Dirk Van de moortel wrote:
"Schoenfeld" wrote in message oups.com...

Dirk Van de moortel wrote:
"Schoenfeld" wrote in message ps.com...

Schoenfeld wrote:
Nth Complexity wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.

Hahahahahahahahahahahahahahahahahahahahahahahahaha hahahahahahahaha!
And you expect to teach OTHERS?!


[IGNORE PREVIOUS POST - SYMBOLS GOT MIXED UP DUE TO CARELESS EDITING]


I'm not certain Dirk is wrong. Most websites (like Wolframs) imply that
0 is neither positive or negative, but I don't think it's possible to
prove this (at least I can't, perhaps someone else can comment).

As others have said, it is a question of defining things. In the part
of the world in which I live, Bourbaki is the standard. So my
statement that "zero is usually taken to be both positive and
negative" is correct - again, in the part of the world in which I
happen to live.

Is there any definition other than the one given by the additive
identity axiom for rings or fields? The remainder of this posts assumes
no.

AXIOM: Additive Identity
There exists y such that for all x, x + y = x

Does that axiom imply that there exists only 1 additive identity? Based
on my understanding of the existential quantifier, it does not.

Indeed, it does not.
So we suppose there are (at least) two such identities, let's not call them
George and Freddy, but Y1 and Y2.

Then we have, thanks to your axiom:
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]
so we also have when we apply [1] to our number Y2:

You are assuming x = Y2.

Y2 + Y1 = Y2 [3]
and likewise, when we apply [2] to our number Y1:

You are also assuming x = Y1. When you assume x = Y2 and x = Y1,
you are also assuming that Y2 = Y1. This is a circular argument.
But what else would we expect from a relativist?

Y1 + Y2 = Y1 [4]

With the axiom of commutativity
for all x, y: x + y = y + x
we then have when we apply it to Y1 and Y2
Y1 + Y2 = Y1 + Y2

You probably meant to write
Y1 + Y2 = Y2 + Y1
which would follow from assuming x = Y1, y = Y2. You must have
made a mistake above when you assumed x = Y2 instead of y = Y2.
But that mistake invalidates your whole argument.

so, with [4] and [3] we conclude
Y1 = Y2.

So, using the commutativy axiom, there is only one additive identity.

[snip]

If I am wrong then I would highly appreciate you pointing out the exact
error in my reasoning.

hope this helps.

Dirk Vdm

Fpm ppfpffmfm pmfmmpmfp ppppfmmfmmfppppffmffmpmf mfmfmffpmmmmfpf Fpm
ppfpffffmfpmpffmffpff mfmfmfpff pfmmpppppmmf Fpm'ffp
mpfmmfpffpmmfpmmmmfmp. Pmmfmfpffmmm pmfmmpmfp mfmmpppmf mfmmmp
pffpmpmmfffmpppfpmmmm mfmmmp ffppff fmfmmppmm fpmmfm fpmmpf pppffmffm
pmmmppmmpmmmfmp, Fpm pmmfpmffmffm ppfpff ffmpppmfpfmpfmffpmmmmfmp
pppmfm pmfmmpmfp. Mmmpffpmpmfm pfmmmpffppffmpf mfmfmfpff
pfmpppmppmfmmmpmmpmmm, mppfpmfmpfmfmfm?

-- Nth Complexity --
-- Have A Nice Day! --
http://www.insurgent.org/~kook-faq/s...query=Uncle+Al
"BWAHAHAHAHAHAHAHAHAHAHA! You complete pillock. To paraphrase
your ****witted claim, 'in some cases, 1 + 1 does not equal 0.'
*PMSL* at you." -- Kadaitcha Man
"The teaching of science and mathematics must be purged of its
authoritarian and elitist characteristics, and the content of these
subjects enriched by incorporating the insights of the feminist,
queer, multiculturalist and ecological critiques." -- A.D.S.





--
Sent by 5 from yahoo within field com
This is a spam protected message. Please answer with reference header.
Posted via http://www.usenet-replayer.com