6
$\begingroup$

Let $S$ be some statement which is unprovable but true in an axiomatic system $T$. If $T$ is consistent, then adding $S$ as an axiom of $T$ keeps the system consistent. But what about adding $\neg S$ as an axiom?

For example, the continuum hypothesis is unprovable in ZFC, and we can add it or its negation as an axiom with no problem.

However, if the Goldbach conjecture is unprovable, it must be true, since if it were false we'd be able to find a counter-example, and check that it is a counter-example. Hence we wouldn't be able to add the negation of the Goldbach conjecture as an axiom if it is unprovable. Does this imply it is provable? Or is this okay?

$\endgroup$
  • $\begingroup$ Why does the Goldbach conjecture have to be true? How do you know there isn't a counter-example that is too big to write down? $\endgroup$ – John Douma Oct 30 '18 at 21:31
  • 2
    $\begingroup$ Yes, we can add the negation of Goldbach conjecture. "True" applies not to the formal system itself (in this case arithmetic), but to its "intended" model. If the Goldbach conjecture is unprovable it means that it has a counterexample in a non-standard model of arithmetic (by Gödel's completeness theorem). If we add the negation our theory will no longer describe "intuitive" integers (assuming there is such a thing) but it will still be a fine theory without contradictions. $\endgroup$ – Conifold Oct 30 '18 at 21:31
  • 2
    $\begingroup$ @JohnDouma: Why do you think I can reason it must be true? I'm reasoning that if it is undecidable in PA, then it is true. But I have no idea whether it is undecidable in PA. $\endgroup$ – Henning Makholm Oct 30 '18 at 21:42
  • 1
    $\begingroup$ @Sambo: Yes: $\varphi$ is "unprovable" if $\varphi$ cannot be proved, and is "undecidable" if neither $\varphi$ nor $\neg\varphi$ can be proved. $\endgroup$ – Henning Makholm Oct 31 '18 at 0:29
  • 1
    $\begingroup$ @Sambo: An undecidable sentence might also "simply be false" in a particular model. An unprovable sentence is either undecidable or disprovable. $\endgroup$ – Henning Makholm Oct 31 '18 at 11:43
6
$\begingroup$

Yes, you can add the negation of any unprovable statement as a new axiom to the theory and still get a consistent theory out of it.

(If the extended theory proved a contradiction, this would directly be a proof of the original "unprovable" statement, which therefore wouldn't be unprovable after all).

If the new axiom happens to be false in the interpretation of the theory you had in mind (such as the actual $\mathbb N$), this interpretation will not be a model any more, of course. The extended theory will have other models that are non-standard models of your original theory.

$\endgroup$
  • $\begingroup$ So in the case of the Goldbach conjecture, if the statement is true but undecidable, adding its negation asserts the existence of an even number which is not the sum of two primes. This number would therefore have to be something which isn't part of what we usually think of as $\mathbb{N}$? $\endgroup$ – Sambo Nov 7 '18 at 14:46
  • $\begingroup$ @Sambo Correct. $\endgroup$ – Henning Makholm Nov 7 '18 at 14:48
  • $\begingroup$ So that means that if the Goldbach conjecture is undecidable, then it must be true in "the actual $\mathbb{N}$", even though we could add its negation as an axiom and end up with a consistent theory. $\endgroup$ – Sambo Nov 7 '18 at 14:54
  • $\begingroup$ @Sambo Again correct. (Note that there still seems to be much more hope of proving Goldbach fair and square than of figuring out a way to prove it independent of say PA) $\endgroup$ – Henning Makholm Nov 7 '18 at 14:57
  • $\begingroup$ Thanks for the clarifications! $\endgroup$ – Sambo Nov 7 '18 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.