Infinitely many transcendental numbers over Q

Question

My previous question was not well-framed so I will ask again:

Can you explicitly produce an infinite set of real numbers which is algebraically independent over $\mathbb Q$?

Answer 1

Yes, using the Lindemann-Weierstrass theorem.

Let $S$ be any infinite set of algebraic real numbers linearly independent over $\mathbb{Q}$, for example $\{\sqrt p : p \text{ is prime}\}$. (A maximal such set is a basis for $\overline{\mathbb{Q}}$ over $\mathbb{Q}$.) Then $\{e^s : s \in S\}$ is algebraically independent.