 4x = 2.


Euclid was irrational!

In a similar vein, I may consider the following two statements.

There is a solution in integers of the equation 2x = 4.

There is no solution in integers of the equation 4x = 2.

or the following two:

There is a solution in rational numbers of the equation x2 = 4.

There is no solution in rational numbers of the equation x2 = 3.

Exercise: Make two similar statements, one positive, of noconsequence, and another, negative, of great significance, leading tothe birth of the complex number field.


Reduction of quadratic to quadratic!

The equations x2 = 4 and x2 = 3 mentioned above are both quadratic.We have known for quite some time how to solve quadratic equations.




The solutions of the equation

ax2 + bx + c = 0 (1)

are given by

x =−b ±

√b2 − 4ac

2a.

Theoretically speaking, taking a square root is only solving a quadraticequation of the form x2 = λ.





ax2 + bx + c = 0 (1)

are given by

x =−b ±

√b2 − 4ac

2a.

Theoretically speaking, taking a square root is only solving a quadraticequation of the form x2 = λ.Q. So what did we achieve ?





ax2 + bx + c = 0 (1)

are given by

x =−b ±

√b2 − 4ac

2a.

Theoretically speaking, taking a square root is only solving a quadraticequation of the form x2 = λ.Q. So what did we achieve ?

A. We reduced the more complicated quadratic equation in (1) to asimpler quadratic equation of the form x2 = λ.


Reduction of quadratic to quadratic !

We showed that the solution of a general quadratic equation can bereduced to the solution of an associated “simplest” quadratic equation.

Can we do something similar with cubic, quartic or equations of higherdegree?


First Step – A General Reduction

It is easy to see that finding the roots of an equation of the form

anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0, an 6= 0, (2)

can be ‘reduced’ to finding the roots of a similar equation in whichan = 1 and an−1 = 0.






Indeed,

Firstly, one can divide the equation by an (since an 6= 0) and so wecan assume that an = 1.






Indeed,


Secondly, we may replace x by y = x − an−1/n and thereby killthe “an−1 term”.






Indeed,


Secondly, we may replace x by y = x − an−1/n and thereby killthe “an−1 term”.

Then α is a solution to the reduced equation (with an = 1 andan−1 = 0) in y if and only if α+ an−1/n is a solution of theequation (2)


Cubic to a Simpler Cubic?

We have described how finding the roots of an equation of the form

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0, an 6= 0



Cubic to a Simpler Cubic?

We have described how finding the roots of an equation of the form

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0, an 6= 0


When n = 3, this procedure reduces finding roots of the cubic equation

a3x3 + a2x2 + a1x + a0 = 0

to finding roots of a simpler cubic equation of the form

y3 + ay + b = 0,

for suitable a and b.


Cubic to a Simpler Cubic ? (contd.)

Continuing, let us take a cubic equation of the (reduced) simpler form

x3 + ax + b = 0. (3)

We will describe how to reduce it yet further. First, find p and q suchthat

a = −3pq and b = −(p3 + q3).

Using the identity

(p + q)3 = p3 + 3p2q + 3pq2 + q3

we can then conclude that x = p + q is a solution of the equation (3).


Cubic to a Simpler Cubic (contd.) ?

Thus, we have reduced the problem of finding roots to

x3 + ax + b = 0. (4)

to the problem of finding p and q that satisfy

a = −3pq and b = −(p3 + q3). (5)

Eliminating q from these two equations, we see that

(a/3p)3 = −q3 = (b + p3),

or, equivalently, that

27(p6) + (27b)p3 − a3 = 0. (6)

This is a quadratic equation in p3, which we know how to solve!And given p3, we also know q3 = −b − p3.


Cubic to a Simpler Cubic (contd.) ?

