On Wednesday, 26 February 2014 16:20:20 UTC+2, Wizard-Of-Oz wrote:
> That means the result is an integer quotient with a zero remainder
Irrelevant bullshit. I explained to you precisely what I meant, more than once. You are an argumentative arsehole.
> I have
No you have not.
> So now you are saying that that is what you mean by 'divides exactly'. I > did ask you what you meant, and you did't say.
My grade 10 students get this and English is their second language. Aren't you ashamed of yourself? You act like a 2 year old.
> YOU need to proof that noone can.
Lesson 2 of the New Calculus. Go read it!!!!!
> You claim what you say is rigorous, so you need to PROVE that for any > function f(x), you can simplify (f(x+n)-f(x-m))/(m+n) and that > simplification will not have any term involving (m+n) in the denominator > Try f(x) = sin(x)
It works for all differentiable functions, including sin(x).