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From: csc@watmath.UUCP (Computer Sci Club)
Newsgroups: net.math
Subject: N dimentional numbers?
Message-ID: <6966@watmath.UUCP>
Date: Tue, 21-Feb-84 13:11:11 EST
Article-I.D.: watmath.6966
Posted: Tue Feb 21 13:11:11 1984
Date-Received: Wed, 22-Feb-84 02:00:26 EST
Organization: U of Waterloo, Ontario
Lines: 32

    The question was asked are there N-dimensional numbers in the
same way that the complex numbers are 2-dimensional. The answer
depends on what you mean by "numbers".  If you mean fields
then the answer is no.
     A field F is a finite extention of the reals R if there
exist a1,a2, ... aN elements of F such that any element f of F
can be written as f= r1*a1 + r2*a2 + ... + rN*aN with
r1, ... rN elements of R.  The only finite extention of the reals
is C the complex numbers. Sketch of proof (see Herstein, Topics
in Algebra, chapter 7, (read chapters 1-6 first))

     If F is a finite extention of R then any element f of F is
a root of a polynomial in R. (basic result of field theory).

     All roots of polynomials in R are complex numbers.

     Therefore f is a complex number, therefore F is contained
in C (and can be shown equal to C).

   There are field extentions of R (fields which contain a copy
of R), for example the field of rational functions p(x)/q(x)
where p and q are real polynomials and q is non zero.  But
these extentions are infinite dimensional.
    If we relax our definitions of numbers to include division rings
(like fields but ab=ba is not necessarily true) then we get one
more finite extention of the reals the real quaternions.  (They
look like a + b*i + c*j + d*k and are four dimensional over the reals)
(This is the only algebraic division ring extention, I am not sure
if it is the only finite division ring extention) 

                                          William Hughes