This triangle is impossible.
If the distance between B and C is 0, B and C are the same points. If that is the case, the distances between A and B and A and C must be the same.
However, i ≠ 1.
If you want it to be real (hehe) the triangle should be like this:
C | \ |i| | \ 0 | \ A---B |1|Drawing that on mobile was a pain.
As the other guy said, you cannot have imaginary distances.
Also, you can only use Pythagoras with triangles that have a 90° angle. Nothing in the meme says that there’s a 90° angle. As I see it, there are only 0° and 180° angles.
Goodbye, I have to attend other memes to ruin.
Context matters. In geometry i is a perfectly cromulent name for a real valued variable.
Oh shit, he used the word cromulent. Every one copy off this guy.
That wouldn’t be cromulent, would it?
Mad mobile drawing!!
As the other guy said, you cannot have imaginary distances.
Incorrect. There are complex valued metric spaces
And even if we assume real valued metrics, then i usually represents the unit vector (0,1) which has distance real 1.
That’s NOT a metric. That’s a measure. Two wholly different things.
It can be a pseudometric
That’s more related to a metric but it still can’t be complex valued and it’s still not a measure.
This is clearly meant to be a right triangle. And the distances between the points are the same (because the squares of the coordinate differences are the same), just the directions are different.
If you move 1 unit forward, turn the correct 90 degrees, and then move i units forward, you will end up back where you started.
You can’t have a distance in a “different direction”. That’s what the |x| is for, which is the modulus. If you rotate a triangle, the length of the sides don’t change.
The vector from one point to another in space has both a distance (magnitude) and a direction. Labeling the side with i only really makes sense if you say we’re looking at a vector of “i units that way”, and not at an assertion that these two points are a directionless i units apart. Then you’d have to break out the complex norms somebody mentioned.
Isnt it fine to assume a 90° angle its just that when u square side AC ur multiplying by i which also represents a rotation by 90° so u now nolonger have a triangle?
It’s not fine to assume a 90° angle. The distance between B and C is 0. Therefore the angle formed by AB and AC is 0°.
If the angle is 90°, then BC should be sqrt(2), not 0. Since the length of both sides is 1. sqrt(|i|2+|1|2) = sqrt(2).
So essentially what ur saying is. The imaginary and real arent 90° or pythagoras is only valid for real numbers?
1 • 1 + i • i = 1 + (-1) = 0 = 0 • 0
Pythagoras holds, provided there’s a 90° angle at A.
this is why it is still a theorem
But that’s not the definition of the absolut value, I.e. “distance” in complex numbers. That would be sqrt((1+i)(1-i)) = sqrt(2) Also the triangle inequality is also defined in complex numbers. This meme is advanced 4-4*2=0 Works only if you’re doing it wrong.
I get it, it’s projected on a comlplex sphere. B and C are the same point
please stop making it make sense
You didn’t really expect an imaginary triangle to behave like a real one, did you?
C and B have a wormhole between them
In the complex plane each of these vectors have magnitude 1 and the distance between them is square root of two as you would expect. In the real plane the imaginary part has a magnitude of zero and this is not a triangle but a line. No laws are broken here.
i = 1is the only logical choiceLiterally, this is one of those questions where they’re testing logic and your understanding that the figures aren’t necessarily representative of physical reality.
It gets worse once you start doing trig on it
I kept seeing this pop up recently, and I finally understand it: it’s an introductory problem in Lorentzian general relativity.
AB is a space-like line, while AC is a time-like line. Typically, we would write AC as having distance of 1, but with a metric such that squaring it would produce a negative result. However it’s similar to multiplying i to the value.
BC has a distance of 0, but a better way of naming this line would be that it has a null interval, meaning that light would travel following this line and experience no distance nor time going by.
I’m sure PBS Spacetime would explain all of this better than me. I just woke up and can’t bother searching for the correct words on my phone.
Maybe the problem is constructing a metric that makes this diagram true. Something like d(x,y) = | |x| - |y| | might work but I’m too lazy to check triangle inequality.
Triangle inequality for your metric follows directly from the triangle inequality for the Euclidean metric. However, you don’t need a metric for the Pythagorean Theorem, you need an inner product and, by definition, an inner product doesn’t allow non-real values.
Wish me luck for I’m doing trig test with radians (2 pi rad = 360 ?)
Radians are the objectivly better way to do angles tho. Just remeber π=180deg and ur right. Btw here is a another brain fuck the units radians/second is just Hz
Thank you for reminding me!
Btw, Radians/sec = Hz? What is this, physics?
Engineering unit maths. Cos angles are unitless so radians/second =1/second=Hz
The easiest way to think about it is that 1 full rotation (2*pi radians) in 1 second makes 1 Hz.
The number of rotations made in a second corresponds to Hz in the same way that the number of sine wave periods that fit in a second also represents Hz. This gif does a really good job of showing how rotation relates to sine/cosine waves, which just so happens to help visualize the rad/s -> Hz <- periods/s relationship:
Radians are the objectivly better way to do angles
Yes, and tau is objectively better than pi. Just remember tau = 360°. Which is a full circle, which easier to work with than half a circle.
You can make something like this properly by defining a different metric. For example with metric dl2 = dx2 - dy2 the vector (1, 1) has length 0, so you can make a “triangle” with sides of lengths 1, -1 and 0.
That’s not a metric. In any metric, distances are positive between distinct points and 0 between equal points
It depends which metric definition are you using. The one I wrote is a pseudo-Riemannian metric that is not positive defined.
Normally physicists use that generalized metric definition because spacetime in most cases has a metric signature of (-1, 1, 1, 1). Points with zero distance are not necessarily the same point, they just are in the same null geodesic.
You’re talking about a metric tensor on a pseudo-Riemannian manifold, I’m talking about a metric space. A metric in the sense of a metric space takes nonnegative real values. If you relax the condition that distinct points have nonzero distance, it’s a pseudometric.
This is true for real-valued metrics but not complex-valued metrics.
Metric, not measure. Metrics are real by definition.
funny Interpretation: in the complex plane, the imaginary axis is orthogonal to the real axis. so instead of the edge marked with i (AC), imagine an edge of length 1 orthogonal to that edge. It would be identical to AB, so
ACCB is 0.But then CB couldn’t also be 0; wouldn’t it be
cos(1 + i)? Or something like that.oh I mixed up the points, I meant to say CB is 0 in the end
I feel violated trying to read that in my brain.
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