"Schoenfeld" wrote in message ups.com...

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.

Ha, so you call the following an admission of your error:
| Well in the case, genius, you just proved your own idiotic error
| demonstrated by the OP. I must therefore concur with that OP,
| "HAHAHAHAHAH".
Nice touch.


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".

And then on top of having tried to sell **** from the beginning, and,
in order to boost his sales, having made a silly glaringly elementary
mistake, that he was not prepared to acknowledge, but now, having
no other option than to indeed do just that, he, as the cherry on the
pie, produces the most transparent lie one could possibly imagine: he
| CONFUSED
| "THERE EXISTS Y FOR ALL X" as
| "FOR ALL X THERE EXISTS Y".

Nice touch, Schoensmeer - Do you really think you can get away
with this? You don't have to admit that you made an error.
I'm waiting for you to admit that you are a dishonest imbecile -
and not just pretending to be one :-)

Dirk Vdm



By the way, congrats on your award.

Dirk Vdm

Robert Low wrote:
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.)

I realize that now. Unfortunately I was thinking
"for all x there exists y such that x + y = x" which doesn't guarantee
uniqueness at all.

For example:
"for all x there exists y such that x + y = 10" is true, but it implies
a different y for every x.

OTOH,
"there exists y for all x such that x + y = 10" is clearly false as no
such y exists for all x.

In article . com "Schoenfeld" writes:
....
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

We have the ring axiom that there is a 0 such that for all x: x + 0 = x.
We have the ring axiom that for each x there is a (-x) such that x + (-x) = 0.
We have also commutativity and associativity of the addition.

Now suppose there is an y such that x + y = x for some x. Add (-x) to both
sides and find that y = 0:
y = // by y + 0 = y
y + 0 = // commutativity
0 + y = // x + (-x) = 0
x + (-x) + y = // commutativity
(-x) + x + y = // by x + y = x
(-x) + x = // x + (-x) = 0
0
--
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:
"Schoenfeld" wrote in message ups.com...

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.

Ha, so you call the following an admission of your error:
| Well in the case, genius, you just proved your own idiotic error
| demonstrated by the OP. I must therefore concur with that OP,
| "HAHAHAHAHAH".
Nice touch.

Have a read of the below two paragraphs of your very own post.

PARAGRAPH 1:

| 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".

PARAGRAPH 2:
| 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".


And then on top of having tried to sell **** from the beginning, and,
in order to boost his sales, having made a silly glaringly elementary
mistake, that he was not prepared to acknowledge, but now, having
no other option than to indeed do just that, he, as the cherry on the
pie, produces the most transparent lie one could possibly imagine: he
| CONFUSED
| "THERE EXISTS Y FOR ALL X" as
| "FOR ALL X THERE EXISTS Y".

Nice touch, Schoensmeer - Do you really think you can get away
with this? You don't have to admit that you made an error.
I'm waiting for you to admit that you are a dishonest imbecile -
and not just pretending to be one :-)

I will afford you the benefit of the doubt, and assume you honestly
believe that:
1. I lied about my question asking you to show "STATEMENT 1" to be
true.
2. I now am lying about my admission of my own error.

To refresh your memory, here was the source my error:

===============================

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

In article om "Schoenfeld" writes:
[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.

In what way? In this part of the world the positive numbers are defined
as all numbers = 0. The negative numbers are defined as all numbers = 0.

Your precise error was that you thought you could split Z in sets P and N
with the following conditions:
1. P union N = Z
2. P intersection N = empty
3. If x in P, -x in N.
But if you wish (3) you have to give up either (1) or (2). Anglosaxon
mathematics give up (1), Bourbaki mathematics gives up (2).
In Anglosaxon mathematics (1) is replaced by:
1a. P union N union {0} = Z
in Bourbaki mathematics (2) is replaced by:
2a. P intersection N = {0}.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

"Schoenfeld" wrote in message oups.com...

