M0PHEAD said:
Maze1125 said:
M0PHEAD said:
You'd still have two left over afterwards though, so the actual answer (and kind of a paradox in itself) to the sum two divided by two is
infinity... remainder two.
Not necessarily.
The answer to 2/0 is infinity exactly, with no need for a remainder because we can specify that it is the infinity that is such that infinity x 0 = 2.
Hang on... you're saying that there's a hypothetical value which when multiplied by zero is equal to two? As I understood it, zero represents nothing- and however many nothings you have it still equates to nothing, right?
As a maths student I find what you're saying, if correct, pretty damn interesting; I don't suppose you've got a link to a page which explains it a bit more thoroughly?
Like I said, I was being imprecise.
The precise way to do such a calculation would be through limits.
Consider the function f(x) = 2x/x.
For all non-zero values of x, f(x) = 2, but f(0) is undefined as it contains 0/0. What you can do is consider the function as 2/x * x which, at x = 0, would be equal to infinity * 0.
This still does not give an answer unless you assume that infinity * 0 has a precise answer, which we cannot do.
So, we need to consider the limit, but for all x =/= 0, f(x) = 2 so it is easy to deduce the limit of the function as x tends to 0 is also 2.
We still do not have justification for claiming f(0) = 2, but what we can do is define a new function g.
Where, for x =/= 0, g(x) = f(x) and g(0) = 2.
g is then a continuous and differentiable function that equals f everywhere except for x = 0 where f did not have a value and g equals 2.
The point at x = 0 is called a removable singularity [http://en.wikipedia.org/wiki/Removable_singularity].
Now, most of that was pretty much rambling for the sake of precision. The point is that, because we are considering the function 2x/x, we know both that, at x = 0, the function is equal to infinity * 0 but also that the function is equal to 2.
So, in this case, infinity * 0 = 2.
(That last paragraph was mildly imprecise again.)