| Topic: |
Science > Physics |
| User: |
"Nth Complexity" |
| Date: |
09 Sep 2005 06:05:41 PM |
| Object: |
NOMINATION: Dirk Van de moortel for VVFWS |
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
-- Nth Complexity --
-- Have A Nice Day! --
http://www.insurgent.org/~kook-faq/search.php?query=Uncle+Al
"The teaching of science and mathematics must be purged of its
authoritarian and elitist characteristics, and the content of these
subjects enriched by incorporating the insights of the feminist,
queer, multiculturalist and ecological critiques." -- A.D.S.
--
Sent by nth_complexity from yahoo included in com
This is a spam protected message. Please answer with reference header.
Posted via http://www.usenet-replayer.com
.
|
|
| User: "Woodchuck Bill" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
09 Sep 2005 06:11:40 PM |
|
|
(Nth Complexity) wrote in
news:l.1126307142.1708282470@[82.96.100.100]:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
-- Nth Complexity --
-- Have A Nice Day! --
http://www.insurgent.org/~kook-faq/search.php?query=Uncle+Al
"The teaching of science and mathematics must be purged of its
authoritarian and elitist characteristics, and the content of these
subjects enriched by incorporating the insights of the feminist,
queer, multiculturalist and ecological critiques." -- A.D.S.
Give it up, George.
--
Bill
.
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|
| User: "odin" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
09 Sep 2005 06:47:10 PM |
|
|
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
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?
.
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|
| User: "The Ghost In The Machine" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 08:00:06 AM |
|
|
In sci.math, odin
<ragnarok@yahoo.com>
wrote
on Fri, 9 Sep 2005 16:47:10 -0700
<Toedne77HJhjg7_eRVn-3Q@whidbeytel.com>:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
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.
.
|
|
|
| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 08:53:09 AM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message news:rhuav2-3l6.ln1@sirius.tg00suus7038.net...
| In sci.math, odin
| <ragnarok@yahoo.com>
| wrote
| on Fri, 9 Sep 2005 16:47:10 -0700
| <Toedne77HJhjg7_eRVn-3Q@whidbeytel.com>:
| >> Dirk Van de moortel wrote:
| >>> By the way, zero is usually taken to be both positive and
| >>> negative.
| >>
| >> Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
| >> 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
.
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|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 11:00:11 AM |
|
|
In sci.math, Androcles
<Androcles@MyPlace.org>
wrote
on Sat, 10 Sep 2005 13:53:09 GMT
<9zBUe.17116$k22.13781@fe2.news.blueyonder.co.uk>:
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message news:rhuav2-3l6.ln1@sirius.tg00suus7038.net...
| In sci.math, odin
| <ragnarok@yahoo.com>
| wrote
| on Fri, 9 Sep 2005 16:47:10 -0700
| <Toedne77HJhjg7_eRVn-3Q@whidbeytel.com>:
| >> Dirk Van de moortel wrote:
| >>> By the way, zero is usually taken to be both positive and
| >>> negative.
| >>
| >> Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
| >> 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.
.
|
|
|
| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 02:18:40 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
message news:oo9bv2-11s.ln1@sirius.tg00suus7038.net...
| In sci.math, Androcles
| <Androcles@MyPlace.org>
| wrote
| on Sat, 10 Sep 2005 13:53:09 GMT
| <9zBUe.17116$k22.13781@fe2.news.blueyonder.co.uk>:
| >
| > "The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in
| > message news:rhuav2-3l6.ln1@sirius.tg00suus7038.net...
| > | In sci.math, odin
| > | <ragnarok@yahoo.com>
| > | wrote
| > | on Fri, 9 Sep 2005 16:47:10 -0700
| > | <Toedne77HJhjg7_eRVn-3Q@whidbeytel.com>:
| > | >> Dirk Van de moortel wrote:
| > | >>> By the way, zero is usually taken to be both positive and
| > | >>> negative.
| > | >>
| > | >>
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
| > | >> 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:
I've just used 32-but signed integers.
| Is 0x7fff positive or negative?
The highest positive number you can represent in 16-bit signed integer.
|
| Is 0x7fff+1 positive or negative?
Neither. It is zero.
| Is 0x7fff+2 positive or negative?
