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I recall reading a comment on reddit that had stated that it is not known if $\pi + e$, (nor $\pi e$) is transcendental over $\mathbb{Q}$, nor even if it is irrational. Is this true? It strikes me as something which is very easy to prove, in my head I could figure out the rough links quite quickly. I'm not necessarily asking $\textit{for}$ a proof, just confirmation that it's not actually some mathematical mystery that people have been trying to solve for centuries, as that comment implied.

EDIT: Okay I've been humbled, there was a gap in my logic, I'd assumed that $\mathbb{Q}(\pi) \neq \mathbb{Q}(e)$, and then realised after reading all the skepticism that that may not be a simple thing to prove....... after a good amount of time spent over it, I realise that it really, really isn't.

I also admit that this sounded really cocky as a question, sorry. I genuinely believed that it couldn't be as difficult a question as the comment indicated, it didn't seem like it could possibly be an open problem.

Nethesis
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    Congratulations. You'll soon be famous. – David Mitra Nov 09 '14 at 17:04
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    Does your argument also imply that the sum of *any* two irrational numbers is irrational? If so, it can't be correct, since the conclusion is false. If not, your argument must be using some special properties of $\pi$ and $e$ beyond the fact that they're irrational. What are those properties? – WillO Nov 09 '14 at 17:06
  • I'd assumed that $e$ was not in $\mathbb{Q}(\pi)$, which I now realise is not actually any simpler a claim to prove. – Nethesis Nov 09 '14 at 18:09
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    Related question: [Why is it hard to prove whether π+e is an irrational number?](http://math.stackexchange.com/questions/159350/why-is-it-hard-to-prove-whether-pie-is-an-irrational-number) – Martin Sleziak May 10 '15 at 07:43

2 Answers2

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It's not known, and so far as I know nobody even has a reasonable plan of attack on the problem. The whole subject of transcendental number theory is just completely intractable with current machinery.

Daniel McLaury
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You might want to consider the following two irrational numbers:

$$\sqrt{2}=1.414213562373095048801688724209698.....$$

$${23481838282\over 245689351}-\sqrt{2}=94.1611061917175085769126386338...$$

Just from a quick look at those two (apparently random) decimal expansions, would you have guessed that the sum of these two irrational numbers is equal to the rational number $23481838281/245689351$?

Does your proof rule out a similar relation between $\pi$ and $e$? How so?

WillO
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