Welcome to Gaia! ::

How so? It seems more logical to me, although I do not pretend to be a mathematician, that the number of zeroes going into a one is undefined, not infinite - I mean, the sum of an infinite number of zeroes is still zero. The question is not valid - zero cannot go into anything.

I hope there's a mathematician around to fix my understanding, since I'm sure it's either wholly inaccurate or incomplete.

HA. I love this. Instead of doing my own math work, I'm contemplating on the equation of 0x0=Infinity.
I don't know what to say. I can could blab about the statement being true, in my own logic. Which could be amusing, but not true.

0x0=0

It's simple elementary math. Any thing times 0 is 0. We all learn it when ever we learn multiplication(I'm horrible because I can't even remember which grade is exactly was. sweatdrop ).
But, what gets me is 0/0 = What? Because any number divided by 0 is undefined.
It could easily be answered by nothing is divided by nothing which results in you still have nothing. Hmmm.... I guess this is a more philosophical question than a math one.

So I did some research on the number 0. Here's what I dug up.

""0 (zero) is both a number — or, more precisely, a numeral representing a number — and a numerical digit. Zero is the last digit to be incorporated in most numeral systems. In the English language, zero may also be called nil when a number, o/oh when a numeral, and nought/naught in either context.""



"" * Multiplication: x · 0 = 0 · x = 0.
* Division: 0 / x = 0, for nonzero x. But x / 0 is undefined, because 0 has no multiplicative inverse, a consequence of the previous rule. For positive x, as y in x / y approaches zero from positive values, its quotient increases toward positive infinity, but as y approaches zero from negative values, the quotient increases toward negative infinity. The different quotients confirms that division by zero is undefined.
* Exponentiation: x0 = 1, except that the case x = 0 may be left undefined in some contexts. For all positive real x, 0x = 0.

The expression "0/0" is an "indeterminate form". That does not simply mean that it is undefined; rather, it means that if f(x) and g(x) both approach 0 as x approaches some number, then f(x)/g(x) could approach any finite number or ∞ or −∞; it depends on which functions f and g are. See L'Hopital's rule.""

I find this last part very interesting, as it suggests that 0/0 is indeed some sort of value. The problem is is that this value has to real definition, but it must exists as according to L'Hopital's rule [which I checked out] The derivative of such a equation must exists in calculus.

""In simple cases, L'Hôpital's rule states that for functions f(x) and g(x):
if or
then
where the prime (') denotes the derivative.

Among other requirements, for this rule to hold, the limit
must exist ""

Getting past the sheer meaning of the calculus, I came to understand that for the derivative of this interger to exist, there must be a value for 0/0. Since the integer can come to be any finite number [as stated in the law], or either of the two infinities.

Cite/Sources

The number 0.
L'Hopital's rule.

EDIT: To link this to the OP's post, the sources I mentioned provide no arguement as to why 0x0 would not = 0. It seems its just a mathematical law.

My head espolode.

All it is sayins is that since the Derivative of 0/0 exists, then 0/0 must have a value. [ L'Hopital's Law ].

Unfortunately, mathematics is a system of logic and only needs to be follow from the axioms and be deductively valid.

The formula's made my head explode, logic is something I believe I have a good grasp of.

Actually, 1/0 is a bit more complicated, because it's having something, and dividing it into no parts. But that would also (technically, and I suppose logically) remove the part that contained the something, so I suppose you'd be left with 0 there as well. On the other hand, where did your something go?

Same follows with any number over 0, except of course 0, since logic should be able to solve that.