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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.

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  • $\begingroup$ What is the question? $\endgroup$
    – user122283
    May 26, 2014 at 22:48
  • $\begingroup$ @SanathDevalapurkar Whether the reconstruction of the argument is correct. Thank you for the edit. $\endgroup$
    – Readingtao
    May 26, 2014 at 22:49
  • $\begingroup$ 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? $\endgroup$
    – mweiss
    May 26, 2014 at 23:54
  • $\begingroup$ @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. $\endgroup$
    – Readingtao
    May 27, 2014 at 0:08
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    $\begingroup$ 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. $\endgroup$
    – mweiss
    May 27, 2014 at 0:33

1 Answer 1

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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.

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