Thus, we have reduced the problem of finding roots to

x3 + ax + b = 0. (4)

to the problem of finding p and q that satisfy

a = −3pq and b = −(p3 + q3). (5)

Eliminating q from these two equations, we see that

(a/3p)3 = −q3 = (b + p3),

or, equivalently, that

27(p6) + (27b)p3 − a3 = 0. (6)

This is a quadratic equation in p3, which we know how to solve!And given p3, we also know q3 = −b − p3.In summary, we first solve the quadratic equation (6) for p3 and thensolve two cubic equations of the “simplest” form

p = 3√

p3, and q = 3√

−b − p3.

to obtain the solution x = p + q to (4).S. Ramanan Distinguished Professor (Retd.) Tata Institute of Fundamental Research ()Negative results are often more interesting! 27.11.11 10 / 34

Quartic Equations ... and so on

Q. What about solutions to quartic equations ?

a4x4 + a3x3 + a2x2 + a1x1 + a0 = 0, a4 6= 0.




a4x4 + a3x3 + a2x2 + a1x1 + a0 = 0, a4 6= 0.

A. A similar solution exists for the quartic equation as well.

One first solves a quadratic equation as above, then a cubic equation,and two more quadratic equations, etc.




a4x4 + a3x3 + a2x2 + a1x1 + a0 = 0, a4 6= 0.

A. A similar solution exists for the quartic equation as well.

One first solves a quadratic equation as above, then a cubic equation,and two more quadratic equations, etc.

Q. What about quintic equations ... and so on .... ?


Here is where negativism enters ...

Paolo Ruffini (1765–1822) gave a proof that it is impossible toproduce an algorithm of this kind for solving any fifth degree equation.Apparently there were mistakes in Ruffini’s arguments and NielsHenrik Abel (1802–1829) gave a definitive proof.


Galois wins the duel!

Évariste Galois (1811–1832) generalized this further ...


Galois wins the duel!

Évariste Galois (1811–1832) generalized this further ...

Galois may be better known for the foolish duel in which he lost his lifethan for his negative result.


Galois’ Result

Galois provedHowever ingenious you are, you cannot find a similar reduction in the

case of all equations of degrees more than 4 ...


Galois’ Result

Galois provedHowever ingenious you are, you cannot find a similar reduction in the

case of all equations of degrees more than 4 ...

• The first difficulty is to make precise what is meant by ‘similar’.S. Ramanan Distinguished Professor (Retd.) Tata Institute of Fundamental Research ()Negative results are often more interesting! 27.11.11 14 / 34

Radical solutions

• The first difficulty is to make precise what is meant by ‘similar’.By this, we mean: start with an equation

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0

and use only the following operations to find the solution.

Let Q be the space of rational numbers (ratios of integers withdenominator non-zero) .

Solve an equation of the form xn = a for some a ∈ Q, integer n.Say α is a solution.


Radical solutions


anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0




Solve another equation of the form xm = f (α) where f is apolynomial with rational coefficients, m is an integer. Say β is thesolution. Solve yet another equation of the form x r = g(β), whereg is a polynomial with rational coefficients. Say γ is the solution ...


Radical solutions


anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0




Solve another equation of the form xm = f (α) where f is apolynomial with rational coefficients, m is an integer. Say β is thesolution. Solve yet another equation of the form x r = g(β), whereg is a polynomial with rational coefficients. Say γ is the solution ...

... go on like this till you are blue in the face!

A solution obtained in this manner is called ‘a solution by radicals’S. Ramanan Distinguished Professor (Retd.) Tata Institute of Fundamental Research ()Negative results are often more interesting! 27.11.11 15 / 34

Galois Leaves His Stamp

Galois’s result restated preciselyFor any integer n > 4, there exist equations of degree n of the form

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0

that do not have a “solution by radicals”


Galois Leaves His Stamp

Galois’s result restated preciselyFor any integer n > 4, there exist equations of degree n of the form

anxn + an−1xn−1 + an−2xn−2 + . . . + a0 = 0

that do not have a “solution by radicals”

• Examples of such equations are not rare or complicated. Theequation

x5 − x − 1 = 0

cannot be “solved by radicals”!S. Ramanan Distinguished Professor (Retd.) Tata Institute of Fundamental Research ()Negative results are often more interesting! 27.11.11 16 / 34

Galois’ Negative Result Gives Birth to Group Theory

Is it so important to solve an equation by radicals?

Can’t we instead use algorithms to approximate square roots,cube roots, · · · for example, on the computer?





