Brainlets BTFO

[math]
0=0^1\\0^1=0^{2-1}\\0^{2-1}=\frac{0^2}{0^1}\\\frac{0^2}{0^1}=\frac{0}{0}\\\frac{0}{0}=0\\\blacksquare[/math]

The real question is, what is 1/0?

In which wheel?

0/0 = 1/100 like how 100% = 1

Open and shut case, Johnson.

Is "division" in a wheel even really division? I mean if x/x != 1 it sounds like they're really forcing the term.

I would argue that "division" is what is defined by the field axioms. Meaning that that thing in a wheel is not division, and it just tries to be profound with misleading terminology.

in the general case brainlet, x/x = 1 most of the time
that's a stupid fucking view, don't you think of eg. euclidean division as division?

[math]0=0^1\\0^1=0^{2-1}\\0^{2-1}=\frac{0^2}{0^1}\\\frac{0^2}{0^1}=\frac{0}{0}\\\frac{0}{0}=0\\[/math]

[math]\frac{0^2}{0^1}\neq\frac{0}{0}\\[/math]

0/0 = 1 nigger

Under field axioms, x/x = 1 whenever it's defined, which is every value except x=0. That's very different than "most of the time". Euclid division isn't so much a binary operation as a theorem proving a unique factoring. And anyways, x = 1*x + 0, which is the closest you can get to x/x = 1.

I will say it's undefined, unknown.

[math]\frac{0}{0} = x[/math]
[math]0 = 0 \cdot x[/math]
[math]\forall x \epsilon \mathbb{R}[/math] 0x=0
Therefore, [math]\frac{0}{0} = \forall x \epsilon \mathbb{R} [/math]
(last theorem is left as an excersize for the reader)

what does the line over infinity mean? repeating infinity?

[math] \displaystyle \lim_{x \rightarrow 0} \frac{0}{x}=0[/math]

you accidentally an x but yep there's your problem

[math]\overline{\mathbb{R}}[/math]

0^(2-1) =/= 0^2/0^1 brainlet

Division is closed over R

[eqn] \lim_{x \rightarrow 0} \dfrac{0}{x} = \dfrac{ \lim_{x \rightarrow 0} 0}{ \lim_{x \rightarrow 0}x }= \dfrac{0}{0}. [/eqn]

(1/inf)/(1/inf)=1
inf/inf=1
inf=inf
//

The sad part is that most calculus courses don't properly formulate the theorem of arithmetic of limits and this shit happens way too often with students.