|Think about Loose Coupling|
Maybe this will help, maybe it will add to the confusion that this post has become. Who knows...
The easy explanation comes from back when i first thought of this as well. i asked a teacher. According to her, zero does not meet the requirements to be in the denominator of any fraction so it should just be avoided. That's the easy-peasy explanation, feel free to avoid the mathematics below if you want...
So, i've been thinking about this for a bit, and eventually i got back to the division algorithm. Let's first take the case of a/b, where a is non-zero and b is zero. According to the division algorithm, there are numbers q and r such that
But in the case of a/0, r will be a. So we have
(which is required in this case) which is against the definition. So we can't divide most things by zero because at a fundamental level it breaks the rules.
Zero over zero is a special case of the above, because r will fall into the appropriate range (it will be zero) but the fraction is still not defined. Why is that? Let's take a closer look...
Let us take the fraction 0/0. Let us attempt to obtain a value for this fraction. It is of the form a/a so it is clearly not reduced (any fraction with common factors in the numerator and denominator can be reduced, and in the case of a/a the fraction is usually reduced by whatever the number a is). We then take some number d in the integers to reduce this fraction by, but we must choose d carefully. When we're done we want the fraction to be in it's most reduced state. So we want to find d such that
Note that no such number exists*. You can continue using larger and larger numbers, but you will never find one (infinity is not a number but a useful concept; it is unusable except in limits). Since we cannot find a number to reduce this fraction by it is irreducable, but it is not reduced. This is a contradiction, so i am forced to conclude that there is no value for this fraction.
This may or may not be the proof,
*: gcd is a function that takes two integer numbers and returns the greatest common divisor of those numbers.
update: put in some formatting and refined some of the definitions. Hopefully this will be readable by non-mathematicians as well...