Your move, brainlet

Impossible because the length of that side is clearly not 0.

Fucking idiot, outside of algebraic geometry you can't have an imaginary length and arbitrarily define a length of i like that.

First, explain what the "i" next to the red line is supposed to represent, otherwise I will assume that two or three of the symbols on the pic are variables

Distance can't be a complexnumber

nothing can physically be "i" in length. No wonder you get an anomalous result

That works on the complex plane just fine

STOP RESPONDING TO THIS BAIT FFS HE POSTS IT EVERY WEEK

It doesn't, because even on the complex plane "i" is not a valid length for a vector

Even assuming [math]i[/math] is a valid length, I think it would be wrong to assume the triangle could be a right angle.

If a side length is imaginary than surely the angles have to be imaginary too.

wrong , idiot.

consider a triangle in C with vertices at 0 + 0i , 1 + 0i and 0 + 1i .
clearly the length of the side between 0 + 0i and 0 + 1i is equal to 0 + 1i -(0 + 0i) = i

what OP's picture demonstrates is that "the square roots of negative numbers" clearly do not exist and lead to mathematics-breaking contradictions.

mofos don't know about my null strut calculus

sci-hub.bz/https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.2042

literally, the distances in are 1+1 space-time geometry. Zero distance = light cones

holy shit you're unironically right, OP's pic is actually a Minkowski Diagram except that no one thought of using the Imaginary axis as the Time axis

woah dude thanks for the nobel prize

This is stupid and not even a good troll. You have to take the difference, then multiply it by its complex conjugate. Not even remotely what the OP is referring to.

Since the triangle inequality is not satisfied, you can't have a triangle with sides 1, 0, i.

triangle inequality only needs to be satisfied in euclidean space

not true, it holds in any metric space, but that's not the case here

>he doesn't know of the complex inner product

clearly you can have a triangle with sides of length 1 , 0 and i . pythagoras' theorem and the triangle 0 + 0i, 1 + 0i, and 0 + 1i demonstrate that you can.

What this actually shows is that " the square roots of negative numbers" are contradictory and invalid.

i cant be length dum dum

[math]\pi a \xi ^2+\pi a (z-\xi )^2+2 \pi a \xi (z-\xi )[/math]

*Ports Tele With Vision*

By that logic, if any three numbers satisfy Pythagorean theorem, then some triangle with those numbers as lengths exists. So what about triangles with negative sides? I think it all comes to what exactly do we mean by a triangle anyway.

Don't respond to the tard i isn't a length. The coordinates ( 0, i) might as well be written as (0 , 1) in an (x, y) plane where x = Re(z) and y = Im(z)

>i was a mistake
yes you were

>Past Tense Intensifies!

good post

>imaginary triangles

>tries to use trig
>violates trig inequality

Lol

you just constructed nothing

althought it even respects the law of triangles:
i + 1 > 0
so does a ''triangle'' with its sides with 1, 1 and 0 of length, yet that doesn't form an actual triangle, now does it?
0 of length forms no triangles, pls leave amerimutts

metrics are functions with codomain [math]\mathbb{R}[/math]
therefore you're not using a metric to label the sides, and so there is no reason for the theorem to apply to your situation

tl;dr kill yourself promptly

Vectors != magnitudes. A vector consists of a magnitude and direction. This is legit babby tier math you should have learned in high school, if not middle school during Geometry class.

>A vector consists of a magnitude and direction
>berates others for "babby tier math"
holy CHRIST I can't stop laughing you are so pitiful

1^2 + 0^2 = i^2
1 = -1

are you retarded? why does this shit 80 IQ bait have 42 replies

i = 1

fuck is wrong with all of you.

1 + i

>distances can be complex
Not with the way "distance" is defined, you brainlet

>complex numbers are totally ordered

I'm not a mathematician, so i might be wrong in this. But isn't i more of a unit than a number?

en.wikipedia.org/wiki/Complex_number

also, units are typically numbers
en.wikipedia.org/wiki/Unit_(ring_theory)

>babby's first hyperbolic geometry

>the complex field is ordered

0 is the hipotenuse though

1 and i are parallel here. i = 1.

I heard OP has imaginary friends.

