I broke maths while lying in bed feeling tired

Oh crap, this is going to require us to reconsider the very foundations of modern mathematics.

>x=-x
>2x=0
>x=0
nice bait

>>8454153
[eqn] e^{2\pi i}=1 [/eqn]
[eqn] ln(e^{2\pi i})=ln(1) [/eqn]
[eqn] 2\pi i=0 [/eqn]
[eqn] \pi=0 [/eqn]

x^2 = -x^2
x^2/x = -x^2/x
x = -x

oh no I broke the universe

OP here, I started from something THAT IS TRUE. Don't shitpost with this starting from x = -x nonsense.

>>8454153
you start with a false premise
sage

>>8454162
Wait, this one actually confuses me.

>>8454184
Can't assume logs work like that for complex numbers

>>8454182
Are you retarded or just haven't taken grade 6 trigonometry yet?

>>8454170
You don't know anything about the arccos function do you?

cos(pi)=cos(10001pi)
arccos(cos(pi))=arccos(cos(10001pi))
pi=10001pi
1=10001
HURR DURR

We all know the area of a circle of radius [math]r_0[/math] is [math]\pi {r_0}^2[/math]. Set this equal to the double integral over the area of a circle:
[eqn]\int_{\theta = 0}^{2 \pi} \int_{r = 0}^{r_0}dr\, d\theta = \pi {r_0}^2[/eqn]
[eqn]\implies \int_{\theta = 0}^{2 \pi} r_0\, d\theta = \pi {r_0}^2[/eqn]
[eqn]\implies 2 \pi r_0 = \pi {r_0}^2 \implies r_0 (r_0 - 2) = 0 \implies r_0 = 0, 2.[/eqn]
The radius of a circle is either 0 or 2!

>>8454212
nice meme

>>8454212
It's been a while since I've taken calc 2, but don't you need to make the integral r dr dtheta?

>>8454221
no his proof is right and he just totally btfo all integration

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46KB, 568x479px

>>8454182
>cos being an even function
>false premise

mods need to label poster IPs like these with permanent tripcodes that say brainlet.

>>8454212
You forgot the Jacobian faggot.

>>8454153
Nice one. The problem is that since cos is not injective cos^-1(cos(a)) = cos^-1(cos(b)) does not imply that a=b.

>>8454184
Take a look at branch cuts in your closest Complex Analysis textbook.

sin(π/2)=1=sin(π5/2)
π/2=π5/2
1=5

>>8454417
>>8454202
>>8454166
>>8454162

DELET THIS

Here's one

[math] \frac{1}{i} \times \frac{1}{i} = \frac{1}{i^2} = -1 \implies \frac{1}{i} = \sqrt{-1} = i [/math]

[math] \frac{1}{i} \times (-1) = \frac{-1}{i} = \frac{i^2}{i} = i \implies \frac{1}{i} = -i [/math]

So, [math] i = \frac{1}{i} = -i [/math]
[math] i=-i [/math]
1=-1
2=0

Haha, mathematics is broken, where's my fields medal?

[clearly a joke, I understand there are 2 square roots]

>>8454153
fuck you shitlords
[math]
\frac{n}{0}=x
[/math]

[math]
\frac{n}{0}-x=0
[/math]

[math]
x\frac{n}{0}=x^2[/math]

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

[math]
x(x-\frac{n}{0})=0[/math]

[math]
x(x-\frac{n}{0})=x-\frac{n}{0}[/math]

[math]
\frac{x(x-\frac{n}{0})}{(x-\frac{n}{0})}=\frac{(x-\frac{n}{0})}{(x-\frac{n}{0})}[/math]

[math]
x=1[/math]

[math]
\frac{n}{0}=1[/math]

>>8454184
he did it wrong, the actual conclusion is i=0

>>8454153
Gj retard

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1MB, 320x240px

>>8454153
>cos^-1(cos(x)) = cos^-1(cos(-x))
>x = -x
Doesn't follow and that's not how acos works.
sage

>>8454455
1/i^2=-1 implies 1/i=+-sqrt(-1), in fact 1/i=-i.

>>8454490
I see the problem, in the 6th line after [/math][math] you divided by 0.

[eqn] 0 = \frac{0}{0} - \frac{0}{0} = \frac{0-0}{0} = \frac{0}{0} = 1[/eqn]

>>8454212
Wrong. In polar coordinates the differential area is proportional to r.

>>8454541
Lrn2division fgt pls

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37KB, 193x300px

>>8454541

I dont know what to think of this thread. Should I post the actual problem with OPs "proof" or should I let you all just shitpost? Not even sure if OP is serious.

>>8454170
>Don't shitpost with this starting from x = -x nonsense.
What a fucking waste of flesh.
x=-x iff x=0
How is that nonsense?

>>8454184
Inverse functions aren't always rigorously defined, as many of the "log rules" you learn about don't hold up for imaginary values.

His post was about as logical as saying sin(2pi) = 0, sin(0)=0, sin(2pi)=sin(0), arcsin(sin(2pi))=arcsin(sin(0)), 2pi=0. This simply doesn't work because we've defined arcsin to only display inverse values of sin from -2pi to 2pi. Go on desmos and enter in sin(y)=x on one line, and arcsin(x) on another, and you'll see what I'm talking about there.