Think we've got dividing and multiplying by zero wrong,

https://en.wikipedia.org/wiki/Wheel_theory

>>8904811
Because your mother dropped you on your head when you were young

[math] 0<n [/math] is true for everything, except 0 under our current rules for <

>>8904823
i clearly meant n^2 btw

>>8904826
>implying i^2 > 0
kys brainlet

>>8904833
it was implicit that i was talking about R

>>8904833
n usually implies an arbitrary element of the set of reals..

>>8904815
how did you get to know this?

>>8904869
n is more often a natural number or maybe an integer, not a real brainlet

>>8904874
i did my thesis on zero division

>>8904811
[math] a \times 0 = a \iff \\ a \times (a - a ) = a \iff \\ a^{-1} \times a \ (a-a) = a^{-1} \times a \iff \\ a-a = 1 \iff \\ 0 = 1 [/math]

Whoops, back to the drawing board.

>>8904840
Why? OP's example works with real and imaginary numbers.

>>8904885
If you followed the rules of PEMDAS, you would get 1 x 0 = 1, which using OP's logic gives 1 =1.

M: NxN => N

1. M(a,0) = 0
2. M(a,S(b)) = P(a,M,(a,b))

>>8904811
a * 0 = a
a * 1 = a
a * (1 + 0) = a
a * 1 + a * 0 = a
a + a = a
2a = 2

you fucked the distributivity and ton of other properties

So...

[math]
\frac{a}{a} = 1 \ or \ 0
[/math]

Doesn't make sense senpai.

>>8904918
did you actually check any example?

File: 1489549129215.jpg (72KB, 1200x740px) Image search: [Google]

72KB, 1200x740px

>>8904885
Your proof is incorrect. Faggot.

File: bb.png (12KB, 1585x456px) Image search: [Google]

12KB, 1585x456px

>>8905050
Just this one.

I'm sure it works for all others too.

[math]|a|\sqrt{(\frac{b}{a})^2 + 1}[/math]

how does that even work?