[snip]

I'll afford you the benefit of the doubt again and assume that your
statement was merely a language issue. But may I ask for just for ONE
reference from "your part of the world" which also follows such
nonstandard nomenclature? (just one)

Schoenfeld, if you don't have the guts to admit on this public
forum that you have been behaving like a dishonest imbecile
again, then you can always use private email - just like you
did last week.

Take care and mind the gap.

Dirk Vdm

"Dik T. Winter" wrote in message
...
| In article . com
"Schoenfeld" writes:
| ...
| 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.

"Androcles" Androcles@ MyPlace.org wrote in message ...

"Dik T. Winter" wrote in message
...
| In article . com
"Schoenfeld" writes:
| ...
| 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
|
| We have the ring axiom that there is a 0 such that for all x: x + 0 =
x.
| We have the ring axiom that for each x there is a (-x) such that x +
(-x) = 0.
| We have also commutativity and associativity of the addition.
|
| Now suppose there is an y such that x + y = x for some x. Add (-x) to
both
| sides and find that y = 0:
| y = // by y + 0 = y
| y + 0 = // commutativity
| 0 + y = // x + (-x) = 0
| x + (-x) + y = // commutativity
| (-x) + x + y = // by x + y = x
| (-x) + x = // x + (-x) = 0
| 0
| --
| dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131
| home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/

Given:
| x + Y1 = x
| x + Y2 = x

How does your argument show Y1 =/= Y2?

Sheesh... 0=0=0=0=0=0=0=0=0=0=0=0=0=0=...=0

Androcles.

Sheesh... limits:
http://users.pandora.be/vdmoortel/di...les/Limit.html
Sheesh... equations:
http://users.pandora.be/vdmoortel/di...SetSolve2.html
http://users.pandora.be/vdmoortel/di...ersuasive.html
http://users.pandora.be/vdmoortel/di...droDistri.html
http://users.pandora.be/vdmoortel/di...ythagoras.html
http://users.pandora.be/vdmoortel/di.../ToothlessBite....
http://users.pandora.be/vdmoortel/di...Competent.html
http://users.pandora.be/vdmoortel/di.../UseTrans.html
http://users.pandora.be/vdmoortel/di...es/Sheesh.html
http://users.pandora.be/vdmoortel/di.../SetSolve.html
http://users.pandora.be/vdmoortel/di...s/DivZero.html
http://users.pandora.be/vdmoortel/di...les/Think.html
Sheesh... square roots:
http://users.pandora.be/vdmoortel/di...les/STILL.html
http://users.pandora.be/vdmoortel/di...anSpecify.html
http://users.pandora.be/vdmoortel/di...es/Nearly.html
http://users.pandora.be/vdmoortel/di...Quadratic.html
http://users.pandora.be/vdmoortel/di...es/GrowUp.html
http://users.pandora.be/vdmoortel/di...Tautology.html
http://users.pandora.be/vdmoortel/di.../Material.html
http://users.pandora.be/vdmoortel/di...les/GIVEN.html
http://users.pandora.be/vdmoortel/di.../PythagoRescue....
http://users.pandora.be/vdmoortel/di...s/SqrtRev.html
http://users.pandora.be/vdmoortel/di...s/NegSqrt.html
http://users.pandora.be/vdmoortel/di...rtAnswers.html
Sheesh... exclusive ors:
http://users.pandora.be/vdmoortel/di...Gibberish.html
Sheesh... partial differential equations:
http://users.pandora.be/vdmoortel/di...rtialDiff.html
http://users.pandora.be/vdmoortel/di...tialDiff2.html
http://users.pandora.be/vdmoortel/di...tialDiff3.html

Dirk Vdm

In sci.math, Androcles

wrote
on Sat, 10 Sep 2005 13:53:09 GMT
:

"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


Assuming 16-bit signed integers, represented in hex:

Is 0x7fff positive or negative?

Is 0x7fff+1 positive or negative?

Is 0x7fff+2 positive or negative?

