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| ``Re: Proof''
by rm50 on 2007-12-28 15:13:59 |
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| I think cam has this one right. The entry on casus irreducibilis talks about polynomials over the reals, not over the rationals. This is incorrect, as the classical meaning of the term refers to the case where a (monic) cubic with integral coefficients has three real roots but is irreducible (over the rationals).
The reason for the term is that, in this case, the roots, while real, cannot be expressed in terms of real radicals since, as you point out, the roots would then be constructible. The casus irreducibilis occurs when the discriminant is positive.
When the discriminant is negative, the cubic has only one real root, so of course there is not any question of being able to represent the roots using real radicals.
Roger |
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