Division by zero is undefined. How would you justify this ?

Thank you all,

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- Jan 26th 2007, 02:57 PMJudiundefined
Division by zero is undefined. How would you justify this ?

Thank you all, - Jan 27th 2007, 12:30 AMearboth
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 then the value of**all**numbers isn't unique anymore.

An example:

. Now divide both sides of this equation by zero:

. According to the assumption you can cancel out the zeros and you'll get

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 - Jan 27th 2007, 02:10 PMThePerfectHacker
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 , we know that is an abelian group. We know that and . Using these definitions we can show that , because and . A ring with unity is a ring such that has identity which we call . And we define a*unit*to be any element that has an inverse. The above theorem shows that if (meaning the identity for addition is not the identity for multiplication) then cannot be a unit (meaning 1/0 does not exist). The only case when this can be is when , 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.