For a successful introduction of a new symbol (e.g. '$\emptyset$') into a mathematical discourse it is necessary and sufficient that the symbol refer to something (e.g. Existence + Specification in ZF) and to nothing else (e.g. Extensionality). I learned of this notion of definability from Halmos (1960).
Luzin (1961) has an introductory section (§8) explaining why division by $0$ isn't allowed. His argument seems new to me so I just want to ask whether I understood it properly. He doesn't say this explicitly, but it seems that he relies on the notion of definability described above. Consider:
$$a\over 0 \tag 1$$
We know that $(1)$ denotes the unique number $b$ s.t. $b \cdot 0 = a$. Now, either $a=0$ or $a \ne 0$.
If $a = 0$, then according to that definition, $(1)$ denotes the unique number $b$ that satisfies the equation $b \cdot 0 = 0$. But all numbers satisfy that equation, so while there is such a number $b$, there is no unique root of that equation, so by the definition of definability, $(1)$ fails to denote a number when $a =0$.
If $a \ne 0$, then according to that same definition, $(1)$ denotes the unique number $b$ that satisfies the equation $b \cdot0 =a$, where $a$ by hypothesis differs from $0$. But no $b$ different from $0$ satisfies that equation, so by the definition of definability, $(1)$ fails to denote a number when $a \ne 0$.
Since in both cases $(1)$ fails to denote a unique number, $(1)$ is said to be ill-defined. That's my reconstruction of Luzin's argument - is it entirely correct?
References
Halmos, P. (1960) Naive Set Theory.
Luzin, N.N. (1961) Differential Calculus.
Yes, your construction is correct.
If $\dfrac{a}{0}$ = c, then $a = 0 \cdot c$. But $0\cdot c = 0$. Hence, if $a$ is not equal to $0$, no value of c can make the statement $a = 0\cdot c$ true, while if $a = 0$, every value of $c$ will make the statement true.
Thus, $\dfrac{a}{0}$ either has no value or is indefinite in value.