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I have a conjecture, but have no idea how to prove it or where to begin. The conjecture is as follows:

A polynomial with all real irrational coefficients and no greatest common factor has no rational zeros.

This conjecture excludes the cases where the polynomial does have a greatest common factor despite having an irrational coefficient, such as $x^3+\pi x^2=0$, as that has rational zero $0$.

I know that not all polynomials with rational coefficients have rational zeros, but I am not sure how to begin. How would I go about beginning to prove this? Has it already been proved - or is there a counterexample that I am missing?

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  • $\begingroup$ what does 'greatest common factor' mean? $\endgroup$ – Trebor Apr 27 at 5:16
  • $\begingroup$ @Trebor, that means that no term shares a factor with any other term except 1 $\endgroup$ – Robert Apr 27 at 5:26
  • $\begingroup$ The problem is that you can't precisely nail down 'factor'. Does $\pi$ and $2\pi$ share one? Obviously. Does $\log 2$ and $2-\log 25$ share one? Not so obvious, but the latter is twice the former. Now, does $\pi$ and $e$ share one? You get prizes for solving this! $\endgroup$ – Trebor May 1 at 15:31
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If one were able to prove that, that would imply that $\pi x-e=0$ has no rational roots. However, it is not known whether $e/\pi$ is rational. So, I think your conjecture as stated would be difficult to prove.

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  • $\begingroup$ Are there any other cases where an operation on two irrational numbers has rationality unknown? $\endgroup$ – Robert Apr 27 at 5:29
  • $\begingroup$ $\pi + e$ is another famous example. I don't know of other examples off of the top of my head, but my guess is that most sums and products of transcendental numbers are not known. $\endgroup$ – Alexander Gruber Apr 27 at 5:46
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    $\begingroup$ @Robert Extension fields are a way of taking a set of numbers $A$ and adding in roots of a polynomial whose coefficients are in $A$. For example, the complex numbers $\mathbb{C}$ are an extension field of the real numbers $\mathbb{R}$ by adding in the roots of the polynomial $x^2+1$ (the roots of which are $\pm i$). $\endgroup$ – Alexander Gruber Apr 27 at 16:41
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    $\begingroup$ What you probably want for this conjecture is look at extension fields of the rational numbers $\mathbb{Q}$ and see how they work. The wikipedia article has some good examples, or you can use any first course book in abstract algebra. $\endgroup$ – Alexander Gruber Apr 27 at 16:44
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    $\begingroup$ I found a counterexample: (x-c)(ax+b) produces a quadratic with all irrational coefficients and a rational zero if c is rational and a and b are both algebraic and irrational. E.g. (x-1)(sqrt(2)x+sqrt(3)) = sqrt(2)x^2 + (sqrt(3)-sqrt(2))x - sqrt(3) ; c = 1, a = sqrt(2), b = sqrt(3). I now know that this conjecture is false for quadratics. I will continue researching and refining. $\endgroup$ – Robert May 1 at 1:06
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For the case of just one irrational coefficient $a_i$, supose by absurd that there is a rational solution $q$. Then: $$(a_i=a_0+...+a_n q^n)\frac{1}{q^i}$$ hence $a_i$ is rational. Therefore for this case there is no rational solutions.

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