Your move, brainlet

>look mom, I posted it again!

>i apparently now belongs to the set of real numbers, that euclidean distance is defined as.

How the fuck is this linear algebra

Hermitian space

>it’s a retard who doesn’t understand what a metric is trying to bait Veeky Forums episode
Sage

>magnitude

[math] i^2=1^2=0^2[/math]

1^2+1^2=0^2*

Sorry to be a brainlet who asks questions, but

Is there any reason why we can't define a geometric space where complex distances exist, and this holds? Does it contain any inherent contradictions?

Distance implies a metric which implies an ordering. You can’t have two solutions to the distance between the same two points. Complex numbers are not well ordered.
Look up what a metric is. You have to have a magnitude and the triangle inequality must hole.

en.wikipedia.org/wiki/Metric_(mathematics)#Definition

Physics fag but if user wanted to create a field with complex distances could he not but itd just be a trivial field?

This is weird, because I never doubted this result. I noticed it in 10th grade and just figured that complex numbers were normally 1-dimensional and it was just kinda cool that the complex plane produced results like e^Xi= sinX +icosX

Like how is this a philsophy question? I don't even know. It seems cool until you think that complex numbers are just usually inputs in 1-dimensional functions, so who gives a fuck? Am I drunk?

by they way are the simpliest 1-dimesnional functions just fucking series ( like specify 1 3 5 7 9 ... ). I am going to shoot someone if this is right

seek help and a girlfriend you autistic fucking creep

Friend, we will win. If you are against me, you are against yourselt.

I don't understand Hitlerspeak, Hans.

Give everyone a gift and never post drunk again, bainlet.

being this much of an undergrad. next time you will claim probabilities must be positive

Why doesn't this make sense if its graphed in an imaginary coordinate plane, which it would have to be?

Why are you replying in Arabic?

But i is strictly larger than 0

Not all metrics are induced by norms.
Metric d is defined as a function with following properties
1) d(x,y)≥0 and d(x,y)= 0 iff x=y
2) d(x,y)+d(y,x)≥d(z,x)
You just gave one specific example of a metric

Drunk again. The only thing wrong is "function". I don't think these fit the defition. Series was also the wrong word. Pattern is better. It isn't 1+3+5+7 it is 1 3 5 7. And we shoudl be able to make any of those points carry more information with Dirac shit. I'm a brainlet, but I am smart as fuck at the same time.

The complex field is not ordinate like the real one.

>laminar flow
>obscure

What's your point ?

But it can be ordered, just like any other set, and assuming AC it can even be well-ordered

>the hypotenuse length is 0

This assumes a two dimensional plane, which is not a good assumption in ALL cases.

maybe there isn't really a triangle there

(i)(i*)+(1)(1*)=(sqrt(2))^2

>i isn't curved

proofwiki.org/wiki/Complex_Numbers_cannot_be_Totally_Ordered

eBbB

The title of that article is misleading, it says:
There exists no total ordering on (C,+,⋅) which is compatible with the structure of (C,+,⋅).

Keypoint being: compatible with the structure.

If i > 0 then i^2 = -1 > 0, no good. If i < 0 then i^2 > 0 which again is no good. Moreover, i is nonzero. Hence the complex numbers cannot be totally ordered

The answer is sqrt(2), try taking a course on complex analysis

t. someone who never took linear algebra

en.m.wikipedia.org/wiki/Well-ordering_theorem
>everysetcan bewell-ordered

>en.m.wikipedia.org/wiki/Well-ordering_theorem
How can this be true? The theorem says that every subset of a set must have a least element, but there are sets of the real numbers with no least element.

See >Keypoint being: compatible with the structure.
The Well-ordering theorem says there exists a well ordering. You're talking about a particular order on the reals, and there's no need for it to be a well-order.

I see. Thanks, user

the hypotenuse can't be shorter than either of the two remaining sides.