# Infinitely many transcendental numbers over Q

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 $\mathbb Q$?

-

## 1 Answer

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.

-
math.stackexchange.com/questions/30687/… discusses the assertion made above about square roots of primes. – Qiaochu Yuan Mar 29 '12 at 4:22
Thanks @QiaochuYuan! – Bruno Mar 29 '12 at 4:23
What if $S=\left\{\ln(2),\ln(3),\ln(5),\ldots\right\}$? The theorem requires $S$ to contain algebraic numbers, not any real numbers linearly independent over $\mathbb{Q}$. – alex.jordan Mar 29 '12 at 4:24
@Bruno: also, to apply Lindemann-Weierstrass $S$ needs to consist of algebraic numbers. – Qiaochu Yuan Mar 29 '12 at 4:25
Haha! Yes, absolutely. Thank you both. I knew that! – Bruno Mar 29 '12 at 4:27
show 3 more comments