Re: Is magnitude more fundamental than the real numbers?


*** T. Winter wrote:
In article <1155061899.245391.199990@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> "Timothy Golden BandTechnology.com" <tttpppggg@xxxxxxxxx> writes:
> *** T. Winter wrote:
...
> > The "identity axis" is nothing new at all. It is well known that when
> > you multiply a number by a zero divisor that the result is a zero
> > divisor. And that is what your "identity axis" states, the only point
> > is that there is not a single "identity axis", there are multiple.
...
> There is only one of these axes. It acts like a real line embedded in
> the even signed P4+ domains. It always has the notation ( 1, 0, 1, 0,
> ... ).

No, there are more of them.

> The term 'zero divisor' is one that I am not yet comfortable with, but
> it is fairly easy to see that in the reals
> ( 0 ) ( +5 ) = 0 .

Yup. But a zero-divisor is a *non-zero* number that can be multiplied by
another non-zero number to give 0. So when a and b are non-zero and
a.b = 0, a and b are both zero divisors.

> This does not make any strong statement about the number +5, just as
> the example in P4
> ( # 1 + 1 ) ( # 1 * 1 ) = 0 .
> does not make a strong statement about ( # 1 * 1 ).

But ( #1 + 1) is non-zero, so it makes a strong statement. Anyhow, you
will not find a solution to:
( #1 *1 +1 )/( #1 *1)
because the first is not a zero-divisor and the second is.

In P4 there are numerous zero-divisors, some basic ones are the following
six (coordinates in order #, *, +, -):
(1, 1, 0, 0) (1, 0, 1, 0) (1, 0, 0, 1) (0, 1, 1, 0) (0, 1, 0, 1) (0, 0, 1, 1)
for *all* six of them you can not find a quotient if you try to divide
(1, 1, 1, 0) by any one of them.

> Also, you've neglected the tail portion of my statement that makes a
> fairly strong argument, particularly the analysis of
> z ( # 1 + 1 ) = c
> for constant c. There is perhaps a misunderstanding.

But z (1, 1, 0, 0) is also on some axis (a different one).
Some linear algebra: to solve
(a1, a2, a3, a4)/(b1, b2, b3, b4) = (p1, p2, p3, p4)
you get at the matrix
( b1 b2 b3 b4 )
( b4 b1 b2 b3 )
( b3 b4 b1 b2 )
( b2 b3 b4 b1 )
we are interested in the determinant, because if it is zero there is in
general no solution. The determinant is zero if either b1+b2+b3+b4 = 0
or b1-b2+b3-b4 = 0. It is however a bit more tricky as we are in
equivalence classes where incrementing each coordinate by the same value
does not change the number: (1, 1, 1, 1) = 0. But whatever increment
we use, the determinant remains zero if b1-b2+b3-b4 = 0. So we can never
find a generic solution for (1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1) and
(1, 0, 0, 1) (or in general for (b1, b2, b3, b4) if b1-b2+b3-b4 = 0).
And, indeed, if b1-b2+b3=b4 = 0, (b1, b2, b3, b4) is a zero-divisor.
In that case: (b1,b2,b3,b4)(1,0,1,0) = (b1,b2,b3,b4)(0,1,0,1) = 0.
You are only seeing that (1, 0, 1, 0) and (0, 1, 0, 1) can only divide
(a1, a2, a3, a4) if a1 = a3 and a2 = a4.

> The identity axis is a dominant feature of the even signs. as is
> demonstrated by

Sorry those pictures do not tell me anything.

> Is this perhaps a counterexample to your claims above?:
> ( # 1 + 1 )( # 1 + 1 ) = # 2 + 2 ;

I see no counterexamples. Try to divide (1, 1, 1, 0) by (5, 3, 1, 3) and
see that it fails. More knowledge about zero-divisors would help you.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/

Hi ***.

I believe that I have resolved our apparent disagreement. I've
scrambled together a low grade reciprocal finder (results with large
error) and sure enough the divisor counterexamples that you provide
have no decent reciprocal. I've read the definition of zero divisor and
it is fine though it does not discuss anything about the underlying
structure of the system it is applied to. I suppose that is up to us
and I am now curious what we'll see above p4 in this context.

The 'other axes' are actually a plane. It is the plane that is
perpendicular to the identity axis that passes through the origin.
We've discussed it a bit before but now I can see it in the context of
division. Just as a multiplication by a point on the identity axis
removes information so does a multiplication by any point in this
plane. Instead of squashing the information into a line it yields two
dimensions of result, but still provides an irreversible effect on any
object. In effect the object has been squashed into a plane and can
never be 'unsquashed' informationally. This is exposed in

http://bandtechnology.com/PolySigned/Deformation/AxisDualDeformStudy.gif
When the sphere crosses over itself (flipping handedness) this is the
place.
This is a product view which explains why the division will not yield
results since the quotient is the reversal of the product. We cannot
get three dimensions back from two. There must be a theorem on
dimensional reduction and zero divisors. It's clear in the
informational context.
Because of this dimensional behavior the only meaningful division on
the zero divisor parts are more values from those parts. The zero
product is composed of one value from each of these parts, those parts
being the identity axis and the plane orthogonal to it on the origin.
There should be a theorem there as well. Upon excluding this plane as
well as the line that I call the identity axis we can have division.

This is quite a structure built in isn't it? The line literally
combined with the plane. Very pretty.
Physics could be here, and in a dimensionally consequential way.
I suppose you are nearing an R x C style equivalent definition of the
P4 product. As I scramble to follow along now I am wondering what you
will find next. And what of P6? Is the zero divisor exclusion the
identity axis and the points orthogonal to it through the origin? What
is a plane in P4 would be a 4D space in P6. This would be a ( 1, n-1 )
type of relation. That is consistent with the zero products, whose
possibility space rises dimensionally.

If the Hopf, Milnor and Kervaire dimensional theorems apply then we
should be able to get a zero divisor structure in P7 as well. But they
suggest that P5 will work perfectly. I'll have a better reciprocal
finder eventually. For now it is slow but could find reciprocals up in
these spaces. If you want me to look for something specific I'll be
happy to try.

I'm not very good at linear algebra but I'll try to follow along. So
far I don't see how you got the matrix in b. Is this a conjugation
style of division? It should be possible to specify a conjugate in P4
so long as we are off of the zero divisor structure but I don't have
it.

-Tim

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