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Nick_Hay comments on Standard and Nonstandard Numbers - Less Wrong
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No we don't.
The model with the natural numbers and a single "integer line" is not a first order model of arithmetic. The reason is this. For a non-standard number "a" large enough there is a (non-standard) natural number that's approximately some rational fraction of "a." This number then has successors and predecessors, so it has an "integer line" around it. But because we can play this game for any fraction, we need lots of integer lines (ordered according to the total ordering on the rationals).
See this for details:
http://web.mit.edu/24.242/www/NonstandardModels.pdf
Very nice. These notes say that every countable nonstandard model of Peano arithmetic is isomorphic, as an ordered set, to the natural numbers followed by lexicographically ordered pairs (r, z) for r a positive rational and z an integer. If I remember rightly, the ordering can be defined in terms of addition: x <= y iff exists z. x+z <= y. So if we want to have a countable nonstandard model of Peano arithmetic with successor function and addition we need all these nonstandard numbers.
It seems that if we only care about Peano arithmetic with the successor function, then the naturals plus a single copy of the integers is a model. If I was trying to prove this, I'd think that just looking at the successor function, to any first-order predicate an element of the copy of the integers would be indistinguishable from a very large standard natural number, by standard FO locality results.
I think whether naturals plus one non-standard integer line is a model of Peano's axioms for the successor only (no addition/multiplication) depends on whether we use second order or first order logic to express induction. (No in second order formulation due to Dedekind's result, yes for any first order formulation).