TITLE: Is this the (real) algebraic numbers?
QUESTION [8 upvotes]: I want to construct the real algebraic numbers from $\mathbb{Q}$ in a manner that sort of "looks like" the construction of the complex numbers from the reals, in a superficial manner. I can't define a single "irrational unit", but I instead tried to start with the set $\{ \sqrt{z} \mid z \in \mathbb{Q} \}$ and build the algebraic numbers from there.
This is what I'm thinking:
Let $A_{0} = \{ q + p \sqrt{z} \mid  q, p, z \in \mathbb{Q} \}$.
In general, let $A_{n} = \{ q + p \sqrt{z} \mid q, p, z \in \mathbb{Q} \cup A_{0} \cup \dots \cup A_{n-1} \}$.
Let $A = \cup_{i=0}^{\infty} A_{i}$.
Is $A = \mathbb{A}$? If not, what's missing, and if so, can this description be simplified in some way?
Thanks all.

REPLY [17 votes]: These are called the constructible numbers, so named because they are the numbers you can construct via Euclidean geometry with a compass and straightedge, starting from a unit interval.
Adjoining all square roots isn't enough to produce the algebraic numbers; you can never produce $\sqrt[3]{2}$ in this fashion.
Adjoining $n$-th roots isn't enough either; for example, there are explicitly known polynomials of degree five whose roots cannot be expressed in that fashion. (i.e. the "insolvability" of the quintic)