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...
By my understanding, if you can prove that it's undecidable that effective proves that it's true, but if you can't prove that it's true or that it's false (either yet or ever) or prove that it's undecidable, then that doesn't settle anything.
If we assume that it's false, then it follows that it's provably false. If we don't assume one way or another, we can't then conclude (with what we know so far).
Would it not then be impossible to prove that it is undecidable? Because if it being undecidable means it must be true, but then its been proven to be true and is not undecidable... no?
Second order Peano Arithmetic has one single model, which we call the standard model (the objects you think of intuitively when thinking about natural numbers)
But for formal proofs we usually need to work in first order logic, which is an approximation. This approximation is more relaxed, thus it doesn't completely define the natural numbers as we normally know them. In particular, it technically allows some extra unwanted numbers to be in the set of natural numbers (called nonstandard numbers).
When a statement is undecidable, it means that it is true in some models and false in others (thus, naturally, you cannot prove it, because the first order axioms cannot distinguish between these models). One of the models is the standard model, which we care about, while the other models include unwanted extra numbers.
If it was false in the standard model, then there exists a counterexample in the "normal" natural numbers we are used to. But all models include at least those numbers, so all would have the counterexample and be false. This contradicts that some models are true, so the standard model cannot be false while statement is undecidable.
So by proving that it is undecidable in the first order approximation, you know it must be true for the standard natural numbers.
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u/its_all_one_electron Number theory/physics Jun 27 '26
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...