# Problem of Dividing By Zero Solved

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• Dec 7th 2006, 06:25 PM
Quick
Quote:

Originally Posted by topsquark
Pffl. If they can teach "intelligent design" in schools why can't they teach this? More garbage for them to have to "unlearn" at some point. :(

-Dan

Pffl. If they can teach "evolutionism" in schools why can't they teach this? More garbage for them to have to "unlearn" at some point. :(

-Quick
• Dec 7th 2006, 06:46 PM
OReilly
Is that "nullity" theory somewhere officialy published?
On that BBC site there is an answer of that proffesor on many posts that he will provide more proofs soon, which left me with doubt whether he actually did fully define that theory or did he just wanted publicity!
• Dec 7th 2006, 06:50 PM
topsquark
Quote:

Originally Posted by Quick
Pffl. If they can teach "evolutionism" in schools why can't they teach this? More garbage for them to have to "unlearn" at some point. :(

-Quick

At least evolution is a scientifically testable theory. I have yet to hear anyone come up with a testable version of intelligent design. (And by its very nature I can't think of a way to do it.)

-Dan
• Dec 7th 2006, 08:01 PM
ThePerfectHacker
Here is something that would help you understand why division by zero is undefined.

$a/b=ab^{-1}$
Where, $b^{-1}$ is the unique solution to,
$bx=xb=1$.
Where, $1$ is the unique element such that,
$c1=1c=c$.

Now,
$a/0=a0^{-1}$
Thus, we $0^{-1}$ is unique solution to,
$0x=1$.
But, that is not possible.

The reason that is not possible is because,
$0x=(1-1)x=x-x=0$
Thus any element multiplied with zero is zero.
Thus there is no solution to the equation above.
• Dec 8th 2006, 12:30 AM
CaptainBlack
Quote:

Originally Posted by OReilly
I was surprised that officials let him do that.

I don't think that theory that hasn't been accepted by all or at least majority of math community should be learned in schools! :mad:

Consensus is an irrelevance in Maths, now consistency ..

RonL
• Dec 8th 2006, 12:32 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Here is something that would help you understand why division by zero is undefined.

$a/b=ab^{-1}$
Where, $b^{-1}$ is the unique solution to,
$bx=xb=1$.
Where, $1$ is the unique element such that,
$c1=1c=c$.

Now,
$a/0=a0^{-1}$
Thus, we $0^{-1}$ is unique solution to,
$0x=1$.
But, that is not possible.

The reason that is not possible is because,
$0x=(1-1)x=x-x=0$
Thus any element multiplied with zero is zero.
Thus there is no solution to the equation above.

Think of it this way nan (or if you must nullity (yuck, yuck ,,))
is a symbol for undefined.

RonL
• Dec 8th 2006, 05:13 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
Think of it this way nan (or if you must nullity (yuck, yuck ,,))
is a symbol for undefined.

That is not a definition.

A "symbol" represents an element from some set. There is no such set.
----
Maybe, what you are saying is useful in computer science. But I do not know.
• Dec 8th 2006, 05:54 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
That is not a definition.

A "symbol" represents an element from some set. There is no such set.

You seem to be lacking in imagination today.

We would be talking here of another version of the extended reals with
three ideal elements added +inf, -inf, nan. That makes nan an element
of the $\mathbb{S} \rm{uper} \mathbb{E} xtended \mathbb{R} eals^{(tm)}$.

Quote:

----
Maybe, what you are saying is useful in computer science. But I do not know.
Why are we interested in number at all if not because of its usefullness
in games and puzzles?

Also, alarm bells should be sounding in your head as you type that statement.
If it were usefull then maths should look at it as usefull tricks in computation
always should be of interest to maths.

RonL
• Dec 8th 2006, 09:21 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
We would be talking here of another version of the extended reals with
three ideal elements added +inf, -inf, nan. That makes nan an element.

Such a defintion, is mathematically acceptable.

However, what are the binary operations on this set?*)
I am willing to bet whatever they do not even turn this set into a ring.

*)That is the only thing you must do for me, otherwise, I cannot accept this defintion.**)

**)And if they are none, for these elements. What is even the purpose in defining such elements, it does not even form a monoid! Or whatever you call it.
• Dec 8th 2006, 09:36 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Such a defintion, is mathematically acceptable.

However, what are the binary operations on this set?*)
I am willing to bet whatever they do not even turn this set into a ring.

The operations are +, -, *, / as normal. With:

$
nan \circ x =nan
$

$
x \circ nan =nan
$

for all $x \in \mathbb{SER}$ where $\circ$ denotes
any of the operations.

It's my job here to point out that this might be a structure of interest
not to show what existing type of structure it is.

Quote:

*)That is the only thing you must do for me, otherwise, I cannot accept this defintion.**)

**)And if they are none, for these elements. What is even the purpose in defining such elements, it does not even form a monoid! Or whatever you call it.
• Dec 8th 2006, 09:48 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
The operations are +, -, *, / as normal.

I do not know what you are using, but - and / are not binary operation.
They represent the inverse operation of + and *.
In the way I defined them above.

So I am going to pretend you did not mention the - and / and work with * and +.

You did not define the binary operation between inf and -inf and its elements.
• Dec 8th 2006, 10:32 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I do not know what you are using, but - and / are not binary operation.
They represent the inverse operation of + and /.
In the way I defined them above.

So I am going to pretend you did not mention the - and / and work with * and +.

You did not define the binary operation between inf and -inf and its elements.

I dont need to these are inherited from the extended reals.

RonL
• Dec 8th 2006, 11:46 AM
TriKri
Quote:

Originally Posted by ThePerfectHacker
$a/b=ab^{-1}$
Where, $b^{-1}$ is the unique solution to,
$bx=xb=1$.
Where, $1$ is the unique element such that,
$c1=1c=c$.

Now,
$a/0=a0^{-1}$
Thus, we $0^{-1}$ is unique solution to,
$0x=1$.
But, that is not possible.

The reason that is not possible is because,
$0x=(1-1)x=x-x=0$
Thus any element multiplied with zero is zero.
Thus there is no solution to the equation above.

But $0\cdot \infty$ is undefined. And if we define $\frac{1}{0}\ =\ 0^{-1}$ to be $+\infty$, then $x - x\ =\ \infty - \infty$ would be undefined too, not zero.
• Dec 8th 2006, 11:53 AM
ThePerfectHacker
Quote:

Originally Posted by TriKri
But $0\cdot \infty$ is undefined. And if we define $\frac{1}{0}\ =\ 0^{-1}$ to be $+\infty$, then $x - x\ =\ \infty - \infty$ would be undefined too, not zero.

You are assuming this algebra structure (and I believe you are reffering to the Extended Reals) is a ring, that is distribution is true. It does not work anymore.

Furthermore, in what I said, all elements have binary operations with each other. Over here not all do (that is some are not defined). So there is no problem.
• Dec 9th 2006, 01:50 PM
The Pondermatic
I just have to go off in defense of intelligent design really fast:

intl. design is NOT:

1. religous. People in all sorts of religions stand with this theory.
2. philisophical. Intl. design proponents are very careful to attatch no description of what intelligence made us to the theory.
3. unscientific. The theory has come up with a way of quantatively telling whether inteeligence made something or not and many proven scientific theories support, through that method, the idea of intl. design.

Is it right? No one knows for sure. Is evolution right? No one knows that, either.

ANYWAY:

It pains me to think about what the proffesor was doing. He is either really, really out of it or is so egotisical he deserves a world record. Either way, he should be put in a mental hospital :D.
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