0 / 0 = 1 theory

damnit, latex didn't work. you're gonna have to render the code in another window or something

>>8863669
Even [math]\LaTeX[/math] finds your """proof""" too idiotic to be worthwhile rendering.

lim x/x, x -> 0 = lim 1 = 1, therefore 0/0 = 1

genius proof, op

>>8863701
thanks

>>8863669
Its retarded.

Learn about continuity BEFORE you make such retarded claims about limits.

Your """"""proof"""""" also shows that 0/0=2, just replace 0.1 with 2*0.1 and you will have the same """"result""""".

But the inverse function is NOT CONTINUOUS. Which invalidates your whole """""""""proof""""""".

>>8863669
that is not how fractions are defined.

>>8863669
Let us consider [math]\mathbb{Z}\times S[/math] where [math]S := \mathbb{Z}-\{0\}[/math]
Consider the following equivalence relation on [math]\mathbb{Z}\times S[/math]
[math](a,s) \sim (b,t) \iff at=bs[/math]
(The reader may feel free to verify this is indeed an equivalence relation.)
Then take the following suggestive notation for an equivalence class:

[math]\frac{a}{s} := \{(b,t): (a,s) \sim (b,t) \}[/math]

Then observe:
[math]\frac{a}{s} = \frac{c}{u} \iff au=cs[/math]
Also define:
[math]\mathcal{Q}(\mathbb{Z}) := \{\frac{a}{s}:a\in\mathbb{Z}, s\in S\}[/math]
Then one can verify [math]\mathcal{Q}(\mathbb{Z})[/math] is a field better known by the notation of [math]\mathbb{Q}[/math]

>this is how the rational numbers are constructed.
>what makes you think 0/0 has any place in it.
If you allow 0 to be in the denomenator, the set wil not be a field anymore.

>>8864047
ofcourse with the traditional operations of addition and multiplication on fractions

>>8864047
Is that because 0/0 would be an element of every equivalence class, which is a contradiction since equivalence classes are disjoint?

Zero is merely a spectrum.