Peano's axioms postulate the existence of some entity $0$ along a successor function which together meet several conditions. I'm wondering whether it possible to axiomatize the natural numbers by starting instead with the prime numbers, along with probably the number $1$ and the operation of multiplication?
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As in have a countably infinite set of constant symbols, one for each prime? – eyeballfrog Oct 08 '19 at 21:16
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You'd need infinitely many "successor" functions, one for each prime number. – 79037662 Oct 08 '19 at 21:16
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1Not sure if this is possible in theory, but such a system would have significant difficulties with defining addition. – 79037662 Oct 08 '19 at 21:18
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@eyeballfrog Yes, I think there's no way around introducing an infinite set of symbols, one for each prime. – pseudosudo Oct 08 '19 at 22:28
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The structure $(\mathbb{N}; \times)$ is much less "contentful" than $(\mathbb{N}; +)$. In particular, $(\mathbb{N}; +)$ is rigid - it has no nontrivial automorphisms. By contrast, $(\mathbb{N};\times)$ has lots of automorphisms, namely one for every permutation of the primes. As a consequence of this, addition is not definable from multiplication alone, so the answer to your question is negative in a strong sense.
Adding constants naming each individual prime doesn't help either: addition is definable in a structure iff it is definable in the structure using only finitely many symbols (since definitions are finite), but fixing finitely many primes leaves lots of nontrivial permutations of the primes which don't respect addition.
Strictly speaking, the second paragraph refers specifically to first-order definability, while the first paragraph applies to any notion of logical definability which is "isomorphism-respecting." (And see this old answer of mine for a discussion of definability; there the focus is on first-order, but the setup is generally applicable.) For example, addition is definable in the natural numbers with multiplication and a constant symbol for each prime using an $\mathcal{L}_{\omega_1,\omega}$ formula.
First-order logic is in many ways quite limited - in particular, it doesn't generally allow "definition by recursion," and we have:
Addition is not first-order definable in the naturals with successor.
Multiplication is not first-order definable in the naturals with successor and addition.
Each of these facts is special to first-order logic - e.g. Peano's original formulation of arithmetic used second-order logic and in that context successor is enough to get addition and multiplication.
In some special situations definition by recursion may be first-order - e.g. the bulk of the proof of Godel's incompleteness theorem amounts to showing that definition by recursion is first-order once we have addition and multiplication - but it shouldn't be taken for granted. (See e.g. the middle part of this answer of mine.)
Noah Schweber
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