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  1. #1
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    undefined

    Division by zero is undefined. How would you justify this ?
    Thank you all,
    Last edited by Judi; January 26th 2007 at 10:09 PM.
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  2. #2
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    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 \frac{0}{0}=1 then the value of all numbers isn't unique anymore.

    An example:

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

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

    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

    EB
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  3. #3
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    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 <R,+,\cdot>, we know that <R,+> is an abelian group. We know that a(b+c)=ab+ac and (b+c)a=ba+ca. Using these definitions we can show that 0x=x0=0, because x0=x(y-y)=xy-xy=0 and 0x=(y-y)x=yx-yx=0. A ring with unity is a ring such that <R,\cdot> has identity which we call 1. And we define a unit to be any element that has an inverse. The above theorem shows that if 1\not = 0 (meaning the identity for addition is not the identity for multiplication) then 0 cannot be a unit (meaning 1/0 does not exist). The only case when this can be is when 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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