Pick a natural number n and apply the following transformation to it: if n is even, the output will be n / 2; otherwise, if n is odd, the output will be 3n + 1. Keep doing this until 1 is reached.
Will this procedure always reach 1? That is, is this a total function? The Collatz Conjecture says yes, but this has not been proved or disproved. It has also not been proved or disproved that it can be proved or disproved.
John Conway argued that the Collatz Conjecture is both true and unsettleable: it cannot be proved, and it cannot be proved that it cannot be proved, and so on. This is not a particularly satisfying situation, but it could very well be the case.
Call the Collatz transformation C. Notice that if C(n) really does reach 1 and terminate for a given n, this is always provable, and indeed the number of iterations required can be explictly calculated. That is, we can exhibit a number k such that Ck(n) = 1. So if the conjecture is true but not provable, then this means that each instance is provable, but the generalization is not. In symbols:
- ⊢ ∃k Ck(1) = 1
- ⊢ ∃k Ck(2) = 1
- ⊢ ∃k Ck(3) = 1
- ⊢ ∃k Ck(4) = 1
- ⊬ ∀n ∃k Ck(n) = 1
This last statement is formalization of the Collatz conjecture, and we’ll refer to it as CC.
But “provable” isn’t an absolute property – provability is always with respect to some theory. What theory are we talking about here? The answer is: take your pick. The conjecture is not known to be decided by any theory in common use: Peano arithmetic (PA), set theory, whatever. For simplicity, we’ll assume PA as our background theory.
Let’s assume that CC is true and also not provable in PA. These assumptions imply that the negation of CC is not provable either, or in other words that CC is independent of PA. This being the case, we are free to add either CC or its negation to PA as new axioms and the resulting theory will be consistent. Adding the negation, however, yields a theory that is 𝜔-inconsistent, or more simply, unsound. It says things about the natural numbers that are not true. (Alternatively, it can be looked at as a theory of number-like objects that are not numbers.)
So we’ll add CC as a new axiom to PA. Call the resulting theory Collatz arithmetic (CA). CA is strictly stronger than PA, since CA ⊢ CC but (by hypothesis) PA ⊬ CC. What other kinds of things can be proved in CA but not PA?
Before getting to that, let’s get C into a more workable form. To say that n is even is to say that has the form 2k, so instead of saying that C(n) = n / 2, we can say that C(2k) = k. Similarly, to say that n is odd is to say that it has the form 2k + 1, and 3(2k + 1) + 1 = 6k + 4. But notice that if n is odd, 3n + 1 will always be even, immediately triggering the even clause. The Collatz function can be “accelerated” slightly by applying this immediately. With a little algebra, this works out to C(2k + 1) = 3k + 2.
Let’s consider a variation of the Collatz function that we’ll call the Half Collatz function. Define a transformation function H to be just like C except that it leaves its even arguments alone – in other words, H(n = 2k) = n and H(2k + 1) = 3k + 2. And instead of applying its tranformation until the argument reaches 1, apply it until its argument is even.
Will the Half Collatz function eventually reach an even number for all inputs? Call this statement the Half Collatz Conjecture (HCC). Like CC, HCC appears to be both true and unprovable, PA ⊬ HCC. But HCC is provable in CA, CA ⊢ HCC. For suppose that n is a number > 1 such that repeated applications of H never reach an even number. By CC, there is a k such that Ck(n) = 1. n > 1, so it must get reduced at some point. That reduction only happens when the even clause of C is triggered, so n must eventually reach an even number. ⊣
Finally, consider one more transformation function: B(n = 3k) = n and B(3k + r) = 5k + 3 + r (with r = 1 or r = 2). We’ll call this the Beaver Collatz function, since this is the function implemented by the 4-state 2-color Beeping Busy Beaver champion. The Beaver Collatz Conjecture (BCC) says that repeated applications of B will eventually reach a number divisible by 3 for all inputs. (For a good time, try starting with 2.) BCC seems to be true, and presumably it is also unprovable in PA. Is BCC provable in CA, CA ⊢ BCC? I conjecture that it is.
We saw that CA ⊢ HCC. But this is equivalent to saying that PA ⊢ CC → HCC, so why bother with a new axiom? Why not just stick with PA and take CC as a hypothesis? Well, if Conway is right, CC is both true and unsettleable. To me this says that CC opens up a whole new method of reasoning, a method of reasoning that is totally inaccessible otherwise. “Unsettleable” is a strong word, and Collatz-based arguments must be correspondingly strong.
But if CA ⊬ BCC, then the new method of reasoning seems somewhat weak. BCC would presumably be unsettleable even in CA, and would therefore have to be added as yet another new axiom. There’s a whole world of Collatz-like functions; are they all co-unsettleable? I suppose that’s possible, but it’s also possible that there are classes of such functions that are co-provable. That is my feeling.