[ 3 / biz / cgl / ck / diy / fa / g / ic / jp / lit / sci / tg / vr / vt ] [ index / top / reports / report a bug ] [ 4plebs / archived.moe / rbt ]

/vt/ is now archived.Become a Patron!

# /sci/ - Science & Math

[ Toggle deleted replies ]
File: 271 KB, 3000x2265, schizo_math.png [View same] [iqdb] [saucenao] [google] [report]

if you extend the Pythagoras theorem to complex numbers you obtain bullshit result a triangle with zero-length hypotenuse, maths are developed by schizos

can /sci/ explain this shit?

 >> Anonymous Sat Jan 23 19:22:22 2021 No.12620287 >>12620268Triangles do not have side numbers, triangles have side $\bold{lengths}$. Lengths are non-negative reals only.
 >> Anonymous Sat Jan 23 19:22:57 2021 No.12620293 >>12620268Incorrect metric. A metric defines how you measure lengths in a space and squaring complex numbers isn't how it's done. You multiply the number with it's conjugate (a+bi => a - bi) so i*(-i)=1, you get sqrt(2).
 >> Anonymous Sat Jan 23 20:28:11 2021 No.12620492 File: 71 KB, 620x675, 1603824455065.jpg [View same] [iqdb] [saucenao] [google] [report] $a = i$$b = 1$$c = 0$So:$A = \frac{ab}{2} = \frac{i}{2}$According to heron's formula, which is $\sqrt{s(s-a)(s-b)(s-c)}$ the area ends up $\frac{i}{2}$ again$a = b\cos\gamma+\sqrt{c^2-b^2\sin^2\gamma}$then$i = 1 * 0 + \sqrt{0 - 1 * 1}$$i = \sqrt{-1}$$c = \sqrt{a^2 + b^2 - 2ab\cos\gamma}$then$0 = \sqrt{-1 + 1 - 0}$$0 = 0$by definition$\tan\alpha = \frac{a}{b}$$\tan\alpha = i$but also$\tan\alpha = \frac{a\sin\gamma}{b-a\cos\gamma}$then$\tan\alpha = \frac{i * 1}{1 - i * 0}$$\tan\alpha = i$There are a couple of divisions by zero but since this is schizo general they're not illegal just undefined.Also the division by zero loses commutativity and associativity.
 >> Anonymous Sun Jan 24 02:43:39 2021 No.12621507 >>12620268This sort of triangle actually occurs in special relativity; the zero-length hypotenuse is what you call a "light-like interval" or "null interval" and it represents how something moving at the speed of light doesn't experience any time.
 >> Anonymous Sun Jan 24 02:49:11 2021 No.12621519 >>12621507god physicists are so fucking stupid
 >> Anonymous Sun Jan 24 02:51:22 2021 No.12621527 >>12621519Its just a function, not a direct observable. Spacetime distance is not space distance.
 >> Anonymous Sun Jan 24 02:52:31 2021 No.12621529 >>12620268Pythagorean theorem only applies on the euclidean metric, which is defined for real numbers. You can construct triangles with imaginary lengths, but Pythagorean theorem is no longer valid and now you need to use something like a Reimann metric to find the hypotenuse.
 >> Anonymous Sun Jan 24 02:54:16 2021 No.12621532 File: 126 KB, 1131x622, fuck math.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Sun Jan 24 02:55:56 2021 No.12621535 >>12621519Physicist here. That guy is an idiot. That triangle doesn't exist in SR. We define a Minkowski Metric. Then you can construct a triangle with legs of values t, and r. The spacetime interval is given by s^2 = t^2 - r^2. Diagrammatically, it's a meaningless triangle used by brainlets who don't understand math.
 >> Anonymous Sun Jan 24 03:00:04 2021 No.12621546 >>12621535>That guy is an idiot.>repeats exactly what I said in different words
 >> Anonymous Sun Jan 24 03:35:31 2021 No.12621610 >>12620492$\tan(ix) \to i$ as $x \to \infty$
 >> Anonymous Sun Jan 24 03:37:26 2021 No.12621614 >>12620268>can /sci/ explain this shit?not every quadratic form is positive definite