Positive overflow.
| Is 0x8000 positive or negative?
That was my question.
Awww....1000 0000 0000 0000 isn't -0 anymore?
|
| Is 0x8000-1 positive or negative?
|
| Is 0x8000-2 positive or negative?
|
| Of course it's not the highest integer; there is no such.
In signed 16-bit representation there is a highest integer, 32767
decimal.
In signed 32-bit representation there is a highest integer,
2,147,483,647 decimal.
How many files can your hard drive hold?
| It's the highest positive integer that can be crammed into
| 16 bits, given current sign conventions.
Exactly, right. Given current sign conventions, there is a -zero.
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.
The computer is never wrong. It is human interpretation that is wrong.
48 = '0' in ASCII.
Androcles.
.
|
|
|
| User: "Dik T. Winter" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 02:54:18 PM |
|
|
In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk> "Androcles" <Androcles@ MyPlace.org> writes:
....
| Assuming 16-bit signed integers, represented in hex:
I've just used 32-but signed integers.
1-s complement or 2-s complement?
| Is 0x7fff positive or negative?
The highest positive number you can represent in 16-bit signed integer.
Right.
| Is 0x7fff+1 positive or negative?
Neither. It is zero.
Wrong. It is overflow.
| Is 0x7fff+2 positive or negative?
Positive overflow.
| Is 0x8000 positive or negative?
That was my question.
Awww....1000 0000 0000 0000 isn't -0 anymore?
It is not on 2-s complement machines, that is on almost *all* machines
currently running. It is -32768.
| It's the highest positive integer that can be crammed into
| 16 bits, given current sign conventions.
Exactly, right. Given current sign conventions, there is a -zero.
Exactly wrong. Given current sign conventions there is no -zero.
When did you retire precisely?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
|
|
|
| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 03:37:32 PM |
|
|
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:IMM9yI.E@cwi.nl...
| In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk>
"Androcles" <Androcles@ MyPlace.org> writes:
| ...
| > | Assuming 16-bit signed integers, represented in hex:
| >
| > I've just used 32-but signed integers.
|
| 1-s complement or 2-s complement?
(Sigh...) I've just used 32-bit SIGNED integers.
If you weren't so anxious to snip you'd see that.
1-s complement is NOT(x).
2-s complement is NOT(x)+1.
'NOT' is unary operator and function that turns 1s into to 0s and 0s
into 1s.
'+' is a binary operator and function that combines x with y
through something we call "addition". You may have heard of it.
| > | Is 0x7fff positive or negative?
| >
| > The highest positive number you can represent in 16-bit signed
integer.
|
| Right.
|
| > | Is 0x7fff+1 positive or negative?
| >
| > Neither. It is zero.
|
| Wrong. It is overflow.
1-s complement or 2-s complement?
|
| > | Is 0x7fff+2 positive or negative?
| >
| > Positive overflow.
| >
| > | Is 0x8000 positive or negative?
| >
| > That was my question.
| > Awww....1000 0000 0000 0000 isn't -0 anymore?
|
| It is not on 2-s complement machines, that is on almost *all* machines
| currently running. It is -32768.
1-s complement or 2-s complement?
| > | It's the highest positive integer that can be crammed into
| > | 16 bits, given current sign conventions.
| >
| > Exactly, right. Given current sign conventions, there is a -zero.
|
| Exactly wrong. Given current sign conventions there is no -zero.
| When did you retire precisely?
I'm going to retire you right now.
*plonk*
Androcles
.
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|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 04:10:46 PM |
|
|
"Androcles" <Androcles@ MyPlace.org> wrote in message news:guHUe.47859$2n6.26635@fe3.news.blueyonder.co.uk...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:IMM9yI.E@cwi.nl...
| In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk>
"Androcles" <Androcles@ MyPlace.org> writes:
| > Exactly, right. Given current sign conventions, there is a -zero.
|
| Exactly wrong. Given current sign conventions there is no -zero.
| When did you retire precisely?
I'm going to retire you right now.
*plonk*
Go ahead, Dik.
This means he pretends he can't read your messages anymore.
