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Timeline for Independence of being an integer

Current License: CC BY-SA 3.0

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Sep 10, 2011 at 22:21 comment added George Lowther As I can't edit comments, I'll just add that obviously I should have said that $e^{79}$, $e^{e^{79}}$, $e^{e^{e^{79}}}$ are algebraically independent. Clearly I shouldn't have had 79 in there too.
Sep 10, 2011 at 15:36 history edited François G. Dorais CC BY-SA 3.0
addendum after comments; deleted 6 characters in body
Sep 10, 2011 at 15:28 comment added François G. Dorais Ah! Never mind, I just figured out the bootstrapping trick...
Sep 10, 2011 at 15:21 comment added François G. Dorais Thanks for confirming my suspicion, Dave and George! What's the easy way to see (the intuitively obvious fact) that $79, e^{79}, e^{e^{79}}$ are linearly independent over $\mathbb{Q}$?
Sep 10, 2011 at 15:18 comment added George Lowther @Dave: Ah, beaten to it.
Sep 10, 2011 at 15:14 comment added George Lowther According to Schanuel's conjecture, $79$, $e^{79}$, $e^{e^{79}}$, $e^{e^{e^{79}}}$ are algebraically independent over the rationals. So it implies that $e^{e^{e^{79}}}$ is not an integer.
Sep 10, 2011 at 15:01 comment added Dave Marker You don't need Macintyre-Wilkie, as Schanuel's Conjecture tells you immediately $\exp(\exp(\exp(79))))$ is not an integer.
Sep 10, 2011 at 14:55 comment added Carl Mummert I'm glad you pointed this out. The only trouble for the purposes of this question, of course, is that Schanuel's conjecture is open. But I have some hope that maybe there is an approach to my question along these lines that doesn't require the full strength of Schanuel's conjecture. For example, the sentence you wrote is quantifier-free, so (naively) we don't need the decidability of the entire theory.
Sep 10, 2011 at 14:37 history edited François G. Dorais CC BY-SA 3.0
fixed wrong adjective
Sep 10, 2011 at 14:09 history answered François G. Dorais CC BY-SA 3.0