Q. What is so great about Galois’ result?






A. The methods are sometimes more important than the result.






A. The methods are sometimes more important than the result.

The key concept used in Galois’ proof was that of a group.

Galois used a certain subgroup of the group of permutations in hisproof.Recall that a permutation σ is a one-to-one mapping from a finiteset {1,2,3, . . . ,n} to itself.

Later abstract groups were defined. Today, the concept of a groupis all pervasive in science.


Symmetry

Symmetry is the notion of invariance under a group – and ispervasive in arts and science.

Translation and Rotational Symmetry


Groups and Symmetries

In elementary dynamics, we have equations of motion, say of aparticle. These are invariant under translations




The equations of “rigid body dynamics” are also invariant underrotations.

The set of all such motions form a group, called the specialEuclidean group.

One says that the special Euclidean group is the group ofsymmetries that leave the motion invariant.




The equations of “rigid body dynamics” are also invariant underrotations.

The set of all such motions form a group, called the specialEuclidean group.

One says that the special Euclidean group is the group ofsymmetries that leave the motion invariant.

There are physical theories which are invariant under other groups(such as, for example, the Lorentz group associated withEinstein’s theory of special relativity.


Back to Galois and his GroupGalois considered the symmetry of the roots of the equation

anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0



anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0

For example, the four roots α, β, γ, δ of the equationx4 + x3 + x2 + 1 = 0 are of the form (with θ = e2πi/5 equal to the5-th root of unity)

α = θ, β = θ2, γ = θ3, δ = θ4.



anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0


α = θ, β = θ2, γ = θ3, δ = θ4.

The transposition t of the first two roots:[t(α) = β, t(β) = α, t(γ) = γ, t(δ) = δ] does not respect therelations among them: e.g.,

α2 = β, but δ = (t(α))2 6= t(β) = α.



anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0


α = θ, β = θ2, γ = θ3, δ = θ4.

The transposition t of the first two roots:[t(α) = β, t(β) = α, t(γ) = γ, t(δ) = δ] does not respect therelations among them: e.g.,

α2 = β, but δ = (t(α))2 6= t(β) = α.

However, the cyclic permutation α 7→ β, β 7→ δ, δ 7→ γ, γ 7→ αpreserves all the relations.


Galois and his Group: A Glimpse into Galois’ Proof

Galois considered the symmetry of the roots of the equation

anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0




anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0

For example, as we just discussed, the (cyclic) permutation groupon 4 symbols given by α 7→ β, β 7→ δ, δ 7→ γ, γ 7→ α is a symmetrygroup (or the Galois group) of the equation

x4 + x3 + x2 + 1 = 0 (7)




anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0


x4 + x3 + x2 + 1 = 0 (7)

The main point is that if a polynomial equation can be solved byradicals, the group of permutations of the equation would be aproper subgroup, as in the case of equation (7).




anxn + an−1xn−1 + an−2xn−2 + . . .+ a0 = 0


x4 + x3 + x2 + 1 = 0 (7)

The main point is that if a polynomial equation can be solved byradicals, the group of permutations of the equation would be aproper subgroup, as in the case of equation (7).On the other hand, the Galois group of the equation

x5 − x − 1 = 0

is the full permutation group and hence cannot be solved byradicals.


Pythagoras: Equations with More Variables

Now consider the quadratic equation in three variables:

x2 + y2 = z2


Pythagoras: Equations with More Variables

Now consider the quadratic equation in three variables:

x2 + y2 = z2

This theorem is of interest togeometers, who think of any solution (x , y , z) as the sides of aright angled triangleas well as algebraists and number theorists, who would like to finda solution in integers for this equation.(Notice that we are not interested in (trivial) solutions such as(x ,0, x) or (0, y , y).


A positive solution

x2 + y2 = z2

Using the fact that x2 = z2 − y2 = (z + y)(z − y) we can easily find allsolutions as follows.


A positive solution

x2 + y2 = z2


Take any two integers p,q, both even or both odd. Then takex = p2 − q2, y = 2pq, z = p2 + q2.


A positive solution

x2 + y2 = z2



It is easy to check that

x2 + y2 = (p2 − q2)2 + (2pq)2 = (p2 + q2)2 = z2

.


