Is the algebraic numbers countable? Prehaps you could prove that for me. The algebraic numbers are fomally defined as... The set of all a such that there exists an m such that a is an element of the union of sets form n eqauls 1 to m of the set of all x such that the sum from i equals 0 to n of c sub i x super i eqauls 0, c sub i is an element of the integers, c subn not equal to 0. However, informally defined as the set of all soulitions of all polynomials of integer coefficents.
There are a countable number of integer coeffecient polynomials of order n, and each such polynomial has n complex roots. Taking only those roots that are real it is clear that the set of real roots is also countable for each n. These sets of real roots can themselves be counted since they are labeled by an integer hence the union of these sets is also countable. Basically the whole thing boils down to using the fact that a countable union of countable sets is countable. That's my semi-rigorous physicist's proof for you.
That "fact" results from the axiom of choice. I vaguely recall a proof similar to the proof of the countability of the rationals (i.e., construct a mapping to the integers) that avoids the axiom of choice.
yes this is all very true however by definition the an algebraic number may be complex. Are reals complex, they must be right. let me amend, an algebraic number may happen not be real. And how is it that there are countabley many nth oder polynomials of integre coeficents?
Hector, That's fine, keeping the complex numbers in the game is no problem. Also, the countability of the integer coeffecient polynomials can be proved inductively if you like. Like I said, it flows from the fact that countable (including therefore finite) unions of countable sets are countable. Edit: D H wants to make sure I don't cheat, so just remember that my fact is related to the axiom of choice. I personally find the axiom "obvious", but quite apart from my personal views you are probably using it anyway in whatever set theory you've got going on.
I suppose you don't need the axiom of choice because you already have the (stronger) property of countability. Use a proof similar to countability of rationals to get countability of N<sup>2</sup> (vector of two natural numbers). Count all elements (m, n) in N<sup>2</sup> and for each one take the mth element of the nth set. That proves the case of a countably infinite union of non-overlapping countably infinite sets, anyway.
