Every odd degree polynomial with real coefficient has at least one real root, that i can prove. but i dont know how to prove that every odd degree polynomial with rational coefficients has at least one rational root?


closed as off-topic by Jack D'Aurizio, Shailesh, Daniel W. Farlow, M. Vinay, user228113 Jun 9 '16 at 1:45

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    $\begingroup$ I'm happy to hear that you can't prove it, because, as M10687. says, this simply isn't true! $\endgroup$ – user247327 Jun 8 '16 at 18:20
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    $\begingroup$ The intermediate value theorem is not true over the rational numbers, see here. So the proof you know will not work. For good reason, because it is no longer true. $\endgroup$ – Dietrich Burde Jun 8 '16 at 18:22
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    $\begingroup$ You may be confused about what the en.wikipedia.org/wiki/Rational_root_theorem says, which is "if a polynomial has a rational root, then it satisfies a certain condition" - but it doesn't say all polynomials must have rational roots. $\endgroup$ – alephzero Jun 8 '16 at 23:23
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    $\begingroup$ I'm voting to close this question as off-topic because it asks for the proof of a non-result. $\endgroup$ – M. Vinay Jun 9 '16 at 1:05

Consider $x^3+ 2$. Does this have a rational root?

  • $\begingroup$ but how do i prove it with a polynomial with rational coefficients like 3/4x^3 +2? $\endgroup$ – lostinmath Jun 9 '16 at 18:27
  • $\begingroup$ That polynomial has rational coefficients, 1 and 2 are rational $\endgroup$ – M10687 Jun 9 '16 at 18:28
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    $\begingroup$ now i get it thank you! $\endgroup$ – lostinmath Jun 9 '16 at 19:34

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