# Question #d0dfe

Mar 25, 2015

When you divide, say, $\frac{3}{0}$ you are trying to find a result such as:
$\frac{3}{0} = a$
But this number $a$ should be a number that multiplied by $0$ gives $3$!
$\frac{3}{0} = a$ so rearranging $3 = 0 \cdot a$
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!

Mar 25, 2015

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

Reason 1:

$\frac{a}{b} = c$ exactly when $b \cdot c = a$

But if $b = 0$, we have

$\frac{a}{0} = c$ exactly when $0 \cdot c = a$

$0 \cdot c = a$ has no solution for $a \ne 0$ because $0 \cdot c = 0$ for all $c$.

(For example: $\frac{5}{0} = c$ would require $0 \cdot c = 5$ 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 $a \cdot x = \text{multiplicative identity}$

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

$0 x + x = 0 x + 1 x = \left(0 + 1\right) x = 1 x = x$
$0 x + x = x$ implies that $0 x = 0$ (Subtract $x$ from both sides.)

(General case)
In any ring whose additive identity is denoted $0$,
we have $0 x = 0$ and $x 0 = 0$ for all $x$.
So the only ring in which $0$ has a reciprocal is the trivial ring: $\left\{0\right\}$.
(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.)