Talk:Liouville's Theorem (Number Theory)

questions
Let $P$ be the minimal polynomial of $x$ over $\Q$. By definition of degree of algebraic number, $\deg P = n$. Since Minimal Polynomial is Irreducible, $P$ does not have a rational root. Are $r_1, \ldots, r_k$ necessary?

In addition, we already assumed $\alpha \notin \set {r_1, \ldots, r_k}$. Why is there Case 3: $\alpha \in \set {r_1, \ldots, r_k}$? --Fake Proof (talk | contribs) 23:59, 27 August 2021 (UTC)


 * Your comment about case $3$ is good. It was a later addition to the original. I confess I didn't look closely, this is not an area of maths I have patience with.


 * As for the rest, the proof does not state that $P$ is minimal, it just has that $P$ is any polynomial that has $x$ as a root.


 * Presumably one can create a proof based on the minimality of $P$, but this one here is not such a proof. --prime mover (talk) 05:08, 28 August 2021 (UTC)