/sci/ - Science & Math » Thread #10830877

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The Imaginary Triangle

Anonymous Tue 23 Jul 19:19:21 2019 No.10830877 View Reply Original Report

Quoted By: >>10830885 >>10830893 >>10830902 >>10830921 >>10831029 >>10831052 >>10831053 >>10831520 >>10832882 >>10832906 >>10832940

So considering all of this:
>>10827994
>>10829333
>>10829418
>>10830299
>>10830428
>>10830477
>>10830832

It turns out that there is no inconsistency with this meme triangle, which means that it should exist. The fact that it exists then proceeds to imply that division by zero is possible (as it is also required for it to exist) and results into its trigonometric functions.
Can some math nerd actually find an inconsistency that disproves the triangle without reverting to autistic screeching about absolute values?

Anonymous

Anonymous Tue 23 Jul 2019 19:22:20 No.10830885 Report

Quoted By: >>10830913

>>10830877
It is a real triangle, in Minkowski space. The length 0 edge is a lightlike vector.

Anonymous

Anonymous Tue 23 Jul 2019 19:24:52 No.10830893 Report

Quoted By: >>10830913

>>10830877
>>10830886

Anonymous

Anonymous Tue 23 Jul 2019 19:28:06 No.10830902 Report

Quoted By: >>10830913

>>10830877
>The fact that it exists then proceeds to imply that division by zero is possible
Nope. You can have division by zero (see point at infinity, one-point compactification, Alexandroff extension) but this triangle has nothing to do with it.

Anonymous

Anonymous Tue 23 Jul 2019 19:30:18 No.10830913 Report

Quoted By: >>10833234 >>10833241 >>10833246

>>10830885
>>10830893
Sure it has an application in physics, but the division by zero part for its sine/cosine is what worries me because it's actually consistent and has huge implications about pretty much everything. Even in physics, that would mean that the density of the singularity, that you derive by dividing by 0 volume, isn't infinite but rather some weird undiscovered number which would imply a new dimension perhaps?

>>10830902
But what about this then >>10830428
How would you calculate the sine and cosine when you have to divide by a zero hypotenuse

Anonymous

Anonymous Tue 23 Jul 2019 19:33:47 No.10830921 Report

Quoted By:

>>10830877
I don't know what a non-real length means. Metric space theory requires that the metric has codomain [0,\infty).

Anonymous

Anonymous Tue 23 Jul 2019 20:21:21 No.10831029 Report

Quoted By: >>10831038 >>10831044 >>10834141

>>10830877
length is a real number so no.

Anonymous

Anonymous Tue 23 Jul 2019 20:24:55 No.10831038 Report

Quoted By: >>10831070

>>10831029
Only in euclidean geometry. There are more geometric possibilities besides euclidean you know. And there's a real possibility the biggest scale objects such as the universe itself has a non-euclidean shape so it has real life value as well.

Anonymous

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Anonymous Tue 23 Jul 2019 20:29:47 No.10831044 Report

Quoted By: >>10831070

>>10831029

Anonymous

Anonymous Tue 23 Jul 2019 20:34:34 No.10831052 Report

Quoted By: >>10831059

>>10830877
0 =/= sqrt(2)

Anonymous

Anonymous Tue 23 Jul 2019 20:35:13 No.10831053 Report

Quoted By:

>>10830877
This is what's known as a Wick rotation. By making one axis imaginary, you're effectively turning Euclidean space into Minkowski space

Anonymous

Anonymous Tue 23 Jul 2019 20:38:39 No.10831059 Report

Quoted By: >>10831913

>>10831052
>sqrt(2)
Anon, i^2 is -1. -1+1 = 0.
It's sqrt(0), not sqrt(2).

Anonymous

Anonymous Tue 23 Jul 2019 20:45:37 No.10831070 Report

Quoted By: >>10831485

>>10831038
>>10831044
>Muh abstract math
Why does it matter if the lines curve and twist if the distance between two points is 0, nothing?

Anonymous

Anonymous Tue 23 Jul 2019 23:58:50 No.10831485 Report

Quoted By:

>>10831070
>Why does it matter if the lines curve and twist if the distance between two points is 0, nothing?
Because their trigonometric functions start behaving weirdly and do shit like tan(x) = i, or sin(x) = i/0

Anonymous

Anonymous Wed 24 Jul 2019 00:16:51 No.10831520 Report

Quoted By: >>10831524 >>10831527

>>10830877
I can't tell if this is bait or not anymore, that triangle does not exists

Anonymous

Anonymous Wed 24 Jul 2019 00:19:06 No.10831524 Report

Quoted By: >>10831532

>>10831520
It's not bait, you're just low IQ and have no awareness of massive swaths of mathematics in general. Maybe try reading a book.

