How are the algebraic and Constructible numbers enumerable? I was reading http://en.wikipedia.org/wiki/Algebraic_number which atates that the algebraic numbers are countable. 
if i am correct this means that there are "only" $\aleph_0 $ of them and that they are orderable.
so i waas wondering how can you enumerate them?
then i was reading that the constructible numbers (http://en.wikipedia.org/wiki/Constructible_number) are an subset of the algebraic numbers. (all constructible numbers are algebraic numbers but not all algebraic numbers are constructable , for example $ \sqrt[3]{2} $ )
Does that mean that also these numbers are  are countable?
and how are they enumerable?  
 A: Are you familiar with the fact that $|\Bbb N \times \Bbb N|=|\Bbb N|$?  You can iterate that to show that the set of finite sequences of naturals (or integers) is countable.  That gives you that the set of polynomials with integer coefficients is countable.  Each of those has only finitely many roots, so the set of algebraic numbers is countable.  As the constructable numbers are a superset of the naturals and a subset of the algebraics, they are countable as well.
A: The way I like to think of these problems is as a "countability chase".
There's countably many integers. So there are countably many $n$-degree integer polynomials. Since there are countably many choices for $n$, there are countably many integer polynomials. Since each of these polynomials has finitely many roots, there are countably many algebraic numbers.
If you wanted to, you could use this chase to construct an ordering, but it would be incredibly messy. If you wanted to avoid double-counting, you'd have to iterate through the irreducible polynomials (because, for example, $x^2 - 6x + 5$ and $x^2 - 7x + 10$  share a root).
And as you said, since the constructable numbers are a subset of the algebraic numbers, they are also countable (you know they can't be finite, because all integers are constructable).
I don't believe there's a nice, "clean" ordering for either set, but I'd love to be wrong.
