Proceedings of the International Congress of Mathematicians 1994, Volume I and II: August 3-11, 1994, Zürich, Switzerland (English and French Edition) [Hardcover]

S.D. Chatterji (Editor)

Book Description

Publication Date: November 12, 1999 | ISBN-10: 3764351535 | ISBN-13: 978-3764351533 | Edition: 1

Since the first ICM was held in Zürich in 1897, it has become the pinnacle of mathematical gatherings. It aims at giving an overview of the current state of different branches of mathematics and its applications as well as an insight into the treatment of special problems of exceptional importance. The proceedings of the ICMs have provided a rich chronology of mathematical development in all its branches and a unique documentation of contemporary research. They form an indispensable part of every mathematical library. The Proceedings of the International Congress of Mathematicians 1994, held in Zürich from August 3rd to 11th, 1994, are published in two volumes. Volume I contains an account of the organization of the Congress, the list of ordinary members, the reports on the work of the Fields Medalists and the Nevanlinna Prize Winner, the plenary one-hour addresses, and the invited addresses presented at Section Meetings 1 - 6. Volume II contains the invited address for Section Meetings 7 - 19. A complete author index is included in both volumes. '...the content of these impressive two volumes sheds a certain light on the present state of mathematical sciences and anybody doing research in mathematics should look carefully at these Proceedings. For young people beginning research, this is even more important, so these are a must for any serious mathematics library. The graphical presentation is, as always with Birkhäuser, excellent....' (Revue Roumaine de Mathematiques pures et Appliquées)

Product Details

• Hardcover: 1605 pages
• Publisher: Birkhäuser Basel; 1 edition (November 12, 1999)
• Language: English, French
• ISBN-10: 3764351535
• ISBN-13: 978-3764351533
• Product Dimensions: 9.1 x 6.6 x 1.1 inches
• Shipping Weight: 1.6 pounds (View shipping rates and policies)
• Average Customer Review: 5.0 out of 5 stars  See all reviews (1 customer review)
• Amazon Best Sellers Rank: #7,008,825 in Books (See Top 100 in Books)

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5.0 out of 5 stars Read Stillwell, January 18, 2007

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Viktor Blasjo - See all my reviews
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This review is from: Proceedings of the International Congress of Mathematicians 1994, Volume I and II: August 3-11, 1994, Zürich, Switzerland (English and French Edition) (Hardcover)

I wish to draw attention to Stillwell's article "Number Theory as a Core Mathematical Dicipline". Stillwell says: "My suggestion is that mathematics, from kindergarten onwards, should be built around a core that is interesting at all levels, capable of unlimited development, and strongly connected to all parts of mathematics. My paper attempts to show that number theory meets these requirements, and that it is natural to build modern mathematics around such a core." The rest of the paper is a wonderful display of beautiful number theory with deep connections with all major areas of mathematics. Naturally there are connections with algebra, and "this is not surprising, because most basic commutative algebra is derived from Gauss's Disquisitiones Arithmeticae via Dirichlet and Dedekind". For example, a "wonderful constellation of results comes from forming the product of elements in an abelian group in two ways", for instance Fermat's little theorem: a^(p-1)=1 mod p when gcd(a,p)=1. This is because the list a,2a,...,(p-1)a mod p must be a permutation of 1,2,...,(p-1) mod p since all elements are nonzero and unequal (since a is invertible), so a*2a*...*(p-1)a=1*2*...*(p-1) mod p, i.e. a^(p-1)*1*2*...*(p-1)=1*2*...*(p-1) mod p, i.e. a^(p-1)=1 mod p. There are also connections with complex numbers. Since (a+bi)(c+di)=(ac-bd)+(ad+bc)i, the multiplicative property o absolute value reads (a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2, which is a classical number theoretic identity apparently know to Diophantus when he said things like "65 is naturally divided into two squares in two ways, namely into 7^2+4^2 and 8^2+1^2, which is due to the fact that 65 is the product of 13 and 5, each of which is the sum of two squares". Now, 13 and 5 in turn are sums of two squares "because" they are primes of the form 4n+1, which is a theorem most easily proved by factorisation in Gaussian integers, i.e. m+ni. The theory of Gaussian integers is largely identical to that of ordinary integers since there is a notion of quotient and remainder from which we get an Euclidean algorithm and from there we prove unique factorisation etc. For ordinary quotient and remainder we need to know, when dividing by x, that multiples of x come within a distance x of any number. With the Gaussian integers we need to know that multiples of m+ni get within abs(m+ni) of any complex number, which is geometrically clear since the multiples of m+ni make a square lattice with side length abs(m+ni). So number theory is apparently also connected with geometry. An even better illustration of this is Diophantus's parametrisation of Pythagorean triples. Primitive Pythagorean triples (a,c,b) correspond to rational points on the unit circle (a/c,b/c) and these in turn correspond to lines through (-1,0) with rational slope t, so we find them by solving y=tx+t and x^2+y^2=1, which gives x=(1-t^2)/(1+t^2) and y=2t/(1+t^2). This rationalisation of the circle also pays of in calculus by showing us how to rationalise integrands like sqrt(1-x^2). Indeed, Bernoulli explicitly credited Diophantus for the substitution he used to turn the integrand for the arc length of a circle into 1/(1+t^2) from where the infinite series of pi follows by geometric series expansion and term-by-term integration. As Bernoulli also recognised, number theory also explains why integrands like sqrt(1-x^4) cannot be rationalised. Assume it can, sqrt(1-x^4)=y, square both sides and multiply up denominators to get Z^4-X^4=Y^2. the impossibility of this when X,Y,Z are integers was essentially proved by Fermat, by infinite descent, and polynomials behave sufficiently like integers for us to be able to mimic his proof in the polynomial case.