>take the vector space of boundary operators in some de rham complex
>make a triangle from orthogonal exterior derivatives in your space
>the hypotenuse is 0
>mathematics is broken

checkmate

nothing. It is imposible since 0 distance cant exist as a triangle side.
also you forgot the |i|

literally oxymoronposting kys
>but what if rain bounced higher than the clouds hehe science btfo

Sorry but your are breaking several rules

yes, several of the rules you can't name

And still mathcucks cant prove why OP is wrong

A^2+b^2=c^2
i^2=-1 1^2=1 1+(-1)=0

Why not? If you can have an axis of units of i, dont they need to be of length i?

...

Your move brainlet
youtu.be/SYGF25asipc

find us we have such secrets to show you desu

...

If 2 angles each equal zero and the remaining angle equals 180*, is this not a triangle by definition? Three angles adding up to 180*

1+0 >1 ?????????????????????? doesn't satisfy shit

haha o wow

Can't have a side with 0 length :)

um sorry but you obviously can, see:

No, you can't. The Pythagorean theorem is defined in euclidean space. You can't have sides of negative or imaginary length. And you can have sides with 0 length.

Brainlet here to save the day. If i is also representative on a point on the Y axis, then going "up i units" sends you to (1, 0). If you calculate the distance from (1, 0) to (1, 0), you have no units.

>The Pythagorean theorem is defined in euclidean space
It's obviously working for complex number as all the values end up being correct

>[math]\color{green}{ a = \sqrt{c^2 - b^2}}[/math]
[math]i = \sqrt{0^2 - 1^2} = \sqrt{-1}[/math]

>[math]\color{green}{ b = \sqrt{c^2 - a^2}}[/math]
[math]1 = \sqrt{0^2 - i^2} = \sqrt{1}[/math]

>[math]\color{green}{ c = \sqrt{a^2 + b^2}}[/math]
[math]0 = \sqrt{i^2 + 1^2} = \sqrt{0}[/math]

It doesnt matter bud. The Pythagorean theorem is defined in EUCLIDEAN space.
Things must have real LENGTH.
You can NOT have negative length or 0 length. These don't exist in real life. Pythagoras made his theorem to apply to real life.

>prove that it works for complex numbers
>"uhh it doesnt matter bud my wikipedia page says it only works for EUCLIDEAN SPACE ok buddy???"

>Sqrt(-1) makes no sense in algebra
>Let's just call it i, and everything just works

>Can I call 1/0 k?
>No, that doesn't make sense in algebra

i is not a length in space, it is a direction

>im a brainlet who never saw how to construct C

And the area of this is [math]\frac{i}{2}[/math] according to both [math]A = \frac{ab}{2}[/math] and Heron's formula

interesting shit OP

When we construct a triangle in [math] \mathbb{C} [/math] and talk about it's lengths, we're obviously talking about norms of the numbers representing the triangle, obviously just as we don't consider negative lenghts in [math] \mathbb{R}^{2} [/math] therefore we can conclude

[math] \nexists a \in \mathbb{C} : \lVert a \rVert = i [/math]

Therefore such triangle doesn't exist

QED

>>Can I call 1/0 k?
>>No, that doesn't make sense in algebra
Yes. Because it leads to all numbers being the same...

>we're obviously talking about norms
Just use a complex measure brah

>negative length

Nothing wrong with that.

>[math]\sqrt{-1}[/math] is negative

KEK

>length of zero

thats just a i and 1 is just a line segment u brainlet

I learned that i is not equal to sqrt(-1), but that i is the solution to i^2+1=0 by my professor

>i is the solution to the equation i^2 + 1 = 0
the academy award goes to your teach this year

>a norm having any value that is not a real number and above 0 except when the vector is zero
found the brainlet

>Yes. Because it leads to all numbers being the same...
No it doesn't, you can call 1/0 "k" just like you can call the negative number under the square root "i"

>No it doesn't, you can call 1/0 "k"
Seriously think before posting.
But then immediately follows that ak=k, for all real numbers a which either means that you have just lost all algebraic structures which you desperately need, or that a=1.

The point is C is a field, whatever you do with k is not, which means that your definition of k is absolute worthless garbage if you want a "good" structure on your numbers.
It is also something which has been explored by mathematicians a long time ago and they have found little use for it, for the above mentioned reasons.

>what is wheel theory en.wikipedia.org/wiki/Wheel_theory

youtube.com/watch?v=2xkdQvavKyY

Nope

i = 1

hypotenuse doesn't exist.

OP's image represents a straight line, drawn very poorly.

You mean |i| = 1

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