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**AlephZero** Recall that a complex number is said to be algebraic provided that it is a root of a polynomial with rational coefficients. To each algebraic number, there is a corresponding minimal polynomial that is both irreducible and monic (that is, its leading coefficient is 1).

It was shown by Cantor that the set of all algebraic numbers is countable. Every proof of this I have ever seen uses an "index" constructed to count the set of all minimal polynomials. This has always seemed sort of artificial to me. I propose the following inductive proof, and it is my hope that someone may be able to verify whether or not it is correct; for some reason, I've always wondered whether or not there is a logical error somewhere in it. Here goes:

Since every algebraic number has a unique minimal polynomial, and to each minimal polynomial there corresponds a finite number of algebraic roots, it is sufficient to show that the set of all minimal polynomials is countable. We proceed by induction. All first degree minimal polynomials have the form $\displaystyle x+r,$ where $\displaystyle r\in\mathbb{Q},$ and thus are clearly equivalent to $\displaystyle \mathbb{Q},$ and therefore countable. Suppose then that the set of all minimal polynomials having degree $\displaystyle n$ or less is countable. Now the $\displaystyle n+1$-th degree minimal polynomials have the form $\displaystyle x^{n+1} + sp(x),$ where $\displaystyle p(x)$ is an $\displaystyle n$-th (or less) degree minimal polynomial, and $\displaystyle s\in\mathbb{Q}.$ Hence as $\displaystyle s$ ranges over the rational numbers, $\displaystyle sp(x)$ yields countably many countable sets, and so the $\displaystyle n+1$-th degree minimal polynomials are countable. QED.

Does that follow?