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.




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. 'y'
is usually called zero and given symbol 0.

Considering the integers Z, you can split Z into two sets P and N such
that:
1. For all x in Z, P contains x iff N contains -x; AND
2. P is closed under addition and multiplication.

Trivially, P is the set of positive integers and N is the set of
negative integers.

Where does 0 lie in here?

If your phrase "split Z into two sets P and N" implictly means
that the union of P and N is Z and that the intersection of P and N
is the empty set, then indeed by design, 0 is not in P and not in N.


The split occurs using only two conditions:
1. For all x in Z, P contains x iff N contains -x; AND
2. P is closed under addition and multiplication.

The only reason 0 is not in P and not in N is because you _assumed_
that 0 does not have a negative. You assumed this most likely because
you assumed that there exists only 1 additive identity. You can't make
this assumption for reasons shown above.

Not only that, but rings and fields demand the additional axiom:

AXIOM: Additive Inverse
For all x there exists y such that, x + y = z where z is the
additive identity of x.

REMARK: The number 'y' is commonly denoted as '-x' and the number 'z'
as 0.

So if you consider the additive inverse of the additive identity you
get the relation:
0 + j = 0 where j = -0

Can you prove that 0 and j are the *same* number from the axioms of the
ring? I don't think you can and this would imply then that there are AT
LEAST two additive identities +0 and -0 and that they are the additive
inverses of each other.

Treating both +0 and -0 as the same would have no arithmetic effect
that I can see, but this does not mean that they are the same. The key
difference of course being their sign.

So unless you explicitly specify what you mean, you can choose,
and the conditions you gave are not sufficient to decide more than
something like
P contains 0 == N contains -0 == N contains 0
i.o.w. 0 is in both or in neither.

The splitting condition I gave in the previous post are the conditions
used to order a set. You will find for example that the complex numbers
are not ordered because if i or -i is in P then P is not closed under
multiplication.


Well if it lies in BOTH P and N there are no contradictions at all. But
this implies that 0 occurs twice in the integers (otherwise it couldn't
be placed in any of P or N).

That depends on what you want the phrase "split Z into two sets"
to mean. As soon as you decide on that, you have your answer.

See above.

[...]

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

The SQUARE ROOT of ZERO is PLUS or MiNUS ZERO, Dimwits.!!
$ Sqrt(0) = + or - Zero(0).!!
You Cracked-pots are so FULL of DOOOooooooooooooooooop.!!
Sincerely, ```Brian

Schoenfeld wrote:
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.



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. 'y'
is usually called zero and given symbol 0.

Considering the integers Z, you can split Z into two sets P and N such
that:
1. For all x in Z, P contains x iff N contains -x; AND
2. P is closed under addition and multiplication.

Trivially, P is the set of positive integers and N is the set of
negative integers.

Where does 0 lie in here?

If your phrase "split Z into two sets P and N" implictly means
that the union of P and N is Z and that the intersection of P and N
is the empty set, then indeed by design, 0 is not in P and not in N.

The split occurs using only two conditions:
1. For all x in Z, P contains x iff N contains -x; AND
2. P is closed under addition and multiplication.

The only reason 0 is not in P and not in N is because you _assumed_
that 0 does not have a negative. You assumed this most likely because
you assumed that there exists only 1 additive identity. You can't make
this assumption for reasons shown above.

Not only that, but rings and fields demand the additional axiom:

AXIOM: Additive Inverse
For all x there exists y such that, x + y = z where z is the
additive identity of x.

REMARK: The number 'y' is commonly denoted as '-x' and the number 'z'
as 0.

So if you consider the additive inverse of the additive identity you
get the relation:
0 + j = 0 where j = -0

Can you prove that 0 and j are the *same* number from the axioms of the
ring? I don't think you can and this would imply then that there are AT
LEAST two additive identities +0 and -0 and that they are the additive
inverses of each other.

Treating both +0 and -0 as the same would have no arithmetic effect
that I can see, but this does not mean that they are the same. The key
difference of course being their sign.

So unless you explicitly specify what you mean, you can choose,
and the conditions you gave are not sufficient to decide more than
something like
P contains 0 == N contains -0 == N contains 0
i.o.w. 0 is in both or in neither.

The splitting condition I gave in the previous post are the conditions
used to order a set. You will find for example that the complex numbers
are not ordered because if i or -i is in P then P is not closed under
multiplication.


Well if it lies in BOTH P and N there are no contradictions at all. But
this implies that 0 occurs twice in the integers (otherwise it couldn't
be placed in any of P or N).

That depends on what you want the phrase "split Z into two sets"
to mean. As soon as you decide on that, you have your answer.

See above.

[...]

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

"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?


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'.

Well... moortel's favourite pastime is supporting the phuckwits
who support relativity, and moortel is the topic.


| 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.
|

c = (c+w)/(1+w/c) - Albert Huckster Einstein.
V = (w+v) /(1+vw/c^2) - Albert Phuckwit Einstein.

moortel is a member of the Holy Church of Relativity.

Androcles

"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:
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, 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

"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.


[repost - small typo corrected]

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:
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 = 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.

Dirk Vdm

"Dirk Van de moortel" wrote in message
...

[snip]

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

typo: Y1 + Y2 = Y2 + Y1 ;-)

Dirk Vdm

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

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


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, 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? :-)

Dirk Vdm

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.

--
#191,
It's still legal to go .sigless.

"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?

Dirk Vdm

In article ,
The Ghost In The Machine wrote:
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.

It's in the IEEE standards because the representation of the
number can have as that many flavors. How hardware defines
zero has nothing to do with how math uses it. It is only
when one is trying to do the math using a computer that
the IEEE standards is used. The reason there had to be
a standard is because there each processor and each
program defined their zeroes differently. Some simply
dropped bits.


However, there were problems with +0 and -0 in some processors,
using one's complement arithmetic. Modern processors all use
two's complement

All? Are you sure about that?

.. for integers, and there's only one representation
for 0 therein.

I would not assume this. Even if hardware were guaranteed
to be pure and perfect and never subject to "Not Invented Here"
syndrome, I would not assume this. Consider programs that have the
bug which concatenates a number when moving it around.

/BAH