Sewing-Machine comments on Edward Nelson claims proof of inconsistency in Peano Arithmetic - Less Wrong

13 Post author: JoshuaZ 27 September 2011 12:46PM

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Comment author: [deleted] 01 October 2011 12:15:52AM *  1 point [-]

I think I can explain that. Nelson doesn't believe in any kind of Platonic number, but he does believe in formal number systems and that they can sometimes accurately describe the world. We have essentially two number systems: the tally-mark system (e.g. SSSSS0 and so on), and the positional system (e.g. 2011 and so on). Both of these are successful at modeling aspects of the world.

Most people regard these two number systems as equivalent in some sense. Specifically, there is a procedure for rewriting a positional number as a tally number, and most people believe that this procedure terminates "at least in principle." (For instance, one can formally prove this in PA using the induction axiom). But Nelson denies it. He believes that the positional number system and the tally number system are different, and that the positional number system is strictly stronger. I think your remark --

This argument applies just as well to addition and multiplication.

-- applies to the tally system, but not to the positional system. But Nelson does not believe that the positional system is inconsistent. So he is happy to adjoin axioms to the tally system that allow for working with addition and multiplication.

Comment author: benelliott 01 October 2011 12:28:46AM *  1 point [-]

My knowledge of complexity theory is very limited, but I thought exponentiation could be done in polynomial time. Is that incorrect, or am I misunderstanding Nelson's argument (please don't tell me he also doubts complexity theory, we must be allowed some fun :P).

EDIT: Yes, I was wrong, the claim of polynomial time in the length of b may well be too strong, but it does seem to use less than 2^b. I think I can use the position system to compute 2^b in O(b^2) steps, which seems to lead us straight from accepting multiplication to accepting exponentiation.

Comment author: [deleted] 01 October 2011 01:14:25AM 1 point [-]

I think so, in positional notation. That's what I was alluding to with my O(log log k) comment. But there is no polynomial time algorithm to reduce 2^n to a tally number (obviously).

I once read something by Nelson about the polynomial hierarchy in complexity theory, but I can't find it now.