This is a natural follow-up to my previous question, here: Can natural numbers with multiplication alone define the order relation?. In the structure $(\mathbb{N};+)$, can one define the relation $div$, that is, $x$ is a divisor of $y$? I believe it cannot, but I don't see how to prove it.
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2Maybe I'm not seeing some nuance here but would this work? $$ x \mid y \stackrel{\text{def.}}{\iff} \exists k \in \mathbb{N} \text{ such that } y = \sum_{i=1}^k x $$ – PrincessEev Aug 28 '23 at 00:29
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3@PrincessEev That's not expressible in first-order logic. – Noah Schweber Aug 28 '23 at 00:50
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You are correct that this cannot be done (in first-order logic, anyways). However, since the structure $(\mathbb{N};+)$ is rigid (= has no nontrivial automorphisms), proving this takes more work than the easier result you mention in the OP.
It turns out that in the structure $(\mathbb{N};+)$ - whose complete theory is Presburger arithmetic - every definable set is eventually periodic (this can be proved by quantifier elimination after passing to a stronger language; see here). This lets us show that $div$ is not definable in $(\mathbb{N};+)$: from $div$ we can define the set of primes, and that set is not eventually periodic.
Noah Schweber
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