# Is $π$ definable in $(\Bbb R,0,1,+,×,<,\exp)$?

(This question is originally from Math.SE, where it didn't receive any answers.)

Is there a first-order formula $\phi(x)$ with exactly one free variable $x$ in the language of ordered fields together with the unary function symbol $\exp$ such that in the standard interpretation of this language in $\Bbb R$ (where $\exp$ is interpreted as the exponential function $x \mapsto e^x$), $\phi (x)$ holds iff $x=\pi$?

(Since a negative answer to this problem would imply that $e$ and $\pi$ are algebraically independent, I cannot expect anyone to give a complete proof that it isn't possible. However, in the case that one suspects strongly that the answer to this question is negative, I would already be pleased if someone could give intuitive arguments why one shouldn't believe in the definability of $\pi$.

However because there are such intricate connections between exponential and trigonometric functions, I don't think that $\pi$ should be undefinable.)

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It seems to me that Schanuel's conjecture (which is a kind of article of faith in transcendental number theory, but of course very far from proven itself) ought to imply that $\pi$ is not definable in this structure. The linked Wikipedia article contains the assertion

• Euler's identity states that $e^{\pi i} + 1 = 0$. If Schanuel's conjecture is true then this is, in some precise sense involving exponential rings, the only relation between $e$, $π$, and $i$ over the complex numbers. [2]

where reference [2] is unfortunately not something I have access to:

• Terzo, Giuseppina (2008). "Some consequences of Schanuel's conjecture in exponential rings". Communications in Algebra 36 (3): 1171–1189. doi:10.1080/00927870701410694.

(link), courtesy of Dominik in a comment below.

See in particular theorem 2.5.1 on page 37, which gives a more precise answer to the OP (under the assumption of Schanuel's conjecture). The bottom line is that a claim that $\pi$ should be definable in this structure would amount to an assertion that SC is false (which would be a pretty big deal, if you know a little about the stature and importance of SC).

Related of course is Wilkie's theorem, which says that any first-order unary formula $\phi(x_1)$ in the language of ordered rings with an exponential function is equivalent to

$$\exists x_{2}\ldots\exists x_n \;\; f_1(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})=\cdots= f_r(x_1,\ldots,x_n,e^{x_1},\ldots,e^{x_n})=0$$

for some exponential polynomials $f_i$ with integer coefficients. So a putative definition of $\pi$ can't be that intricate in principle, at least not in terms of logical complexity.

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I downvoted this answer. The OP's question is very precise, and this response is wishy-washy. In my own view of MO function, the content of this answer deserves only to be a comment under the original post. –  J. Martel Jul 14 '13 at 3:43
@J.Martel The OP explicitly said, if there is strong suspicion that $\pi$ is not so definable, that he wouldn't expect a complete proof but would be happy with some intuitive reasons. I gave a precise conditional response, with a presumably helpful reference, based on something widely believed to be true. It's sort of like if someone asked a question about distribution of primes, and someone responded by saying it would follow from RH; that would be an intuitive reason for believing it. That's an example of a conditional statement, and such statements have good pedigree in mathematics. –  Todd Trimble Jul 14 '13 at 4:02
Here is a link to the paper: fedoa.unina.it/2845 –  Dominik Jul 14 '13 at 10:55
"Tolerable." Thanks so much! Seriously, I think there is room in MO answers to share "mere" intuitions on occasion, when they are useful and relevant and helpful, without being able to pin down every last detail. Normally I am much more precise in my answers, but felt emboldened on this occasion by the explicit invitation to share intuition -- this is actually a topic I've thought about, and my speculations were seriously considered before being tendered, and not half-assed. Since the observations were relevant, we were able to pin down details relatively quickly, much as happens IIRW. –  Todd Trimble Jul 14 '13 at 19:10
If you look at the message that appears when you hover over the downvote button, you will see "this is not helpful". I think usefulness, helpfulness, and relevance should be the benchmarks in deciding whether to downvote -- not whether the answer has reached a definitive state. Downvoting definitely has a chilling effect on people who are trying to be helpful and professional. I hope you will consider this in the future. Cheers. –  Todd Trimble Jul 14 '13 at 19:15