Anonymous

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Anonymous Wed 24 Jul 2019 00:19:38 No.10831527 Report

Quoted By: >>10831539

>>10831520
>that triangle does not exists

Anonymous

Anonymous Wed 24 Jul 2019 00:20:41 No.10831532 Report

Quoted By: >>10831537 >>10831554

>>10831524
Pythagorean theorem only holds in Euclidean spaces where there are not complex valued lenghts

Anonymous

Anonymous Wed 24 Jul 2019 00:21:55 No.10831537 Report

Quoted By:

>>10831532
By standard definition to avoid specifically this. Roots only held for positive numbers as well some time ago.

Anonymous

Anonymous Wed 24 Jul 2019 00:22:36 No.10831539 Report

Quoted By: >>10831545

>>10831527
ok, now i know this is bait

Anonymous

Anonymous Wed 24 Jul 2019 00:24:14 No.10831545 Report

Quoted By: >>10831565

>>10831539
Disprove the consistency between the trigonometric functions. Or the fact that even integrating for the area of the triangle still gives the exact same result as the ancient formulas do

Anonymous

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Anonymous Wed 24 Jul 2019 00:25:00 No.10831548 Report

Quoted By: >>10831566 >>10831587

imaginary time is just a trick though. this doesn't really appear in physics. lightlike geodesics have 0 spacetime interval really because the metric has (-+++) signature and not because time is imaginary. that only appears when you want to make life easier with wick rotations or whatnot, but imaginary time stuff is purely a shortcut. you don't get complex numbers in the metric tensor, unlike in quantum mechanics where you have legit complex components of eigenstates. the i is just there to rewrite the signature.

that memeyness of taking the i seriously from the imaginary time trick is exactly what one of the bogdanovs wrote their thesis on. i'm not shitting you, that is true. their theory was about how the signature was somehow quantum and the metric somehow rotated in the complex plane from euclidean (++++) signature to normal (-+++) signature. so basically, thinking this way is bog level, literally. look it up.

Anonymous

Anonymous Wed 24 Jul 2019 00:27:12 No.10831554 Report

Quoted By: >>10831576

>>10831532
>Pythagorean theorem
Is just a special case for the law of cosines you brainlet.

Anonymous

Anonymous Wed 24 Jul 2019 00:31:13 No.10831565 Report

Quoted By: >>10831568 >>10831577

>>10831545
I don't care if there is consistency with trigonometric functions, if you want this fucking triangle fulfil the Pythagorean theorem
then it MUST be in the euclidean space which it clearly doesn't. Ther's your proof, it doesnt matter with how many formulas it is consistent with, it simply can't exisit in an euclidean space

Anonymous

Anonymous Wed 24 Jul 2019 00:32:03 No.10831566 Report

Quoted By:

>>10831548
kek its actually real
https://en.wikipedia.org/wiki/Bogdanov_affair

Anonymous

Anonymous Wed 24 Jul 2019 00:33:20 No.10831568 Report

Quoted By: >>10831576

>>10831565
Pythagorean theorem is a special case for law of cosines you retard. Law of cosines can be extended to non-Euclidean geometry.

Anonymous

Anonymous Wed 24 Jul 2019 00:38:01 No.10831576 Report

Quoted By:

>>10831554
>>10831568
including non metric spaces?

Anonymous

Anonymous Wed 24 Jul 2019 00:38:09 No.10831577 Report

Quoted By: >>10831589

>>10831565
>I don't care that it fits everything perfectly, our definitions (that we ourselves put to specifically avoid cases like this) forbid imaginary lengths so therefor it doesn't exist and our arbitrary rules should triumph over the natural functionality of every trigonometric function complimenting the rest in perfect sync

Anonymous

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Anonymous Wed 24 Jul 2019 00:45:30 No.10831587 Report

Quoted By: >>10831592

>>10831548
>because the metric has (-+++) signature and not because time is imaginary.
I think you are getting the horse and the buggy backwards there. The reason (dx_0)^2 has a minus sign in the matrix representation of the metric is because dx_0 is an imaginary quantity. The metric takes the form
$g_{\mu\nu}=\sum_{\mu} \sum_{\nu} dx_\mu dx_\nu$
and the matrix representation having a diagonal signature is not fundamental at all. The minus sign comes from
$dx_0=ic dt$

cool hat, btw.

