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# 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

$\displaystyle P(x)\;=\;0$ | (1) |

where

$P(x)\;:=\;a_{0}x^{n}\!+\!a_{1}x^{{n-1}}\!+\!\ldots\!+\!a_{{n-1}}x\!+\!a_{n}$ |

is an irreducible and primitive polynomial with integer coefficients $a_{j}$ and $a_{0}>0$. Each algebraic number satisfies exactly one such equation (see the minimal polynomial). For every integer $N=2,\,3,\,4,\,\ldots$ there exists a finite number of equations (1) such that

$n\!+\!a_{0}\!+\!|a_{1}|\!+\ldots+\!|a_{n}|\;=\;N$ |

(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 roots of these equations. These algebraic numbers may be ordered to a finite sequence $S_{N}$ using a fixed ordering system, for example by the magnitude of the real part and the imaginary part. When one forms the concatenated sequence

$S_{2},\,S_{3},\,S_{4},\,\ldots$ |

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

# References

- 1 E. Kamke: Mengenlehre. Sammlung Göschen: Band 999/999a. – Walter de Gruyter & Co., Berlin (1962).

## Mathematics Subject Classification

11R04*no label found*03E10

*no label found*

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