I broke maths while lying in bed feeling tired

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!

nice meme

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

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

>cos being an even function
>false premise

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

You forgot the Jacobian faggot.

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.

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

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]

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]

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

Gj retard

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

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

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]

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

Lrn2division fgt pls

...

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.

>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?

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.