A positive solution

x2 + y2 = z2



It is easy to check that

x2 + y2 = (p2 − q2)2 + (2pq)2 = (p2 + q2)2 = z2

.

Every solution is obtained this way!


Diophantus

Looking for solutions in integers for equations with integral coefficientswas apparently first studied by the Greek, Diophantus.


Sulva Sutras

But the ancient Hindus also were interested in such problems. Duringthe construction of the Vedis they needed to be right about rightangles! The measuring was done by a rope with knots (called Sulva)and the theory was called sulva sootra.


Fermat’s Last Theorem and its Lost Proof

Fermat asked the question whether there are similar (non-trivial)solutions for equations of the form

x3 + y3 = z3.




x3 + y3 = z3.

He proved that there were none!




x3 + y3 = z3.

He proved that there were none!

History was made when he asked: Is it true that for every n > 3

xn + yn = zn

also has no (non-trivial) solutions?

He wrote on the margin of a book that there were none and thathe had a nice proof of this fact. He apologized that the margin wastoo small to accommodate a proof.You bet!


Three centuries later . . .

There were attempts to settle this "negative result" for centuries.These gave birth to whole new fields!!




Finally, in the 1990‘s, Andrew Wiles came up with a proof of thisclaim.

True to its history, his argument which was substantially correct,had a lacuna! Thankfully it was soon stitched together by him andTaylor and wrapped up.

The proof of Fermat’s Theorem involved new ideas and theories,which could not have been imagined by Fermat,

marginally or otherwise !




Finally, in the 1990‘s, Andrew Wiles came up with a proof of thisclaim.

True to its history, his argument which was substantially correct,had a lacuna! Thankfully it was soon stitched together by him andTaylor and wrapped up.

The proof of Fermat’s Theorem involved new ideas and theories,which could not have been imagined by Fermat,

marginally or otherwise !Did Fermat have a simpler method to prove it? Perhaps this century

will provide the answer, perhaps in the NEGATIVE!!!


Hilbert’s Tenth Problem

David Hilbert, one of the great mathematicians of the last century,gave a list of unsolved problems as worthy of working on in the20th century.




Hilbert’s tenth problem, roughly speaking, states:Devise an algorithm involving a finite number of operations tosolve a finite set of Diophantine equations.




Hilbert’s tenth problem, roughly speaking, states:Devise an algorithm involving a finite number of operations tosolve a finite set of Diophantine equations.

Note that solving f1 = 0, f2 = 0, . . . , fr = 0 is the same as solvingthe single equation

∑

f 2i = 0.


MRDP

The collaborative effort of four mathematicians

Matiyasevic, Robinson, Davis and Putnam

settled the problem.


MRDP

The collaborative effort of four mathematicians

Matiyasevic, Robinson, Davis and Putnam

settled the problem.

Yuri Matiyasevic (Professor at the Steklov Institute)

Julia Robinson (1919–1985)

Martin Davis (Professor Emeritus at Courant Institute)

Hilary Putnam (Professor Emeritus at Harvard University)

from 1940–1970 showed that there can be NO such algorithm!


A problem of current interest

The proof unites notions from number theory and computabilitytheory.

• There is a documentary on this topic called “Julia Robinson andHilbert’s Tenth Problem”. There are still several important openproblems related to extensions of Hilbert’s Tenth Problem.

Julia Robinson was the first woman to be elected President of theAmerican Mathematical Society (1983-1984)


Life as a Mathematician is Hard ...

You have to spend a lot of time lost in thought



Sometimes you’re forced to go to conferences in exotic locations!

Then you have to meet and make friends with an internationalgroup of people



Finally ....

You get paid for having flexible hours, being your own boss andsolving your favourite puzzles ...

and



Finally ....

You get paid for having flexible hours, being your own boss andsolving your favourite puzzles ...

and

THE OPPORTUNITY TO INTERACT WITH YOUNG INQUIRINGMINDS ALL ONE’S WORKING LIFE (AND EVEN PAST IT!)


Thank you for your attention!


s. ramanan distinguished professor (retd.) tata institute...

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