Is 0x8000 positive or negative?

Is 0x8000-1 positive or negative?

Is 0x8000-2 positive or negative?

Of course it's not the highest integer; there is no such.
It's the highest positive integer that can be crammed into
16 bits, given current sign conventions. There are larger
unsigned integers (0xffff = 65535 = USHRT_MAX) but that's
a different convention, represented in C as 'unsigned int'
as opposed to 'signed int' or just plain 'int'.

0x8000 is the smallest (most negative) 16-bit integer
under those conditions, and has some odd properties;
for starters 2 * 0x8000 = [overflow] 0, as you've already noticed.
Of course 2 * 0x7fff = 0xfffe = [overflow] -2.

And then 0xffff + 0x0001 = -1 + 1 = [carry/borrow] 0;
it's not really considered overflow as such.

I'm not sure 0x8000 ever was -0 in two's complement arithmetic;
in contemporary code it's SHRT_MIN in /usr/include/limits.h ,
and might have been INT_MIN in older model computer units (PDP/11
comes to mind) which squished an int into 16 bits.

Of course the real problem is that a computer models the thinking
processes of humans, and imperfectly at that. To the computer
0x8000 is just a bunch of voltage pulses, pushed down one side
of an arithmetic logic unit in an attempt to get a result fast.

Sometimes, the result is wrong.

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

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

[snip]

I will take that [snip] as indicating now that you finally realized I
was not lying at all, but rather just made an elementary error which I
eventually realized and admitted to over 10 times in this entire
thread.

I'll afford you the benefit of the doubt again and assume that your
statement was merely a language issue. But may I ask for just for ONE
reference from "your part of the world" which also follows such
nonstandard nomenclature? (just one)

Schoenfeld, if you don't have the guts to admit on this public
forum that you have been behaving like a dishonest imbecile
again, then you can always use private email - just like you
did last week.

I see. Well here is the email I sent you last week for the public
record (since this matters to you).

========================
Dirk,

As you know I requested some time ago for you to withdraw certain pages
about me from your personal ridicule site. You've recently decided to
put them up again. Unfortunately for your subjects, your site gets high
google page-rank (because it is linked to by various other sites) and
searches for those peoples names gives your pages first. This may not
be an issue to some people, but for me it is a very big issue. I'm not
sure how much you understand about e-commerce, but online consumers are
by default skeptical of [4 WORDS REMOVED]. Such consumers usually
perform quick [2 WORDS REMOVED] to validate whether or not such an [1
WORD REMOVED] is legitamit and whether or not to commit to the purchase
of that [1 WORD REMOVED].

Now your site, by association to header page article links, implies
your subjects are mentally retarded, psychopaths, autistic, ignorant,
incompetent, etc. Your site obviously has the potential to act as a
*massive* deterant to a [1 WORD REMOVED] I plan to [2 WORDS REMOVED]
very soon. I am not making this up, this is a legitamit issue. The
point of this email is primarily to make you AWARE of the impliciations
of your defamatory pages. Obviously, it would benefit me greatly for
you to take down all these pages about me as soon as possible - but in
doing so, the fair thing is to request you remove that entire section.

Please don't make me waste my time or your time or your ISP's time with
this. At the risk of sounding threatening, which I'm not trying to be,
if you don't withdraw such pages I'll need to pester your ISP until
these arguably defamatory and fiscally damaging pages are withdrawn.

PLEASE WITHDRAW SUCH PAGES!


Regards
==============================
NOTE: In the interest of maintaining ones business/geographic identity
anonymous from the ever increasing parade of internet lunatics and
psychopaths, certain key words removed.


I still hold to that e-mail, and am pleased you decided to withdraw
your pages ridiculing me (you aren't as gutless as I put you for).
Though, I still recommend that you withdraw all those pages, as IMO, it
only acts to strengthen the crackpot usenet mentality you so vehemetely
oppose.


Take care and mind the gap.

Dirk Vdm