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Possible Duplicate:
Division by $0$

I've always been inclined to believe that x/0 = NaN is a placeholder for a character or constant that no one has created yet.

I know assume that none of you can tell the future, but is there an expectation that someone will eventually (successfully) define division by zero?

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    $\begingroup$ "NaN": there, I just created it. The real question is not to "define" division by zero, but to define it in a way that makes the result useful. $\endgroup$ – Robert Israel May 13 '12 at 20:26
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    $\begingroup$ It already has been, when it is a useful concept. For example, as the point at infinity in the real projective plane or in the extended complex plane. But one still has to be careful because it does not obey all the laws of arithmetic for finite numbers. One has to take care when making such exceptions! $\endgroup$ – bgins May 13 '12 at 20:26
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    $\begingroup$ Or read the answers here $\endgroup$ – MJD May 13 '12 at 20:27
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    $\begingroup$ Ooh, here it is again! $\endgroup$ – MJD May 13 '12 at 20:29
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    $\begingroup$ There is an algebraic structure called a wheel where division by 0 is perfectly well-defined. $\endgroup$ – Zev Chonoles May 13 '12 at 20:29
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Division by zero can be defined. It is called Wheel Theory. It's not a very popular set of mathematics and the paper that it originates from is a little difficult to find. Division by zero is left undefined in modern mathematics because it causes a loss of many useful statements. For example, $\frac{a}{b}=c \Rightarrow cb=a$. This is not true when $b=0$ and $a$ is nonzero. ($cb=0\neq a$) So, we lose generality by allowing division by zero to be defined.

Allowing division by zero also leads to proofs such as this which are valid:

$$a=b$$ $$a^2=ab$$ $$a^2-b^2=ab-b^2$$ $$(a+b)(a-b)=b(a-b)$$ $$a+b=b$$ $$2b=b$$ $$2=1$$

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    $\begingroup$ I downvoted because division by zero does not really have anything to do with this style of erroneous proof, and the claim that it is undefined because it is the source of these errors seems to me mistaken. The error in this proof is to deduce $x=y$ from $f(x) = f(y)$. In this case the function is $f(x) = 0\cdot x$, but any other noninjective function will work, and some are common. For example, one often sees reasoning like this: $(-1)^2 = 1^2$; take the square root of both sides, thus $-1 = 1$. Or similarly $e^0 = e^{2\pi i}$; take logarithm of both sides, therefore $0=2\pi i$. $\endgroup$ – MJD May 13 '12 at 20:32
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    $\begingroup$ @MarkDominus, that very well may be a perspective that I cannot follow very well. It seems to me that you're generalizing the error when the transition from line $4$ to line $5$ involves division by zero, which was what I meant when I said It is the source of many erroneous proofs that you see online. It is very possible I do not fully understand your viewpoint, however. $\endgroup$ – 000 May 13 '12 at 20:37
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    $\begingroup$ Here is my reasoning: You can produce a proof that $0=1$ that works in essentially the same way without dividing by zero, say instead by taking square roots. Yet we don't say that taking square roots is undefined. In what way is dividing by zero different from taking square roots? Your reply doesn't get to the essence of why it was left undefined. $\endgroup$ – MJD May 13 '12 at 20:42
  • $\begingroup$ I see a huge difference between dividing by zero and taking square roots: Taking square roots has an obvious and meaningful interpretation, dividing by zero does not. There are certain cases where it would seem that dividing by zero would make sense to be defined a certain way, but the fact that there are cases rather than a case makes it unsatisfactory to define division by zero. I guess it may be more sensible to look at it from the perspective of, "What can I gain from defining this and what do I lose from defining this?" In the case of division by zero, we lose a lot. What do you think? $\endgroup$ – 000 May 13 '12 at 20:46
  • $\begingroup$ Okay, but that is not what you said in your post, which I think obscures the real issues with irrelevant ones. $\endgroup$ – MJD May 13 '12 at 20:48
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My understand of arithmetic division over R is this: a/b = {x in R| b*x = a}. In case of division by 0, the answer would be none unless a = 0.

By studying the limits of the function 1/x we can tell that 1/0 is the biggest number ever(which is known to not exists :))

What I wanna say, is that 1/0 doesn't exists and thus cannot be defined. Unless..who knows :)

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