Re: Is magnitude more fundamental than the real numbers?

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/~***/
.