Originally Posted by

**kingwinner** **Prove that the set of algebraic numbers is countable.**

__Proof:__

Let A be the set of all algebraic numbers.

Let $\displaystyle A_n$={x E R : p(x) = 0 for some p a polynomial of degree n with integer coefficients}

∞

U $\displaystyle A_n$ = A

n=1

Consider the n-th degree polynomial

$\displaystyle a_nx^n+...+a_1x+a_0$, $\displaystyle a_i$ E Z, $\displaystyle a_n$≠0

|{$\displaystyle (a_n,...,a_1,a_0)$: $\displaystyle a_i$ E Z, $\displaystyle a_n$≠0}| = |$\displaystyle Z^n$| = |Z| = |N|

Each such polynomial has at most n roots.

Therefore, $\displaystyle A_n$ is countable.

A countable union of countable sets is countable, thus A is countable.

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1) I don't understand why |{$\displaystyle (a_n,...,a_1,a_0)$: $\displaystyle a_i$ E Z, $\displaystyle a_n$≠0}| = |$\displaystyle Z^n$|. $\displaystyle a_n$ cannot be 0, so the set on the LHS is a little bit different from $\displaystyle Z^n$. How can we formally prove that it has the same cardinality as $\displaystyle Z^n$?

2) Assuming the above, I understand why the set of all polynomials with integer coefficients is countable, but I don't understand why the set of all "roots" to these polynomials, |$\displaystyle A_n$|, is also countable. I know each polynomial with degree n has at most n roots, but I don't see how it rigorously follows that the set of all "roots" are countable. How can we prove it formally?

These are the two fine points that I don't understand in this proof.

I hope someone can clarify these. Thanks for any help!

[note: also under discussion in math links forum]