[FOM] Large cardinals and finite combinatorics
Timothy Y. Chow
tchow at alum.mit.edu
Mon Jun 18 14:25:05 EDT 2007
Joe Shipman wrote:
>For a given natural number n, there is a well-defined operation * on
>the integers mod 2^n which can be calculated recursively using 2 rules:
>
>p*1 = (p+1) mod 2^n
>
>p*(q*r)=(p*q)*(p*r) [self-distributive law]
>
>The 2^n x 2^n "multiplication table" is called the "nth Laver table".
>
>It can be shown with large cardinals (using an embedding from a rank
>into a rank) that for any k there exists n such that, in the nth Laver
>table, (1*1) does not equal (1*(2^k)).
Do you mean 1*(2^k+1) here?
This is interesting...I had not heard of it before. Does anyone suspect
that this is unprovable in ZFC? Comparing it with Friedman's examples, I
would be surprised if this required a rank-into-rank.
Here's a simple observation. In the nth Laver table, relabel every
integer i with the label "2^n - i." Thus rows and columns are reversed
and the entries are replaced with their distance from 2^n. Then the top
half of the nth Laver table consists of two copies of the (n-1)st
relabeled table. I don't know if this observation is worth anything---it
might mess up all the nice algebraic properties---but I didn't see it
explicitly mentioned when I briefly glanced at a couple of papers on the
subject.
Tim
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