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