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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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Abhishek Parab
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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.

Bruno Joyal
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    http://math.stackexchange.com/questions/30687/the-square-roots-of-the-primes-are-linearly-independent-over-the-field-of-ration discusses the assertion made above about square roots of primes. – Qiaochu Yuan Mar 29 '12 at 04:22
  • 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 $\mathbb{Q}$. – 2'5 9'2 Mar 29 '12 at 04:24
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    @Bruno: also, to apply Lindemann-Weierstrass $S$ needs to consist of algebraic numbers. – Qiaochu Yuan Mar 29 '12 at 04:25
  • Haha! Yes, absolutely. Thank you both. I knew that! – Bruno Joyal Mar 29 '12 at 04:27
  • We can certainly produce a (necessarily uncountable) transcendence base for the reals. But to apply Lindemann-Weierstrass, we need the elements of $S$ to be algebraic. – André Nicolas Mar 29 '12 at 04:27
  • My brain malfunctions in mysterious way - I used the theorem correctly, all while making this surprising claim... hmm! – Bruno Joyal Mar 29 '12 at 04:28
  • Thanks @Bruno, this answers my question. $\sqrt p$ is such a set. – Abhishek Parab Mar 29 '12 at 04:46