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You're right that if ZFC is consistent then it has a model in which this polynomial has an integer root. The issue is that the "integers" in the model are diffe
by OscarCunningham 7y ago
You're right that if ZFC is consistent then it has a model in which this polynomial has an integer root. The issue is that the "integers" in the model are different from the actual integers. So you can't take the root out of the model and use it to prove ZFC inconsistent.
- joe_the_user 7y agoThat is certainly how I would understand the situation. However, it seems intuitively you construct the notion of finite, could engage in the ordinary computation and have a way of distinguishing these "weird roots" from regular roots. I suppose if you make the position of (something like) "every polynomial whose root cannot be found by finite calculation does not, in fact, have one" an axiom but then you would have discarded the notion of finite axioms (you'd have a second order system). Still, might actually have a way to specify "truth", something I'd thought was a bit beyond logic at this point.
- OscarCunningham 7y agoYou might be interested in https://en.wikipedia.org/wiki/Tarski%27s_undefinability_theorem https://en.wikipedia.org/wiki/Tarski%27s_undefinability_theo....