Re: Is magnitude more fundamental than the real numbers?


*** T. Winter wrote:
In article <1155047137.861451.128940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> "Timothy Golden BandTechnology.com" <tttpppggg@xxxxxxxxx> writes:
> *** T. Winter wrote:
...
> > Nope. It will not work with either #1+1, #1*1 and +1-1. Because they
> > are all three zero-divisors.
>
> Could you please explain the #1*1 instance?
> I don't understand why you have that value in this list.
> I understand that
> ( # 1 + 1 ) ( # 1 * 1 ) = 0 .

And that means that *both* are zero divisors. It also means the inverse
of #1*1 does not exist, so you can not divide by it in general. It also
mean that when you take an arbitrary number, the product with #1*1 is
a zero divisor. And so on.

> But this is an artifact of the general behavior that any value
> multiplied by #1+1 will land on the axis.

Any value multiplied by #1*1 will land on another axis.
And BTW, all numbers that have exactly two 1's in their representation
and 0's otherwise, are zero divisors.

> argument arises. To impose the old math rules on this new construction
> may not be appropriate. The identity axis is a new beast. The old cage
> will not necessarily hold it in.

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.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/

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,
.... ).
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 .
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 ). It is one of many
values that will provide this zero result just as +5 can be replaced
for the example in the reals.

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.

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

http://bandtechnology.com/PolySigned/Deformation/DeformationUnitSphereP4.html
When the sphere turns into a line that is when the red dot has taken a
value of the identity axis. Information is destroyed. The only similar
behavior downward in sign is multiplication by zero. Here there is a
slight difference due to the fact that a real value remainder exists so
it is not exactly a zero concept, but very similar. The phrase
'identity axis' perhaps should be rephrased 'null axis'. The existing
field definition for the reciprocal uses a strictly zero exception. The
polysign system suggests that there ought to be more excepted;
particularly all values along the identity axis provide multiple
solutions.

Is this perhaps a counterexample to your claims above?:
( # 1 + 1 )( # 1 + 1 ) = # 2 + 2 ;
Therefor
( # 2 + 2 ) / ( # 1 + 1 ) = # 1 + 1 .
Do you see the real line embedded behavior?

-Tim

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