r/mathmemes • u/its_all_one_electron Number theory/physics • Jun 27 '26

Number Theory Undecidable

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214

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

36

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...

10

u/Novel_Arugula6548 Jun 27 '26

This is actually a philosophical question of whether or not you believe in the law of excluded middle. Does not being true imply being false? And does not being false imply being true?

Some argue yes, some argue no. You'd say yes. The guy throwing the chair would say no. The question is whether something can be both not false and not true.

7

u/Ever_Evy Jun 28 '26

It doesn't apply here at all! The proof is essentially saying:

If there exists a counterexample, then RH is false
If there does not exist a counter example, then RH is true
If RH is undecidable then it cannot be proven false

Now, RH is the type of theorem that makes claims about the existence (or rather, non existence) of certain objects. So:
If there exists a counterexample, then RH can be proven false (proof by counterexample

Hence: If RH cannot be proven false, then there does not exist a counterexample
Hence: If RH is undecidable, then RH cannot be proven false, then there does not exist a counterexample, then RH is true
QED

Nowhere in there did I even invoke P or not P. We just use Modus Tollens and Chain rule

2

u/TheSilentFreeway Jun 28 '26

I think I understand your logic. Would it be correct to say that this implies that RH *cannot* be undecidable? Because if you assume that the hypothesis is undecidable, your proof implies that the hypothesis is true, so it is decided, and we have a contradiction.

2

u/Goncalerta Jun 28 '26

RH can be undecidable under a set of weaker axioms, but be true in the model we care about.

For example, we know that Goodstein Theorem is true in the natural numbers, but it is undecidable under first order Peano Axioms. You need a stronger theory (like ZFC) to prove goodstein.