Download e-book for iPad: Proofs from THE BOOK (4th Edition) by Martin Aigner, Günter M. Ziegler

By Martin Aigner, Günter M. Ziegler

ISBN-10: 3642008550

ISBN-13: 9783642008559

Amazon: http://www.amazon.com/Proofs-THE-BOOK-Martin-Aigner/dp/3642008550

This revised and enlarged fourth version beneficial properties 5 new chapters, which deal with classical effects reminiscent of the "Fundamental Theorem of Algebra", difficulties approximately tilings, but in addition fairly fresh proofs, for instance of the Kneser conjecture in graph thought. the hot variation additionally offers extra advancements and surprises, between them a brand new evidence for "Hilbert's 3rd Problem".

From the Reviews:

"... within [this ebook] is certainly a glimpse of mathematical heaven, the place smart insights and gorgeous principles mix in marvelous and excellent methods. there's mammoth wealth inside of its pages, one gem after one other. ..., yet many [proofs] are new and tremendous proofs of classical effects. ...Aigner and Ziegler... write: "... all we provide is the examples that we have got chosen, hoping that our readers will proportion our enthusiasm approximately remarkable principles, smart insights and lovely observations." I do. ... " AMS Notices 1999

"... the extent is just about straight forward ... the proofs are excellent. ..." LMS e-newsletter 1999

Show description

Read Online or Download Proofs from THE BOOK (4th Edition) PDF

Similar graph theory books

Download PDF by Jean-Claude Fournier: Graph Theory and Applications: With Exercises and Problems

Content material: bankruptcy 1 uncomplicated innovations (pages 21–43): bankruptcy 2 bushes (pages 45–69): bankruptcy three shades (pages 71–82): bankruptcy four Directed Graphs (pages 83–96): bankruptcy five seek Algorithms (pages 97–118): bankruptcy 6 optimum Paths (pages 119–147): bankruptcy 7 Matchings (pages 149–172): bankruptcy eight Flows (pages 173–195): bankruptcy nine Euler excursions (pages 197–213): bankruptcy 10 Hamilton Cycles (pages 26–236): bankruptcy eleven Planar Representations (pages 237–245): bankruptcy 12 issues of reviews (pages 247–259): bankruptcy A Expression of Algorithms (pages 261–265): bankruptcy B Bases of Complexity concept (pages 267–276):

Get Theory and Application of Graphs PDF

Within the spectrum of arithmetic, graph conception which reviews a mathe­ matical constitution on a suite of parts with a binary relation, as a well-known self-discipline, is a relative newcomer. In contemporary 3 a long time the fascinating and quickly starting to be region of the topic abounds with new mathematical devel­ opments and important functions to real-world difficulties.

Extra resources for Proofs from THE BOOK (4th Edition)

Sample text

A for every n ≥ 0. But this cannot be true, because on the right-hand side we have an integer, while the left-hand side with e = 1+ 1 1 1 1 1 1 + +. + + + + + ... 1! 2! n! (n + 1)! (n + 2)! (n + 3)! decomposes into an integral part bn! + 1! 2! n! and a second part b 1 1 1 + + + ... n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3) M. M. 718281828... ex := 1 + x 1 xk = k! k≥0 + x2 2 + x4 6 1 24 + ... + x4 24 + ... 36 Some irrational numbers Geometric series For the infinite geometric series Q = 1 q + 1 q2 + 1 q3 + ...

Regiae sci. Göttingen XVI (1808), 69; Werke II, 1-8 (contains the 3rd proof). [4] C. F. G AUSS : Theorematis fundamentalis in doctrina de residuis quadraticis demonstrationes et amplicationes novae (1818), Werke II, 47-64 (contains the 6th proof). [5] F. L EMMERMEYER : Reciprocity Laws, Springer-Verlag, Berlin 2000. For p = 3, q = 5, G2 = (ζ − ζ 2 )2 = ζ 2 − 2ζ 3 + ζ 4 = ζ 2 − 2 + ζ = −3 = 3−1 (−1) 2 3, since 1 + ζ + ζ 2 = 0. 30 The law of quadratic reciprocity What’s up? I’m pushing 196 proofs for quadratic reciprocity Every finite division ring is a field Chapter 6 Rings are important structures in modern algebra.

In the three-dimensional case the end of a segment may also be given by a crossing of two edges. However, in any case all the interior points of a segment belong to the same set of edges of pieces. Hilbert’s third problem: decomposing polyhedra P = P ∪ P1 ∪ · · · ∪ Pm 55 Q = Q ∪ Q1 ∪ · · · ∪ Qm P1 Q1 P Q P2 Q2 P = P1 ∪ · · · ∪ Pn P1 P2 P3 P4 Q = Q1 ∪ · · · ∪ Qn Q2 Q1 Q4 Q3 For a parallelogram P and a non-convex hexagon Q that are equicomplementary, this figure illustrates the four decompositions we refer to.

Download PDF sample

Proofs from THE BOOK (4th Edition) by Martin Aigner, Günter M. Ziegler


by Anthony
4.1

Rated 5.00 of 5 – based on 48 votes