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From: jkcender on 5 Aug 2010 01:49
On Aug 4, 7:26 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote:
>
> > On Aug 4, 6:50 pm, Tim Little <t...(a)little-possums.net> wrote:
> >> Just to clarify, (1 + (-1)) + z = 2z, but 1 + ((-1) + z) = z right?
> >> So this form of addition does not obey the associative law?
>
> > I think you are referring to 4. If so, no. 1 + ((-1) + z) = (1 + (-1))
> > + z = 2z.
>
> I'm referring to both 3 and 4.
>    1 + ((-1) + z)
>  = 1 + (-1)   ... by (3)
>  = z          ... by (4).
>
> If associativity holds, then
>    1 + ((-1) + z)
>  = (1 + (-1)) + z
>  = z + z     ... by (4)
>  = 2z        ... by (3).
>
> Since you already stated that 2z =/= z, associativity cannot hold.
>
> - Tim

1 + (-1)= z therefore 1 + (-1) + z = 1z + 1z = 2z also 1 + ((-1) +
z) = (1 + (-1)) + z = 2z
Associativity holds.
From: Tonico on 5 Aug 2010 02:21
On Aug 5, 8:19 am, jkcender <jkcm...(a)yahoo.com> wrote:
> > 5. Additive inverse. n – n = nz,  and nz – nz = (nz)^2
>
> Tim called my attention to 5. To avoid possible confusion, only the z
> part is squared. The n is not.



It's impossible to avoid confusion if you still write it that way.
Write it n*z^2 or at least nz^2.

Tonio
From: jkcender on 5 Aug 2010 15:31
On Aug 4, 8:21 pm, Tonico <Tonic...(a)yahoo.com> wrote:
> On Aug 5, 8:19 am, jkcender <jkcm...(a)yahoo.com> wrote:
>
> > > 5. Additive inverse. n – n = nz,  and nz – nz = (nz)^2
>
> > Tim called my attention to 5. To avoid possible confusion, only the z
> > part is squared. The n is not.
>
> It's impossible to avoid confusion if you still write it that way.
> Write it n*z^2 or at least nz^2.
>
> Tonio

It is written the way you have suggested (equivalent to nz^2) in my
paper. I goofed when transcribing it. Do you suggest removing the
original post and reposting a corrected version, or just leaving it
with this post as a correction?
From: jkcender on 5 Aug 2010 16:06
On Aug 4, 7:26 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-08-05, jkcender <jkcm...(a)yahoo.com> wrote:
>
> > On Aug 4, 6:50 pm, Tim Little <t...(a)little-possums.net> wrote:
> >> Just to clarify, (1 + (-1)) + z = 2z, but 1 + ((-1) + z) = z right?
> >> So this form of addition does not obey the associative law?
>
> > I think you are referring to 4. If so, no. 1 + ((-1) + z) = (1 + (-1))
> > + z = 2z.
>
> I'm referring to both 3 and 4.
>    1 + ((-1) + z)
>  = 1 + (-1)   ... by (3)
>  = z          ... by (4).
>
> If associativity holds, then
>    1 + ((-1) + z)
>  = (1 + (-1)) + z
>  = z + z     ... by (4)
>  = 2z        ... by (3).
>
> Since you already stated that 2z =/= z, associativity cannot hold.
>
> - Tim

