On 10 Sep 2005 05:35:47 -0700, "Schoenfeld"
wrote:


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

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

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.

Fine.

Then we have, thanks to your axiom:
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]

That looks reasonable, but there is a subtely here which I can't deduce
from the axioms.

The axiom stated:
There exists y such that for all x, x + y = x

Does this mean that,
STATEMENT 1:
For all y in set of additive identities, for all x in Z, x + y = x.

You make this assumption when you state that:
x + Y1 = x
x + Y2 = x

Dirk doesn't use an axiom here, just the definition of an additive
identity: y is an additive identity iff for all x, x+y=x

--
Wim Benthem

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

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.

Fine.

Then we have, thanks to your axiom:
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]

That looks reasonable, but there is a subtely here which I can't deduce
from the axioms.

The axiom stated:
There exists y such that for all x, x + y = x

No. That was not what the axiom stated.
It stated:
AXIOM: Additive Identity
There exists y such that for all x, x + y = x
This way, a thing y that satisfies the above condition, is an
additive identity by definition.


Does this mean that,
STATEMENT 1:
For all y in set of additive identities, for all x in Z, x + y = x.

You make this assumption when you state that:
x + Y1 = x
x + Y2 = x

I did not state that.
When you are doing mathematics, try to be precise.
I stated that if Y1 and Y2 are additive identities, then
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]
because that is how the axiom defines additive identities.
For every AdditiveIdentity the axiom allows us to say:
for all x: x + Additive Identity = x.
So your statement is trivially true - by definition.
The axiom says that the set of additive identities is not empty.
Below I prove that this set can only have one element.



so we also have when we apply [1] to our number Y2:
Y2 + Y1 = Y2 [3]
and likewise, when we apply [2] to our number Y1:
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

So you didn't even spot the typo.
That should be
Y1 + Y2 = Y2 + Y1

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.

Everything you said relies on "STATEMENT 1" being true. Can show you
this statement true? :-)

Yes, but not to an imbecile.
Did my "pointing out the exact error in your reasoning" help you,
or are you an imbecile, or perhaps merely pretending to be one?

Well in the case, genius, you just proved your own idiotic error
demonstrated by the OP. I must therefore concur with that OP,
"HAHAHAHAHAH".

Dirk Vdm

"The Ghost In The Machine" wrote in
message ...
| In sci.math, odin
|
| wrote
| on Fri, 9 Sep 2005 16:47:10 -0700
| :
| 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?!
|
| The IEEE Floating-Point Arithmetic Standard (IEEE 754) defines zero
| representions as positive zero and negative zero. Just about every
CPU on
| the planet uses this standard. If zero is not positive or negative,
then
| what do you figure it is?
|
|
| Personally, I think a modified Law of Trichotomy might apply:
| a real is either positive, zero, or negative. Therefore,
| zero is neither one or the other. Terms such as "nonnegative"
| or "nonpositive" are occasionally used in proof descriptions,
| if one needs to be able to allow or select 0 from a set of reals
| during a proof.
|
| However, there were problems with +0 and -0 in some processors,
| using one's complement arithmetic. Modern processors all use
| two's complement for integers, and there's only one representation
| for 0 therein.

Awww....1000 0000 0000 0000 0000 0000 0000 0000 isn't -0 anymore?

It's not an overflow, I'd double it for that.
1000 0000 0000 0000 0000 0000 0000 0000 * 10 =
1 0000 0000 0000 0000 0000 0000 0000 0000

It's not the highest integer, I can add one to it.
1000 0000 0000 0000 0000 0000 0000 0001

1111 1111 1111 1111 1111 1111 1111 1111 is -1,
I can add one to that and get zero:
1111 1111 1111 1111 1111 1111 1111 1111 +
0000 0000 0000 0000 0000 0000 0000 0001 =
1 0000 0000 0000 0000 0000 0000 0000 0000

So just what is
1000 0000 0000 0000 0000 0000 0000 0000
in modern processors with only one representation for 0?

Androcles

"Wim Benthem" wrote in message ...
On 10 Sep 2005 05:35:47 -0700, "Schoenfeld"
wrote:


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

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

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.

Fine.

Then we have, thanks to your axiom:
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]

That looks reasonable, but there is a subtely here which I can't deduce
from the axioms.

The axiom stated:
There exists y such that for all x, x + y = x

Does this mean that,
STATEMENT 1:
For all y in set of additive identities, for all x in Z, x + y = x.

You make this assumption when you state that:
x + Y1 = x
x + Y2 = x

Dirk doesn't use an axiom here, just the definition of an additive
identity: y is an additive identity iff for all x, x+y=x

Actually, knowing Schoenfeld, he's just trying to find a
way to avoid ackowledging the fact that he was selling
**** from the beginning, and that, in order to boost his
sales, he made a silly glaringly elementary mistake.
It's a Schoenfeld Trademark. I love it :-)

Dirk Vdm

Schoenfeld wrote:

If you consider commutative rings (e.g. integers) or ordered fields
(e.g. reals) there is the additive identity axiom:
There exists y such that for all x, x + y = x
This is entirely insufficient to imply a single unique additive
identity y, although this seems to be the universal interpretation.

The axiom as stated there does not immediately state that
the additive identity is unique; that is a consequence of
this together with other axioms. It is easy to see that
commutativity of addition implies that the additive inverse
is unique.

For suppose that y is an additive identity and
y' is an additive identity.

Then since y is an additive identity,

y' + y = y'

(adding y to anything leaves that thing unchanged)

and, since y' is an additive identity,

y + y' = y

(adding y' to anything leaves that thing unchanged).

