# Halmos on Definability and Luzin on Division by 0

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 $$a \ne 0$$.

1. 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 $$a =0$$.

2. 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.

• What is the question?
– user122283
May 26 '14 at 22:48
• @SanathDevalapurkar Whether the reconstruction of the argument is correct. Thank you for the edit. May 26 '14 at 22:49
• That argument is (as far as I know) the standard explanation for why division by 0 isn't allowed. I am curious that you describe this as "new to you". Is there some other explanation that you are more familiar with? May 26 '14 at 23:54
• @mweiss All of mathematics is new to me, so that probably was a rather misleading thing to say. I have heard people say that it's meaningless, by the definition of rational numbers: a rational number is an expression of form $a / b$ where $b \ne 0$. Therefore, $x / 0$ is not a rational number for any $x$. As you know, there are many such explanations. Luzin's argument is the first one that I feel satisfactorily addresses the question. May 27 '14 at 0:08
• Anybody who's given you that explanation ("It's meaningless by definition") is dodging the issue. The question is, why is it defined that way? Definitions are instrumental; any time you see a definition with a clause in it excluding some case (like a stipulation that $b \neq 0$ it usually signals that something goes wrong in that case. May 27 '14 at 0:33

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.