The Ghost In The Machine 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

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

"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.



Take care and mind the gap.

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

Dirk Vdm

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.






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

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

"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?
Androcles.

"The Ghost In The Machine" wrote in
message news | 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:

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.

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?

| 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/

In article "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/

"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.


| | 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