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January 26th 2007, 02:57 PM
Judi

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Division by zero is undefined. How would you justify this ?
Thank you all,
January 27th 2007, 12:30 AM
earboth

Quote:

Originally Posted by Judi

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 :cool:

EB
January 27th 2007, 02:10 PM
ThePerfectHacker

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.