Anonymous

Anonymous Wed 24 Jul 2019 00:46:06 No.10831589 Report

Quoted By: >>10831590

>>10831577
>let me disregard this one arbitrary rule and abide and use this other arbitrary rules to prove my point

Anonymous

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Anonymous Wed 24 Jul 2019 00:48:40 No.10831590 Report

Quoted By: >>10831607 >>10831627

>>10831589
>the trigonometric functions are an arbitrary rule

Anonymous

Anonymous Wed 24 Jul 2019 00:51:24 No.10831592 Report

Quoted By: >>10833153

>>10831587
no jon, you are the one who has it backwards. first of all, i never said anything about a diagonal structure for the metric. and the metric is not a matrix, it's a tensor, smartypants. and writing down an equation and stating something that basically represents the notation associated with the imaginary time trick doesn't elevate it from being a trick. the metric is a real thing. that is obvious in general relativity -- it basically is the field through which physical gravitational waves propagate. how you choose to write it down your signature convention is not

here's another example, in some schools of relativity the signature is actually (+---). that's a convention. you could just as well do some trickery to make space imaginary but keep time real. in other words, notational conventions don't mean shit

Anonymous

Anonymous Wed 24 Jul 2019 00:58:47 No.10831607 Report

Quoted By: >>10831636

>>10831590
pls tell me you are not thinking about trigonometry on the plane

Anonymous

Anonymous Wed 24 Jul 2019 01:12:42 No.10831627 Report

Quoted By: >>10831636 >>10833234

>>10831590
>trigonometry
Oh yes, im sure have a side with length 0 will not fuck up things

Anonymous

Anonymous Wed 24 Jul 2019 01:19:38 No.10831636 Report

Quoted By: >>10831670

>>10831607
Please disprove any of the initial posts

>>10831627
>im sure have a side with length 0 will not fuck up things
That's exactly the point - it actually doesn't and everything holds together, in the same way that having a negative number under the root didn't fuck math up and complemented it

Anonymous

Anonymous Wed 24 Jul 2019 01:37:17 No.10831670 Report

Quoted By: >>10831705 >>10833234

>>10831636
>Please disprove any of the initial posts

What about every fucking sine and cosine would involve dividing by 0

Anonymous

Anonymous Wed 24 Jul 2019 01:49:49 No.10831705 Report

Quoted By: >>10831748

>>10831670
Did you read the posts? Division by zero within that framework ends up producing either sine or cosine and that remains the exact same definition throughout every single formula reliant upon sine/cosine. The only indeterminate part is the angle itself that produces those functions, i.e the angles around the hypotenuse

Anonymous

Anonymous Wed 24 Jul 2019 02:11:39 No.10831748 Report

Quoted By: >>10831756

>>10831705
if you allow contradictions then you can prove everything you want, so go ahead

Anonymous

Anonymous Wed 24 Jul 2019 02:16:40 No.10831756 Report

Quoted By: >>10831840

>>10831748
I knew that this was about to be your reply so I ran most of the formulas for a triangle with sides of (i, 0, 0), (i, -1, 0), (i, 1, 1) and (i, 1, sqrt(2)) - every single one of them produced inconsistencies everywhere, with (i, 1, 0) being the only triangle that cannot produce an inconsistency.

Anonymous

Anonymous Wed 24 Jul 2019 03:12:48 No.10831840 Report

Quoted By:

>>10831756
>(i, 0, 0), (i, -1, 0), (i, 1, 1) and (i, 1, sqrt(2))
because non of them are fucking triangles

>with (i, 1, 0) being the only triangle that cannot produce an inconsistency
except of course division by zero

Im done, have fun

Anonymous

Anonymous Wed 24 Jul 2019 03:37:41 No.10831913 Report

Quoted By: >>10833040 >>10833160

>>10831059
The modulus of 1+i is still sqrt(2). Imagine a point at (1,i) in the Argand plane. Calculate its absolute distance to the origin. Realise you just calculated the hypothenuse of that 'imaginary triangle'.

Anonymous

Anonymous Wed 24 Jul 2019 14:17:39 No.10832862 Report

Quoted By: >>10832865

0^2 - 1^2 = i

Anonymous

Anonymous Wed 24 Jul 2019 14:20:13 No.10832865 Report

Quoted By:

>>10832862
It equals $i^2$ which is true, no contradiction, anon.

Anonymous

Anonymous Wed 24 Jul 2019 14:28:13 No.10832882 Report

Quoted By: >>10833237

>>10830877
>there is no inconsistency with this meme triangle
>which means it should exist
No it doesn't, get some better scientific understanding

Anonymous

Anonymous Wed 24 Jul 2019 14:44:49 No.10832906 Report

Quoted By:

>>10830877
Instead of considering this triangle, why wouldn't you first consider (i ; 2 ; sqrt(3)) for example ? You wouldn't have to divide by zero and you'd be able to see if imaginary lengths make sense.

Then, try to see how these evolve when the third side tends towards zero.

