In article .com "Schoenfeld" writes:
Dik T. Winter wrote:
In article .com "Schoenfeld" writes:
Dik T. Winter wrote:
...
The statement,
Axiom: Additive Identity
"for all x there exists y such that x + y = x"
defines an additive identity for all x.

No it does not. Consider the following addition table:
+ a b c
a a c b
b c b a
c b a c

For each 'x' there is an 'y' such that 'x + y = x'. But there is not
an additive identity.

It is an additive identity by definition, the definition was for each x
there is y.

What is the additive identity in the addition table above? In what
way does the addition table above fail the requirement: "for all x
there exists y such that x + y = x"?

Your table has no relation to what I said. In fact, it doesn't even
have a relation to how additive identities are defined for
rings/groups/fields (another elementary error you've just made).

Your axiom does not talk about rings/groups/fields or whatever. You
just state an axiom and tell us that it defines an additive identity.
I provide an addition table that satisfies that axiom, and it does
*not* have an additive identity.

I'll repeat again, the AXIOM:
"Axiom: Additive Identity, for all x there exists y, x + y = x"
guarantees that an object 'y' exists for every single object 'x',
however the object 'y' is not an number.

Yes, there is an object 'y' for every single object 'x', just as in my
addition table, but it is not sure that there is a single object 'y'
that is valid for *all* objects 'x', so it is not sure that there is
an additive identity. I have no idea what you are meaning with:
"however the object 'y' is not an number".
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Schoenfeld:


I DEFINED the 'additive identity':
Axiom: Additive Identity
"for all x there exists y such that x + y = x"

There DOES exist additive identities for all x by DEFINITION. This

You are missing the point. What you've written means that if,

x_1 + y_1 = x_1

x_2 + y_2 = x_2

Bilge wrote:
Schoenfeld:


I DEFINED the 'additive identity':
Axiom: Additive Identity
"for all x there exists y such that x + y = x"

There DOES exist additive identities for all x by DEFINITION. This

You are missing the point.

No, I am making the SAME point you just made. That IS my point - the
axiom, defined with for-all quantifier first guarantees existence but
such additive identity is NOT a number. Originally, this was how I was
thinking the axiom was defined for rings in general and why I couldn't
understand Dorks proof, and why I asked him to elaborate (he simply
decided to unleash a set of personal attacks for obvious reasons
(reasons related to a cracked lids on a teapots)).


What you've written means that if,

x_1 + y_1 = x_1

x_2 + y_2 = x_2
.
.

your definition does not require y_1 = y_2, etc.

Thanks for repeating what I said 5 times already and for agreeing with
me.

Obviously, I've
enumerated an element, y, for all x. By contrast, ``There exists
_a_ y, such that...'' implies y_1 = y_2, etc. The seconf definition
implies the identity is unique.

Thanks for repeating what I said 5 times already and for agreeing with
me.

Your definition doesn't.

Thanks for repeating what I said 5 times already and for agreeing with
me.

It might seem
like nitpicking, but you decided that nitpicking was important. One
way to make sure everyone picks nits, is to insist that it will
validate your position and shame your detractors.

Thanks for repeating what I said 5 times already and for agreeing with
me.

Schoenfeld wrote:

understand Dorks proof, and why I asked him to elaborate (he simply

Typo: Dork = Dirk

In article .com,
Schoenfeld wrote:

I DEFINED the 'additive identity':
Axiom: Additive Identity
"for all x there exists y such that x + y = x"

That was dumb of you.

-- Richard

"Richard Tobin" wrote in message
...
| In article .com,
| Schoenfeld wrote:
|
| I DEFINED the 'additive identity':
| Axiom: Additive Identity
| "for all x there exists y such that x + y = x"
|
| That was dumb of you.
|
| -- Richard

Actually it is dumb of you. Obviously you have no idea what a group is.
Rubik's cube is solved when you find the identity.
Androcles

In article "Androcles" Androcles@ MyPlace.org writes:
"Richard Tobin" wrote in message
...
....
| I DEFINED the 'additive identity':
| Axiom: Additive Identity
| "for all x there exists y such that x + y = x"
|
| That was dumb of you.
|
| -- Richard

Actually it is dumb of you. Obviously you have no idea what a group is.
Rubik's cube is solved when you find the identity.

I think not. The identity of the group of Rubik's cube is making no turn
at all. I do not think that that solves the cube.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Androcles wrote:
"Richard Tobin" wrote in message
...
| In article .com,
| Schoenfeld wrote:
|
| I DEFINED the 'additive identity':
| Axiom: Additive Identity
| "for all x there exists y such that x + y = x"
|
| That was dumb of you.
|
| -- Richard

Actually it is dumb of you. Obviously you have no idea what a group is.
Rubik's cube is solved when you find the identity.

Keep looking, Androcles. Eventually you may find it.

Androcles wrote:
"Richard Tobin" wrote in message
...
| In article .com,
| Schoenfeld wrote:
|
| I DEFINED the 'additive identity':
| Axiom: Additive Identity
| "for all x there exists y such that x + y = x"
|
| That was dumb of you.
|
| -- Richard

Actually it is dumb of you. Obviously you have no idea what a group is.
Rubik's cube is solved when you find the identity.

Androcles is still looking for it.

In article ,
Androcles Androcles@ MyPlace.org wrote:
Actually it is dumb of you. Obviously you have no idea what a group is.
Rubik's cube is solved when you find the identity.

Just when you think they can't get any stupider, they do.

-- Richard