Re: Is magnitude more fundamental than the real numbers?

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/~***/
.

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