algebraic numbers are countableAlgebraicNumbersAreCountable2005-05-02 15:26:002009-09-30 08:58:23Theoremalgebraic number% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}\begin{thmplain}
The set of (a) all algebraic numbers, (b) the real algebraic numbers is countable.
\end{thmplain}
{\em Proof.}\, Let's consider the algebraic equations
\begin{align}
P(x) \;=\; 0
\end{align}
where
$$P(x) \;:=\; a_0x^n\!+\!a_1x^{n-1}\!+\!\ldots\!+\!a_{n-1}x\!+\!a_n$$
is an \PMlinkname{irreducible}{IrreduciblePolynomial2} and primitive polynomial with integer coefficients $a_j$ and\, $a_0 > 0$.\, Each algebraic number \PMlinkescapetext{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 \PMlinkescapetext{roots} of these equations.\, These algebraic numbers may be ordered to a \PMlinkname{finite sequence}{OrderedTuplet} $S_N$ using a \PMlinkescapetext{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 $\mathbb{Z}_+$.
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\bibitem{EK}{\sc E. Kamke:} {\em Mengenlehre}.\, Sammlung G\"oschen: Band 999/999a.\, -- Walter de Gruyter \& Co., Berlin (1962).
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