Zero Model

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December 8th 2006, 10:02 AM
ThePerfectHacker

Zero Model

I am going to try to define numbers which will permit division by zero. Furthermore, I am going to try to keep the numbers behaving as well as possible. I was working on this problem. So far I have addition of these numbers complete. Here they are.

Consider the group $<\mathbb{R},+>$ , we will "add" two numbers to this group and redefine the addition operation to keep it a group.

Definition: "Hackerian space" is the set $\mathcal{H}=\mathbb{R}\cup \{-nan,+nan\}$

Definition: The binary operation,
$*:\mathcal{H}\times \mathcal{H}\to \mathcal{H}$ will be as follows:
$\forall a,b\in \mathbb{R} \, \, \, \, a*b=a+b$
$\forall a\in \mathcal{H} \not = \mp nan \, \, \, \, a*\pm nan=\pm nan*a=\pm nan$
$\forall a\in \mathcal{H} \, \, \, \, \, 0*a=0*a=0$
$\pm nan*\mp nan=\mp nan*\pm nan=0$

Theorem: Hackerian space is a group.
Proof: Trivial

We do obtain something interesting with this construction.
The "Blank set" $=\{-nan,0,nan\}$ is a normal subgroup of Hackerian space. And the interesting property is that,
$\boxed{\mathcal{H}/\mathbb{B}\simeq \mathbb{R}}$

Now there is a chance that $\mathcal{H}\simeq \mathbb{R}$ because they have the same cardinality. But the boxed property shows that it is isomorphic by a non-zero factor group. I hope that shows that they are distinct algebra structures but I am not sure.*)

*)Conjecture.

We have constructed a new group which is distinct by isomorphism from the reals if the conjecture is true. We can now try to procede to define multiplication.
December 8th 2006, 11:31 AM
topsquark

Careful! If you keep going like this you might even justify renormalization at some point. :p

-Dan
December 9th 2006, 01:50 PM
ThePerfectHacker

Quote:

Originally Posted by topsquark

Careful! If you keep going like this you might even justify renormalization at some point. :p

-Dan

What does that mean? Something with Quantum physics?
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I have solved my problem.
If you were reading what I wrote, I mentioned it would be quite stupid if Hackerian space is the same as the Real numbers (up to isomorphism). But the answer is simple. Hackerian space has a subgroup order 3,i.e. Blank space. However, R does not!
So we are truly dealing with a different algebraic structure.
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But the sad thing is that we cannot turn R by introducing new made up numbers (inf, nan, ... ) into an algebriac structure where division is permitted. Because it needs to be a division ring. Which is violated by "division by zero". Thus, it is not possible to extend the reals into numbers division where zero divison is permitted.