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Topic: Two Finite Arithmetics
Replies: 19   Last Post: Apr 9, 2014 9:27 PM

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 William Elliot Posts: 2,637 Registered: 1/8/12
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 S|N -> 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.