Is $\frac{0}{0}$ a set?

Question

Say we want to find a number $x$ such that: $$x^2-5x+6=0$$ So, there's no unique value of $x$ satisfying that but we still say $x={2,3}$

Is the situation with $\frac{0}{0}$ the same? I mean, finding the solution of $x\cdot 0=0$. And, since all values of $x$ are satisfying that so $x=1,1.5,1.29,\sqrt{2},e,\pi,5+6i...$ or $x=C$, the set of complex numbers.

Answer 1

Assuming that you formally defined the notation

$$\frac pq$$

to represent the solution set of the equation

$$q\cdot x=p,$$

just like

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

would denote the solutions of

$$ax^2+bx+c=0^*$$ for any $a,b,c$, then yes, $\dfrac00=\mathbb C$.

Such a notation is never adopted. In particular, it conflicts with the well-established symbol $\dfrac00$ that designates an undeterminate form; an undeterminate form is no particular value but a way to state that the expression at hand must be evaluated as a limit (with a specific pattern).

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$^*$In particular $(5\pm\sqrt{5^2-4\cdot1\cdot6})/(2\cdot1)$, equivalent to $(5\pm1)/2$. More generally,

$$\frac{p\pm q}r$$ would denote the solution set of

$$\frac r2x^2-px+\frac{p^2-q^2}{2r}=0$$ for $r\ne0$ and

$$\frac{p\pm q}0$$

would be undefined or undeterminate.

Answer 2

If you define $\frac{0}{0}$ as the set of $x$ such as $0x=0$ of course you will get all of R but the point about $\frac00$ as it arises in the calculus is to consider limits of fractions of quantities getting smaller and smaller. Such a situation is summarized by the phrase "indeterminate form of type $\frac00$". With that interpretation of $\frac00$ you can't define it as the set of $x$ such that $0x=0$ because that's imcompatible with the analysis of the concept of limit.