Prev: how does Infinity = negative numbers, affect Algebra? #486 Correcting Math
Next: Second doubt
From: William Elliot on 4 Mar 2010 06:47
Peano's axioms with a set N and a unary function S.
1 Some 1 in N.
2 For all n in N, Sn in N.
3 For all n,m in N, if Sn = Sm, then n = m.
4 For all n in N, Sn /= 1.
5 P(1), for all n in N, if P(n) implies P(Sn),
then for all n in N, P(n).

Two ways that N can be finite.
Remove 4 and have integers modulus n with
.. . 2 = S1, 3 = S2,... 1 = Sn
in a circular order. Addition and multiplication
are addition and multiplication modulus n.

Remove 3 and have integers 1,2,.. n with Sn = n
in a linear order. Addition and multiplication are
r plus s = min(n, r + s), r times s = min(n, r*s).

Both of those can be revised by recruiting zero to be the someone in N.

Any resemblance to a sci.logic post or posts,
past or present, is purely coincidental. ;->

From: Arturo Magidin on 4 Mar 2010 12:41
On Mar 4, 5:47 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> Peano's axioms with a set N and a unary function S.
> 1       Some 1 in N.
> 2       For all n in N, Sn in N.
> 3       For all n,m in N, if Sn = Sm, then n = m.
> 4       For all n in N, Sn /= 1.
> 5       P(1), for all n in N, if P(n) implies P(Sn),
>                 then for all n in N, P(n).
>
> Two ways that N can be finite.
> Remove 4 and have integers modulus n with
> . . 2 = S1, 3 = S2,... 1 = Sn
> in a circular order.  Addition and multiplication
> are addition and multiplication modulus n.
>
> Remove 3 and have integers 1,2,.. n with Sn = n
> in a linear order.  Addition and multiplication are
> r plus s = min(n, r + s), r times s = min(n, r*s).
>
> Both of those can be revised by recruiting zero to be the someone in N.
>
> Any resemblance to a sci.logic post or posts,
> past or present, is purely coincidental. ;->

Both are special cases of a cyclic semigroup, which is all you need.

Cyclic semigroups are determined by pairs of nonnegative integers
(m,n); the pair (m,n) corresponds to the multiplicative semigroup
generated by x and subject to the relations x^m = x^{m+n}.

The cyclic group of order n is (0,n): <x : x^0 = x^n>; your second
example is (m,1), with m>0: <x : x^m = x^{m+1}>. But you can have,
say, x, x^2, x^3, x^4, x^5, and x^6 = x^3.

--
Arturo Magidin
From: abo on 4 Mar 2010 15:02
On Mar 4, 12:47 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
> Peano's axioms with a set N and a unary function S.
> 1       Some 1 in N.
> 2       For all n in N, Sn in N.
> 3       For all n,m in N, if Sn = Sm, then n = m.
> 4       For all n in N, Sn /= 1.
> 5       P(1), for all n in N, if P(n) implies P(Sn),
>                 then for all n in N, P(n).
>
> Two ways that N can be finite.
> Remove 4 and have integers modulus n with
> . . 2 = S1, 3 = S2,... 1 = Sn
> in a circular order.  Addition and multiplication
> are addition and multiplication modulus n.
>
> Remove 3 and have integers 1,2,.. n with Sn = n
> in a linear order.  Addition and multiplication are
> r plus s = min(n, r + s), r times s = min(n, r*s).
>
> Both of those can be revised by recruiting zero to be the someone in N.
>
> Any resemblance to a sci.logic post or posts,
> past or present, is purely coincidental. ;->

Actually, subtracting any of 1/, 2/, 3/, or 4/ means that a model need
not be finite.

PA \ 1 : Possible model of N is the empty set.
PA \ 2 : Possible model of N is {1,2,...,n} for any n, where Sn is
not in N.
PA \ 3 : Possible model of N is {1,2,...,n,...,m} for any n, m (with
m >= n > 1), where Sm = n
PA \ 4 : Possible model of N is {1,2,..,n} for any n, where Sn = 1

If you're interested you might look at General Arithmetic, here:

http://www.andrewboucher.com/papers/ga.pdf

Leon Henkin looked at this well before, in his paper “On Mathematical
Induction,” The American Mathematical Monthly, Vol. 67, 0), pp.
323-338.
From: abo on 4 Mar 2010 15:04
On Mar 4, 9:02 pm, abo <dkfjd...(a)yahoo.com> wrote:
> On Mar 4, 12:47 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
>
>
>
> > Peano's axioms with a set N and a unary function S.
> > 1       Some 1 in N.
> > 2       For all n in N, Sn in N.
> > 3       For all n,m in N, if Sn = Sm, then n = m.
> > 4       For all n in N, Sn /= 1.
> > 5       P(1), for all n in N, if P(n) implies P(Sn),
> >                 then for all n in N, P(n).
>
> > Two ways that N can be finite.
> > Remove 4 and have integers modulus n with
> > . . 2 = S1, 3 = S2,... 1 = Sn
> > in a circular order.  Addition and multiplication
> > are addition and multiplication modulus n.
>
> > Remove 3 and have integers 1,2,.. n with Sn = n
> > in a linear order.  Addition and multiplication are
> > r plus s = min(n, r + s), r times s = min(n, r*s).
>
> > Both of those can be revised by recruiting zero to be the someone in N.
>
> > Any resemblance to a sci.logic post or posts,
> > past or present, is purely coincidental. ;->
>
> Actually, subtracting any of 1/, 2/, 3/, or 4/ means that a model need
> not be finite.

Agh, should be "might be finite" or "need not be infinite."
 | 
Pages: 1
Prev: how does Infinity = negative numbers, affect Algebra? #486 Correcting Math
Next: Second doubt