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Yeah, that's not quite what I meant. What I was trying to get at was more analogous to the ambiguity of i and -i. Suppose I wanted to create a system of math

by lisper 2y ago

Yeah, that's not quite what I meant. What I was trying to get at was more analogous to the ambiguity of i and -i. Suppose I wanted to create a system of math whose intended model was the negative naturals rather than the positive naturals. Let's call this new system NPA. NPA has the exact same axioms as PA except with predecessor instead of successor as the core axiomatic construct. So zero has a predecessor, but no successor. Everything works as before, except with the letter P instead of the letter S. Addition still works, you're just adding negative numbers and getting negative numbers as the result, which is what you would expect. It doesn't break down until you get to multiplication, where, if you follow the same definition as standard PA you get the "wrong" result because you want the product of two negative numbers to be a positive number, and there are no positive numbers in NPA. So AFAICT multiplication is required to distinguish PA from NPA. Which raises an interesting question: is there a non-trivial operation you can define to distinguish i from -i the way that multiplication allows you to distinguish 1 from -1?

btilly 2y ago
If you start with NPA it is identical as PA. You can apply all the same logic and all the first theorems. The problem with multiplication is not that the proofs fall apart, it is that multiplication does not mean what you want it to mean. When you go to the complex numbers, you have a similar situation. The only thing that distinguishes i from -i is what you want the principal branch of sqrt to be. Otherwise nothing would change if you flip them.
- lisper 2y ago
  > The problem with multiplication is not that the proofs fall apart, it is that multiplication does not mean what you want it to mean. Well, yeah, that's kind of the point: the positiveness of the naturals is not just a convention, it is actually a logical consequence of something in the axioms. Specifically, it is a logical consequence of the definition of multiplication. I find that interesting.
  - btilly 2y ago
    That ties to one of the standard axioms of the real numbers in classical mathematics. There exists a set P which is closed under addition and multiplication, such that for every real number x, one of these three things is true: x is in P, -x is in P, x = 0. P is, of course, the positive numbers. This axiom makes the reals into an ordered field. Add in completeness, and the reals are unique (up to choice of set theory).
Tainnor 2y ago
In every model of a theory, you can just consistently relabel elements in order to get another model. But that doesn't really tell you anything because that model is just isomorphic to the old one (that term has a precise meaning, but I think the "relabeling" intuition captures it). So your NPA is just a consistent relabeling of elements n to elements -n - it's fully isomorphic to PA. There's nothing particularly interesting about it. Notice that "the product of two negative numbers is positive" doesn't help because in your alternative model, "positive" (or rather, "nonnegative") just means something different (or more precisely, it doesn't even make sense to talk about "nonnegative numbers" because that's all of them). It only becomes interesting when you have different non-isomorphic models of a theory - that then means that there are (distinguishable) properties of the model that the theory can't talk about. In the case of the complex numbers, if you just exchange i and -i, I believe that you get a field which is fully isomorphic to C.
- Tainnor 2y ago
  > In the case of the complex numbers, if you just exchange i and -i, I believe that you get a field which is fully isomorphic to C. To add to this and expand a little, there are only two isomorphisms between C and itself which keep all the real numbers fixed: the identity and the conjugate map (which sends a+ib to a-ib, and in particular i to -i). If you don't fix the real numbers there are other such isomorphisms but I believe you can only construct them via the axiom of choice (so you can't really "write them down"). In the case of the real numbers there's really only one such isomorphism, the identity.
  - lisper 2y ago
    Right. So in those terms, NPA is isomorphic to PA -- until you introduce multiplication, at which point the isomorphism goes away becauase S(0)S(0) is S(0) but P(0)P(0) is not P(0). That seems like an interesting fact to me, and feels like it might be a clue pointing to some deeper truth.
    - Tainnor 2y ago
      You can't define a model "NPA" in which all numbers are nonpositive but then define multiplication such that it can yield positive numbers - that just wouldn't be a function. It's pretty easy to prove that S(0)*S(0) = S(0) with the axioms of PA. That has to be true no matter what interpretation you give "S" in your model, so in your proposed model "NPA", in which the domain are the nonpositive integers and S is the predecessor function, of course it has to be true that the "product" of two -1 and -1 is -1. It just means that in such a model, multiplication isn't what you think it is. One should really be careful not to mix up logical systems and their models. The way you're using notation, it's really easy to get confused.

by lisper 2y ago

btilly 2y ago

If you start with NPA it is identical as PA. You can apply all the same logic and all the first theorems. The problem with multiplication is not that the proofs fall apart, it is that multiplication does not mean what you want it to mean. When you go to the complex numbers, you have a similar situation. The only thing that distinguishes i from -i is what you want the principal branch of sqrt to be. Otherwise nothing would change if you flip them.

lisper 2y ago
> The problem with multiplication is not that the proofs fall apart, it is that multiplication does not mean what you want it to mean. Well, yeah, that's kind of the point: the positiveness of the naturals is not just a convention, it is actually a logical consequence of something in the axioms. Specifically, it is a logical consequence of the definition of multiplication. I find that interesting.
- btilly 2y ago
  That ties to one of the standard axioms of the real numbers in classical mathematics. There exists a set P which is closed under addition and multiplication, such that for every real number x, one of these three things is true: x is in P, -x is in P, x = 0. P is, of course, the positive numbers. This axiom makes the reals into an ordered field. Add in completeness, and the reals are unique (up to choice of set theory).

Tainnor 2y ago

In every model of a theory, you can just consistently relabel elements in order to get another model. But that doesn't really tell you anything because that model is just isomorphic to the old one (that term has a precise meaning, but I think the "relabeling" intuition captures it). So your NPA is just a consistent relabeling of elements n to elements -n - it's fully isomorphic to PA. There's nothing particularly interesting about it. Notice that "the product of two negative numbers is positive" doesn't help because in your alternative model, "positive" (or rather, "nonnegative") just means something different (or more precisely, it doesn't even make sense to talk about "nonnegative numbers" because that's all of them). It only becomes interesting when you have different non-isomorphic models of a theory - that then means that there are (distinguishable) properties of the model that the theory can't talk about. In the case of the complex numbers, if you just exchange i and -i, I believe that you get a field which is fully isomorphic to C.

Tainnor 2y ago
> In the case of the complex numbers, if you just exchange i and -i, I believe that you get a field which is fully isomorphic to C. To add to this and expand a little, there are only two isomorphisms between C and itself which keep all the real numbers fixed: the identity and the conjugate map (which sends a+ib to a-ib, and in particular i to -i). If you don't fix the real numbers there are other such isomorphisms but I believe you can only construct them via the axiom of choice (so you can't really "write them down"). In the case of the real numbers there's really only one such isomorphism, the identity.
- lisper 2y ago
  Right. So in those terms, NPA is isomorphic to PA -- until you introduce multiplication, at which point the isomorphism goes away becauase S(0)S(0) is S(0) but P(0)P(0) is not P(0). That seems like an interesting fact to me, and feels like it might be a clue pointing to some deeper truth.
  - Tainnor 2y ago
    You can't define a model "NPA" in which all numbers are nonpositive but then define multiplication such that it can yield positive numbers - that just wouldn't be a function. It's pretty easy to prove that S(0)*S(0) = S(0) with the axioms of PA. That has to be true no matter what interpretation you give "S" in your model, so in your proposed model "NPA", in which the domain are the nonpositive integers and S is the predecessor function, of course it has to be true that the "product" of two -1 and -1 is -1. It just means that in such a model, multiplication isn't what you think it is. One should really be careful not to mix up logical systems and their models. The way you're using notation, it's really easy to get confused.