That would be a much more sensible approach

Anonymous

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Anonymous Wed 24 Jul 2019 14:46:33 No.10832908 Report

Quoted By: >>10833331

Anonymous

Anonymous Wed 24 Jul 2019 15:21:28 No.10832940 Report

Quoted By:

>>10830877
You are treating complex numbers as if they were real, real numbers are a poin in the real line while complex numbers are points in a plane, multiplication in the complex plane is extansion/contration and rotation, and the sum of complex numbers if vector addition.

i^2 + 1^2 = (0,1)^2 + (1,0)^2 = (-1,0) + (1,0) = (0,0)

there's nothing strange there, you are just thinking about complex numbers the wrong way

Anonymous

Anonymous Wed 24 Jul 2019 16:31:20 No.10833040 Report

Quoted By:

>>10831913
It's a presupposition of that conclusion that the nature of the hypotenuse acts in accordance with the rules of an argand space

Anonymous

Anonymous Wed 24 Jul 2019 17:49:56 No.10833153 Report

Quoted By:

>>10831592
You can embed spacetime in a complex space and have dx_0 = idt. This is every bit as real as the signature of the metric, it says literally the same thing. That's the whole point of the Wick rotation, it's just a different way of saying the same thing, like the Fourier transform.

Anonymous

Anonymous Wed 24 Jul 2019 17:52:56 No.10833160 Report

Quoted By: >>10833267

>>10831913
Only if you take the modulus first. If you work in the geometry where you don't, the triangle in OP is correct. As has been pointed out already in this thread, that corresponds to working in Minkowski space.

Anonymous

Anonymous Wed 24 Jul 2019 18:31:22 No.10833234 Report

Quoted By: >>10833259

>>10830913
>>10831627
>>10831670
The fact that some formulas yield division by zero in certain cases just means such formulas don’t apply in those cases. It doesn’t mean you’ve broken math.

Consider the “formula” x/x = 1. If x=0, you get division by 0. You haven’t made math inconsistent or anything like that, it’s just that the formula doesn’t apply at x=0.

For a more closely related example, consider a right-angled triangle. Then the length of one of the legs is given by a = H sin t where H is the length of the hypotenuse and sin t is the sine of the angle between this hypotenuse and the other leg. You can rearrange this formula to get the length oft he hypotenuse in terms of the height of the leg: H = a / sin t. Now what happens as t goes to 0?

Anonymous

Anonymous Wed 24 Jul 2019 18:32:27 No.10833237 Report

Quoted By:

>>10832882
>math
>scientific understanding
Yikes!

Anonymous

Anonymous Wed 24 Jul 2019 18:34:34 No.10833241 Report

Quoted By:

>>10830913
>but the division by zero part for its sine/cosine is what worries me because it's actually consistent and has huge implications about pretty much everything
what the fuck. no it doesn’t you retard.

Anonymous

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Anonymous Wed 24 Jul 2019 18:36:18 No.10833246 Report

Quoted By: >>10833338

>>10830913
>Even in physics, that would mean that the density of the singularity, that you derive by dividing by 0 volume, isn't infinite but rather some weird undiscovered number which would imply a new dimension perhaps?

Anonymous

Anonymous Wed 24 Jul 2019 18:45:09 No.10833259 Report

Quoted By:

>>10833234
>The fact that some formulas yield division by zero in certain cases just means such formulas don’t apply in those cases.
Exactly, having a triangle with length 0 is one of such cases, so even considering calculating sine and cosine is reterded

Anonymous

Anonymous Wed 24 Jul 2019 18:49:47 No.10833267 Report

Quoted By: >>10833356

>>10833160
>If you work in the geometry where you don't
which one is that ?

Anonymous

Anonymous Wed 24 Jul 2019 19:20:38 No.10833331 Report

Quoted By:

>>10832908
this is the most beautiful triangle in all of mathematics

Anonymous

Anonymous Wed 24 Jul 2019 19:26:34 No.10833338 Report

Quoted By: >>10833342

>>10833246
i love the funny inverted brain pictures they crack me up

Anonymous

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Anonymous Wed 24 Jul 2019 19:28:53 No.10833342 Report

Quoted By: >>10833348

>>10833338
>i love the funny inverted brain pictures they crack me up

Anonymous

Anonymous Wed 24 Jul 2019 19:31:00 No.10833348 Report

Quoted By:

>>10833342
what a cool way to respond without much effort good thinking anon

Anonymous

Anonymous Wed 24 Jul 2019 19:33:22 No.10833356 Report

Quoted By: >>10834103

>>10833267
Minkowski geometry. The invariant line-element is given by $ds^2 = -x_0^2+x_1^2+x_2^2+x_3^2$ . Massless particles i.e. photons travel along geodesics where $ds^2 = 0$ , at least in flat space.

Anonymous

Anonymous Wed 24 Jul 2019 23:30:12 No.10834103 Report

Quoted By:

>>10833356
>at least in flat space
Not just flat space.

Anonymous

Anonymous Wed 24 Jul 2019 23:38:54 No.10834141 Report

Quoted By:

>>10831029
This

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