Go back
The Nullity

The Nullity

Posers and Puzzles

R

Joined
30 Oct 05
Moves
3072
Clock
10 Dec 06
1 edit
Vote Up
Vote Down

0^0 = 0^(1-1)

0^0 = (0^1)(0^-1) = (0/1)(1/0) = (0/0) = nullity


Here's the paper:
http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf


And here are its supposed ramifications for calculus:
http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf


So, it's a new number lying outside the real number line that supposedly answers a nagging mathematical question, but is it sound? And also importantly, is it trivial? Taking a cursory look at both papers, and the BBC news report of it, this "breakthrough" seems only to be attaching a name to what was previously undefined. Is there something I'm missing?

s
Slappy slap slap

Under your bed...

Joined
22 Oct 05
Moves
70042
Clock
10 Dec 06
Vote Up
Vote Down

Originally posted by Ramiri15
0^0 = 0^(1-1)

0^0 = (0^1)(0^-1) = (0/1)(1/0) = (0/0) = nullity


Here's the paper:
http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf


And here are its supposed ramifications for calculus:
http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf


So, it's a new number lying outside the real number line that supposedly ans ...[text shortened]... y to be attaching a name to what was previously undefined. Is there something I'm missing?
I was thinking about this at work today. Thanks for posting the papers. I hope to read them soon.

F

Joined
11 Nov 05
Moves
43938
Clock
10 Dec 06
Vote Up
Vote Down

Originally posted by Ramiri15
0^0 = 0^(1-1)

0^0 = (0^1)(0^-1) = (0/1)(1/0) = (0/0) = nullity


Here's the paper:
http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf


And here are its supposed ramifications for calculus:
http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf


So, it's a new number lying outside the real number line that supposedly ans ...[text shortened]... y to be attaching a name to what was previously undefined. Is there something I'm missing?
Does it include the possibility to divide by zero?
Then it is not sound math.

T
Kupikupopo!

Out of my mind

Joined
25 Oct 02
Moves
20443
Clock
10 Dec 06
Vote Up
Vote Down

Originally posted by Ramiri15
0^0 = 0^(1-1)

0^0 = (0^1)(0^-1) = (0/1)(1/0) = (0/0) = nullity


Here's the paper:
http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf


And here are its supposed ramifications for calculus:
http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf


So, it's a new number lying outside the real number line that supposedly ans ...[text shortened]... y to be attaching a name to what was previously undefined. Is there something I'm missing?
By naming it they can make a new set of axioms, to deal with previously undefined things.

It'll have no effect on normal calculations, only on calculations on sets with nullity in them.

I think it's nice, but it has no use aside from being able to write everything down in one set of axioms (and no exceptions).

D

Joined
21 Sep 05
Moves
75006
Clock
10 Dec 06
Vote Up
Vote Down

Part of the TV clip I saw included the axiom that 1/0=+oo and -1/0=-oo

It appears that all he has decided to do is to allow calculations with infinity and to give 0/0 a name. 'The nullity'.

I checked the date. Doesn't seem like the start of April. Perhaps he's from the southern hemisphere 😉

R

Joined
30 Oct 05
Moves
3072
Clock
11 Dec 06
Vote Up
Vote Down

Originally posted by Diapason
Part of the TV clip I saw included the axiom that 1/0=+oo and -1/0=-oo

It appears that all he has decided to do is to allow calculations with infinity and to give 0/0 a name. 'The nullity'.

I checked the date. Doesn't seem like the start of April. Perhaps he's from the southern hemisphere 😉
He's a professor from Reading 😞.

I've been checking this a bit more and it turns out the operation 0/0 has already been defined in numerous ways, among them simply NaN (not a number). British humor is supposed to be quite absurdist, no? Perhaps it's all a joke...

F

Joined
11 Nov 05
Moves
43938
Clock
11 Dec 06
Vote Up
Vote Down

Originally posted by Ramiri15
He's a professor from Reading 😞.

I've been checking this a bit more and it turns out the operation 0/0 has already been defined in numerous ways, among them simply NaN (not a number). British humor is supposed to be quite absurdist, no? Perhaps it's all a joke...
Given a function f(x)=x/x, then f(x) = 1 wherever f is defined. However it is not defined at x=0 where 'f(0)=0/0' which is impossible because of the division by zero impossibility.

However, if we form a limit where x tends to go towards 0 from right or from right we get a limit value wich is 1. So if we define f(0) to be equal to 1 then we have a continous function throug the whole R.

So '0/0' can consider to have a value but in reality it has not. Sometimes it makes semse, sometimes not. Wherever we talk about 0/0 we have to assume a limit value of some kind.

