# Proving the non-existence of rational zeros in a polynomial with irrational coefficients

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?

• what does 'greatest common factor' mean? – Trebor Apr 27 at 5:16
• @Trebor, that means that no term shares a factor with any other term except 1 – Robert Apr 27 at 5:26
• 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! – Trebor May 1 at 15:31

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.
• $\pi + e$ 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. – Alexander Gruber Apr 27 at 5:46
• @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 $\pm i$). – Alexander Gruber Apr 27 at 16:41
• 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. – Alexander Gruber Apr 27 at 16:44
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.