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

"Dik T. Winter" wrote in message
...
| In article
"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

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

In sci.math, Androcles

wrote
on Sat, 10 Sep 2005 19:01:44 GMT
:

"The Ghost In The Machine" wrote in
message ...
| In sci.math, Androcles
|
| wrote
| on Sat, 10 Sep 2005 15:42:30 GMT
| :
|
| "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?
|
| The above proof can be tightened considerably. If we assume
| that there are two elements 0 and y in the ring
| such that y + x = x + y = x and 0 + x = x + 0 = x
| for all x in the ring, then:
|
| y (initiate)
| = y + 0 (arith identity ring axiom)
| = 0 + y (additive commutativity ring axiom; UI^2)
| = 0 (hypothesis)
|
| therefore the additive element is unique. This proof means
| that the identity is also unique within commutative monoids,
| which need not have inverses. See
http://mathworld.wolfram.com/Monoid.html .
|
|
| Sheesh... 0=0=0=0=0=0=0=0=0=0=0=0=0=0=...=0
|
| Androcles.
|
|
|
|

Sheesh... 0+0+0+0+0+0+0+0+0+0+0= 0
Why don't you have an infinity of identities?
How does your argument show Y1 =/= Y2?

Given what?

If one assumes that for some y1 and some y2, not necessarily distinct,
that

for all x, x + y1 = x [1] (identity #1)

and

for all x, x + y2 = x [2] (identity #2)

and

for all x and y, x + y = y + x [3] (commutativity)

then one can conclude the following:

y2 + y1 = y2 (substituting x=y2 in [1])
y1 + y2 = y1 (substituting x=y1 in [2])

By commutativity,

y2 + y1 = y2
y2 + y1 = y1

therefore y1 = y2 since they're both equal to y2 + y1.

In any event, I'm not trying to *show* y1 != y2. I'm trying
to show !(y1 != y2) or y1 = y2.

Androcles.


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

"The Ghost In The Machine" wrote in
message ...
| In sci.math, Androcles
|
| wrote
| on Sat, 10 Sep 2005 19:01:44 GMT
| :
|
| "The Ghost In The Machine" wrote in
| message ...
| | In sci.math, Androcles
| |
| | wrote
| | on Sat, 10 Sep 2005 15:42:30 GMT
| | :
| |
| | "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?
| |
| | The above proof can be tightened considerably. If we assume
| | that there are two elements 0 and y in the ring
| | such that y + x = x + y = x and 0 + x = x + 0 = x
| | for all x in the ring, then:
| |
| | y (initiate)
| | = y + 0 (arith identity ring axiom)
| | = 0 + y (additive commutativity ring axiom; UI^2)
| | = 0 (hypothesis)
| |
| | therefore the additive element is unique. This proof means
| | that the identity is also unique within commutative monoids,
| | which need not have inverses. See
| http://mathworld.wolfram.com/Monoid.html .
| |
| |
| | Sheesh... 0=0=0=0=0=0=0=0=0=0=0=0=0=0=...=0
| |
| | Androcles.
| |
| |
| |
| |
|
| Sheesh... 0+0+0+0+0+0+0+0+0+0+0= 0
| Why don't you have an infinity of identities?
| How does your argument show Y1 =/= Y2?
|
| Given what?
|
| If one assumes that for some y1 and some y2, not necessarily distinct,
| that
|
| for all x, x + y1 = x [1] (identity #1)
|
| and
|
| for all x, x + y2 = x [2] (identity #2)
|
| and
|
| for all x and y, x + y = y + x [3] (commutativity)
|
| then one can conclude the following:
|
| y2 + y1 = y2 (substituting x=y2 in [1])
| y1 + y2 = y1 (substituting x=y1 in [2])
|
| By commutativity,
|
| y2 + y1 = y2
| y2 + y1 = y1
|
| therefore y1 = y2 since they're both equal to y2 + y1.
|
| In any event, I'm not trying to *show* y1 != y2. I'm trying
| to show !(y1 != y2) or y1 = y2.

Unsuccessfully.

y2 + y1 = y2 = y2 + y1 = y1

Androcles.

Robert Low wrote:
Schoenfeld wrote:
Robert Low wrote:
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.
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.

Indeed, it doesn't even guarantee the existence of an
additive identity.

I'm glad you could read what I said (and then snip it).

Yep, there's nothing to beat understanding
the things you're trying to talk about.

Yeah, like constructing a sentence which carries meaning.

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

[snip]

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.

As usual you got it all wrong.
I don't oppose crackpot usenet mentality. I study and enjoy it.
If these forums didn't have people like you, I woulnd't even
be here.

hint: if you wanted to study crackpots, you need only study a mirror.



Take care and mind the gap.

Or don't - and twist your ankle again.
Your choice.

See.

Dirk Vdm

In article "Androcles" Androcles@ MyPlace.org writes:
"Dik T. Winter" wrote in message
...
| In article
"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/

Dirk Van de moortel wrote:

[snip]

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.

[snip more crap]

Idiot imbecile. Next thing you tell us is that you met Bourbaki and he
taught you personally this garbage.



Mike





Dirk Vdm

In article David Kastrup writes:
"Dik T. Winter" writes:
In article "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/

Bilge wrote:
Schoenfeld:

========================
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.

Or possibly it's popular for the same reason that websites like
rotten.com are popular. People find the contents hard to imagine without
seeing it for themselves. I know that I found it hard to believe that
anyone as logically challenged as androsleaze or stowe could be anything
but anomalies until I started reading this newsgroup.

No, Google's page rank algorithm treats the web as a graph and weighs
each node (web page) as function of the number of incoming edges (among
various other factors). I don't know if 'rotten.com' is popular, but
that's your business not mine. Although if you wanted my advice I would
say that if such material pleases you, seek psychiatric help ASAP.

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].

There is an obvious solution to your dilemna: Think before
you post.

That's poetic.

Now your site, by association to header page article links, implies
your subjects are mentally retarded, psychopaths, autistic, ignorant,
incompetent, etc.

I'm afraid I don't really see the problem.

Well primarily because the e-mail was not directed to you, and
secondarily, because you're too busy stuffing your mind with the
mind-poison found at 'rotten.com'.

[...]


``For today's plaintiff to prevail in a defamation action, he must
prove publication of the defamatory statement, identification of
the plaintiff, falsity, defamatory content, injury and fault.''

http://www.law.duke.edu/journals/dlt...2dltr0004.html

See number (4) above. Rule number (5) Never ask for favors using
an implied and empty threat. It's likely to recieve the same sort
of consideration that cryptome.org gives that sort of thing: ``No
court order has ever been served; any order served will be published
here -- or elsewhere if gagged by order. Bluffs will be published
if comical but otherwise ignored.''

I never made any threat, I specifically stated that I was not trying to
make threat. Also I don't think US law applies here.

You're a real man of principle, john... Your interest only extends
as far as maintaining your own anonymity. When it comes to anyone
else, your interest is in whether or not you think you can halt
your detractors by posting their personal information. Hypocrisy is
not the best way to convince anyone that your motivations are based
on principles.


Make a mental note (you'll need to get some paper first), in that email
I specifically stated the 'fair thing to do' was to remove all pages
even the ones not referring to me. Your statements are mere lies.


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.

I fail to see how posting some of the more outrages lapses in
logical thinking, common sense, etc., could possibly encourage
emulation. I find it hard to believe that lots of people will
read dirk's fumbles page and then say to themselves, ``I hope
I'm that smart someday.'' If so, then perhaps they should be
relocated to antarctica until evolution catches up.

David, some people say you are the perfect idiot. I say that you are
not perfect but you are doing alright.