# What is Mathematical Argument? What is Mathematical Argument? Arash Rastergar Department of mathematical sciences Sharif University of Technology What are possible perspectives Experimental approach Philosophical approach Intuitionistic approach What are assumptions of each perspective? What do we believe about each perspective? What are possible perspectives in our point of view? Tasting the truth, enlightenments, wisdom, revelation, inspiration, mind and thought. Possible approaches to the question Humanistic: Aristotle Hierarchized: Plato Hierarchy of wisdom, soul, body

Hierarchy of several layers of darkness and light Hierarchy of several layers of light Hierarchy of several layers of existence What is our approach to the question? Hierarchy of personality, light, wisdom, spirit, heart, soul and body Possible approaches to communication with audience Intuitionistic Top-down: from theology to mathematics and then to physics: approach of Einstein Down-top: from physics to mathematics and then to theology: approach of Newton Limit the realm of study to inside of mathematics: which is the welcome approach in our time. What an argument could be? Aristotelian point of view: Thurston, Hirsh

Platonic point of view: Hilbert, Erdsh Leibnizian point of view Kantian point of view Fregeian point of view Hilbertian point of view Lingual point of view Social sciences point of view The information age point of view Advices to a problem solver 1) Writing neat and clean 2) Writing down the summary of arguments 3) Clarifying the logical structure 4) Drawing big and clean figures 5) Recording the process of thinking 6) Deleting irrelevant remarks and explanations 7) Writing down side results 8) Putting down the full proof after finishing the arguments 9) Notifying important steps in form of lemmas 10) Considering the mind of reader Decisions to be made 11) Where to start

12) Listing different strategies to attack the problem 13) Mathematical modeling in different frameworks 14) Using symbols or avoiding symbols 15) Deciding what not to think about 16) Organizing the process of coming to a solution 17) How to put down the proof Habits to find 18) Tasting the problem 19) Gaining personal view towards the problem 20) Talking to oneself 21) Considering all the cases 22) Checking special cases 23) Performing a few steps mentally 24) Thinking simple Personality of good problem solvers 25) Patience 26) Divergent thinking 27) Criticizing conjectures 28) Looking for equivalent formulations 29) Fluency in working with ideas and concepts

30) Looking for simpler models Intuition 31) Geometric imagination 32) Recognizing simple from difficult 33) Decomposition and reduction to simpler problems 34) Jumps of the mind 35) Estimating how much progress has been made 36) Finding the trivial propositions quickly 37) Formulating good conjectures 38) Being creative and directed in constructions 39) Understanding an idea independent of the context 40) Imagination and intuition come before arguments & computations The role of arguments in problem solving Arguments clarify the logical structure Notifying important steps in form of lemmas simplifies arguments Arguments organize the process of coming to a solution Arguments promote divergent thinking Arguments criticize conjectures

Looking for equivalent formulations help coming up with a new argument Looking for simpler models simplifies arguments Decomposition and reduction to simpler problems is performed by arguments Arguments help us estimate how much progress has been made Formulating good conjectures using the perspective given by arguments The role of arguments in making assumptions Arguments Arguments There are test assumptions generalize assumptions to wider scopes natural barriers to generalization of assumptions revealed by arguments. Sometimes one can not unite two given theories. Arguments do surgery on assumptions in order to repair implications. Surgery and repair performed by arguments could

lead to unification of assumptions. Strength and weakness of assumptions are assessed by fluency and naturality of implications. The role of arguments in development of theories Arguments test theories Arguments generalize theories to wider scopes. By generalization one can unite the realms of two theories. Recognition of relations between assumptions via arguments usually leads to unification of theories. Recognition of relations between theories via arquments forms a paradigm. One is interested to find relations between two theories for the further development of mathematics. Search for the truth Truth in mathematics is understood by

analogies. which are revealed by arguments. In mathematics one compares two or three theories and find dictionaries between them in order to look for background truth. In mathematics concepts are just a model of the truth. So arguments discuss relations between the models of truth. Psychology of problem solving Problem Problem solvers solvers race against time. take advantage of a clean mind. Problem solvers manage the process of thought. Problem solvers manage the language. Problem solvers manage the feelings and

mathematical behavior. Problem solvers develop intuition. All these aspects show up in the nature of arguments Psychology of theorization Theorizers Theorizers Theorizers Theorizers race against truth. take advantage of a clear wisdom. manage wisdom. manage the process of formulation. Theorizers manage the personality and mathematical qualities. Theorizers develop Intuition. All these aspects show up in the nature of arguments Dual pairs of arguments

