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It is well known that there's no conclusion now whether $\pi+e$ is rational or not. What would happen if we knew that $\pi+e$ is rational? Specifically, are there related open problems that would be settled?

Bill Dubuque
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zhangwfjh
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    Probably nothing; usually, its the techniques that lead to a new result that are the exciting part, rather than the result itself. – goblin GONE Mar 28 '14 at 10:51
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    Human sacrifice, dogs and cats living together... mass hysteria! –  Mar 28 '14 at 11:07
  • @Mike: I like it! +1 – MPW Mar 28 '14 at 11:19
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    Depends. If $\pi+e$ turned out to be rational, I would be inclined to believe that God is joking with us. – Jyrki Lahtonen Mar 28 '14 at 11:43
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    If $\pi + e$ turned out to be rational then there will be a much more simpler and natural relation between $e$ and $\pi $ than the traditional $e^{i\pi} + 1 = 0$. – Paramanand Singh Mar 28 '14 at 11:47
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    @ParamanandSingh, I'm not so sure. What if the denominator of $\pi+e$ has 2 million digits? Will $q\pi+qe=p$ be simpler than $e^{i\pi}+1=0$, if $q$ is that big? – Gerry Myerson Mar 28 '14 at 12:06
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    @GerryMyerson: I had in my mind that the relationship would be conceptually simple rather than visibly simple. In fact the existing Euler's relation between $e$ and $\pi$ seems to be the most impressive. – Paramanand Singh Mar 28 '14 at 12:10
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    Restating the fact given by @ParamanandSingh, if $\pi + e$ was rational, or even algebraic, Schanuel's hypothesis (currently an open problem in transcendence theory) would fall off, since $\pi + e = k$ for some $k \in \Bbb Q$ and $e^{i\pi} + 1 = 0$ are two different expressions in the exponential ring, and Schanuel's conjecture essentially implies that Euler's identity is the unique $\mathcal E$-algebraic relation between $e$, $\pi$ and $i$. – Balarka Sen Mar 28 '14 at 13:39
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    So I think it's not really true that absolutely nothing would happen, considering the importance of Schanuel's conjecture in transcendental number theory. – Balarka Sen Mar 28 '14 at 13:45
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    Well, we'd know that $\pi e$ was irrational, so there's that. – Milo Brandt Jan 04 '15 at 05:08

2 Answers2

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This is a comment that's too long for the usual format. If $\alpha=e+\pi$ can be written as a fraction of integers, both numerator and denominator must be $\geq 2 \times 10^{32}$. To see this, let $A,B,C,D$ be integers defined by

$$ \begin{array}{lcl} A &=& 3063742572717320569341511991159738 \\ B &=& 522834163445445988434458010516405 \\ C &=& 9765222175513935643148512770417523 \\ D &=& 1666455861030599542832067804101203 \\ \end{array} $$

Then any good formal computing system will confirm to you that $\frac{A}{B} < \alpha < \frac{C}{D}$ and $BC-AD=1$. If $\alpha$ is rational, $\alpha=\frac{p}{q}$ with $p,q \in {\mathbb N}_{>0}$, then $u=pB-qA$ and $v=qC-pD$ must be positive integers. But then $p=Cu+Av\geq A+C \geq 2\times 10^{32}$ and similarly $q=Du+Bv\geq B+D \geq 2\times 10^{32}$.

Ewan Delanoy
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It would imply in particular that $e$ is a period, which it conjecturally isn't. There are deep reasons why $e$ is conjectured not to be a period, stemming (I believe) from Deligne's theory of motivic weights. I recommend taking a look at Kontsevitch and Zagier's fantastic article Periods.

Bruno Joyal
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