Algebraically independent numbers and Archimedean field

Question

What is the cardinality of the set of all algebraically independent numbers in $\mathbb{R}$?

Can this be related to the total number of Archimedean fields possible as rational extensions of sets of algebraically independent numbers?

Answer 1

Let $L/K$ be a field extension. Let $(x_i)_{i \in I}$ be a family of elements of $L$.

This family is algebaically independent over $K$ iff for every pairwise distinct indices $i_1,\ldots ,i_n$ and every polynomial $P \in K[X_1, \ldots, X_n]$, $P(x_{i_1}, \ldots,x_{i_n})=0$ implies $P=0$.

So saying that a number is algebraically independent has not much of a meaning.

If you are interested in numbers $x$ such that the family $(x)$ is algebaically independent over $\mathbb{Q}$, the standard name is that $x$ is transcendent over $K$.

Note that this is abstract algebra that has no direct relationship with Archimedean properties.