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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?

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    $\begingroup$ Probably nothing; usually, its the techniques that lead to a new result that are the exciting part, rather than the result itself. $\endgroup$ – goblin Mar 28 '14 at 10:51
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    $\begingroup$ Human sacrifice, dogs and cats living together... mass hysteria! $\endgroup$ – user98602 Mar 28 '14 at 11:07
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    $\begingroup$ Depends. If $\pi+e$ turned out to be rational, I would be inclined to believe that God is joking with us. $\endgroup$ – Jyrki Lahtonen Mar 28 '14 at 11:43
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    $\begingroup$ 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 $e^{i\pi} + 1 = 0$. $\endgroup$ – Paramanand Singh Mar 28 '14 at 11:47
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    $\begingroup$ 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 $i$. $\endgroup$ – Balarka Sen Mar 28 '14 at 13:39
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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 $q=Du+Bv\geq B+D \geq 2\times 10^{32}$.

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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.

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  • $\begingroup$ Interesting ref, thanks! +1 $\endgroup$ – Markus Scheuer Jan 15 '15 at 13:38
  • $\begingroup$ @Markus My pleasure! $\endgroup$ – Bruno Joyal Jan 15 '15 at 13:45

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