You can now calmly say what you have to say and he will not
be able to respond anymore - until he forgets in about 3 weeks
or so... He did it 4 (or was it 5 or 6?) times with me ;-)
Dirk Vdm
.
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|
|
| User: "Daryl McCullough" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 04:40:55 PM |
|
|
Dirk Van de moortel says...
"Androcles" <Androcles@ MyPlace.org> wrote in message
I'm going to retire you right now.
*plonk*
Go ahead, Dik.
This means he pretends he can't read your messages anymore.
You can now calmly say what you have to say and he will not
be able to respond anymore
Yes, I've been calling Androcles an idiot recently to see if he is really
reading my posts or not. So far he hasn't responded.
Androcles, you are an idiot.
--
Daryl McCullough
Ithaca, NY
.
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| User: "Dik T. Winter" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 07:29:13 PM |
|
|
In article <guHUe.47859$2n6.26635@fe3.news.blueyonder.co.uk> "Androcles" <Androcles@ MyPlace.org> writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message
news:IMM9yI.E@cwi.nl...
| In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk>
"Androcles" <Androcles@ MyPlace.org> writes:
| ...
| > | Assuming 16-bit signed integers, represented in hex:
| >
| > I've just used 32-but signed integers.
|
| 1-s complement or 2-s complement?
(Sigh...) I've just used 32-bit SIGNED integers.
If you weren't so anxious to snip you'd see that.
1-s complement is NOT(x).
2-s complement is NOT(x)+1.
'NOT' is unary operator and function that turns 1s into to 0s and 0s
into 1s.
'+' is a binary operator and function that combines x with y
through something we call "addition". You may have heard of it.
Yes, so what?
| > | Is 0x7fff positive or negative?
| >
| > The highest positive number you can represent in 16-bit signed
integer.
|
| Right.
|
| > | Is 0x7fff+1 positive or negative?
| >
| > Neither. It is zero.
|
| Wrong. It is overflow.
1-s complement or 2-s complement?
Both.
| > | Is 0x7fff+2 positive or negative?
| >
| > Positive overflow.
| >
| > | Is 0x8000 positive or negative?
| >
| > That was my question.
| > Awww....1000 0000 0000 0000 isn't -0 anymore?
|
| It is not on 2-s complement machines, that is on almost *all* machines
| currently running. It is -32768.
1-s complement or 2-s complement?
2-s complement.
| > | It's the highest positive integer that can be crammed into
| > | 16 bits, given current sign conventions.
| >
| > Exactly, right. Given current sign conventions, there is a -zero.
|
| Exactly wrong. Given current sign conventions there is no -zero.
| When did you retire precisely?
I'm going to retire you right now.
*plonk*
Ah, when you do not know the answer just plonk.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
|
|
|
|
|
|
| User: "Dik T. Winter" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 02:55:05 PM |
|
|
In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk> "Androcles" <Androcles@ MyPlace.org> writes:
....
| 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.
The computer is never wrong.
I thought you had been doing quality control? I start to doubt that.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
|
|
|
| User: "Bilge" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
11 Sep 2005 08:57:59 AM |
|
|
Dik T. Winter:
In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk>
"Androcles" <Androcles@ MyPlace.org> writes:
...
| 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.
The computer is never wrong.
I thought you had been doing quality control? I start to doubt that.
His contrtibution was to not show up.
.
|
|
|
|
| User: "David Kastrup" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
11 Sep 2005 01:03:43 AM |
|
|
"Dik T. Winter" <Dik.Winter@cwi.nl> writes:
In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk> "Androcles" <Androcles@ MyPlace.org> writes:
...
| 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.
The computer is never wrong.
I thought you had been doing quality control? I start to doubt
that.
I once debugged a certain problem in a program of mine until my hairs
were close to falling out. The computer worked fine, just that one
program delivered rubbish. I finally traced it down to an arithmetic
right shift command which failed to shift in the sign bit consistently
when it was set: sometimes it shifted in 0 instead (I did not figure
out the exact required conditions). The only obvious failure had been
that one program. Replacing the CPU fixed it.