Geometric versus algebraic arguments Continuous versus discrete arguments Local versus global arguments Categorical versus interstructural arguments Rigid versus deforming arguments Pairs of formulations Lagrangian Differential versus Hamiltonian mechanics forms versus vector formulation of electromagnetism Differentials versus finite differences Infinitesimals versus limits Hopf-algebra versus geometric formulation of affine algebraic groups Geometric versus arithmetic formulation of algebraic curves Paradigms back up arguments Paradigm Paradigm

integral The role of language Language is the ultimate tool for forming paradigms. question: Could one conceptually relate two theories, without joining their

lingual formulation? Answer: Even physicists do not believe that! This means that people trust language but not intuition. Because they can pretend to speak but they cant even pretend to communicate mental images directly. Therefore arguments are governed by language. Marriage of arguments Relativity-quantum marriage give Real-imaginary marriage give rise to field theory rise to complex numbers Geometric versus arithmetic thinking give rise to algebraic thinking Continuous versus discrete thinking give rise to fundamental theorem of calculus Local versus global thinking give rise to

superposition Boundary versus differential give rise to Poincare duality Fruit of marriage Role of father argument: Provides ideas and intuitions. Management of relations with other arguments. Provides the global structure. Determines how to generalize. Furnishes the soul. Role of mother argument: Provides appropriate formulation and language.

Management of internal relations between subarguments. Provides the local structures. Determines how to solve problems. Furnishes the body. Fundamental theorem of algebra Peter Rothe, in Arithmetica Philosophica 1608: A polynomial equation of degree n (with real coefficients) may have n solutions. Albert Girard, in L'invention nouvelle en l'Algbre 1629: A polynomial equation of degree n has n solutions unless the equation is incomplete, By incomplete he meant that no coefficient is equal to 0. He actually believes that his assertion is always true; For instance, he shows that the equation x44 = 4x 3, although incomplete, has four solutions (counting multiplicities): 1 (twice), 1 + i2, and 1 i2.

In 1702 Leibniz said that no polynomial of the type x4 + a4 (with a real and distinct from 0) can be written in such a way. Later, Bernulli made the same assertion concerning the polynomial x4 4x3 + 2x2 + 4x + 4. But Bernulli got a letter from Euler in 1742 in which he was told that his polynomial happened to be equal to where is the square root of 4 + 27. Also, Euler mentioned that Attempts to prove the theorem d'Alembert in 1746, but his proof was incomplete. Among other problems, it assumed implicitly a theorem (now known as Puiseux's theorem) which would not be proved until more than a century later (and furthermore the proof

assumed the fundamental theorem of algebra). Euler (1749), de Foncenex (1759), Lagrange (1772), and Laplace (1795). These last four attempts assumed implicitly Girard's assertion; to be more precise, the existence of solutions was assumed and all that remained to be proved was that their form was a + bi for some real numbers a and b. In modern terms, Euler, de Foncenex, Lagrange, and Laplace were assuming the existence of a splitting field of the polynomial p(z). At the end of the 18th century, two new proofs were published which did not assume the existence of roots. One of them, due to James Wood 1798 and mainly algebraic, and it was totally ignored. Wood's proof had an algebraic gap. The other one was published by Gauss 1799 and it was mainly geometric, but it had a topological gap, filled by Alexander Ostrowski in 1920, as discussed in Smale 1981. A rigorous proof was published by Argand in 1806; it was here that, for the first time, the fundamental theorem of algebra was stated for polynomials with complex

coefficients, rather than just real coefficients. Gauss produced two other proofs in 1816 and another version of his original proof in 1849. Cauchy's Cours d'analyse de l'cole Royale Polytechnique 1821 contained Argand's proof, although Argand was not credited for it. Constructive proofs It was Weierstrass who raised for the first time, in the middle of the 19th century, the problem of finding a constructive proof of the fundamental theorem of algebra. He presented his solution, that amounts in modern terms to a combination of the DurandKerner method with the homotopy continuation principle, in 1891. Another proof of this kind was obtained by Hellmuth Kneser in 1940 and simplified by his son Martin Kneser in 1981. Without using countable choice, it is not possible to constructively prove the fundamental theorem of algebra for complex numbers based on the Dedekind real numbers (which are not constructively equivalent to the Cauchy real numbers without countable choice). Fred Richman proved a reformulated version of the theorem that does work.

