# Is every model of modular arithmetic either even or odd?

Modular Arithmetic (MA) has the same axioms as first order Peano Axioms (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. (http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic)

MA has arbitrarily large finite models based on modular arithmetic. All finite models of MA have either an even or odd number of elements. I call a model of MA "even" if it satisfies either of these two sentences:

E1) $\exists x(x \ne 0 \land x+x = 0)$

E2) $\forall x(x+x \ne S0)$

A model of MA is odd if it satisfies either of:

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

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

We can use compactness to prove MA has infinite "even" size models by adding the even definitions above as axioms. We can similarly prove there are infinite "odd" size models of MA. Some infinite sets, like the integers, are both even and odd. The integers are not the basis for a model of MA. For example, the four square theorem (every number is the sum of at most four squares) is a theorem of both MA and PA. The four square theorem is false in the integers. It has been conjectured the complex numbers are a basis for a model of MA. If so, the complex numbers would be an "odd" model of MA.

My question is whether every model of MA must be exclusively even or exclusively odd? Are the following statements theorems of MA?

$$\exists x(x \ne 0 \land x+x = 0) \ \overline{\vee}\ \exists x(x+x = S0)$$

$$\forall x(x+x \ne S0) \ \overline{\vee}\ x(x = 0 \lor x+x \ne 0)$$

I included the ring theory tag because all of the axioms of ring theory can be derived from the axioms of MA. Every model of MA is a commutative ring with unity. I have found that the 1-element model of MA (the trivial ring) can cause a lot of problems in proofs. I would be happy to prove these statements are true for all models with two or more elements.

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+1, nice question! But I don't understand your statement "Some infinite sets, like the integers, are both even and odd." The integers don't fulfil your axiom E1 -- in what sense do you mean that they're even? Also, it seems to me that a somewhat clearer formulation of your question "whether every model of MA must be exclusively even or exclusively odd" would be "whether E1 and E2 are equivalent in MA, or equivalently whether O1 and O2 are equivalent in MA". – joriki Oct 15 '12 at 5:02
The integers satisfy E2 and O1, so they are both even and odd. Or neither even or odd if you prefer. Proving E1 and E2 are equivalent in MA would answer my question. They are not equivalent in PA. Both are independent of the axioms of MA. – Russell Easterly Oct 16 '12 at 2:13
Posted to MO, mathoverflow.net/questions/119375/even-xor-odd-infinities where there was rather more discussion than there has been here. – Gerry Myerson Feb 7 '13 at 2:16
The MO link posted by Gerry Myerson gives the answer, no. The ring of $p$-adic integers $\mathbb{Z}_2$, satisfying O1 but not O2, and E2 but not E1, is a counterexample. – HTFB Aug 8 '14 at 9:15
I tried to say that ℤ is a model of MA, but Emil Jeřábek immediately pointed out that it is not, because the induction axiom fails miserably (so I deleted the “answer”). On the third approach, I think that in PA-without-∀_x_(_Sx_≠0) one can use induction to prove that if E2 (there are no numbers giving “1” after doubling), then any number is either even (equals to some number doubled) or odd, and an even number is always followed by an odd and vice versa. – Incnis Mrsi Aug 11 '14 at 12:27