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.
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.
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.
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.
It gets worse once you start doing trig on it
I feel violated trying to read that in my brain.
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.
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Uhh 21?
