Question #d0dfe

2 Answers
Mar 25, 2015

When you divide, say, 3/030 you are trying to find a result such as:
3/0=a30=a
But this number aa should be a number that multiplied by 00 gives 33!
3/0=a30=a so rearranging 3=0*a3=0a
But this is not possible!
So, it is not possible to divide by 00.

On the other hand have a look at what happens if you get "near" to zero but not zero.
Try 0.010.01, 0.00010.0001, 0.0000010.000001 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:

a/b = cab=c exactly when b*c=abc=a

But if b=0b=0, we have

a/0 = ca0=c exactly when 0*c=a0c=a

0*c=a0c=a has no solution for a!=0a0 because 0*c=00c=0 for all cc.

(For example: 5/0=c50=c would require 0*c=50c=5 which cannot happen.)

Reason 2:

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

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

For any number, xx, we can show that 0*x=00x=0 So 00 has no multiplicative inverse (no reciprocal).

0x+x=0x+1x=(0+1)x=1x=x0x+x=0x+1x=(0+1)x=1x=x
0x+x=x0x+x=x implies that 0x=00x=0 (Subtract xx from both sides.)

(General case)
In any ring whose additive identity is denoted 00,
we have 0 x=00x=0 and x0=0x0=0 for all xx.
So the only ring in which 00 has a reciprocal is the trivial ring: {0}{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.)