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From: Jonathan Cender on 10 Aug 2010 22:02
On Aug 6, 5:20 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote:
>
> > rewritten with all zero numbers we can see that it does not equal z.
> > 1z^0 + ((-1z^0) + z^1) =/= z. I hope it is clear here that, just like
> > in any polynomial, terms with different powers cannot be added.
>
> So your rule (3) is actually false?  It is not true that x + nz = x?
>
> - Tim

What you may have teased out here is that it is possible to have more
than one arithmetic with the new zero. Rule 4 as originally given
mushes too many things together. Two different arithmetics could
result from separating out the two different parts of rule 4. If x +
nz is to be in final form, a different rule 4 will be necessary in
order for 1 + ((-1) + z) to always equal 2z. In that case, as you say,
"It is not true that x + nz = x." The zero number never drops off (or
perhaps not until the very end.) We can have one or the other, but not
both at the same time. [yes, this mostly repeats part of a reply to
another post from Tim.]
-Jonathan
From: Jonathan Cender on 10 Aug 2010 22:04
On Aug 6, 5:22 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote:
>
> > The associative property fails here in the same way that the
> > associative property of addition fails when adding Reals and
> > imaginaries.
>
> The associative property of addition works perfectly well when adding
> reals and imaginaries.  The difference is that nobody working in
> complex arithmetic claims that imaginaries are additive identities for
> reals.
>
> - Tim

I stand corrected. It was a poorly chosen example.
-Jonathan
From: Tim Little on 11 Aug 2010 00:38
On 2010-08-11, Jonathan Cender <jkcmsal(a)yahoo.com> wrote:
> OK, for arithmetic with just rule 4a, I've begun to wonder if
> separating it out like this is a key to a number of issues I've
> struggled with. By following the steps you have questioned, nz and
> –nz can always be reduced to z and –z because nz = z + (n-1)z.

If the arithmetic of ordinary (nonzero) integers holds, then they can
both be reduced to just z:

-z = -(1 - 1)
= (-1) + 1
= 1 + (-1)
= z.

Though of course you're free to posit that in your system the nonzero
integers are not the ordinary ones, and do not behave the same way.
It would make it more difficult to determine what should be the value
of fairly ordinary expressions like 2+2 though, or whether (1+2)+1
should have the same value as 1+(2+1).


> The result of this is that z is a lot more like 0 except that
> division is defined for z.

What do you get if you multiply 2z by (1/z)? If associativity holds
then you would get 2, but then 2z=z so 2=1 which would be undesirable.
So we may have to lose associativity of multiplication as well as
addition.


> I hope the above response in this post also answers this part.
> Associativity doesn't hold. Let me know if there's something still
> not clear or I'm missing something.

It does clarify at least what sort of properties you're aiming for in
your system.


- Tim
From: Transfer Principle on 11 Aug 2010 00:46
On Aug 10, 6:50 pm, Jonathan Cender <jkcm...(a)yahoo.com> wrote:
> On Aug 6, 5:18 pm, Tim Little <t...(a)little-possums.net> wrote:
> > >> 1 + ((-1) + z)
> > >> = 1 + (-1) ... by (3)
> > >> = z ... by (4).
> > You haven't addressed this derivation.  Which step is incorrect?  It
> > should be easy, there are only two and I've labelled both of them with
> > the numbering of rules you provided.
> Rule 5, since it is not commutative as Transfer Principle showed at
> the end of his first post, is now
> n-n= nz and –(n-n)= -nz

Here's what I've come up with so far: The equivalence relation
for ordered pairs of polynomials (P,Q) and (R,S) is:

(P,Q)~(R,S) iff P+S(1-z) = R+Q(1-z)

Using this, we can prove the following:

~ is an equivalence relation.
(1,1)~(z,0), since 1+0(1-z) = z+1(1-z) = 1
(n,n)~(nz,0), since n+0(1-z) = nz+n(1-z) = n
(z,z)~(z^2,0), since z+0(1-z) = z^2+z(1-z) = z

If (P,Q)~(R,S), then (P+T,Q+U)~(R+T,S+U), so that the relation
~ respects (componentwise) addition.

But unfortunately, ~ doesn't respect multiplication. But first
we have to define multiplication. We assume that it is defined
similarly to how it is defined in the construction of the
integers Z from N. We recall that since (P,Q) really means
P-Q, we have:

(P,Q)(R,S) = (P-Q)(R-S)
= PR-PS-QR+QS
= PR+QS-PS-QR
= (PR+QS,PS+QR)

But ~ doesn't respect this multiplication. To see this, we
recall that (1,1)~(z,0), and then we have:

(1,1)(1,1) = (2,2) (yes, the product equals the sum here)
(z,0)(1,1) = (z,z)

Yet (2,2) isn't equivalent to (z,z), since the former is
equivalent to (2z,0) and the latter to (z^2,0).

Rewriting this in Cender's notation, we have:

1-1 = z
(1-1)(1-1) = z(1-1)
1-1 -1+1 = z-z

Now RHS is still z-z = z^2, but what about LHS? We want to
be able to write:

1-1-1+1 = 2-1-1 = 2-2 = 2z

But we're not allowed to combine the positive 1's. Instead
we must write:

1-1-1+1 = z-1+1 = z-z = z^2

so that LHS finally matches RHS. The multiplication
(P,Q)(R,S) = (PR+QS,PS+QR) assumes that PR and QS can be
added by reordering in PR-PS-QR+QS, but we can't.

And so it appears that addition isn't commutative.

> > You're assuming that associativity holds to show that associativity
> > holds.  If you're stating that associativity holds as an *axiom*, you
> > should say so and accept that your system is then inconsistent.
> I hope the above response in this post also answers this part.
> Associativity doesn’t hold. Let me know if there’s something still not
> clear or I’m missing something.

It appears that Cender is now resolved that addition is
neither commutative nor associative. We can now go back
and redefine both addition and multiplication so that ~
respects them both. In particular, we can no longer use
componentwise addition (which is both commutative and
associative) and replace it with anothe operation.
From: Jonathan Cender on 11 Aug 2010 19:35
On Aug 10, 6:38 pm, Tim Little <t...(a)little-possums.net> wrote:
>
> What do you get if you multiply 2z by (1/z)?  If associativity holds
> then you would get 2, but then 2z=z so 2=1 which would be undesirable..
> So we may have to lose associativity of multiplication as well as
> addition.
>
> - Tim


(1/z) = z(inv) so 2z * (1/z) = 2z * z(inv) = 2*z*z(inv) = 2 * 1 = 2

(1/z) = z(inv) by the first part of rule 3.
z*z(inv) = 1 by rule 1.

At least that was the way it was _supposed_ to work as you obviously
recognized:) Now i'll have to rethink.
One thought that occurred to me after sending the posts yesterday is
whether something from Calculus might apply? Might it apply here as
well? Specifically, I'm referring to the order in which taking the
limit to zero occurs. If I have my terminology correct, the limit
can't be taken to zero just any old time. It has to be after at least
some other arithmetic operations are done.
Seems rather arbitrary, but along these lines would it be acceptable
to say that "taking" 2z to z needs to wait until multiplication is
finished? But for addition it's OK?
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