Proofs All proofs below involve some analysis, at the very least the concept of continuity of real or complex functions. Some also use differentiable or even analytic functions. This fact has led some to remark that the Fundamental Theorem of Algebra is neither fundamental, nor a theorem of algebra. Some proofs of the theorem only prove that any nonconstant polynomial with real coefficients has some complex root. Lemma: Given a non-constant polynomial p(z) with complex coefficients, the polynomial has only real coefficients and, if z is a zero of q(z), then either z or its conjugate is a root of p(z). Growth lemma A large number of non-algebraic proofs of the theorem use the fact (sometimes called growth lemma) that Lemma: (Estimate of growth) An n-th degree polynomial function

p(z) whose dominant coefficient is 1 behaves like zn when |z| is large enough. A more precise statement is: there is some positive real number R such that: when |z| > R. Complex-analytic proofs Proof 1: Find a closed disk D of radius r centered at the origin such that |p(z)| > |p(0)| whenever |z| r. The minimum of |p(z)| on D, which must exist since D is compact, is therefore achieved at some point z00 in the interior of D, but not at any point of its boundary. The minimum modulus principle implies then that p(z00) = 0. In other words, z00 is a zero of p(z).

Another analytic proof Proof 2: Since |p(z)| > |p(0)| outside D, the minimum of | p(z)| on the whole complex plane is achieved at z00. If |p(z00)| > 0, then 1/p is a bounded holomorphic function in the entire complex plane since, for each complex number z, |1/p(z)| |1/p(z00)|. Applying Liouville's theorem, which states that a bounded entire function must be constant, this would imply that 1/p is constant and therefore that p is constant. Hence a contradiction, and thus p(z00) = 0. A variation of analytic proof Does not require the minimum modulus principle and so no use of Cauchy's integral theorem. Lemma: For the special case of a polynomial function, the minimum modulus principle can be proved directly using elementary arguments. More precisely, if we assume by contradiction that then, expanding p(z) in powers of z z00 we can write

Here, the cjj's are simply the coefficients of the polynomial and we let k be the index of the first coefficient following the constant term that is non-zero. An asymptotic proof Proof 3: For z sufficiently close to z0 this has behavior asymptotically similar to the 0 simpler polynomial q(z) = a + ckk(z z00)kk, in the sense that (as is easy to check) the function is bounded by some positive constant M in some neighborhood of z00. Therefore if we define 0 = (arg(a) + arg(ck)) / k and let , then for any 0 k sufficiently small positive number r (so that the bound M mentioned above holds), using the triangle inequality we see that When r is sufficiently close to 0 this upper bound for | p(z) | is strictly smaller than | a | , in contradiction to the definition of z00. (Geometrically, we have found an explicit direction 00 such that if one approaches z00 from

that direction one can obtain values p(z) smaller in absolute value than | p(z00) | .) Proof using the argument principle Proof Proof 4: 4: Let Let R R be be a a positive positive real real number number large large enough enough so so that that every every root root of

of p(z) p(z) has has absolute absolute value value smaller smaller than than R; R; such such a a number number must must exist exist because because every every non-constant non-constant polynomial polynomial function function of of degree degree n

n has has at at most most n n zeros. zeros. For For each each rr > > R, R, consider consider the the number number where where c(r) c(r) is is the the circle circle centered centered at

at 0 0 with with radius radius rr oriented oriented counterclockwise; counterclockwise; The The argument argument principle principle says says that that this this number number is is the the number number N N of of zeros zeros of of p(z) p(z) in

in the the open open ball centered at 0 with radius r, which, since r > R, is the number of zeros of p(z). ball centered at 0 with radius r, which, since r > R, is the number of zeros of p(z).

On On the the other other hand, hand, the the integral integral of of n/z n/z along along c(r) c(r) divided divided by by 2i 2i is is equal equal to to n. n. But But the the difference between

the two numbers is difference between the two numbers is The The numerator numerator of of the the rational rational expression expression being being integrated integrated has has degree degree at at most most n n 1 1 and

and the the degree degree of of the the denominator denominator is is n n+ + 1. 1. tends to +. But the number is also equal tends to +. But the number is also equal

Therefore, Therefore, the the number number above above tends tends to to 0 0 as as rr to to N N n n and and so so N N= = n. n. Proof 5:

Combining linear algebra with the Cauchy theorem. To establish that every complex polynomial of degree n > 0 has a zero, it suffices to show that every complex square matrix of size n > 0 has a (complex) eigenvalue. The proof of the latter statement is by contradiction. Let A be a complex square matrix of size n > 0 and let Inn be the unit matrix of the same size. Assume A has no eigenvalues. Consider the resolvent function , which which is is a a meromorphic meromorphic function function on on the the complex complex plane plane with with values values in in the

the vector space of matrices. The eigenvalues of A are precisely the poles of R(z). Since, by assumption, A has no eigenvalues, the function R(z) is an entire function and Cauchy's theorem implies that On On the the other other hand, hand, R(z) R(z) expanded expanded as as a a geometric geometric series series gives: gives: This formula is valid outside the closed disc of radius ||A|| (the operator norm of A).

