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Now, I understand that dividing by zero in any case is undefined. However, in math, there are always exceptions. I'm just really curious...what are the different cases for different answers? For most controversial arguments, I've looked up both sides, but I can't seem to find that many sources for this one.

Personally, I think it $\frac 00$ is 0 because for younger kids, division can often be a word problem such as the following:

If John has 20 apples, and he divides his apples up between his 5 friends, how many apples did he give each of his friends?

Obviously, the answer is 4 apples. Now, reword the problem to say:

If John has 0 apples, and he divides his apples up between his 5 friends, how many apples did he give each of his friends?

Now the answer, given this context, is 0. Of course, once the decimal system is introduced in a person's education, and the "remainder" system is useless (eg, $\frac54 = 1 $, R $ 1$), this proof no longer works.

TL;DR: If there is nothing there, and you don't divide it, there's still nothing there. You can't magically create 1 to satisfy the $\frac aa = 1$ rule.

Antonio Maria Di Mauro
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piticent123
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    If you define ${a \over b} = c$ such that $c \times b =a$; then ${0 \over 0}$ has infinite solutions since for any $c$ it's true that $c\times 0 = 0$. – Felix Marin Feb 04 '14 at 01:32
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    In your second scenario (John has 0 apples and divides them to 5 friends), the mathematical expression that would model the situation would be $0/5$, which is indeed $0$. However, if you are trying to model $5/0$, try a problem like this: "John has 5 apples, and divides them evenly to 0 friends. How many apples does each friend get?" The answer is surely not 0. It's undefined. – chharvey Feb 04 '14 at 01:33
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    If you divide $0$ apples between $0$ friends, then each of your $0$ friends has received $n$ apples, where $n$ is any real number. – T.J. Gaffney Feb 04 '14 at 01:34
  • @Gaffney only if you're working in certain types of algebra systems. In high school algebra though, $0/0$ is undefined, not an arbitrary real number. – chharvey Feb 04 '14 at 01:43
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    @Harveybars Yes, dividing 5 apples among no friends wouldn't make sense to the average person, but if you instead said "John has 5 apples, divides them evenly to his friends, giving 0 apples to each friend. How many friends does John have?", someone without much mathematical sophistication might say "infinity", and they might even think that an uncountable sum of zeroes can equal finite number, because an uncountable union of measure-zero sets can have finite measure, and they may not recognize that measures are only countably additive. – Keshav Srinivasan Feb 04 '14 at 01:48
  • @Gaffney, it's easier to say "if you divide 0 apples between your friends, then each of your $n$ friends has received 0 apples". – Keshav Srinivasan Feb 04 '14 at 01:50
  • @Harveybars $0/0$ is said to be undefined precisely because there is no single value that can be assigned. $5/0$ is undefined as well but for a different reason: As you said, there isn't any value that can be assigned. Nowadays in K-12 education (at least in my exposure) they are careful to make this distinction because it otherwise is confusing to a student seeing it for the first time. – T.J. Gaffney Feb 04 '14 at 05:30
  • @KeshavSrinivasan wow, you lost me on that last part. What did you mean by "measures are only countably additive?" – chharvey May 02 '14 at 17:19
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    @chharvey I don't know how familiar you are with measure theory, but measures obey a property called countable additivity, which states that if you have countably many disjoint sets, then the measure of their union is the sum of the measure of each of the sets. But they don't have the property of uncountable additivity: if you have uncountably many disjoint sets, then the measure of the union need not equal the sum of the measure of each of the sets. In particular, the union of uncountably many measure-zero sets can have positive measure (e.g. uncountably many points make up an interval). – Keshav Srinivasan May 02 '14 at 22:42

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I'm trying really hard to figure out exactly what you are asking, but here's my best answer.

For any real $c$,

Case 1: denominator approaches zero from the right $$\lim_{x\to 0^+}\frac{c}{x}=\infty$$

Case 2: denominator approaches zero from the left $$\lim_{x\to 0^-}\frac{c}{x}=-\infty$$

Case 3: numerator approaches zero from both directions $$\lim_{x\to 0}\frac{x}{c}=0$$

Case 4: numerator and denominator approach zero from both directions $$\lim_{x\to 0}\frac{x}{x}=1$$

Maybe you could rephrase your question to be more specific. In the mean time, do some light reading so you can refine your question.

chharvey
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