algebraic numbers are countable

Theorem.

The set of (a) all algebraic numbers, (b) the real algebraic numbers is countable.

Proof.  Let’s consider the algebraic equations

$P(x)=\mathrm{\hspace{0.33em}0}$(1)

where

$$P(x):=a_0{x}^n+a_1{x}^{n-1}+\mathrm{\dots }+a_{n-1}x+a_n$$

is an irreducible (http://planetmath.org/IrreduciblePolynomial2) and primitive polynomial with integer coefficients $a_j$ and  $a_0>0$.  Each algebraic number exactly one such equation (see the minimal polynomial).  For every integer  $N=2,\mathrm{\hspace{0.17em}3},\mathrm{\hspace{0.17em}4},\mathrm{\dots }$  there exists a finite number of equations (1) such that

$$n+a_0+|a_1|+\mathrm{\dots }+|a_n|=N$$

(e.g. if  $N=3$,  then one has the equations  $x-1=0$  and $x+1=0$) and thus only a finite set of algebraic numbers as the of these equations.  These algebraic numbers may be ordered to a finite sequence (http://planetmath.org/OrderedTuplet) $S_N$ using a system, for example by the magnitude of the real part and the imaginary part.  When one forms the concatenated sequence

$$S_2,S_3,S_4,\mathrm{\dots }$$

it comprises all algebraic numbers in a countable setting, which defines a bijection from the set onto ${\mathbb{Z}}_{+}$.