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Is there a bijective function $f(n)$, where $n \in \mathbb{N}$, which enumerates all algebraic numbers? Is it possible to define such function?

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@GitGud: that doesn't deal with each polynomial having a number of roots, some of them being the same, and different polynomials having the same roots. – Ross Millikan Apr 17 '13 at 19:33
@RossMillikan I had just realised that. – Git Gud Apr 17 '13 at 19:34
up vote 1 down vote accepted

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...

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Unless I'm overlooking something, you're providing a hint to the proof that there is a bijection between the set of algebraic numbers and $\Bbb N$, but not actually a hint to how to find such a bijection. $\textbf{Edit:}$ I just noticed the OP also asked if there is such a bijection. I'll leave my comment here anyway so it's clear that this is a partial answer. – Git Gud Apr 17 '13 at 19:35
Can't one construct such a bijection by a bruteforce algorithm? Finding an explicit bijection seems more of a programming exercise than math. – genepeer Apr 17 '13 at 21:48

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