# Why Not Define $0/0$ To Be $0$?

For every number $x$, $x\times 0=0$, hence $\dfrac{0}{0}$ can be any number! So $\dfrac{0}{0}$ "is knows as indeterminate" [1]. But what if we define it to be $0$? I already have an answer, but don't know how convincing it is:

$1=\dfrac{1}{1}=\dfrac{1}{1}+0=\dfrac{1}{1}+\dfrac{0}{0}=\dfrac{1\times 0}{1\times 0}+\dfrac{0\times 1}{0\times 1}=\dfrac{0}{0}+\dfrac{0}{0}=0+0=0$, a contradiction.

Is there any better explanation why not to define $\dfrac{0}{0}$ to be $0$ (or any other number)?

Thanks.

• Why do you need a better explanation? what's wrong with this one? – Dennis Gulko Nov 8 '13 at 15:21
• If $\mathbb{Q}$ is viewed as a meadow, then the reciprocal of $0$ is zero. So $x/0 = x \cdot 0^{-1} = x \cdot 0 = 0$. – goblin Nov 8 '13 at 15:22
• Related thread: math.stackexchange.com/q/527613/264 – Zev Chonoles Nov 8 '13 at 15:31
• That $0$ is a multiple of any number by $0$ is already a flawless, perfectly satisfactory answer to why we do not define $0/0$ to be anything, so this question (which is eternally recurring it seems) is superfluous. Your answer is not convincing because it assumes $\frac{a}{b}=\frac{ac}{bc}$ holds for $c=0$, but haven't justified this (even for $c\ne0$ it has to be justified in the beginning). – anon Nov 8 '13 at 15:31