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Thread: undefined

  1. #1
    Junior Member
    Jun 2006


    Division by zero is undefined. How would you justify this ?
    Thank you all,
    Last edited by Judi; Jan 26th 2007 at 10:09 PM.
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  2. #2
    Super Member
    earboth's Avatar
    Jan 2006
    Quote Originally Posted by Judi View Post
    Division by zero is undefined. How would you justify this ?
    Thank you all,
    Hello, Judi,

    normally you are told that you get the result 1 if you divide a number by itself.

    If you permit this result at $\displaystyle \frac{0}{0}=1$ then the value of all numbers isn't unique anymore.

    An example:

    $\displaystyle 3 \cdot 0 = 8 \cdot 0$. Now divide both sides of this equation by zero:

    $\displaystyle \frac{3 \cdot 0}{0} = \frac{8 \cdot 0}{0}$. According to the assumption you can cancel out the zeros and you'll get

    $\displaystyle 3 \cdot 1= 8\cdot 1$

    For me as a customer this method has some advantages: My next car costs only $1.00, because $1.00 is the same as $32,499.00

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  3. #3
    Global Moderator

    Nov 2005
    New York City
    There is nothing wrong with what Earboth said, but I presume you want a more algebraic answer.

    When I was a young boy a long time ago it used to bother me as well. Until I learned what a ring was, and I was happy. I will assume you are familar with the terms I am about to use.

    Given a ring $\displaystyle <R,+,\cdot>$, we know that $\displaystyle <R,+>$ is an abelian group. We know that $\displaystyle a(b+c)=ab+ac$ and $\displaystyle (b+c)a=ba+ca$. Using these definitions we can show that $\displaystyle 0x=x0=0$, because $\displaystyle x0=x(y-y)=xy-xy=0$ and $\displaystyle 0x=(y-y)x=yx-yx=0$. A ring with unity is a ring such that $\displaystyle <R,\cdot>$ has identity which we call $\displaystyle 1$. And we define a unit to be any element that has an inverse. The above theorem shows that if $\displaystyle 1\not = 0$ (meaning the identity for addition is not the identity for multiplication) then $\displaystyle 0$ cannot be a unit (meaning 1/0 does not exist). The only case when this can be is when $\displaystyle 1=0$, that is in a trivial ring of a single element (excerise). Thus, in general, for a general ring with more than one element, it cannot contain all units. The best we can do are all non-zero elements are units, that is called a division ring.
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