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Note that I am talking about rational roots not rational coefficients. I know that Galois theory can tell you but I want to know if knowing whether all the roots of a polynomial are rational can also tell you. Note also that the roots and their properties are usually unknown, but I'm talking about when you only know that they are rational but don't know their values.

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By definition, a polynomial is solvable if and only if its roots can be expressed using rational numbers, addition/subtraction, multiplication/division, and radicals (squareroot symbols, cuberoot symbols, etc.) If the roots are rational, they certainly can be expressed in the above form.

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Thank you, that is exactly what I wanted to know. So if you know the roots are rational, then Galois theory can help you find them, right? You are talking about THAT "solvable" aren't you? – Kenny Mar 7 '12 at 17:15
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@Kenny: Well, Galois theory doesn't really give you any information in this case, since the Galois group is trivial. But if the roots are rational, you don't need such powerful tools. See Tib's post. – Brett Frankel Mar 7 '12 at 17:18
What do you mean by "the Galois group would be trivial?" So even if you don't know the roots but do know they are rational, Galois theory doesn't tell you anything about the roots? – Kenny Mar 7 '12 at 17:34
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Yes. Galois theory also explains, in abstract terms, why some equations are solvable by radical and others aren't. And of course there are myriad applications to number theory, algebraic geometry, group theory, etc. – Brett Frankel Mar 7 '12 at 18:12
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@BlueRaja-DannyPflughoeft: It is solvable over $\mathbb{Q}(\pi)$, but this post is about polynomials with rational coefficients. – Brett Frankel Mar 7 '12 at 22:40
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You can always find all the rational roots of a polynomial using the rational root theorem. See http://en.wikipedia.org/wiki/Rational_root_theorem

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(1) This does not answer OP's question about whether solvability equivalent to having all roots rational. (2) OP is interested in polynomials whose all of its roots are rational rather than finding all the rational roots of a polynomial. – user2468 Mar 7 '12 at 17:21
@J.D. True, but this does answer OP's follow-up comment. Perhaps it would be better as a comment than a solution, but I don't think it deserves a downvote. – Brett Frankel Mar 7 '12 at 17:24
@BrettFrankel Well, then this is more of a comment. Answers should address the question title & body. – user2468 Mar 7 '12 at 17:26
BTW I did not downvote! I only posted my remarks. – user2468 Mar 7 '12 at 17:27
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He didn't ask whether it's equivalent to being solvable; he asked whether it implies that it's solvable. Clearly there are some cases where it's solvable and none of the roots are rational. – Michael Hardy Mar 7 '12 at 18:08
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