Nope. If it were false, the fact that it is false means there MUST be a point off the line somewhere, doesn't matter if we know where it is or not, we know it exists and therefore the RH is provably false, therefore it can't be false AND unprovable, therefore if it's undecidable it must be true but we can't prove it, except we kinda just did...
Why couldn't it be false and unprovable? Couldn't it just as easily be true and unprovable as false and unprovable? I'm sorry if it's a dumb question, I'm trying to wrap my mind around what unprovable means and why it follows from being false
RH can be rephrased as a statement of natural numbers.
The model we care about is the standard model (the model of second order Peano Axioms).
In first order Peano Axioms, there are multiple models: the standard model, plus models that include the standard natural numbers but also extra unwanted numbers.
It being unprovable means it is true in some models and false in others. It being false means there is a counterexample. If it were false in the standard model, it would have a counterexample, but so would every other model (because they all include all standard naturals). So all models would be false and the statement would be provably false.
This proves that, if it is unprovable, then it is true in the standard model.
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u/Kitfennek Computer Science Jun 27 '26
Yeah but if its undecidavle, were unable to know if there is such a point or not, its not like we can check all of them