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

Nope. The axiom is "there is an y such that for all x: x + y = x".
Something slightly different.

Is that your way of dodging an error you made but tacticly did not
explicitly write?

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.

Your error was assuming that no such y existed because no such number
exists, but to get numbers you need a ring and the axiom above is not
compatable with that of a ring.

I am assuming nothing about numbers at all. I only look at the axiom
and sees that it does not define an additive identity.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

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

Nope. The axiom is "there is an y such that for all x: x + y = x".
Something slightly different.

Is that your way of dodging an error you made but tacticly did not
explicitly write?

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.

Your error was assuming that no such y existed because no such number
exists, but to get numbers you need a ring and the axiom above is not
compatable with that of a ring.

I am assuming nothing about numbers at all. I only look at the axiom
and sees that it does not define an additive identity.

Of course not :-)
The statement
There is a y such that for all x: P(x,y)
merely implies the statement
For all x, there is a y such that P(x,y)
but not the other way around.
Finding a counterexample is always trivial.

But that is not at stake here.
You are now arguing with Schoenie about the definition
of "additive identity" because that is the only way he's got
left to save that little bit of face he thinks he surely must
have got left - somewhere down there ;-)

Dirk Vdm

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

Nope. The axiom is "there is an y such that for all x: x + y = x".
Something slightly different.

Is that your way of dodging an error you made but tacticly did not
explicitly write?

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. That definition is not the same as there is y for each x.
If you DEFINE it the proper way (there is y for each x) then the
statement (for each x there is y) is a lemma NOT an AXIOM.

Your error was assuming that no such y existed because no such number
exists, but to get numbers you need a ring and the axiom above is not
compatable with that of a ring.

I am assuming nothing about numbers at all. I only look at the axiom
and sees that it does not define an additive identity.

It is the definition of the additive identity used in this case.

Will you admit your error ?

--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Dirk Van de moortel wrote:
"Dik T. Winter" wrote in message ...
In article . com "Schoenfeld" writes:
Dik T. Winter wrote:
...
Axiom: Additive Identity
for all x there exists y such that x + y = x
guarantees an additive identity.

Nope. The axiom is "there is an y such that for all x: x + y = x".
Something slightly different.

Is that your way of dodging an error you made but tacticly did not
explicitly write?

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.

Your error was assuming that no such y existed because no such number
exists, but to get numbers you need a ring and the axiom above is not
compatable with that of a ring.

I am assuming nothing about numbers at all. I only look at the axiom
and sees that it does not define an additive identity.

Of course not :-)
The statement
There is a y such that for all x: P(x,y)
merely implies the statement
For all x, there is a y such that P(x,y)
but not the other way around.
Finding a counterexample is always trivial.

But that is not at stake here.
You are now arguing with Schoenie about the definition
of "additive identity" because that is the only way he's got
left to save that little bit of face he thinks he surely must
have got left - somewhere down there ;-)

I explained to you already, i consider everyone here crackpots. I don't
care for your opinions (i mean seriously, you don't contribute anything
other than derision of others, at least I tried asking a question to
learn something and perhaps teach others as well). As I already told
you, I plan to leave these groups (after finishing of threads). I hope
I won't impact your crackpot study.

Dirk Vdm

Richard Tobin wrote:
In article .com,
Schoenfeld wrote:

The statement,
Axiom: Additive Identity
"for all x there exists y such that x + y = x"

defines an additive identity for all x.

It defines a thing like an additive identity for *each* x, but it
doesn't require that it's the same for each one, which is what is
meant by "additive identity".

That's exceedingly obvious

So why did you make the false claim above?


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
however is not a number since the axiom is different for
rings/groups/fields. You can deduce the the as a LEMMA from
rings/groups/fields, but this requires a different definition for the
Additive Identity.


You're the 3rd person to make this error. Will you run away like a
coward too?


But strictly speaking, with that definition, the additive
identity DOES exist but it is obviously not a number.

That doesn't sound like "strictly speaking" to me. On the contrary, it
sounds like you're making it u p as you go along.

note:

Sorry, not interested.

-- Richard

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"?

Will you admit your error ?

When are you going to?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

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).

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.


Will you admit your error ?

When are you going to?

You have made the error, and in your attempt to cover it up, you have
made another error.


--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

"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).

I'll repeat again, the AXIOM:
"Axiom: Additive Identity, for all x there exists y, x + y = x"

You got quantifier dyslexia, right?

guarantees that an object 'y' exists for every single object 'x',
however the object 'y' is not an number.

The word "number" as you use it is meaningless. Anyway, additive
identities are not something dependent on x. You got your quantifiers
wrong.

If you were able to read a simple table, you'd notice the difference.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum

David Kastrup wrote:
"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).

I'll repeat again, the AXIOM:
"Axiom: Additive Identity, for all x there exists y, x + y = x"

You got quantifier dyslexia, right?

You're as useless as rubber lips on a wood pecker.

guarantees that an object 'y' exists for every single object 'x',
however the object 'y' is not an number.

The word "number" as you use it is meaningless. Anyway, additive
identities are not something dependent on x. You got your quantifiers
wrong.

You're two letters off from being an asset to this thread.

If you were able to read a simple table, you'd notice the difference.

A guy with your IQ should have a low voice too.

hint: read properly.


--
David Kastrup, Kriemhildstr. 15, 44793 Bochum

"Schoenfeld" writes:

David Kastrup wrote:
"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).

I'll repeat again, the AXIOM:
"Axiom: Additive Identity, for all x there exists y, x + y = x"

You got quantifier dyslexia, right?

You're as useless as rubber lips on a wood pecker.

guarantees that an object 'y' exists for every single object 'x',
however the object 'y' is not an number.

The word "number" as you use it is meaningless. Anyway, additive
identities are not something dependent on x. You got your quantifiers
wrong.

You're two letters off from being an asset to this thread.

If you were able to read a simple table, you'd notice the
difference.

A guy with your IQ should have a low voice too.

Oh, I have. When I am not singing alto in my choir for stylistic
reasons, I am singing bass. Easier on the voice than tenor. One
should just realize when shouting from the chest does not help in
making a particular approach go the distance.

Anyway, we all feel honored for you offering those pearls of wisdom
which your teachers must never have tired of telling you. You could
try calling your axiom something differently than "additive identity"
since that name is already taken, maybe something like "indifferent
companion". On the other hand, changing the name might make your
attempt to bluff your way out of the hole you are digging for yourself
more apparent on other fronts.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum