/sci/ - Science & Math

[Anonymous No.1931584]

Here is my proof that complex numbers are BS

e^(2*Pi*i)=1
e^0=1
hence
0=2*Pi*i
i=0

Problem 300 years of mathematics?

[Anonymous No.1931593]

OP you forgot this
>Pic related

[Anonymous No.1931591]

>mfw I would like a mathfag to explain how this isn't true

[Dahammer No.1931594]

>>1931584
a problem? Yes, your mathematics.
Go troll /b/. You seem to b more at their level.

[Anonymous No.1931601]

>>1931591
This isnt true becuase exp() function is not really a function on complex plain.
Diffrent Inputs can give same output.

[Anonymous No.1931605]

>>1931591
it is e^(pi*i), and it equals -1.

[Anonymous No.1931609]

10/10
Much better than those 1=2 proofs

[Anonymous No.1931614]

bitches don't know 'bout my periodic functions

[Anonymous No.1931612]

f(x) = (x-1)(x-2)

f(2) = 0
f(1) = 0
1 = 2

what now faggots?!

[Anonymous No.1931611]

A related but true result is i^i = e^{-pi/2}.

[Anonymous No.1931615]

Two can play that game, OP!

Here is my proof that trigonometry is BS
sin(0) = 0
sin(Pi) = 0
hence
0 = Pi

[Anonymous No.1931622]

I can support this kindoff trolling. It will provide some highschool fags and 1st yearfags useful lesson in periodic functions and complex plane.

[Anonymous No.1931624]

Hey guys dert
1^23 = 1
1^684853 = 1
23 = 684853

[Anonymous No.1931626]

6/6 = 1
5/5 = 1
hence
6=5

[Anonymous No.1931629]

>>1931626
Wtf is that
That means
6/6=5/5
lrn2troll

[Anonymous No.1931640]

>>1931629

[Anonymous No.1931642]

>>1931615
That's exactly the same trick he used, except that cos(0) = 1, cos(pi) = -1, so 0 != pi.
However, cos(2pi) = 1, so 0 = 2pi

[Anonymous No.1931655]

>>1931642
Touche, But my trick was superior.