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Cubic with three real irrational roots.

  1. May 27, 2006 #1

    uart

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    Considering the case of cubic polynomials with integer coefficients and three real but irrational roots. Is it true that it's impossible that all three roots can be in the form of simple surd expressions like [tex]r+s \sqrt{n}[/tex] (where r and s are rational and sqrt(n) is a surd). The arguement is that if [tex]r+s \sqrt{n}[/tex] is a solution then you can show that [tex]r-s \sqrt{n}[/tex] must also be a solution.

    Does anyone have any extra info or links to theorems or other useful info related to this.

    Thanks.
     
    Last edited: May 27, 2006
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  3. May 27, 2006 #2

    Hurkyl

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    It sounds like you have a reasonable proof of this. Do you know about Galois theory?
     
  4. May 27, 2006 #3

    HallsofIvy

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    No, it's not true if by "surd" you mean the usual definition: a root of a real number. Of course, any cubic equation can be solved by the cubic formula, giving the root as a combination of cube roots of an expression involving a square root. However, the result of the square may be imaginary and so the cube root is not a "surd".
     
  5. May 27, 2006 #4

    uart

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    No I don't know about Galois theory but I just looked at a quick overview at http://en.wikipedia.org/wiki/Galois_theory.

    So would it be true to say that any cubic root could be expressed as a something like a nested surd then?
     
  6. May 28, 2006 #5

    HallsofIvy

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    I'm beginning to wonder if I know anything! My answer above was "no" if by surd you mean the "usual definition: a root of a real number". In order to give support for my contention, I "googled" on surd and found that MathWorld defines "surd" as "any irrational number"!
    Wikpedia, however, defines "surd" as "an unfinished expression used in place of resolving a number's square, cube or other root, usually because the root is an irrational number.", my definition.

    With the first of those definitions, "any cubic root could be expressed as a something like a nested surd" would be false because two of the cube roots might be complex rather than irrational numbers, the MathWorld definition of "surd".

    With the second of those definitions, "any cubic root could be expressed as a something like a nested surd" is still false because the nested roots might be roots of complex numbers rather than real numbers, the Wikpedia definition of "surd".

    So my answer to your question is still NO, even though the roots of every cubic equation can be written as combinations of nested second and third roots, those roots or the numbers in the roots might be complex rather than real and, thus, not surds!
     
  7. May 29, 2006 #6

    uart

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    Yeah that's interesting Halls. So Mathworld just considers the term "Surd" to be an antiquated term for an irrational. Note however that they do refer to the term "Quadratic Surd" as having the meaning that I (and probably yourself) normally think of as a Surd. http://mathworld.wolfram.com/QuadraticSurd.html
     
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