P
Bananarama

False berry

Joined
14 Feb 04
Moves
28719
Clock
11 Dec 06
Vote Up
Vote Down

As I understand it, the whole point of this fix is to stop computers from throwing up when they encounter singularities like infinity and nullity. By using arithmetic that includes definite values for these objects, computer programmers may be able to save themselves a bunch of grief. However, right now I don't see much use for them outside the computer realm.

u
The So Fist

Voice of Reason

Joined
28 Mar 06
Moves
9908
Clock
11 Dec 06
1 edit
Vote Up
Vote Down

Originally posted by PBE6
As I understand it, the whole point of this fix is to stop computers from throwing up when they encounter singularities like infinity and nullity. By using arithmetic that includes definite values for these objects, computer programmers may be able to save themselves a bunch of grief. However, right now I don't see much use for them outside the computer realm.
Good...now when I start up Diablo my female character can begin with "Nullity Armour"

m

Joined
07 Sep 05
Moves
35068
Clock
12 Dec 06
1 edit
Vote Up
Vote Down

Originally posted by FabianFnas
Given a function f(x)=x/x, then f(x) = 1 wherever f is defined. However it is not defined at x=0 where 'f(0)=0/0' which is impossible because of the division by zero impossibility.herever we talk about 0/0 we have to assume a limit value of some kind.
This is only because of the axioms that define the real numbers.

What the paper is doing is defining a set of axioms to construct a complete arithmetic where actions such as division by zero are possible. Your argument is based on the real number system, but this specifically isn't the real number system (although presumably it behaves like where it would make sense, or it's a bit of a waste of time). Number systems are defined by their axioms and nothing else.

I only glanced at it, so I've no idea if there are any real problems with it, but there's no absolute reason why it can't work.

F

Joined
11 Nov 05
Moves
43938
Clock
13 Dec 06
1 edit
Vote Up
Vote Down

Originally posted by mtthw
This is only because of the axioms that define the real numbers.

What the paper is doing is defining a set of axioms to construct a complete arithmetic where actions such as division by zero are possible. Your argument is based on the real number system, but this specifically isn't the real number system (although presumably it behaves like where i a if there are any real problems with it, but there's no absolute reason why it can't work.
I think there is good reasons *not* to include division by zero to any of the existing number systems we have.
Higher mathematical education explains quite thoroughly why it if of no value to do this but I trust them completely in this matter.

I was once a fan of the idea of bringing in 1/0 as a value as one brings in the the concept of square root of minus one and call it 'i', hence constructing the complex number system. But when doing so one has also to invent a totally new set of operations and arithmetic rules.

But constructing a brand new number system by making 1/0 legal is not so easy.

m

Joined
07 Sep 05
Moves
35068
Clock
13 Dec 06
Vote Up
Vote Down

Originally posted by FabianFnas
I was once a fan of the idea of bringing in 1/0 as a value as one brings in the the concept of square root of minus one and call it 'i', hence constructing the complex number system. But when doing so one has also to invent a totally new set of operations and arithmetic rules.

But constructing a brand new number system by making 1/0 legal is not so easy.
No, it's not easy. Which is why when somebody manages to do it they get noticed, which appears to be what has happened. And they've had to invent another number (this "nullity" concept) to make it work consistently.

u
The So Fist

Voice of Reason

Joined
28 Mar 06
Moves
9908
Clock
13 Dec 06
Vote Up
Vote Down

we had a similar debate here just after I signed up last year. I forget the thread but I remember I was roundly thrashed for suggesting something similar to what this guy has done.

Anyone remember the thread?

g
Wayward Soul

Your Blackened Sky

Joined
12 Mar 02
Moves
15128
Clock
13 Dec 06
1 edit
Vote Up
Vote Down

Originally posted by Ramiri15
0^0 = 0^(1-1)

0^0 = (0^1)(0^-1) = (0/1)(1/0) = (0/0) = nullity


Here's the paper:
http://www.bookofparagon.com/Mathematics/PerspexMachineVIII.pdf


And here are its supposed ramifications for calculus:
http://www.bookofparagon.com/Mathematics/PerspexMachineIX.pdf


So, it's a new number lying outside the real number line that supposedly ans y to be attaching a name to what was previously undefined. Is there something I'm missing?
is 0^0 not 1?

if 0/0=nullity, what would that mean for when you are posed with finding the limit of a sequence which involves dividing by zero?

for instance, lim(x->0) of sin(x)/x = lim(x->0)cos(x)/1=1 (by l'hopitals rule), but it also equals the nullity. or am i just confused? i don't have time to look at this properly, but i'll read the paper sometime...maybe...it sounds...interesting...

P
Bananarama

False berry

Joined
14 Feb 04
Moves
28719
Clock
13 Dec 06
2 edits
Vote Up
Vote Down

Originally posted by genius
is 0^0 not 1?

if 0/0=nullity, what would that mean for when you are posed with finding the limit of a sequence which involves dividing by zero?

for instance, lim(x->0) of sin(x)/x = lim(x->0)cos(x)/1=1 (by l'hopitals rule), but it also equals the nullity. or am i just confused? i don't have time to look at this properly, but i'll read the paper sometime...maybe...it sounds...interesting...
Nullity is supposed to represent the actual answer you get when you divide 0/0, not the limit of some function h(x) = f(x)/g(x) as f(x) and g(x) tend toward 0. Limits of sequences would remain largely unchanged, except in this system you could report the actual answer as positive infinity, negative infinity, or nullity, not just the limit.

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.