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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 $\mathbb Q$?

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up vote 6 down vote accepted

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.

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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 Joyal 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
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@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 Joyal Mar 29 '12 at 4:27
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