
Re: Two Finite Arithmetics
Posted:
Apr 8, 2014 11:47 PM


On Mon, 7 Apr 2014, Dan Christensen wrote: > > > > Naive Finite Arithmetic > > > > Let N be a set, 0,m two elements and SN > N a function. > > > > Axioms for naive finite arithmetic: > > > > 0, m in N; Sm = m > > > m should have no successor. Perhaps a partial function S on N? > > Naive finite arithmetic is not your finite arithmetic. > > S is a full function to banish the pests of your system. > Inventing a successor for the maximum number seems a bit weird to me. It fits much better than not having one. > An elementary result of ordinary number theory is that no number can be it's > own successor. You can probably obtain the same result with your axioms, so > you may need another axiom to get around this potential contradiction. I doubt it. The theorem would likely be for all x /= m, x /= Sx
> > > > for all x, Sx in N > > for all x, Sx /= 0. > > > > for all x,y /= m. (Sx = Sy implies x = y) > > > > For all A subset N, if > > > > 0 in A, (for all x in A implies Sx in A) > > > > then N subset A > > > > Definition of addition by induction. > > > > 0 + y = y > > Sx + y = x + Sy > > > Is this where the "naive" comes in? The use of the infix '+' really > > > needs to be justified, i.e. the sum of a pair of a numbers should be > > > formally proven to be unique, however you may define sums. > > That definition is a binary function over N^2. > > That it works for double induction is common knowledge. > So, you are defining your way out of the problem. Why prove sums are unique > when you can just assert that it is true? It's, ahem... "common knowledge" > after all, right?
It's a problem of your creation.
> > > As with your function S, perhaps '+' should be a partially function. > > No. That's a mess for the partiallity is difficult to describe. > > This way, once you get to m, then adding more is not more. > > > Also, shouldn't you have something have something like Sx + y = S(x+y)? > > Hey, that's a better idea, avoiding double induction > > as now the inductive definition is the add y function. > > > > > > Definition of mulplication by induction. > > > > 0 * y = 0 > > > > Sx * y = x*y + y > > > > Sx + y = x + Sy > > > > Is this a consistent set of axioms with the model of a finite > > > > set of integerss { 0,1,.. m } and addition defined by > > > > a + b = max{ m, a+b }? > > For the model work, I should have carified that + is usual addition > > ++ the naive finite arithmetic and write, by definition > > a ++ b = max( m, a+b } > > > Again, you probably want a partial function. The way I see it, m + 1 should > > > be undefined. > > No way!. You have been able to finish your system. > > Mine is done and complete, corrected and imporved. > It may need a bit more work.
Futile effort.
> > > There may be a good reason that finite arithmetic has never been > > > successfully formalized. It may be impossible. > > The max ceiling prevents cancelation > Can you prove this, or this yet another new axiom? It's obvious from the axiom Sm = m. > > a + b = a + c implies b = c > > At best, if a + b = a + c /= m, then b = c. > > Also it not very useful. For very large m, it can > > suffice for most mathematics, but the needed m seems to > > keep increasing with time. Would m = googleplex suffice? > m could be 0, and no number would have a successor. Not very interesting, > but a possibility. No, that contradicts for all x, Sx /= 0.

