From jay.kominek@colorado.edu Thu May 02 17:17:33 2002
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Date: Thu, 2 May 2002 18:17:31 -0600 (MDT)
To: lojban@yahoogroups.com
Subject: Re: Fw: [lojban] cipja'o
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From: Jay Kominek
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On Thu, 2 May 2002, G. Dyke wrote:
> so transcendental numbers are those which cannot solve poynomial equation=
s?
>
> what does that make them??
Nono, they can be the roots of polynomial equations (geez, can't believe I
screwed that up in my previous email), but not of polynomial equations
with integer coefficients. For instance, if your polynomial has got a
transcendental coefficient, then you can (and likely will) have
transcendental roots.
However, polynomials with integer coefficients can have irrational roots,
for instance, x^2-2=3D0, has roots of +sqrt(2) and -sqrt(2), which are
both irrational. (Meaning you can't express them as p/q where p and q are
integer.)
- Jay Kominek
Plus =C3=A7a change, plus c'est la m=C3=AAme chose