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/sci/ - Science & Math

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Anonymous Wed Aug 14 07:32:00 2019 No.10887102 [Reply] [Original]

Easy question.

>>	Anonymous Wed Aug 14 07:32:58 2019 No.10887103 >>10887102 2/3

>>	Anonymous Wed Aug 14 07:43:55 2019 No.10887115 File: 72 KB, 1080x1020, 1543836533346.jpg [View same] [iqdb] [saucenao] [google] >>10887102 >nononooNOONONONONO MY PHD NOOOOOOOOOOOO STOOOOOOOP please please PLEASE DONT SAY ZERO PLEASE SAY SQRT(2) PLEASEEEE COMPLEX NOT ORDERED UGHHH AAAAAHHHHHH

>>	Anonymous Wed Aug 14 07:48:43 2019 No.10887118 >>10887102 1.4

>>	Anonymous Wed Aug 14 07:59:26 2019 No.10887136 c = 1 + i

>>	Anonymous Wed Aug 14 08:05:48 2019 No.10887145 >>10887115 hAhA hE poStUd le EBIN fAcE mAymAy aGaiN rofbl MATH LIBSHITS BTFOOOOOO

>>	Anonymous Wed Aug 14 08:41:36 2019 No.10887193 >>10887102 sqrt(2), obviously "The combined area of the squares on the legs of a right triangle is equal to the area square on the hypotenuse." In this case, the combined area is 2.

>>	Anonymous Wed Aug 14 08:43:05 2019 No.10887195 >>10887102 -1/12

>>	Anonymous Wed Aug 14 08:57:26 2019 No.10887211 >>10887195 This

>>	Anonymous Wed Aug 14 08:59:21 2019 No.10887215 >>10887193 Hello, non-math person here, how can the length of a side of a triangle be complex?

>>	Anonymous Wed Aug 14 09:02:08 2019 No.10887219 >>10887215 it can’t, this is one troll anon trying to force his meme >inb4 spacetime intervals no

>>	Anonymous Wed Aug 14 09:08:14 2019 No.10887226 >>10887215 length is never complex, that's the whole point. the length of the side labelled i is actually 1.

Anonymous Wed Aug 14 09:08:26 2019 No.10887229

>>10887219
Oh I just realized the entire triangle is meant to represent a complex number and C is a modulus. That brings an interesting question, can we assume those factors just going by the picture? Surely it needs to be defined as complex space for the answer sqrt 2 to be valid?

>>	Anonymous Wed Aug 14 09:23:47 2019 No.10887268 >>10887193 the area of the square on the i leg is -1.

>>	Anonymous Wed Aug 14 09:36:11 2019 No.10887298 >>10887215 It can't be, length can't be complex. Kind of. All real numbers belong to the set of complex numbers. All complex numbers have a real component. That length of i has a real component of 1, so its length is one.

>>	Anonymous Wed Aug 14 09:39:32 2019 No.10887301 Zero? Sqrt(1^2+i^2)=sqrt(1-1)=0 right?

>>	Anonymous Wed Aug 14 09:42:37 2019 No.10887304 >>10887195 kek

>>	Anonymous Wed Aug 14 09:43:39 2019 No.10887308 >>10887268 wrong

>>	Anonymous Wed Aug 14 13:12:52 2019 No.10887702 File: 34 KB, 1180x793, memetrianglesolution.png [View same] [iqdb] [saucenao] [google] >>10887102 I solved it for you

>>	Anonymous Wed Aug 14 13:51:43 2019 No.10887791 great, but u could have used paint

>>	Anonymous Wed Aug 14 15:07:41 2019 No.10887934 File: 38 KB, 1180x793, revisededition.png [View same] [iqdb] [saucenao] [google] >>10887702 I'm retarded, pic related is the actual solution

>>	Anonymous Wed Aug 14 15:10:37 2019 No.10887939 File: 11 KB, 639x364, idontgetit.png [View same] [iqdb] [saucenao] [google] >>10887102 why is this wrong? no bully pls

>>	Anonymous Wed Aug 14 15:30:10 2019 No.10887965 >>10887939 Because length is by definition a real number. This is an ill-posed "problem" meant to trick retards that only plug shit into formulas without understanding them, aka (You).

>>	Anonymous Wed Aug 14 15:36:00 2019 No.10887973 >>10887102 Doesn't matter i or 1, because it's a triangle. triangles can only be flat. So it's just the absolute value of each side: sqrt(1^2+1^2) = sqrt(2) = ~1.4

>>	Anonymous Wed Aug 14 15:53:08 2019 No.10887996 >>10887229 >Surely it needs to be defined as complex space for the answer sqrt 2 to be valid? It is already defined as a complex space by having imaginary numbers. The length i is actually 1, but expressed in i. The hypotenuse is sqrt2

>>	Anonymous Wed Aug 14 15:58:03 2019 No.10888003 it doesn't exist. "i" is to geometry what a number divided by zero is to aritmethics

>>	Anonymous Wed Aug 14 16:11:44 2019 No.10888031 >>10887934 Well... at least you're right about you being retarded.