Anyway, "quality control" nowadays can amount to telling people to
maintain paperwork while they blow it.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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| User: "Dik T. Winter" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
11 Sep 2005 06:59:55 PM |
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In article <851x3wjg3k.fsf@lola.goethe.zz> David Kastrup <dak@gnu.org> writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> writes:
In article <kkGUe.47765$2n6.9258@fe3.news.blueyonder.co.uk> "Androcles" <Androcles@ MyPlace.org> writes:
....
The computer is never wrong.
I thought you had been doing quality control? I start to doubt
that.
I once debugged a certain problem in a program of mine until my hairs
were close to falling out. The computer worked fine, just that one
program delivered rubbish. I finally traced it down to an arithmetic
right shift command which failed to shift in the sign bit consistently
when it was set: sometimes it shifted in 0 instead (I did not figure
out the exact required conditions). The only obvious failure had been
that one program. Replacing the CPU fixed it.
I have a few similar expeeriences. One was a program that was used in
a large number of long runs on a Cyber 205. On occasion the results would
be bogus. The reason was that apparently one of the fans was connected
the wrong way. And it appears Androcles never has heard of the Intel FPU
bug.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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| User: "David Kastrup" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 11:10:31 AM |
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The Ghost In The Machine <ewill@sirius.tg00suus7038.net> writes:
In sci.math, Androcles
So just what is
1000 0000 0000 0000 0000 0000 0000 0000
in modern processors with only one representation for 0?
Assuming 16-bit signed integers, represented in hex:
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.
Not quite.
0x8000 is the smallest (most negative) 16-bit integer
under those conditions,
You are not exactly being consistent.
and has some odd properties; for starters 2 * 0x8000 = [overflow] 0,
as you've already noticed. Of course 2 * 0x7fff = 0xfffe =
[overflow] -2.
That's all pretty harmless. The really appalling thing is that you
can't negate the number without causing overflow: the number stays the
same.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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| User: "" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 08:18:55 AM |
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In article <rhuav2-3l6.ln1@sirius.tg00suus7038.net>,
The Ghost In The Machine <ewill@sirius.tg00suus7038.net> wrote:
In sci.math, odin
<ragnarok@yahoo.com>
wrote
on Fri, 9 Sep 2005 16:47:10 -0700
<Toedne77HJhjg7_eRVn-3Q@whidbeytel.com>:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
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
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| User: "Robert J. Kolker" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 11:14:11 AM |
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odin wrote:
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?
It could be taken as neither. Define postive to mean (strictly) greater
than zero and negative to mean (strictly) less than zero. In this
definition, zero is neither positive nor negative.
Bob Kolker
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| User: "Virgil" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
09 Sep 2005 10:46:28 PM |
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In article <l.1126307142.1708282470@[82.96.100.100]>,
(Nth Complexity) wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
Depends. According to Bourbaki both, but by the common English standard,
neither.
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| User: "Uncle Al" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
09 Sep 2005 08:02:28 PM |
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Nth Complexity wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
Idiot.
-- Nth Complexity --
-- Have A Nice Day! --
http://www.insurgent.org/~kook-faq/search.php?query=Uncle+Al
"The teaching of science and mathematics must be purged of its
authoritarian and elitist characteristics, and the content of these
subjects enriched by incorporating the insights of the feminist,
queer, multiculturalist and ecological critiques." -- A.D.S.
Fucking imbecile.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "Schoenfeld" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 01:57:27 AM |
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Nth Complexity wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
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).
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 R is the set of
negative integers.
Where does 0 lie in here?
Well if it lies in BOTH P and Q 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 Q). Again, this is not strictly prohibited by
additive identity axiom. You could say there are two additive
identities 0+ and 0-, positive and negative respectively. A cursory
analysis their arithmetic reveals no axiomatic contradictions.
Based on this, I would need to say that 0 by itself does not exist.
Rather you have positive 0+ and negative 0-.
Is there an error here?
REMARK: The widely used terminology related to all this is:
"Positive integers" = 1,2,3, ...
"nonnegative integers" = 0, 1, 2, 3, ...
"negative integers" = -1, -2, -3, ...
"nonpositive integers" = 0, -1, -2, -3, ....
-- Nth Complexity --
-- Have A Nice Day! --
http://www.insurgent.org/~kook-faq/search.php?query=Uncle+Al
"The teaching of science and mathematics must be purged of its
authoritarian and elitist characteristics, and the content of these
subjects enriched by incorporating the insights of the feminist,
queer, multiculturalist and ecological critiques." -- A.D.S.