Tim, thanks for your fresh pair of eyes and my apologies. I'm so used
to thinking in terms of the new zero and therefore only of how the
associative property holds with the new zero. So yes, you are correct,
the new zero is not associative in the same way the current zero is.
Clarification is in order to show how it is associative in familiar
math language. For the new zero, combining like terms takes precedence
over associativity the way it is done with the Real number zero. And
upon reviewing my paper, this is dealt with too cryptically there as
well on page 41. I hope looking at the issue from the standpoint of
polynomials helps. By this I mean the usual, that terms of a
polynomial are combined according to the power of each term. How does
this apply to the new zeros? Just as the Real numbers can be written
as a Complex number with a 0 term (a+0i), the Reals can be written as
a zero number with a 0 term. Rule 7 says that z^0=1. So a real can be
written x = xz^0. Your original equation was 1 + ((-1) + z) = z. When
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
when like terms are added, the equation does equal 2z. In this sense
the associative property is preserved.
From: Transfer Principle on 5 Aug 2010 18:33
On Aug 4, 7:30 pm, jkcender <jkcm...(a)yahoo.com> wrote:
> 1. nz * nz(inv) = 1, and xz * yz(inv) = xy    [note 3]
> 2. Zero times anything else is still a zero number. x * nz = xnz,  and
> xz * nz = xn(z)^2,
> 3. Division defined. x/nz = x/n * z(inv) = (x/n)z(inv), and xz/yz = x/
> y, and xz(inv)/yz(inv) = x/y
> 4. Additive identity.  x + nz  = x  or x + nz  may be considered in
> final form, but z + z = 2z.
> 5. Additive inverse. n – n = nz,  and nz – nz = (nz)^2
> 6.  xz is not equal to yz, rather xz + yz = (x+y)z
> 7. (nz)^0 = 1 (other zeros such as the exponent zero still use the
> symbol “0”)

As usual, let me see whether there's a way to make this
more rigorous.

We know that there's a rigorous construction of the set R
of real numbers. Typically, we begin with the set N of
naturals, then form the set Z of integers via a certain
equivalence class on NxN, then form the set Q of rationals
via a certain equivalence class on ZxZ, then form the set
R of reals via either Cauchy sequences or Dedekind cuts.

Now the OP, jkcender, differs from this construction at
the step from N to Z. For in standard Z, we define 0 to be
the equivalence class {(n,n)|neN}, but not only does
jkcender avoid standard 0, but he wants to identify each
ordered pair (n,n) with a different value, namely nz.

But even though jkcender differs from the standard
construction fairly early, we can postpone this difference
by using a trick recently mentioned by David Ullrich. We
go not from N -> Z -> Q -> R, but N -> Q+ -> R+ -> R. Then
Q+ (equivalence classes in NxN) and R+ (D-cuts in R+) can
be defined as in Ullrich, and then we can work with R+ to
form jkcender's system.

After Little points out that associativity of addition
fails in jkcender's system, jkcender decides to use
polynomials in z to define his system. I point out that
there is a similar trick to make Tim Golden's polysigned
numbers work, so we may try the same thing here.

We begin by considering R+[z], the set of polynomials in z
with coefficients in R+. Then (R+[z])^2 is the set of all
ordered pairs of polynomials with positive coefficients,
and just as in forming R from (R+)^2 (or Z from N^2), the
first element of the ordered pair is positive while the
second element is negative.

Notice that although the coefficients are positive, zero
coefficients are allowed, since a polynomial 4z+0z^2+5z^3
is really just the polynomial 4z+5z^3. Indeed, the
polynomial 0 is allowed -- (P,0) is a polynomial with all
positive coefficients, and (0,P) is a polynomial with all
negative coefficients. But what we can't have is (0,0),
since this would be the forbidden zero polynomial.

So what we need is (R+[z])^2\{(0,0)}. Notice that this
set is closed under componentwise addition since R+ itself
is closed under addition.

Now we need an equivalence relation ~ on this set. And the
relation should have the following properties:

-- If P,Q,R,S,T are polynomials and (P,Q)~(R,S), then
(T+P,U+Q)~(T+R,U+S). (This allows + to be welldefined.)
-- If k is a scalar in R+ and (P,Q)~(R,S), then
(kP,kQ)~(kR,kS).
-- If T is a polynomial and (P,Q)~(R,S), then
(TP,TQ)~(TR,TS).
-- (1,1)~(z,0).

Before I leave, let me point out that I'm not sure whether
jkcender's addition is commutative, either. For we know
that 1+(-1) = z, but what is (-1)+1? We see that:

(-1)+1 = -1z^0 + 1z^0
= -1(1z^0 + -1z^0)
= -1z^1
= -z

a problem which doesn't go away by using polynomials.

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