>>	Anonymous Wed Aug 14 23:28:24 2019 No.10889054 >>10887934 the second square's area is 1 the squares you have built are the actual square around the triangle, not the squares that are supposed to represent squaring

>>	Anonymous Wed Aug 14 23:34:15 2019 No.10889070 >>10887102 If you somehow extend the meaning of length to include nonreals, you'd have to come up with a new way of measuring linear distance between points. The pythagorean theorem is only defined for the euclidean plane. tl;dr c=your mom

>>	Anonymous Thu Aug 15 06:13:29 2019 No.10889589 >>10887102 C=0 obviously

>>	Anonymous Thu Aug 15 06:36:06 2019 No.10889604 Are you all retarded? It's written that C = ?.

>>	Anonymous Thu Aug 15 08:56:21 2019 No.10889730 >>10889604 Pretending to be retarded is still retarded.

>>	Anonymous Thu Aug 15 09:07:01 2019 No.10889745 >>10887102 i isn't a quantity or a number, it's a symbol. same goes for pi, sqrt2 etc

>>	Anonymous Thu Aug 15 09:10:26 2019 No.10889748 File: 188 KB, 1920x1080, 1564349726813.jpg [View same] [iqdb] [saucenao] [google] >>10889745 >same goes for pi, sqrt2 etc

>>	Anonymous Thu Aug 15 09:12:59 2019 No.10889752 >>10887102 sqrt(1/2) + sqrt(1/2)*i

>>	Anonymous Thu Aug 15 09:29:50 2019 No.10889783 >>10889745 /thread

Anonymous Thu Aug 15 09:31:51 2019 No.10889788

>>10887102
whether the answer is [math]\sqrt{2}[/math] or 0 depends on your interpretation of the question. 0 IS a valid answer if you think of the complex argument as a rotation in [math]\mathbb{R}^{2}[/math], though there are a number of reasons why that isn't considered acceptable - one of which is displayed right here because it implies the triangle is actually a flat line, which is why the hypotenuse equals 0.

>>10887965
trivial and p-adic norms btfo

>>	Anonymous Thu Aug 15 09:35:34 2019 No.10889795 sqrt(i^2 + 1)

>>	Anonymous Thu Aug 15 10:29:02 2019 No.10889859 >>10887102 >I will assign as length of this edge something literally called 'imaginary number' >hehe I am so smart math ppl btfo

>>	Anonymous Thu Aug 15 12:53:26 2019 No.10890073 File: 15 KB, 480x360, hqdefault.jpg [View same] [iqdb] [saucenao] [google] >>10887298 >kind of you know you're not retarded enough to say all real numbers are complex so shove that "ACSHULLY" up your ass sperglord

Anonymous Thu Aug 15 21:04:46 2019 No.10891322
File: 5 KB, 548x82, wtf.gif [View same] [iqdb] [saucenao] [google]

https://www.wolframalpha.com/input/?i=ipi+as+continued+fraction

What's going on here? This is just the computational engine running out of bits without doing the full symbolic expansion, right?

Can we construct a continued fraction that unpacks according to [math]0=1 + e^{i\pi}[/math]?

>>	Anonymous Thu Aug 15 22:00:47 2019 No.10891559 0?

>>	Anonymous Thu Aug 15 22:03:57 2019 No.10891569 >>10887102 Sqrt(2)

>>	Anonymous Thu Aug 15 22:08:22 2019 No.10891592 Does nobody know how to take a complex modulus?

Anonymous Fri Aug 16 02:29:06 2019 No.10892131
File: 2.89 MB, 262x300, 1408390925944.gif [View same] [iqdb] [saucenao] [google]

>>10887965
I don't remember a lot from vector calculus but can't you plot imaginary and real values simultaneously and use the vector magnitude formula to find the length of the hypotenuse?

>>	Anonymous Fri Aug 16 02:58:25 2019 No.10892174 >>10887102 Can't you just put a protractor on your screen to measure the angle between c and the side with length 1, then just divide 1 by the cosine of that angle to find c?

>>	Anonymous Fri Aug 16 03:06:04 2019 No.10892188 Man, fuck this rabbit hole. https://www.youtube.com/watch?v=NaL_Cb42WyY I'm pretty sure there are infinite ways to express pi now.

>>	Anonymous Fri Aug 16 03:23:07 2019 No.10892218 >>10892188 read a book.

>>	Anonymous Fri Aug 16 03:38:10 2019 No.10892245 >>10887102 C=0 it's a triangle with only 2 angles and each angle is 90 degrees

>>	Anonymous Fri Aug 16 03:40:45 2019 No.10892249 >>10887115 God what a retard lmao

>>	Anonymous Fri Aug 16 04:16:14 2019 No.10892300 You guys do realize that 1 and i are numbers, not lines in the complex plane, right? c is a set of complex numbers, not any one particular number. But root2 and 0 are not in c.

>>	Anonymous Fri Aug 16 05:22:47 2019 No.10892407 >>10887102 A^2+B^2=C^2 1^2+1^2=2^2 To find C: √2=1,4=C

>>	Anonymous Fri Aug 16 05:39:25 2019 No.10892430 >>10887102 C=((i^2)+1)^(1/2)

>>	Anonymous Fri Aug 16 06:17:33 2019 No.10892473 Are people really so retarded they need complex numbers to solve this.

>>	Anonymous Fri Aug 16 06:21:58 2019 No.10892476 >>10892473 Are you so retarded that you don't realise this is a troll thread?

>>	Anonymous Fri Aug 16 11:35:55 2019 No.10892900 >>10887102 ans = sqrt(abs(i)^2 + abs(1)^2) //ans = 1

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