# Undecidable conjectures

Suppose we work with a conjecture saying that something is true for any natural $n$. For each $n$ there exists an algorithm of finite length allowing one to decide whether it is true or false for this $n$, and suppose that we even have an explicit formula how the length depends on $n$. It is proved that the conjecture is true for all numbers until, say, $10^{1000}$; no counterexample is found. Is it possible that this conjecture is undecidable?

As far as I understand, the answer is yes, but I do not understand the following. If the conjecture is false, then we can take the corresponding $n$ and hence we get a finite proof of that fact. If it is undecidable, than we may add the axiom that it is false, and then we seem to arrive at a contradiction with my preceding phrase. Could anyone explain what is wrong?

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Hi all, thanks for your comments. I will read them carefully and after that I'll get back. –  frog Aug 13 '11 at 8:51

We will show where your intuitive argument breaks down. Call the conjecture $\varphi$, and suppose that $\varphi$ is undecidable. Then, as you observed, under your very strong assumptions, $\varphi$ is true in the natural numbers, but not provable.

Not provable in what theory? By undecidable we always mean undecidable in a particular theory. Say that theory is PA, first-order Peano Arithmetic. But for the rest of this post, PA could be replaced by any strong enough theory that has the natural numbers as a model.

Let us add to PA the axiom $\lnot\varphi$ as you specified. Then the theory $T$ with axioms the axioms of Peano Arithmetic, together with $\lnot\varphi$, is consistent, and therefore has a model $M$. In $M$, the conjecture $\varphi$ is false.

This model $M$ is not (isomorphic to) $\mathbb{N}$, since $\varphi$ is true in $\mathbb{N}$. The object $\omega\in M$ that "witnesses" the falsity of $\varphi$ in $M$ is therefore not a natural number. Your algorithm will not be applicable to $\omega$, which is, in the informal sense, greater than $1$, $1+1$, $1+1+1$, and so on. (We can without loss of generality assume that $\mathbb{N}\subset M$). Informally, under your assumptions, any $\omega$ that witnesses the falsity of $\varphi$ in $M$ must be an "infinite" integer.

If a certain argument breaks down, it is still possible that the result one is after is true! However, we can find explicit examples of sentences $\varphi$ of the type you described that turn out to be undecidable in PA.

In particular, one can construct an explicit Diophantine equation $E$ (in $k$ variables, where $k$ is not very large, last I heard about $20$) which has no solutions in $\mathbb{N}$, but such that this fact is not provable in PA. Let $\varphi$ be the assertion that this Diophantine equation has no solutions. For any specific $k$-tuple $(a_1,a_2, \dots, a_k)$ of natural numbers, the fact that $(a_1,a_2,\dots,a_k)$ is not a solution of $E$ is easily checkable, by a computation of predictable length. By suitably encoding the $k$-tuples of natural numbers, we can make sure that $\varphi$ satisfies your conditions.

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+1 for a very clear explanation, but this brings up a confusing point to me. It seems to assert the reality of $\mathbb N$ as a structure that's not defined using axioms. Is this a Platonist perspective or is something more subtle going on? –  Grumpy Parsnip Aug 12 '11 at 14:03
I used a semantic argument (models) partly because of a wish for clarity. Also the OP's question itself was framed in terms of truth. And we were doing classical number theory, or functional analysis, we would not bat an eye at taking for granted the "reality" of $\mathbb{N}$, and indeed of far more exotic objects. So why not apply the same standard to undecidability results, which are, after all, just plain old theorems. For my answer, I could have taken a far more syntactic line, at the likely price of intelligibility. –  André Nicolas Aug 12 '11 at 14:18
The Platonist point of view is certainly the easiest to digest. I'm not criticizing your presentation, just trying to get a sense of the issues involved. –  Grumpy Parsnip Aug 12 '11 at 14:58
@Jim: The second-order theory of arithmetic is categorical, so if you're willing to believe that our universe of sets is real, then there is a unique $\mathbb{N}$ in it. –  Zhen Lin Aug 13 '11 at 2:24
@Jim: the natural numbers, as a set in ZF(C), constitute a model of Peano arithmetic, and it is this model that people often implicitly refer to when they talk about $\mathbb{N}$. By the way, first-order Peano arithmetic is much weaker than you think it is: the first-order axiom schema of induction only allows induction to be applied to subsets of $\mathbb{N}$ that can be described by a first-order formula. In particular, there are only countably many such subsets. –  Qiaochu Yuan Aug 13 '11 at 3:26

Assume you are working on the conjecture of consistency of PA (denote it $con(PA)$) in PA itself. $con(PA)$ essentially says for each $n$, it is not the case that $n$ is the Godel number of a proof of a contradiction - i.e. $con(PA)$ denotes the formula $\forall x \lnot Prov(x, 0 = 1)$. So $con(PA)$ is a type of conjecture you are looking at.

We know that for each natural number $n$, ${\rm PA} \vdash \lnot Prov(n, 0 = 1)$. The universal quantification (i.e. $con(PA)$) on the other hand, is not provable in PA. If you add $\lnot con(PA)$ to PA, then you end up with a theory that is consistent, but not $\omega$-consistent. The later means that the theory proves $\exists x p(x)$ and $\lnot p(0), \lnot p(1), ...$ for some $p$.

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Goldbach's conjecture (which states that if $n\ge4$ is even then it is a sum of two primes) is of this type. If it's false, then there's a counterexample, and so a proof that it's false; hence, if it's undecidable (within some particular axiom system), then it's true.

The twin prime conjecture (that there are infinitely many primes $p$ such that $p+2$ is also prime) is a different kettle of fish. The argument above does not apply, so it could be undecidable in a stronger sense.

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"...if [Goldbach] is undecidable then it's true." I believe that would be "in the standard model". –  Arturo Magidin Aug 13 '11 at 2:45
@Arturo, yes, thanks. –  Gerry Myerson Aug 13 '11 at 6:50