Given first-order logic with equality and the real field $\mathbb{R} = (R, 0, 1, <, +, \cdot)$, is $\pi$ first-order definable?

By first-order definable, I mean a sentence of the form $\exists x \;\phi(x)$ such that $\pi$ is the only element in $R$ satisfying $\phi$.

  • 1
    $\begingroup$ I doubt it as $\pi$ is transcendental and cannot be expressed as the root of a polynomial. $\endgroup$
    – user65203
    Jun 24, 2016 at 9:27
  • $\begingroup$ Maybe this would be helpful? If you could convert each binary digit into the actual value by multiplying by the appropriate power of two, and add all such values, you would get $\pi$. I don't know whether this could be stated as a first-order logic statement. $\endgroup$ Jun 24, 2016 at 9:30
  • 2
    $\begingroup$ The conventional wording for your "pinpoint" would be to ask whether $\pi$ is first-order definable in the structure $(\mathbb R,0,1,{<},{+},{\cdot})$. $\endgroup$ Jun 24, 2016 at 9:47
  • $\begingroup$ @shardulc It cannot, in fact. $\endgroup$ Jun 24, 2016 at 10:23
  • $\begingroup$ Depending on the fine details of the question, there is a cheat: if you've given me $R$ to use as constant symbols, then I can use $\phi(x) := (x = \pi)$. But the answer is "no" if $0$ and $1$ are the only constant symbols you give me. $\endgroup$
    – user14972
    Jun 24, 2016 at 11:23

1 Answer 1


No, you cannot. Although $\pi$ has "reasonably concrete" definitions in terms of $+, \times, <$ (e.g. via infinite series), none of them can be made first-order. This follows, e.g., from the fact that:

  • The algebraic reals form a real closed field.

  • The theory of real closed fields is complete, and in fact if $F_1, F_2$ are real closed fields with $F_1\subseteq F_2$, then $F_1\preccurlyeq F_2$.

  • $\pi$ is not algebraic.

  • 3
    $\begingroup$ In somewhat more detail, the second bullet point implies that any first-order statement true of $\mathbb{R}$ is also true of the real algebraic numbers, so if there were a first-order statement picking out $\pi$ the statement "$\pi$ exists" would be first-order, hence true of the real algebraic numbers, which would mean that $\pi$ is algebraic. $\endgroup$ Jun 24, 2016 at 18:21

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