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This question already has an answer here:

Would this even assist math in the way that $i$ did? Or is this just outright pointless and/or too exclusive to call for a definition?

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marked as duplicate by Will Jagy, Michael Albanese, user236182, Noah Schweber, BLAZE Dec 3 '15 at 22:22

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    $\begingroup$ From the analysis side, there is no way to continuously extend $f(x,y)=\frac{x}{y}$ to $(0,0)$. That is a problem. It's not catastrophic; for instance, even though $g(x,y)=x^y$ cannot be continuously extended to $(0,0)$, $0^0$ is still often treated as being $1$. But it is enough of a problem that it is not useful to define $\frac{0}{0}$ in analysis. Similarly, from the algebraic side, $\frac{0}{0}$ should be "the unique solution $x$ to the equation $0\cdot x=0$". But there is no such unique solution. So it is completely nonsensical to define $\frac{0}{0}$ from the algebraic perspective. $\endgroup$ – Ian Dec 3 '15 at 21:23
  • $\begingroup$ Assuming that 0 is the unity element for the + operation, and also assuming that we work in a field where thus dividing by 0 is undefined, as $0^{-1}$ does not exist, the suggestion is indeed pointless. $\endgroup$ – Maestro13 Dec 3 '15 at 21:23
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It would require us to give up many of the usual rules of algebra -- for example the cancellation rule which says that $$ \frac{a\times b}{a} = b $$ whenever both sides exist.

However, taking $(a,b)$ first to be $(0,1)$ and $(0,2)$ we would get $$ \frac00 = \frac{0\times 1}{0} = 1 \qquad\text{and}\qquad \frac00 = \frac{0\times 2}{0} = 2 $$

Since $1$ is manifestly not the same as $2$ (and if we make them the same, all of arithmetic ceases to be useful), your extended system would need to be one that doesn't satisfy the cancellation rule.

A lot of other such rules go the same way, so in the end you may have something you call $\frac00$, but the fraction bar in that expression would no longer denote anything resembling division anyway.

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