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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.

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    $\begingroup$ Congratulations. You'll soon be famous. $\endgroup$ Nov 9, 2014 at 17:04
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    $\begingroup$ 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? $\endgroup$
    – WillO
    Nov 9, 2014 at 17:06
  • $\begingroup$ I'd assumed that $e$ was not in $\mathbb{Q}(\pi)$, which I now realise is not actually any simpler a claim to prove. $\endgroup$
    – Nethesis
    Nov 9, 2014 at 18:09
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    $\begingroup$ Related question: Why is it hard to prove whether π+e is an irrational number? $\endgroup$ May 10, 2015 at 7:43

2 Answers 2

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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.

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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?

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  • $\begingroup$ Series for $e$ and $\pi$ however have regular structures, so $e +\pi$ should also be somehow regular.. $\endgroup$
    – Widawensen
    Jul 21, 2016 at 5:26

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