October 17 Information

Can the Standard Model of natural numbers be defined, as the model, satisfying Peano system whose ">" relation well-orders all elements of that model?

In other words, do the natural numbers have, any non-standard model whose elements are well-ordered by the ">" relation of Peano system? HOOTmag ( talk) 17:30, 17 October 2015 (UTC) reply

Yes to 1, no to 2. The natural numbers are the unique wellfounded model of Peano arithmetic.

(Quibble: You mean does the theory of the natural numbers have a nonstandard model etc. The natural numbers are a model; they don't have a model.) -- Trovatore ( talk) 18:48, 17 October 2015 (UTC) reply

Thank you; Actually, that's what I thought, but I wanted to be sure. Btw when I posted my question, the person I thought about was you :-)

Btw, I'd thought about a simple way to prove this: On one hand: every number - contained in any model of Peano Arithmetic - is even, or its successor - contained in that model - is even (as one can easily prove from Peano system); On the other hand: had any non-standard model of Peano Arithmetic been well orderable by ">" relation (of Peano system), it would have contained "the smallest infinite number", which (of course) wouldn't have been even, nor would its successor have been even. HOOTmag ( talk) 20:15, 17 October 2015 (UTC) reply

I think that works, if you define "even" as "equal to twice some element". Maybe slightly easier, any nonzero element has an immediate predecessor, but the least infinite one would not. -- Trovatore ( talk) 20:24, 17 October 2015 (UTC) reply

Of course "even" means (in Peano arithmetic) "equal to twice some element", can there be any other definition of "even" (in the context of Peano arithmetic)? However, you're right, your alternative proof is easier... HOOTmag ( talk) 20:32, 17 October 2015 (UTC) reply

By the way, much more can be said. Nonstandard models of arithmetic (doesn't matter much whether we mean recursively axiomatizable Peano arithmetic or the stronger, not recursively axiomatizable true arithmetic) all have a copy of the natural numbers at the bottom (in the < ordering). Above that, they have a bunch of copies of the order type of the integers. Those copies are linearly ordered, in a dense linear order without endpoints.

All countable dense linear orders without endpoints are isomorphic; they have the order-type of the rationals. So putting all this together, you can see that the order type of any countable nonstandard model of arithmetic is N+Z×Q. We actually have an article on nonstandard models of arithmetic, which I just glanced at briefly. -- Trovatore ( talk) 19:07, 17 October 2015 (UTC) reply