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
EB
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.