Your move, brainlet

Your move, brainlet

Distances can't be negative.

NIGGERS DESHRTOYIN MI RAYCE!!11

nigger what negative value do you see there?

>i is negative

hellooooooo brainlet

How do you know it's a right-angled triangle?

>The nonzero leg length that would result in a zero hypotenuse is imaginary

Sounds about right, I don't see the issue.

Distances can only be positive and real.

mod(1)=mod(-1)=mod(i)=mod(-i)=1

As an electrical engineer, this is just stupid. If you make a drawing like that, one dimension (here x) has to be real numbers, the other one (here y) has to be imaginary numbers. The blue line therefore is 1+j ( as said, distances can't be negative, so you need to add the absolute value of 1^2 to the absolute value of i^2), or a length of sqrt(2) with a phi of pi/2.

However, either your x-dimension is positive facing to the left, or the drawing needs to be flipped.

>HURR DURR, can you brainlets find the length of one of the sides of this two sided triangle?

Fucking imbecils, have you even finished middle school?

If i is negative then what's -i?

>distance measured in 'i'.
There's your problem, brainlet.

i is a tool used (usually by brainlet engineers) to think through a problem, not an actual number.

You are the brainlets.

Propose you have a square with side length 'i'.
i^2 = -1. You now have a square with negative area.

i isn't negative, it is a complex number of magnitude (positive) 1 in the imaginary axis

areas can be negative you dumb fuck, the introduction of complex numbers gives enough ground to assume that.

it's called vectors you whore

This is an incorrect problem, you troglodyte.

>the problem is wrong
the absolute state of maths

Correct me for my brainletism but
>[math]sin = \frac{a}{c} = \frac{i}{0}[/math]
>[math]cos = \frac{b}{c} = \frac{1}{0}[/math]
>[math]tan = \frac{a}{b} = \frac{i}{1}[/math] but also [math]tan = \frac{sin}{cos}[/math] and since [math]tan = \frac{i}{1}[/math] then [math]sin = i[/math] and [math]cos = 1[/math]

Now here comes the interesting part, since [math]sin[/math] and [math]cos[/math] were initially [math]\frac{i}{0}[/math] and [math]\frac{1}{0}[/math] wouldn't considering the above mean that [math]\frac{i}{0}[/math] = [math]i[/math] and [math]\frac{1}{0}[/math] = [math]1[/math]?
i.e [math]\frac{x}{0} = x[/math]?

a:b=A:B does not imply a=A, b=B.
Yes, i does strange things outside the safe Gauss plane.
Consider the line r of equation y=ix. Then r is perpendicular to itself, because i=-1/i

I'd use pyhtagorhas rtheorrem
or even better I'd study gommorrha's manual or kamasutra for this angle, so I can know in what position to satisfy better.

retarded sin = ik and cos = 1k for any value k
basic trignometrik

Hypotenuse shld be longest side, 0 is less than 1, its not right triangle

[math]\sqrt{-1}^{2} + 1^{2} = 0^{2} = 0[/math] It follows pythagorean theorem which is probably why user thinks he is smart.

It's a shitty point though because distance can't be negative and the hypotenuse can't be 0, the it may as well be a point.

the square of i is -1 but its norm is still 1

The problem with this proof is that i is on the imaginary axis and not an X,Y plane. In which cannot represent distance in the same way as a -1. So the pythagorean theorem cannot apply to imaginary numbers

1.
The triangle folds in on itself and becomes a line, obviously.

Oh shit i thought we were solving for i.

>distances can't be negative

Minkowski space?

What does this have to do with Minkowski Space?

Minkowski space?

Minkowski space?

Light path in Minkowski space!

>As an electrical engineer
Stop right there

The triangle is spaced in the same plane. It means by default this cannot be true, since the imaginairy space needs to be perpendicular to the real space. Or else you get a contradiction

0^2 = 1^2 + i^2?

You mean the 45 degrees light path ? Why would it be 0 tho

>Everyone complaining about a negative area when the diagram implies a zero distance.

It's fucking degenerate.

Your side denoted by ā€œiā€ is through a higher dimension. Imaginary numbers can be represented on a grid when all non-imaginary numbers are on the x-axis.
This triangle is possible if you are measuring the hypotenuse in a lower dimension than side ā€œiā€.

Checkmate atheismo.

more like distances can't be zero Lol

zero is the absence of number not a number itself

therefore it is the absence of distance

if you replace [math]\tau = it [/math] then Minkowski space has (++++) signature. What I mean is instead of calculating distance in spacetime as [math]ds^2 = dx^2 + dy^2 + dz^2 - dt^2[/math] you have [math]ds^2 = dx^2 + dy^2 + dz^2 + d\tau^2[/math], since [math]\tau^2 = (it)^2 = -t^2[/math].

its akshually spics and whytes deserve it for turning on their own gods to worship a Kikeonastick

i = 1

It's root 2 senpai.

Just plot it on the complex plane and see yourself.

this.

op's picture is a really shitty drawing of a straight line.

by Allah you are truly disgusting

>tfw so inept at drawing my straight lines come out looking like triangles