1. ## undefined

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

2. 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

EB

3. 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 $a(b+c)=ab+ac$ 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 $$ 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.