Martin Aigner, Günter M. Ziegler, Karl H. Hofmann's Proofs from THE BOOK PDF

By Martin Aigner, Günter M. Ziegler, Karl H. Hofmann

This revised and enlarged 5th version beneficial properties 4 new chapters, which include hugely unique and pleasant proofs for classics corresponding to the spectral theorem from linear algebra, a few more moderen jewels just like the non-existence of the Borromean earrings and different surprises.

From the Reviews

"... within PFTB (Proofs from The booklet) is certainly a glimpse of mathematical heaven, the place shrewdpermanent insights and lovely principles mix in incredible and wonderful methods. there's mammoth wealth inside its pages, one gem after one other. ... Aigner and Ziegler... write: "... all we provide is the examples that we've got chosen, hoping that our readers will percentage our enthusiasm approximately wonderful principles, shrewdpermanent insights and lovely observations." I do. ... "

Notices of the AMS, August 1999

"... This publication is a excitement to carry and to examine: plentiful margins, great images, instructive photographs and lovely drawings ... it's a excitement to learn to boot: the fashion is apparent and pleasing, the extent is just about simple, the required history is given individually and the proofs are remarkable. ..."

LMS e-newsletter, January 1999

"Martin Aigner and Günter Ziegler succeeded admirably in placing jointly a vast number of theorems and their proofs that might unquestionably be within the ebook of Erdös. The theorems are so basic, their proofs so stylish and the remainder open questions so exciting that each mathematician, despite speciality, can take advantage of interpreting this booklet. ... "

SIGACT information, December 2011.

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Additional info for Proofs from THE BOOK

Example text

But, since F is a field, xd − 1 has at most d roots, and so the elements b, b2 , . . , bd = 1 are precisely these roots. In particular, every element of order d is of the form bi . d On the other hand, it is easily checked that ord(bi ) = (i,d) , where (i, d) denotes the greatest common divisor of i and d. Hence ord(bi ) = d if and only if (i, d) = 1, that is, if i and d are relatively prime. Denoting Euler’s function by ϕ(d) = #{i : 1 ≤ i ≤ d, (i, d) = 1}, we thus have ψ(d) = ϕ(d) whenever ψ(d) > 0.

B. The multiplicative group F ∗ = F \ {0} is cyclic of size q p−1 − 1 (see the box on the next page). Since by Fermat’s little theorem p is a divisor of q p−1 − 1, there exists an element ζ ∈ F of order p, that is, ζ p = 1, and ζ generates the subgroup {ζ, ζ 2 , . . , ζ p = 1} of F ∗ . Note that any ζ i (i = p) is again a generator. Hence we obtain the polynomial decomposition xp − 1 = (x − ζ)(x − ζ 2 ) · · · (x − ζ p ). Now we can go to work. Consider the Gauss sum p−1 G := i=1 i i ζ ∈ F, p where ( pi ) is the Legendre symbol.

Move For k = p− 1 ≡ −1 (mod p) this gives ( −1 p )(p− 1), since ζ k = p − 1 in front and write G2 = −1 (p − 1) + p p−2 k=1 k p p−1 Euler’s criterion: ( −1 ) = (−1) p ζ (1+k)i . p−1 2 i=1 Since ζ 1+k is a generator of the group for k = p − 1, the inner sum equals p−1 i i=1 ζ = −1 for all k = p − 1 by our first observation. Hence the second p−2 summand is − k=1 ( kp ) = ( −1 p ) by our second observation. It follows 2 that G2 = ( −1 p )p and thus with Euler’s criterion G = (−1) completes the proof.

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