Now we wheel out the commutativity of addition.

y' = y' + y = y + y' = y

So any two objects which have the defining property
of an additive identity must be equal. Or, to put
that in plainer language, the additive identity
is unique.

(You might even see presentations with uniqueness
in there are part of the definition. It's a matter
of taste.)

"Robert Low" wrote in message ...
Schoenfeld wrote:

If you consider commutative rings (e.g. integers) or ordered fields
(e.g. reals) there is the additive identity axiom:
There exists y such that for all x, x + y = x
This is entirely insufficient to imply a single unique additive
identity y, although this seems to be the universal interpretation.

The axiom as stated there does not immediately state that
the additive identity is unique; that is a consequence of
this together with other axioms. It is easy to see that
commutativity of addition implies that the additive inverse
is unique.

For suppose that y is an additive identity and
y' is an additive identity.

Then since y is an additive identity,

y' + y = y'

(adding y to anything leaves that thing unchanged)

and, since y' is an additive identity,

y + y' = y

(adding y' to anything leaves that thing unchanged).

Now we wheel out the commutativity of addition.

y' = y' + y = y + y' = y

So any two objects which have the defining property
of an additive identity must be equal. Or, to put
that in plainer language, the additive identity
is unique.

(You might even see presentations with uniqueness
in there are part of the definition. It's a matter
of taste.)

He seems not to be prepared to admit that he understand this.
He's trying to avoid losing his face, which must be a painful
process for someone who already lost it ;-)

Dirk Vdm

Dirk Van de moortel wrote:
Actually, knowing Schoenfeld, he's just trying to find a
way to avoid ackowledging the fact that he was selling
**** from the beginning, and that, in order to boost his
sales, he made a silly glaringly elementary mistake.
It's a Schoenfeld Trademark. I love it :-)

Speaking of elementary mistakes, I hope now you have taught yourself
why:

[Dirk Van de moortel wrote:]
"By the way, zero is usually taken to be both positive and
negative. "

is a "silly glaringly elementary mistake" by your own efficient
demonstration.

To be precise, the exact error in my reasoning was to confuse:
"There exists y such that for all x, x + y = x" as
"For all x there exists y such that x + y = x".

So I guess the OP was right. Congratulations! (on your award, that is).


Dirk Vdm

Dirk Van de moortel wrote:
"Robert Low" wrote in message ...
Schoenfeld wrote:

If you consider commutative rings (e.g. integers) or ordered fields
(e.g. reals) there is the additive identity axiom:
There exists y such that for all x, x + y = x
This is entirely insufficient to imply a single unique additive
identity y, although this seems to be the universal interpretation.

The axiom as stated there does not immediately state that
the additive identity is unique; that is a consequence of
this together with other axioms. It is easy to see that
commutativity of addition implies that the additive inverse
is unique.

For suppose that y is an additive identity and
y' is an additive identity.

Then since y is an additive identity,

y' + y = y'

(adding y to anything leaves that thing unchanged)

and, since y' is an additive identity,

y + y' = y

(adding y' to anything leaves that thing unchanged).

Now we wheel out the commutativity of addition.

y' = y' + y = y + y' = y

So any two objects which have the defining property
of an additive identity must be equal. Or, to put
that in plainer language, the additive identity
is unique.

(You might even see presentations with uniqueness
in there are part of the definition. It's a matter
of taste.)

He seems not to be prepared to admit that he understand this.
He's trying to avoid losing his face, which must be a painful
process for someone who already lost it ;-)

I admitted the error I made.

To be precise, the exact error in my reasoning was to confuse:
"There exists y such that for all x, x + y = x" as
"For all x there exists y such that x + y = x".

The difference is important, because the second statement (which is not
how the axiom is stated) implies that all y are not necessarily
additive identities for all x. But I'll admit this error for a third
time, to extinguish your claims that i'm trying to 'save face'. I
CONFUSED "THERE EXISTS Y FOR ALL X" as "FOR ALL X THERE EXISTS Y".

By the way, congrats on your award.

Dirk Vdm

Androcles says...

Is there any definition that gives the additive identity for time?
This one doesn't cut it:
[quote]
we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Einstein can prove nothing can go faster than a turtle.

Oops!... Did I say 'a turtle'? Sorry...'light'.

You are an idiot, Androcles.

--
Daryl McCullough
Ithaca, NY

Wim Benthem wrote:
On 10 Sep 2005 05:35:47 -0700, "Schoenfeld"
wrote:


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

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

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.

Fine.

Then we have, thanks to your axiom:
for all x: x + Y1 = x [1]
for all x: x + Y2 = x [2]

That looks reasonable, but there is a subtely here which I can't deduce
from the axioms.

The axiom stated:
There exists y such that for all x, x + y = x

Does this mean that,
STATEMENT 1:
For all y in set of additive identities, for all x in Z, x + y = x.

You make this assumption when you state that:
x + Y1 = x
x + Y2 = x

Dirk doesn't use an axiom here, just the definition of an additive
identity: y is an additive identity iff for all x, x+y=x

Yes, that wasn't immediately obvious to me because I mixed up the order
of the existential quantifers in that axiom (when i was thinking about
it, that is).

Dirk's malicious claims of me somehow lying to not admit this error are
characteristically false. However, they do serve to explicitly
demonstrate that he is the perfect candidate for the OP's award. Not
necessarily for the error he made (and won't admit to), but rather for
his malicious arrogance and transitive ignorance.


--
Wim Benthem