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An algebraic number is a number which is a root of some non-zero polynomial equation with rational coefficients.

A transcendental number is a number which is not a root of any non-zero polynomial equation with rational coefficients.

Are there any real numbers that are not the roots of any polynomial equation with algebraic coefficients?

I'm guessing that somebody must have asked this before somewhere, though I have not been able to find it, so please excuse me for the possible duplicate...

Thanks

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  • $\begingroup$ Any trancendental number $p$ is the root of the polynomial $x - p$, which is a polynomial with irrational coefficients. $\endgroup$
    – Arthur
    Jul 1, 2014 at 15:40
  • $\begingroup$ @Arthur: Thank you for pointing that out. I "tuned up" the question accordingly... $\endgroup$ Jul 1, 2014 at 15:42

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The complement of the transcendentals in $\mathbb{R}$ is the real algebraic numbers. The algebraic numbers are algebraically closed. Therefore every polynomial with non-transcendental coefficients has algebraic coefficients and thus algebraic roots.

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  • $\begingroup$ So what you're saying is - we will not "get" transcendental roots even if we use irrational algebric coefficients? $\endgroup$ Jul 2, 2014 at 4:26
  • $\begingroup$ Correct. Make a polynomial with algebraic coefficients, get algebraic roots. Since "algebraic" = "not transcendental", you can't get to the transcendentals using only the algebraics (in polynomials). $\endgroup$ Jul 2, 2014 at 5:16
  • $\begingroup$ Gotcha, thanks. $\endgroup$ Jul 2, 2014 at 5:54
  • $\begingroup$ A nitpick: complement of transcendentals in $\Bbb R$ are real algebraic numbers which aren't algebraically closed, since $x^2+1$ has no roots, algebraic or not. $\endgroup$
    – Wojowu
    Dec 27, 2015 at 10:25
  • $\begingroup$ @Wojowu : Right you are. Corrected. $\endgroup$ Dec 27, 2015 at 18:57
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"Are there real numbers ..." Yes, there are, and in fact those numbers coincide with the transcendental numbers. In fact, if $x$ is a root of a polynomial with non-transcendental(=algebraic) coefficients, then $x$ is itself algebraic.

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  • $\begingroup$ So no transcendental number is a root of some polynomial equation with algebric coefficients? $\endgroup$ Jul 2, 2014 at 6:03
  • $\begingroup$ @barakmanos exactly. $\endgroup$ Jul 2, 2014 at 21:05
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Now that you've added non-trancendental (i.e. "algebraic") to the question, the answer is yes.

The roots of a polynomial with algebraic coefficients are also roots of some polynomial (possibly of higher degree) with rational coefficients.

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  • $\begingroup$ That's an interesting notion (about the roots of a polynomial with algebraic coefficients being the roots of some polynomial with rational coefficients). Thanks. $\endgroup$ Jul 2, 2014 at 5:53
  • $\begingroup$ Nevertheless, I think that the answer is in fact - yes. Even according to your own description (if I understand it correctly), not only there are real numbers that are not the roots of any polynomial equation with algebric coefficients, but there are many of them - all the transcendental numbers (an uncountable set). $\endgroup$ Jul 2, 2014 at 6:01
  • $\begingroup$ The answer is yes, you're right. I just had the wrong question in mind when I wrote the answer. $\endgroup$
    – Arthur
    Jul 2, 2014 at 8:11

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