From: mtrott@wolfram.com (Michael Trott)
Newsgroups: sci.math.research
Subject: Re: Real-Radicals field.
Date: 2 Sep 1997 15:03:34 GMT
In sci.math.research article <5udh2h$t8s$1@cantuc.canterbury.ac.nz> you
wrote:
> Statement:
> ------------
> For any cubic with three different irrational real roots,
> the roots are NOT elements of the real-radical field.
> ------------
>
> Is that true?
Yes, see
H\"older. Math. Annalen 38, 307 (1891)
I. M. Isaacs. Am. Math. Monthly 92, 571 (1985)
--
Michael Trott
Wolfram Research, Inc.
==============================================================================
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Complex Representation
Date: 20 Feb 1998 06:21:34 GMT
In article , SCN User wrote:
>
>One root of the cubic equation x^3 - 3x + 1 = 0
>is known to be about - 1.8793 ...actually all of
>its three roots are in fact real. My problem is,
>I have only been able to obtain, for example,
>
>-[e^(i*pi/9) + e^(-i*pi/9)] as the representation
>
>for the root I gave above
Right; that's not the fault of your algebra. It is known that if all
three roots of a rational cubic are real and irrational, they _do_ lie
in a radical extension of Q but they do _not_ lie in a radical
extension of Q which is contained in the real line. In English: you
_can_ express the roots using + - * / sqrt() and cbrt() but you
must at some point take the square root of a negative number. Back in the
days when all educated persons spoke Latin this was called the
"casus irreducibilis" of cubic polynomials.
Of course you _can_ express the roots using only the (real) trigonometric
functions, but that's not "algebra".
dave