3 ms·
You actually do not have to remove any axioms from Peano arithmetic to prove that a model of PA with such a 'bleem' exists. It is a straightforward consequence
by metashurick 15y ago
You actually do not have to remove any axioms from Peano arithmetic to prove that a model of PA with such a 'bleem' exists. It is a straightforward consequence of the compactness theorem: just enrich PA with a new constant c and an infinite series of axioms 'c != N' for every numeral N.
For any finite set of these new axioms a model exists (just set c to a large enough number), therefore by the compactness theorem there exists a model which satisfies all our new axioms.
See also http://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic http://en.wikipedia.org/wiki/Non-standard_model_of_arithmeti...
- cperciva 15y agoSorry, I should have been clearer -- I meant arithmetic with Peano's original axioms (i.e., with 2nd-order induction).
- shasta 15y agoExcept the number you get this way is not between 3 and 4. It's essentially infinite with no functions in the model that separate it from the finite numbers (and this can be proven in PA: if a number isn't 0..N, then it's greater than N). Generally, I don't think it's fair to characterize this story as something studied by mathematicians. Really, it's an exercise in reasoning about nonsense.
- cperciva 15y agoDepends what your definition of < is. Some definitions leave unspecified what the relationship between two values not in {0, 1, 2, ...} is.
- shasta 15y agoInequality can be defined in Peano arithmetic in a total way. Maybe you're talking about defining some relation on the model? But what justifies calling such a relation an inequality?