 >> Anonymous Sun Jan 24 03:39:53 2021 No.12621617 >>12621535Brainlet. Let (t, r) be (0, 0), (0, 1), and (1, 0) for three vertices of a triangle and compute s for all three sides.
 >> Anonymous Sun Jan 24 03:41:19 2021 No.12621620 >>12621535what the fuck are you smoking, that guy is 100% right and you haven't said anything different than him
 >> Anonymous Sun Jan 24 06:40:10 2021 No.12621894 File: 19 KB, 800x536, imaginenotknowingcomplex.png [View same] [iqdb] [saucenao] [google] [report] I was writing an elaborate response when realized I was just being jebaited by sum dumbass. Lemme just leave here a quate from Gauss about the subject and an image, both from Wikipedia."If one formerly contemplated this subject from a false point of view and therefore found a mysterious darkness, this is in large part attributable to clumsy terminology. Had one not called +1, −1, √−1 positive, negative, or imaginary (or even impossible) units, but instead, say, direct, inverse, or lateral units, then there could scarcely have been talk of such darkness." ~ Gauss (1831)Lastly fuck you OP.
 >> Anonymous Sun Jan 24 07:41:23 2021 No.12622016 File: 7 KB, 250x250, 1603321180172.jpg [View same] [iqdb] [saucenao] [google] [report] >>12620492let $r$ be the radius of the inscribed circle, $r = \sqrt{\frac{1}{s}(s-a)(s-b)(s-c)}$, where $r = \frac{1}{2}+\frac{i}{2}$and $s$ the semi-perimeter $s = \frac{a+b+c}{2}$, where $s = \frac{1}{2}+\frac{i}{2}$From the law of cotangents we have$\cot \frac{\alpha}{2} = \frac{s - a}{r}$So we get$\cot \frac{\alpha}{2} = \frac{\frac{1}{2}+\frac{i}{2} - i}{\frac{1}{2}+\frac{i}{2}}$$\cot \frac{\alpha}{2} = -i$Extending the law further:$\cot \frac{\beta}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 1}{\frac{1}{2}+\frac{i}{2}}$$\cot \frac{\beta}{2} = i$$\cot \frac{\gamma}{2} = \frac{\frac{1}{2}+\frac{i}{2} - 0}{\frac{1}{2}+\frac{i}{2}}$$\cot \frac{\gamma}{2} = 1$We can then conclude with the last identity which is$\frac{\cot \frac{\alpha}{2}}{s - a} = \frac{\cot \frac{\beta}{2}}{s - b} = \frac{\cot \frac{\gamma}{2}}{s - c} = \frac{1}{r}$which becomes$\frac{-i}{s - a} = \frac{i}{s - b} = \frac{1}{s - c} = \frac{1}{\frac{1}{2}+\frac{i}{2}}$which is proven to be true herehttps://www.wolframalpha.com/input/?i=%28-i%2F%28s-a%29%29%3D%28i%2F%28s-b%29%29%3D%281%2F%28s-c%29%29%3D%281%2F%281%2F2%2Bi%2F2%29%29+where+a%3Di%2C+b%3D1%2C+c%3D0%2C+s%3D%281%2F2%2Bi%2F2%29
 >> Anonymous Sun Jan 24 17:37:37 2021 No.12624014 As you know$\sin(ix) = i \sinh(x)$$\cos(ix) = \cosh(x)$$\tan(ix) = i \tanh(x)$$\sin(x+iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)$$\cos(x+iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)$$\tan(x+iy) = \displaystyle \frac{\tan(x) + i \tanh(y)}{1 - i \tan(x) \tanh(y)}$cos is finite for finite inputs, but it approaches complex infinity as the imaginary part of the angle approaches $\pm \infty$, with which direction it goes in the complex plane dependent on the real part of the angle. Similarly for sin.$\tan(x + iy) \to i$ as $y \to \infty$ and $\tan(x + iy) \to -i$ as $y \to -\infty$.We can't assign finite angle values to the two non-right angles of this triangle, but we do approach its trigonometric ratios when the angle opposite the side of length i has an imaginary part tending towards positive infinity and the angle opposite the side of length 1 has an imaginary part tending towards negative infinity.
>>