Your move, brainlet

Your move, brainlet

i = 1

But then the hypotenuse would be square root of 2, not 0

No. It's not a right triangle.

works in the complex plane

Can't have a triangle with a side distance of zero

should be ||a|| + ||b|| = ||c|| but that's too complex for brainlets

> ||a|| + ||b|| = ||c||
> 3 + 4 = 5
Hmm.

i= root -1
i^2 + 1^2 = 0
root -1 ^2 + 1^2 = 0
-1 + 1 = 0
0 = 0
but also this

Sure you can, all the math works out. Just because you get two strait line segments atop one another does not mean you cant define a triangle that describes such a thing.

This shit is so retarded. Stop making this thread

|i|^2 + |1|^2 = 1 + 1 = 2 /= 0

Doesn't work OP

| a + ib |^2 = a^2 + b^2, where a and b are real.

And?

Can't have a triangle with negative side-length in Euclidean geometry.

>Can't have a triangle with negative side-length in Euclidean geometry.
True, but how is that related to this thread?

Not that guy, but what's the mathematical significance of Euclidean geometry (specifically, the Euclidean inner product) anyway?
We can't even say that "it's the geometry of the real world" anymore.

Hardcore shitpost

Pythagorean theorem actually says:

||a||^2+||b||^2 = ||c||^2

And ||i||=1

you labeled the wrong length as 0, user
the hypotenuse of the "triangle" between 0+i and 1+0i is 0
see pic related

Your move

>Non-Euclidean geometries are geometries in which the parallel axiom does not hold. An example of this is the geometry of the sphere surface. There, the theorem of Pythagoras no longer applies, since in such geometries the interior angle theorem does not apply, that is, the sum of angles of a triangle of 180 ° is different.

Now does the parallel axiom hold? I can draw multiple lines through a point that do not meet a line drawn in the complex space, can't I?

It should be RxR where the line is drawn usually. If you adopt this for complex numbers you'd get CxC or in practical terms: RxRxImaginary part at least. I think.

>3