Question: If $x$ is an algebraic number, then it occurs as a root of some polynomial with integer (or rational) coefficients. Is there also a polynomial with integer (rational) coefficients such that $x$ is the smallest root of this polynomial (w.r.t. the usual ordering of the reals)?
The question is inspired by a "solution" handed in by a student. The problem was to show that the set of algebraic numbers is countable. The solution started by the observation that $\mathbb Q[x]$ is countable. And then it continued more-or-less like this: "We take an enumeration of polynomials in $\mathbb Q[x]$. Then we omit the polynomials without real roots. We assign to each of the remaining polynomials the smallest root. If we omit the duplicates from this new list, we get an enumeration of the set of all algebraic numbers." (I will add that it is not difficult to fix this proof so that it works even without relying on using smallest root. We have countably many polynomials, each of them has only finitely many roots. Union of countably many finite sets if countable. But this is a digression to topic of other posts on this site, such as Prove that the set of all algebraic numbers is countable or How do we prove the existence of uncountably many transcendental numbers?)
Unless I have missed something, I can shown the answer to the question is No. (Simply by taking some irreducible polynomial with more than two real roots. Unless somebody else does this, I can expand this into an answer and post it.)
Still I consider the problem interesting enough to post it here. (And it is often the case that even when I ask a question to which I know a solution, I am pleasantly surprised by interesting alternative solutions I get as answers.)
