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How can we show that the addition relation $\{\langle m,n,p\rangle\mid p=m+n\}$ is not definable in $(\mathbb{N};\space \cdot)$ (structure with universe $\mathbb{N}$ and the usual multiplication)?

Graham Kemp
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    Hint: Can you find an automorphism of $(\mathbb{N}; \cdot)$ which does not preserve addition? – Alex Kruckman Sep 10 '19 at 21:47
  • How do you define multiplication without first defining addition? – Dan Christensen Sep 10 '19 at 21:47
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    @DanChristensen Your question is completely irrelevant to the OP's problem. We know what addition and multiplication of natural numbers is. The question is whether there is a first-order formula just involving the multiplication operation, which defines the graph of addition on $\mathbb{N}$. – Alex Kruckman Sep 10 '19 at 21:49
  • @AlexKruckman So, you can't define multiplication without first defining addition. Thanks. – Dan Christensen Sep 10 '19 at 21:52
  • @DanChristensen It's also the case that the graph of multiplication is not definable in the structure $(\mathbb{N}; +)$. From the point of view of first-order logic, neither addition nor multiplication is definable from the other. – Alex Kruckman Sep 10 '19 at 21:53
  • It seems that exchanging two chosen primes whenever they are a divisor of some x works. Thank you for help. –  Sep 10 '19 at 22:11
  • @AlexKruckman I've always thought that multiplication on N was defined as a binary function $+$ on $N$ such that (1) for all $x \in N$, we have $x\cdot 0=0$, and (2) for all $x, y \in N$, we have $x\cdot (y+1)=x\cdot y+x$. – Dan Christensen Sep 10 '19 at 22:16
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    @DanChristensen That's not a first-order definition. Definability in the context of model theory is a technical condition, and doesn't have to do with where the operations/relations "come from," merely their (relative) intrinsic complexities as measured in a very particular way. – Noah Schweber Sep 10 '19 at 22:18
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    @ranger281 That's right! I suggest you write an answer to your own question and accept it - this way the problem won't remain on the "unanswered questions" queue. – Alex Kruckman Sep 11 '19 at 01:01

1 Answers1

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Consider the automorphism $h:\mathbb{N} \rightarrow\mathbb{N} $ that is the identity map except for multiples of some distinct prime numbers a, b, where, in any such $n$, we replace each a by b, and each b by a (e.g., if $n=2*a*a*b$, then $ h(n)=2*b*b*a$).

For any m-ary relation R definable in $\mathbb{N}$, and any $a_{1}, ..., a_{m}\in\mathbb{N}$, we have $\langle a_{1}, ..., a_{m}\rangle\in R \iff \langle h(a_{1}), ..., h(a_{m})\rangle\in R$. By assuming that addition is definable, we arrive at a contradiction.

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    +1. It's worth noting that more broadly the primes "independently generate" (like a basis in linear algebra) the whole structure: any permutation of the primes induces an automorphism of the structure (indeed, automorphisms of the structure correspond exactly to permutations of the primes). – Noah Schweber Sep 12 '19 at 05:02