Let r > ||A||. Then in which only the summand k = 0 has a nonzero integral: a contradiction. Topological proofs Proof6: Let z00 C be such that the minimum of |p(z)| on the whole complex plane is achieved at z00; It was seen at the proof which uses Liouville's theorem that such a number must exist. We can write p(z) as a polynomial in z z00: there is some natural number k and there are some complex numbers ckk, ckk ++ 11, ..., cnn such that ckk 0 and that It follows that if a is a kth th root of p(z0)/ck and if t is positive 0 k and sufficiently small, then |p(z00 + ta)| < |p(z00)|, which is impossible, since |p(z00)| is the minimum of |p| on D.

Proof 7: Suppose that p(z) has no zeros. Choose a large positive number R such that, for |z| = R, the leading term znn of p(z) dominates all other terms combined; in other words, such that |z|nn > |ann 11znn 1 1 + + a0|. 0 As z traverses the circle given by the equation |z| = R once counterclockwise, p(z), like znn, winds n times counter-clockwise around 0. At the other extreme, with |z| = 0, the curve p(z) is simply the single (nonzero) point p(0), whose winding number is clearly 0. If the loop followed by z is continuously deformed between these extremes, the path of p(z) also deforms continuously. We can explicitly write such a deformation as H(Rei i,t) = p((1 t)Rei i) where t is greater than or equal to 0 and less than or equal to 1. If one views t as time, then at time zero the curve is p(z) and at time one the curve is p(0). Clearly at every point t, p(z) cannot be zero by the original assumption, therefore during the deformation, the curve never crosses zero. Therefore the winding number of the curve around zero should never change. However, given that the winding number started as n and ended as 0, this is absurd. Therefore, p(z) has at least one zero.

Algebraic proofs These proofs only use the intermediate value theorem: every polynomial with odd degree and real coefficients has some real root; every non-negative real number has a square root. The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if R is any real-closed field, then its extension is algebraically algebraically closed. closed. Proof 8: As mentioned above, it suffices to check the statement every non-constant polynomial p(z) with real coefficients has a complex root. This statement can be proved by induction on the greatest non-negative integer k such that 2kk divides the degree n of p(z). Let a be the coefficient of znn in p(z) and let F be a splitting field of p(z) over C; in other words, the field F contains C and there are elements

elements z z11,, z z22,, ..., ..., z znn in in F F such such that that If k = 0, then n is odd, and therefore p(z) has a real root. Now, suppose that n = 2kkm (with m odd and k > 0) and that the theorem is already proved when the degree of the polynomial has the form 2kk 11m with m odd. For a real number t, define: Then the coefficients of qt(z) are symmetric polynomials in the zi's with t i real coefficients. Therefore, they can be expressed as polynomials with

real coefficients in the elementary symmetric polynomials , that is, in a11, a22, ..., (1)nnann. So qtt(z) has in fact real coefficients. Furthermore, the degree of qtt(z) is n(n 1)/2 = 2kk 11m(n 1), and m(n 1) is an odd number. Using the induction hypothesis, qt has at least one complex root; in other t words, zii + zjj + tziizjj is complex for two distinct elements i and j from {1,...,n}. Since there are more real numbers than pairs (i,j), one can find distinct real numbers t and s such that zii + zjj + tziizjj and zii + zjj + sziizjj are complex (for the same i and j). So, both zii + zjj and ziizjj are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that zii and zjj are complex numbers, since they are roots of the quadratic polynomial z22 (zii + zjj)z + ziizjj. Another algebraic proof Proof 9: We use Galois theory. It suffices to show that C has no proper finite field extension. Let K/C be a finite extension. Since the normal closure of K over R still has a finite degree over C (or R), we may assume without loss of generality that K is a normal extension of R (hence it is a

Galois extension, as every algebraic extension of a field of characteristic 0 is separable). Let G be the Galois group of this extension, and let H be a Sylow 2-group of G, so that the order of H is a power of 2, and the index of H in G is odd. By the fundamental theorem of Galois theory , there exists a subextension L of K/R such that Gal(K/L) = H. As [L:R] = [G:H] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L = R, thus [K:R] and [K:C] are powers of 2. Assuming for contradiction [K:C] > 1, the 2-group Gal(K/C) contains a subgroup of index 2, thus there exists a subextension M of C of degree 2. However, C has no extension of degree 2, because every quadratic complex polynomial has a complex root.

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