Cubic with three real irrational roots.

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Discussion Overview

The discussion revolves around the nature of cubic polynomials with integer coefficients that have three real but irrational roots. Participants explore whether all three roots can be expressed in the form of simple surd expressions, such as r+s√n, and the implications of this regarding Galois theory and definitions of surds.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant suggests that if a root can be expressed as r+s√n, then r-s√n must also be a root, implying a restriction on the form of the roots.
  • Another participant questions the validity of the initial claim, stating that cubic equations can be solved using the cubic formula, which may involve cube roots of expressions that include square roots, potentially leading to imaginary results.
  • A participant expresses uncertainty about the definitions of "surd," noting discrepancies between sources like MathWorld and Wikipedia, which may affect the interpretation of whether cubic roots can be expressed as nested surds.
  • It is argued that depending on the definition of "surd," the claim about cubic roots being expressible as nested surds may not hold true, particularly if the roots are complex rather than real.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether all three roots of a cubic polynomial can be expressed as simple surd expressions. Multiple competing views remain regarding the definitions of surds and the implications for cubic roots.

Contextual Notes

Participants highlight the ambiguity in the definition of "surd," which affects the discussion. There is also mention of the potential for roots to be complex rather than real, which complicates the claims about their forms.

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 argument 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.
 
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It sounds like you have a reasonable proof of this. Do you know about Galois theory?
 
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".
 
Hurkyl said:
It sounds like you have a reasonable proof of this. Do you know about Galois theory?

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

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...

So would it be true to say that any cubic root could be expressed as a something like a nested surd then?
 
uart said:
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?

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!
 
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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