I'm reading Ross' "Elementary Analysis". He shows the rational zeros theorem and a proof of it. And then there are some examples such as the following ones:
The basic idea is that testing all the candidates for rational roots, if none of them is a root, then the number we are testing for irrationality is not rational because as there are no rational roots, then all roots must be irrational or complex.
But I thought about the following: Can't we have a case where we are testing for the irrationality of some $\alpha$ and:
- Some of the candidates for rational roots are roots;
- Some of the candidates for rational roots aren't roots?
I suspect that if $\alpha$ is an irrational number, that is not possible. I recognize the procedure of finding the polynomial $x^6-6x^4 +12x^2-13=0$ given the number $\alpha=\sqrt{2+\sqrt[3]{5}}$ as computing the minimal polynomial of $\alpha$. And it seems to me that the case I pointed out can't happen because of something related to field extensions but I can't figure out what is it nor if it is actually true.
The main doubt I have is the truth or falsity of the following proposition: If $\alpha$ is an irrational number, then the minimal polynomial of $\alpha$ over $\Bbb{Q}$ does not have a rational root. Is this proposition true? If so, why?