--
Sent by nth_complexity from yahoo included in com
This is a spam protected message. Please answer with reference header.
Posted via http://www.usenet-replayer.com
.
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| User: "David Kastrup" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 02:01:40 AM |
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"Schoenfeld" <schoenfeld1@gmail.com> writes:
Nth Complexity wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
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).
You don't prove definitions.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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| User: "Schoenfeld" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 02:05:15 AM |
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David Kastrup wrote:
"Schoenfeld" <schoenfeld1@gmail.com> writes:
Nth Complexity wrote:
Dirk Van de moortel wrote:
By the way, zero is usually taken to be both positive and
negative.
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
And you expect to teach OTHERS?!
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).
You don't prove definitions.
Please READ what I said properly. I referred to the definitions (i
re-posted my post correcting symbol misplacements). Address queries in
other thread branch.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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| User: "Robert Low" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 09:01:38 AM |
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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.)
.
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| User: "Dirk Van de moortel" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 09:03:44 AM |
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"Robert Low" <mtx014@coventry.ac.uk> wrote in message news:3og780F5pdodU1@individual.net...
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
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| User: "Schoenfeld" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 09:10:58 AM |
|
|
Dirk Van de moortel wrote:
"Robert Low" <mtx014@coventry.ac.uk> wrote in message news:3og780F5pdodU1@individual.net...
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
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| User: "Dirk Van de moortel" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 10:08:45 AM |
|
|
"Schoenfeld" <schoenfeld1@gmail.com> wrote in message news:1126361458.593179.53700@g14g2000cwa.googlegroups.com...
Dirk Van de moortel wrote:
"Robert Low" <mtx014@coventry.ac.uk> wrote in message news:3og780F5pdodU1@individual.net...
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
.
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| User: "Schoenfeld" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 10:32:47 AM |
|
|
Dirk Van de moortel wrote:
"Schoenfeld" <schoenfeld1@gmail.com> wrote in message news:1126361458.593179.53700@g14g2000cwa.googlegroups.com...
Dirk Van de moortel wrote:
"Robert Low" <mtx014@coventry.ac.uk> wrote in message news:3og780F5pdodU1@individual.net...
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.
You make this assumption when you state that:
x + Y1 = x
x + Y2 = x
[YOU]
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.
===================
What I called "STATEMENT 1" may have been obvious to you, but it was
not obvious to me at all. The reason it was not obvious to me was
because I kept thinking of additive identity axiom as follows:
"For all x there exists y such that x + y = y"
I didn't write it that way, but that is how I was thinking about it.
And if you think about it like that (which I will admit once again that
it is wrong) then it is not possible for y to be the same for all x.
For example,
"For all x there exists y such that x + y = 10" does not give the same
'y' for all x. Rather, it gives a DIFFERENT y for each x. This was the
source of my error which I'll admit for the final time.
Now, getting back to the whole point raised by the OP on your
statement:
"By the way, zero is usually taken to be both positive and negative."
Here is YOUR explanation of this statement:
"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. "
AND
"In my part of the world:
"Positive integers" = { 0, 1, 2, 3, ... }
"Strictly positive integers" = "Nonnegative integers" = { 1, 2, 3,
.... }
"Negative integers" = { 0, -1, -2, -3, ... }
"Strictly negative integers" = "Nonpositive integers" = { -1, -2, -3,
.... }"
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)
Dirk Vdm
By the way, congrats on your award.
Dirk Vdm
.
|
|
|
| User: "Dirk Van de moortel" |
|
| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 10:41:24 AM |
|
|
"Schoenfeld" <schoenfeld1@gmail.com> wrote in message news:1126366367.771433.180890@g47g2000cwa.googlegroups.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
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| User: "Schoenfeld" |
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| Title: Re: NOMINATION: Dirk Van de moortel for VVFWS |
10 Sep 2005 11:03:47 AM |
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Dirk Van de moortel wrote:
"Schoenfeld" <schoenfeld1@gmail.com> wrote in message news:1126366367.771433.180890@g47g2000cwa.googlegroups.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
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