No. 0/0 is completely undefined. For example, consider limit as x approaches 0 of (-x)/x. This approaches (-0)/0 = 0/0 but simplifying the limit given an answer of -1. So 0/0 is completely dependent on the characteristics of the convergence and had no meaning in isolation. In fact, the field of real (and complex) numbers does not even include 0 in the multiplicative group. Therefore, dividing by 0 is nonsensical as it has no inverse as it is not in the multiplicative field. According to the basic axioms of groups theory, given a field F with a the identity of the additive group in the field and b some element of the multiplicative group, it can be proven that mul(a,b) = a where mul is the multiplication operation. This explains why 0 times any other number is 0. If you would like a copy of the proof, just drop me an email.

Sorry for the rant, math major.

I'd love to see the proof. But you did not post your email address. Could you send it to mine?

oops. Sorry about that. Fixed.

PS. Your email address is not posted either.

Isn't it something like this?

0=a-a, for all a

(where a-a is actually, a+(-a))

so: mul(b,0)=mul(b,(a-a))=mul(b,a)-mul(b,a)=z-z=0

pretty much.