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This question already has an answer here:

I asked my teacher what is the real meaning of $\cfrac{0}{0}$, and the answer I got was "nobody knows". I can't leave this subject "as is". I need a decent explanation, at least an explanation to why "nobody knows". I'm sure you'll come up with a few good answers.

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marked as duplicate by Alexander Gruber Jan 12 '15 at 4:05

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I really hope your calculus teacher didn't say "nobody knows": $\frac{0}{0}$ is pretty well understood. It's just that it's undefined (without further context).

What I mean by "context" is: it's possible to give meaning to limits of the form $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ where $\lim_{x \to a}f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, but this depends on your choice of $f$ and $g$. For instance $$\lim_{x \to 0} \dfrac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \dfrac{e^{kx}-1}{x} = k, \quad \lim_{x \to 0} \dfrac{x}{x^3} = \infty, \quad \cdots$$ These are all different, despite the fact they're all of the form $\frac{0}{0}$.

So it's not that nobody knows, it's just that it's meaningless! But we can assign meaning if we know how the $0$s in the fraction came to be.

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    $\begingroup$ They're only "of the form $0/0$" if we pretend that the limit of the quotient is the quotient of the limits, which it isn't. $\endgroup$ – Trevor Wilson Oct 15 '13 at 20:46
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    $\begingroup$ @TrevorWilson: Hence "of the form" rather than "equal to". I think this language usage is standard, cf. http://en.wikipedia.org/wiki/Indeterminate_form. $\endgroup$ – Clive Newstead Oct 15 '13 at 20:47
  • $\begingroup$ I didn't mean to pick on you, because as you say the language is standard, but in my opinion this standard language is almost as silly as the teacher's saying "nobody knows." I tried to express this idea better in my answer. $\endgroup$ – Trevor Wilson Oct 15 '13 at 20:58
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The expression $0/0$ is meaningless because the operation of division $(a,b) \mapsto a/b$ is not defined for pairs $(a,b)$ where $b$ is zero, just as it is not defined for pairs $(a,b)$ where $b$ is an elephant.

Sometimes you will hear that $0/0$ is an "indeterminate form" that can equal different things depending on the context. This is a horribly imprecise way of speaking. What it really means is that if you are evaluating a limit and you do the computation

$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 0} f(x)}{\lim_{x \to 0} g(x)} = \frac{0}{0}$$

Then you have made a mistake in the first step and you have to evaluate the limit in a different way, e.g. with l'Hospital's rule, to get a valid answer (which could be anything depending on the particulars of the problem, hence "indeterminate.") This reason that the first step in the displayed calculation is a mistake is that the rule "the limit of a quotient is the quotient of the limits" is not true in general—it is only true when the limit of the denominator is nonzero.

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Math is logic.

$12/3=4$ because if you have twelve oranges and three kids, each gets four.

However, dividing by zero is illogical; what does it mean? It is no longer a math question, but a philosophical one.

Everything you do in math needs to make sense!

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  • $\begingroup$ Exactly. So when I divide nothing to no one (as giving no-apples to nobody) is like I'm doing something (or nothing) infinitely. And as much as it doesn't make any sense, it kinda sounds like something that happens all the time. $\endgroup$ – yonyon100 Oct 15 '13 at 20:50
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    $\begingroup$ I resent the implication that philosophy is not logical. $\endgroup$ – leonbloy Oct 15 '13 at 20:59
  • $\begingroup$ Correction: math is applied logic, in the same way physics is applied math, chemistry is applied physics, biology is applied chemistry...it goes on forever..! $\endgroup$ – zerosofthezeta Oct 15 '13 at 21:07
  • $\begingroup$ Some infinities being larger than others also doesn't make sense, yet is accepted as math. $\endgroup$ – Kaz Oct 16 '13 at 0:38
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    $\begingroup$ Does taking the square root of $-1$ make sense in terms of oranges? $\endgroup$ – Trevor Wilson Oct 16 '13 at 6:00
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Well, if $0/0$ is well-defined then it should play nicely with our definition of division. Specifically, $a/b = c$ if $a = cb$. So there's gotta be some c such that $0c = 0$. Let's try 1. $0\cdot 1 = 0$, so $0/0 = 1$.

Not too bad. Now let's try two. $0\cdot 2 = 0$, so $0/0 = 2$. Since equality is transitive, $1=2$.

Aaaaaand we just broke math. We resolve this by saying that $0/0$ is not defined, which keeps everything running nicely with a minimum of fuss.

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