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In the thread Even Xor Odd Infinities I defined odd models of Modular Arithmetic (MA) as models satisfying the axioms of MA and two first order statements. Even XOR Odd Infinities?

$O1) \forall x(x=0 \lor x+x \ne 0)$

$O2) \exists x(x+x=S0)$

MA has the same axioms as first order Peano Arithmetic (PA) except $\forall x(Sx \ne 0)$ is replaced with $\exists x(Sx=0)$.

Ashutosh proved $O2 \to O1$. Assume $x+x=0$ and $y+y=1$.


Ashutosh's proof doesn't use induction or the axiom $\exists x(Sx=0)$. This means it works even in theories weaker than PA and MA. Emil Jerabek proved $O1$ does not imply $O2$ in MA by proving the 2-adic numbers are a counter example.

The standard models of MA are the rings $Z/nZ$. Any infinite model of MA must be non-standard. Several people have stated any algebraically closed field is a model of MA. If so, Ashutosh's proof shows algebraically closed fields are odd models of MA. The complex numbers satisfy every definition for odd I have been able to come up with. This seems to suggest there are an odd number of complex numbers.

I am looking for first order arithmetic statements true in rings $Z/nZ$ if and only if n is odd. Showing how the statement implies or is implied by other definitions of odd would be an added bonus. Finding a statement independent of others would be interesting. Finding a definition of oddness that is not true in the complex numbers would be even more interesting. Some examples:

$O3) \exists x(x+Sx=0)$

$O4) \forall x \exists y(x=y+y)$

$O5) \forall x \forall y(x=0 \lor y=0 \lor Sx \ne 0 \lor x*y \ne y)$

$O5$ is a complicated way of saying $\forall x(-x \ne x)$

All of these statements are independent of the axioms of MA and require the axiom $\forall x(Sx \ne 0)$ to prove (or disprove) in PA.

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Over MA, O3 and O4 are both equivalent to O2, and O5 is equivalent to O1, so these give you nothing new. – Emil Jeřábek Feb 14 '13 at 11:00
Thanks. So, all of the statements false in PA are equivalent and all the statements true in PA are equivalent. I am still curious if there are definitions of odd that aren't equivalent to one of these two classes. If not, then the complex numbers satisfy every definition of odd. – Russell Easterly Feb 14 '13 at 17:46
Well, you can cook up any number of examples by taking a conjunction of O1 or O2 with a sentence which is valid in all finite models of MA, but not in $\mathbb C$ (such as $\forall x\exists y(y^2=x)\to0=2$). I suppose these are not terribly interesting. – Emil Jeřábek Feb 14 '13 at 20:40
$O2 \land ( \forall x \exists y(y^2=x) \to 0=2)$ is false in PA yet is not equivalent to O2, O3, and O4. Doesn't this prove there must be more than two classes of definitions for oddness? I think a conjunction with any statement true in all finite models works. – Russell Easterly Feb 15 '13 at 1:51
Isn’t that what I just wrote? Anyway, I thought you are interested in more natural examples. – Emil Jeřábek Feb 19 '13 at 11:24

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