Of course the natural numbers can be put into 1-1 correspondence with the rationals -- how to show that is the classic Cantor diagonal argument.
At a tea before a seminar, I asked Max Zorn what Paul Cohen had just proved. Zorn didn't explain it and instead just loaned me his copy of Cohen's paper -- I lost it in a move! If Zorn wasn't strongly interested in Cohen's proof, then neither should I be.
For any set of axioms, there will be statements we can neither prove nor disprove. Surprising, interesting, got it.
Zermelo-Fraenkel set theory with the axiom of choice seem fine to me -- now back to the real work!
So if were the case that ZFC+¬Con(ZFC) was sound then what it proves would be true, and "ZFC+¬Con(ZFC) is inconsistent" would be true. But that would mean ZFC+¬Con(ZFC) is actually inconsistent, and thus ZFC+¬Con(ZFC) would actually be unsound after all.
But to be a bit more specific, ¬Con(ZFC) says that the Turing machine that searches for a contradiction in ZFC does indeed halt. However (in all likelihood) such a machine does not actually halt, in the sense that it does not halt in 1 step, and it does not halt in 2 steps, and it does not halt in 3 steps, etc., and indeed (in all likelihood) for each numeral n, ZFC even proves that the machine does not halt in n steps.
(Now there is a small possibility that maybe such a machine does halt in some particular number of steps. If it does actually halt, that means it has found a proof that ZFC is inconsistent. But this scenario is even worse, because it means that ZFC itself is not only unsound, but inconsistent, (and hence ZFC+¬Con(ZFC) is also inconsistent). In particular ZFC+¬Con(ZFC) being inconsistent means it proves that every Turing machine halts, which is even more wrong in general, even if it happens to be right about this particular machine.)
Er, no. The fact is that there is no fact of the matter as to whether those particular Turing machines either a: do not halt at all, or b: halt after a (colloquially) infinite number of steps. (A implication of there being no fact of the matter is that, empirically, we can't tell the difference by running them, but we can't tell the difference for a machine that halts in 2^(2^(2^256)) steps (or not at all) either, so that's not very interesting on it's own.)
(As you note, we don't actually know that (for example) a Turing machines looking for contradictions in ZFC is one of those particular machines; indeed I'm not sure offhand if we actually know of any specific example of such a machine. But that's presumably not the issue here.)
So again, we have a proof in some sketchy axiom system that the machine that searches for an inconsistency in ZFC halts. The question is whether that conclusion is a true statement or not, whether that machine actually does or does not halt. You can start up that program on your laptop while we consider it. Maybe it will halt during our discussion. I also want to point out that whether that conclusion is a true statement or not has no bearing on whether the deduction system was sound to begin with. Unsound systems can prove true statement, simply by accident rather than by design.
Case (1) that machine doesn't really halt, then ZFC+¬Con(ZFC) proves an untrue statement, and we thus the system is unsound.
Case (2) that machine does really halt. Then this particular conclusion wasn't untrue, but we have another problem. We can take the output of that machine, decode it, and we have a proof of an inconsistency in ZFC. Again this is just a deduction, it doesn't mean that math is wrong or the world ends. Just that the ZFC axiom system is so unsound that it is inconsistent, and it has some untrue axioms. Nevertheless we can use such a proof and derive a proof that ZFC prove `while(true)` halts, and in turn transform that into a proof in ZFC+¬Con(ZFC) that `while(true)` halts. And it is definitely the case that `while(true)` doesn't halt. So in this case we have found a different but still untrue statement that ZFC+¬Con(ZFC) proves. Our conclusion is the same: ZFC+¬Con(ZFC) is unsound.
So in either case `ZFC+¬Con(ZFC)` proves some machine halts that doesn't. I haven't concluded which of the two machines it is wrong about though I have my personal suspicions, but it is definitely wrong about one of them.
Axioms are true by definition; that's what "axiom" means. (They might also be false, but (if and) only if your system is inconsistent.)
> Case (1) that machine doesn't really halt
This is not a state of affairs: all you know at any given time is that the machine hasn't halted yet (after some specific finite number of steps), or that it has aquired some termination-precluding invariant like a infinite loop (which you can detect only strictly less than all of).
Regardless, the particular technical definition of truth I'm using is Tarski's definition of truth[1]. We don't have to agree whether Tarski's definition is "correct", and I'm happy to qualify my use of the word "truth" here, but I'm referring to a specific technical definition here.
[1] https://en.wikipedia.org/wiki/Semantic_theory_of_truth#Tarsk...
Where can I read more about that? Because, both are infinite, but there still should be more rational numbers, than natural numbers?
Your intuition about there being more rational numbers might be based on viewing the rationals as a proper superset of the natural numbers. You might similarly consider that there are more natural numbers than even natural numbers. However, by "renaming" every even number, and that shouldn't change how many there are, to half its value, we obtain the natural numbers. Formally, there exists a bijection between N and 2N, as between N and Q, and this is what mathematicians mean when they say that sets have the same size, or cardinality.
In this case, at most one of these two a priori quite reasonable statements can be true:
- if set B is a strict superset of A, B is larger than A.
- if you can map the times in A to the items in B in 1:1 fashion, A and B have the same size.
Giving up the first is deemed less problematic than giving up the second (probably because that means giving up comparing sizes of infinite sets at all, but I’m not familiar enough with that to be sure about that)
for a in range(1, Infinity):
for b in range(a):
print(a-b, "/" , b+1)
1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ...
QED
Google ”counting rational numbers” and ”different sizes of infinity” to learn more.
I don’t see why that should imply it wouldn't be "okay" to add ¬Con(ZFC).
It may be highly counterintuitive, but the history of mathematics is full of counterintuitive results that nowadays are accepted as true in mainstream mathematics.
Well-known examples are the existence of irrational numbers, the claim that the set of natural numbers has the same size as that of the rational numbers, the existence of hyperbolic geometry, and the Banach-Tarski paradox.