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The answer to the question "Could it be that Goldbach conjecture is undecidable?" claims that it is possible for something such as the Goldbach conjecture to be undecidable, meaning that assuming that it is true and assuming that it is false would both lead to no contradiction.

But if it is undecidable, then, if we assume that it is false, it would mean that exists an even number that cannot be written as the sum of two primes. If a natural number exists, then it can be written down using a finite number of digits (any natural number is definable). This means that that number exists whether or not the conjecture is true, so if we assumed that it was true, there would be a contradiction, so it therefore can't be undecidable.

What is the flaw in what I just said?

Community
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abcd8888
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    Even if Goldbach's conjecture is true, it might not be possible to deduce it from the set of axioms we are working with. In that case if you assume it is false you will not obtain a contradiction, even if you don't have a specific integer that can't be written as a sum of two primes. There's a bit of a disconnect between the syntax and semantics of It. – Matt Samuel Nov 30 '16 at 05:52
  • @MattSamuel: But how is it possible for it to be false not to have an integer that contradicts it? Saying "there is no integer that cannot be written as the sum of two primes" is exactly the same as saying that "any even integer can be written as the sum of two primes"... – abcd8888 Nov 30 '16 at 06:06
  • Right, but if that's true and it's not possible to prove that it's true, then it is consistent to assume it is false even without having a counterexample to back it up. There is nothing behind the scenes making sure that we can back up everything we are able to prove with evidence. We can prove that an integer with more digits than the number of particles in the universe exists logically, but we can't write down all of its digits. – Matt Samuel Nov 30 '16 at 06:14
  • @MattSamuel: So you're saying that even though Goldbach's conjecture is false implies that there is a number for which it fails, there does not necessarily exist a way to represent that number...? Aren't all natural numbers definable (you can define a finite number n as "1 + 1 + ... + 1 + 1 n times"). – abcd8888 Nov 30 '16 at 06:35
  • See Qiaochu Yuan's answer. It is articulated more clearly there. – Matt Samuel Nov 30 '16 at 06:38

2 Answers2

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What is the flaw in what I just said?

Your argument correctly shows that the Goldbach conjecture cannot be false and undecidable. However, it can be true and undecidable.

This is equivalent to saying that it can be true in the standard model of, say, Peano arithmetic, but false in some nonstandard model: in a nonstandard model there may be a nonstandard counterexample to Goldbach's conjecture which is not an ordinary integer, and so which cannot be written down in the usual sense.

The reason it can be true and yet assuming it's false doesn't lead to a contradiction is that if it's undecidable, Peano arithmetic doesn't know it's true, so Peano arithmetic can't use the fact that it's true in a proof.

Qiaochu Yuan
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  • I don't understand why you agree with the OP that the Goldbach conjecture cannot be false and undecidable. There is no reason why the conjecture might fail but not be disprovable in PA. – Rob Arthan Dec 02 '16 at 23:40
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    @Rob: yes there is, and the OP exactly describes why. If Goldbach's conjecture is false (in the standard model), then there is a standard counterexample, and that counterexample constitutes a disproof of it in PA. – Qiaochu Yuan Dec 03 '16 at 01:11
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    @RobArthan: More generally, PA is $Σ_1$-complete, meaning that it proves every $Σ_1$-sentence that is satisfied by $\mathbb{N}$ (from the viewpoint of the meta-system), simply because the witness can be represented in PA by the standard numeral (depending on the variant, of the form "$1+1+\cdots+1$" or "$S(S(\cdots(0)))$", and PA proves any $Δ_0$ sentence satisfied by $\mathbb{N}$. Thus PA disproves every $Π_1$-sentence that is not satisfied by $\mathbb{N}$. However, PA is essentially $Π_1$-incomplete, because of the incompleteness theorem. Sadly, none of this is stated in the linked answer. – user21820 Apr 04 '17 at 13:15
  • What if we talk of ZFC instead? Can it be undecidable in ZFC but true in the external universe isn’t that a contradiction if the external universe is ZFC itself? – Vivaan Daga Feb 07 '24 at 13:20
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if we assume that it is false, it would mean that exists an even number that cannot be written as the sum of two primes. If a natural number exists, then it can be written down using a finite number of digits

Ah not quite. If the proof the negation of Goldbach's conjecture is constructive, and if all the assumptions of the proof are constructive, then what you say is true. Converting a nonconstructive proof into a constructive proof is easy, just convert every nonconstructive inference step into an axiom. But converting a nonconstructive set of axioms into a constructive set of axioms is not trivial. Here are some examples of what would have to be true about every single assumption in the proof for the result to be constructive:

  • For every assumption of the form $A \lor B$, you must be able to actually compute which of $\{A,~B\}$ is true. As a corollary:
  • For every predicate in the proof $P$, where the assumption $P(x) \lor \lnot P(x)$ is assumed, you must be able to actually compute which of $\{P(x),~\lnot P(x)\}$ is true
  • Every assumption of the form $A(x) \implies B(x)$, whenever $A(x)$ is computable, $B(x)$ is also computable
  • Every assumption of the form $A(x)$, $A(x)$ must be computable

And there's more. Even 1 assumption could make the axiom set non-constructive. Let $G(m)$ be the predicate that holds when $m$ is less than 3, odd, or the sum of 2 primes. What if the (dis)proof of Goldbach's conjecture defines a predicate:

$$Q(n) = \forall m > n ~:~ G(m)$$

and made the assumption $Q(x) \lor \lnot Q(x)$. Can you write a computer program that, inputting $x$, outputs true if $Q(x)$ and false otherwise? If not, then the assumptions are not constructive (wrt your computing ability, that's a separate consideration though), and then you can't necessarily infer that the digits of the Goldbach counterexample are computable.

DanielV
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    Sorry, your answer is wrong; it conflates truth with provability, which is invalid. Also, the natural numbers are always expressible as a finite sum of ones. So if Goldbach's conjecture is false then the counter-example can be proven in extremely weak systems including PA$^-$. See this post for a detailed explanation of the conventional meaning of "true" when talking about an arithmetic sentence. – user21820 Apr 04 '17 at 12:57