[FOM] arithmetic without product

[Shahab Tasharrofi]

I think the following is an axiomatization of that theorem using only
addition. I might have of course made

a mistake in axiomatizing it but I don't think that it would be a mistake
that is not easily fixable.


Assumptions:

For all n, m: m < n implies A(n, m, 0) = m + 1

For all n, m: m >= n implies A(n, m, 0) = 0


For all n, m: S(n, m, 0) = A(n, m, 0)

For all n, m, k: S(n, m, k+1) = S(n, m, k) + A(n, m, k+1)


For all n, m, k: A(n, m, k+1) =/= 0 if and only if S(n, m, k) < n+1

For all n, m, k: A(n, m, k+1) =/= 0 implies A(n, m, k+1) = A(n, m, k) + 1


For all n, m, k: (A(n, m, k) =/= 0 & A(n, m, k+1) = 0) implies T(n, m) =
S(n, m, k)

For all n, m: A(n, m, 0) = 0 implies T(n, m) = 0


For all n: C(n, 0) = 0  or  C(n, 0) = 1

For all n: C(n, 0) = 1 if and only if T(n, 0) = n+1

For all n, m: C(n, m+1) = C(n, m)  or  C(n, m+1) = C(n, m) + 1

For all n, m: C(n, m+1) = C(n, m) + 1 if and only if T(n, m+1) = n+1


Claim:

For all k > 0, there exists n: C(n) >= k


Theorem: Assumptions implies Claim

BR, Shahab


On Mon, Jul 23, 2018 at 9:00 AM, morgorod at gmx.net <morgorod at gmx.net> wrote:

>
>
> > On 21. Jul 2018, at 10:20, José Manuel Rodriguez Caballero <
> josephcmac at gmail.com> wrote:
> >
> > Consider the following theorem:
> >
> > For any positive integer k, there is a positive integer n having at
> least k representations as sum of consecutive positive integers.
> >
> > For example, for k = 3 we have n = 9 = 4+5 = 2+3+4.
> >
> > For the statement of this theorem we need the addition.
>
>
> It is not so clear that you only need addition for stating the theorem.
>
> You have to state that there are k-many numbers x_1, … x_k such that
> adding some succesors to these numbers will sum up to n.
>
> - One possibilty would be to speak about the set of such numbers and
> demanding that the cardinality of such a set is greater than k. It is not
> possible to formulate this in first order language of PA.
> - Another option would be to demand the existence of the Goedel number of
> the sequence x_0, … x_n in the theorem; the Goeddell number encodes all the
> necessary informations. But we need the beta-function to encode sequences
> of arbitrary lengths. And this function is defined in terms of
> multiplication.
> - Finally, we could formulate a separate version of this theorem for all n
> in N; this way we can demand a fixed number of natural numbers x_k, and
> multiplication can be avoided. But the resulting statements are weaker than
> the original theorem, and it seem not to be the desired approach to the
> question.
>
> There could be other formal representations of the theorem; but I do not
> see a version not using multiplication at all.
>
> Best, Rene G.
>
> > If we also have the multiplication, then our theorem can be proved using
> a well-known argument due to J. J. Sylvester. On the other hand, if we have
> not multiplication and we restrict ourselves to first order logic, the
> existence of a proof of this result is not so clear. Is there an
> "arithmetic with just addition" where this theorem can be stated but it
> cannot be proved?
> >
> > Kind Regards,
> > Jose M.
> > _______________________________________________
> > FOM mailing list
> > FOM at cs.nyu.edu
> > https://cs.nyu.edu/mailman/listinfo/fom
>
> _______________________________________________
> FOM mailing list
> FOM at cs.nyu.edu
> https://cs.nyu.edu/mailman/listinfo/fom
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: </pipermail/fom/attachments/20180725/a7247d53/attachment.html>