Question

Question #d0dfe

Answer 1

When you divide, say, `3/0` you are trying to find a result such as:
`3/0=a`
But this number `a` should be a number that multiplied by `0` gives `3`!
`3/0=a` so rearranging `3=0*a`
But this is not possible!
So, it is not possible to divide by `0`.

On the other hand have a look at what happens if you get "near" to zero but not zero.
Try `0.01`, `0.0001`, `0.000001` and see what happens!

Answer 2

You can't do it.
(Any attempt to define division by zero will "break arithmetic" somewhere.)

Reason 1:

`a/b = c` exactly when `b*c=a`

But if `b=0`, we have

`a/0 = c` exactly when `0*c=a`

`0*c=a` has no solution for `a!=0` because `0*c=0` for all `c`.

(For example: `5/0=c` would require `0*c=5` which cannot happen.)

Reason 2:

I am an algebraist, I define division to be multiplication by a reciprocal.

A reciprocal of `a` is a multiplicative inverse. That is, it is a solution to `a*x="multiplicative identity"`

For any number, `x`, we can show that `0*x=0` So `0` has no multiplicative inverse (no reciprocal).

`0x+x=0x+1x=(0+1)x=1x=x`
`0x+x=x` implies that `0x=0` (Subtract `x` from both sides.)

(General case)
In any ring whose additive identity is denoted `0`,
we have `0 x=0` and `x0=0` for all `x`.
So the only ring in which `0` has a reciprocal is the trivial ring: `{0}`.
(The trivial ring has one thing in it. That thing is the additive and multiplicative identities. In non-trivial rings, it is not possible for both identities to be the same.)