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Math Help - Zero Model

  1. #1
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    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.
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    Forum Admin topsquark's Avatar
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    Careful! If you keep going like this you might even justify renormalization at some point.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Careful! If you keep going like this you might even justify renormalization at some point.

    -Dan
    What does that mean? Something with Quantum physics?
    ----
    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.
    ----


    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.
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