Re: A MOD B NO MATH HAHAHAHA !!

On Jan 13, 4:37 pm, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<7b6561e8-de9b-4bc3-8ccc-06b73267a...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Tim BandTech.com" <tttppp...@xxxxxxxxx> wrote:

I already know that my own thinking is askance to your thinking here.
But I have another means by which the mod 1 system is of importance.
Upon generalizing sign the real number takes the form
s x
where s is sign and x is magnitude. For the real number there are two
signs and the modulo arithmetic takes its place in the arithmetic
product of two values
( s1 x1 )( s2 x2 )
whose resultant is
(s1 + s2) x1 x2
where this sign summation is mod 2.

Isn't ( s1 x1 )( s2 x2 ) = (s1 s2) (x1 x2) even simpler and a good deal
less artificial?

As you substitute for s1 and s2 above the product will not give the
correct results. For instance in P2(the reals):
(-)(+) = - :=: (1 + 2)%2 = 1
which is the mod 2 sum of the sources.
We can literally substitute a 1 for '-' and a 2 for '+' and get the
behavior in mod 2 summation, though the zero sign has to be taken to
mean 2, which is actually consistent as the identity element being
left unsigned as meaning positive.

If we tried your product in this regard we'd get
(-)(+) = + :=: (1)(2) = 2
which is not consistent with the reals.

The sign s is a new data type. It requires a bit of room and it
determines that room, not us. In this regard the usage of s here is
somewhat a linguistic addition, yet it can be expressed in terms of
traditional language. This is a problem of stating notation clearly
and so your representation could work, yet when you define the product
s1s2 you will wind defining the mod 2 sum when s1,s2 are signs in P2.
So you could quibble with my initial refutation, but upon filling out
the dipsute I would think that we would come to perfect agreement.

- Tim
.

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