TITLE: Bijection for algebraic numbers
QUESTION [5 upvotes]: Is there a bijective function $f(n)$, where $n \in \mathbb{N}$, which enumerates all algebraic numbers? Is it possible to define such function?

REPLY [2 votes]: HINT: Given an equation of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0=0$, where each of the $a_i$ are integers, define the "index" of this equation to be $|a_n| + |a_{n-1}| + \cdots +|a_1|+|a_0|+n$.
We can count the algebraic numbers by counting the solutions of equations with "index" one. Then by counting the solutions of "index" two that we haven't already counted. An so on...