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> What do we actually mean by saying "statement S is unprovable"? Do we mean "there is no [edited:] proof for S" (which includes the case that S is provably fal

> What do we actually mean by saying "statement S is unprovable"? Do we mean "there is no [edited:] proof for S" (which includes the case that S is provably false), or do we mean "there is no proof for S and there is no proof for its converse"? By “converse” I think you mean “negation”. A statement being unprovable means we will never know whether it is this or false. > Because if you show that the converse of Goldbach's conjecture is not provable, you have actually proven Goldbach's conjecture (since you have shown that there is no counterexample!). Nope: demonstrating that (the opposite of Goldbach’s conjecture) is unprovable is logically equivalent to demonstrating that (Goldbach’s conjecture) is unprovable. It means we’ll never know either way.

trashtester 4y ago
> Nope: demonstrating that (the opposite of Goldbach’s conjecture) is unprovable is logically equivalent to demonstrating that (Goldbach’s conjecture) is unprovable. It means we’ll never know either way. I think this is false. If you find a number N that is not the sum of two primes, you did disprove Goldbach's conjecture. Any such number would be smaller than infinity, and so there would only be a finite set of primes smaller it to check for. So basically, if Goldbach's conjecture turns out to be false it IS going to be PROVABLY false. Only if Goldbach's conjecture is actually true will it be the case that it is impossible to prove its negation. But if it is ALSO unprovable (but STILL true) you will NOT be able to prove that the negation is unprovable, because that would mean that there doesn't exist ANY number N that disproves the conjecture, so would prove the unprovable original conjecture.... Consider the statement "All swans are white", but you live in a universe with an infinite number of swans. Lets assume that there exists at least one black swan. Proving the statement false is trivial once you find the first black swan. However, if all swans in the given universe ARE white, and you have no way of inspecting every one, you can never PROVE that they are all white. Also you can NOT prove that it is impossible to prove the negation.
- kubanczyk 4y ago
  > if all swans in the given universe ARE white, and you have no way of inspecting every one, you can never PROVE Yes, you can. Math is not physics. Pythagoras was able to prove his theorem about every triangle in the universe without inspecting every triangle in the universe. The trick is that "the universe" is not the physical universe, it's just a choice of axioms.
  - trashtester 4y ago
    > Pythagoras was able to prove his theorem about every triangle in the universe without inspecting every triangle in the universe. My understanding is that statements about members of infinite sets can be decidable if there exists a method to set up use recoursion/induction to cover all of them. For the set of right triangles, this is relatively straight forward (if we ignore the complexity introduced by real numbers). For statements on infinite sets where it is not possible to reduce a proof to such recoursion/induction, you quickly end up needing an infinite number of steps to cover all cases, meaning the problem is undeciable/uncomputable. See the Church-Turing thesis: https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis

trashtester 4y ago

> Nope: demonstrating that (the opposite of Goldbach’s conjecture) is unprovable is logically equivalent to demonstrating that (Goldbach’s conjecture) is unprovable. It means we’ll never know either way. I think this is false. If you find a number N that is not the sum of two primes, you did disprove Goldbach's conjecture. Any such number would be smaller than infinity, and so there would only be a finite set of primes smaller it to check for. So basically, if Goldbach's conjecture turns out to be false it IS going to be PROVABLY false. Only if Goldbach's conjecture is actually true will it be the case that it is impossible to prove its negation. But if it is ALSO unprovable (but STILL true) you will NOT be able to prove that the negation is unprovable, because that would mean that there doesn't exist ANY number N that disproves the conjecture, so would prove the unprovable original conjecture.... Consider the statement "All swans are white", but you live in a universe with an infinite number of swans. Lets assume that there exists at least one black swan. Proving the statement false is trivial once you find the first black swan. However, if all swans in the given universe ARE white, and you have no way of inspecting every one, you can never PROVE that they are all white. Also you can NOT prove that it is impossible to prove the negation.

kubanczyk 4y ago
> if all swans in the given universe ARE white, and you have no way of inspecting every one, you can never PROVE Yes, you can. Math is not physics. Pythagoras was able to prove his theorem about every triangle in the universe without inspecting every triangle in the universe. The trick is that "the universe" is not the physical universe, it's just a choice of axioms.
- trashtester 4y ago
  > Pythagoras was able to prove his theorem about every triangle in the universe without inspecting every triangle in the universe. My understanding is that statements about members of infinite sets can be decidable if there exists a method to set up use recoursion/induction to cover all of them. For the set of right triangles, this is relatively straight forward (if we ignore the complexity introduced by real numbers). For statements on infinite sets where it is not possible to reduce a proof to such recoursion/induction, you quickly end up needing an infinite number of steps to cover all cases, meaning the problem is undeciable/uncomputable. See the Church-Turing thesis: https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis