@@Garfield_Minecraft i ≠ "verticle one" "verticle" is just one of interpretations, but technically i = "such something whose square is -1" Does it really make sense with "verticle" so that "vertical" times "vertical" is "horizontal"? In the context of complex numbers yes. Generally-I don't think so
Missing the joke sooooo much 💀 i is the magnitude of the vector. "Vectors can't have imaginary magnitude" yes that's the joke. It doesn't make sense to say you're i meters away from something.
Who said they can't? You could define a normalising operation that maps your vector space to the imaginary axis specifically (albeit leaving the reals/"regular" complex numbers behind to achieve that). The real problem is that vector magnitudes within a vector space must be comparable, and 1 isn't comparable with i due to a lack of ordering.
11 дней назад
@@longlivethe9989 the quote I took from a reply this channel made to a comment here, stating "magnitudes are always real numbers"
I'm not sure if I agree with this. The diagram shows "i" as the value for the length of the side, not as a vector itself. If you treated the side like a vector, then the vector should have imaginary magnitude (i.e. magnitude = i), not a magnitude equalling 1 as you described in your explanation. I'm not sure if imagninary magnitude is a well-defined concept but if we were to extend mathematics to allow such a concept, it doesn't seem unreasonable that we could extend Pythagoras' theorem to allow us to claim 1^2 + i^2 = 0^2 as an answer to the problem.
Imaginary magnitude is a well-defined concept. If you want to use only the magnitude of the vector, then you are saying that one of the sides of the triangle is equal to the module of 𝔦 Module of complex number: z=a+b𝔦 ⇒ |z|=sqrt(a²+b²). In this case, when z=𝔦, a=0 and b=1 So the magnitude of the vector 𝔦 is equal to 1 Therefore the side x is equal to sqrt(2) There is no complex number with magnitude equal to 𝔦. Hope you can agree with the video now :)
Magnitudes are always real numbers so I'm afraid that I cannot agree, i on its own is just a displacement of 1 unit in the positive vertical direction of the complex plane. If you draw a line of length 1 in the complex plane and move or rotate it, it is still of length 1 since the complex plane is flat and the distance is defined in a particular way as mentioned in the other replies.
@@addwithad Let's say we tried to solve a similar problem, we are trying to add 2 vectors together. The first vector has a magnitude of 1 in the x-direction, the second vector has a magnitude of i in the y-direction. Would this problem have a solution?
The vector reasoning is incompatible with other right triangles. Why can't I look at a 3-4-5 triangle and say, well, 3 and 4 are vectors that go in the same direction, so the remaining side is just 1? You're assuming the triangle is drawn to scale, and supposing that the triangle actually does point in the imaginary direction. I'm not saying you're wrong with √2 but there's absolutely more nuance to what you did.
A triangle requires at least two dimensions, in the case of a 3,4,5 triangle these can both be real dimensions so it does not follow that 3,4 are in the same direction
For the length to be a complex number, it must be in the complex plane, on the complex plane numbers are defined by their displacement from the origin and displacement is a vector. An example of displacement is driving to work/school, you go in a certain direction and a certain distance(same as length), this is a vector since it has a direction and a magnitude, the magnitude being the distance
@@hiccupwarrior89 The trouble is that it's a poorly posed question, that's why people re-post it so much. If we take it as an axiom that it is a right angle triangle then clearly the hypotenuse x is not less than the base 1, and therefore cannot be 0. Treating i as a real number in this case and squaring it contradicts this. I think the only sensible solution is to take i in its well defined setting as being a displacement in the complex plane rather than length, if you put points in a plane then you can define the displacement relative to the origin. I'm more than happy to discuss it if there's a method which produces a different real-valued length for x greater than or equal to 1.
As far as I know, that is only assuming that the triangle is in the complex plane and not an actual triangle. Though I guess that isn't too far fetched as an actual triangle with a side length of i would be impossible.
Yeah I think the problem is that this length x can only be interpreted in the complex plane, otherwise this is not a triangle since i=0 on the real number line. But many peoples' instincts are to apply Pythagoras' theorem as if i is a real number and overlook the extra subtlties
I think that it boils down to how one interprets a mathematical diagram. When one first looks at the diagram, he or she is probably inclined to think that he or she is looking at a triangle and will assume that it is a right one and therefore use the Pythagoras theorem to try to find x. If there were axis in the background, he or she might think about complex numbers and vectors if he or she had studied those things before. The other problem is that one can't assume that that triangle is a right one since there is no indicator that any of the triangle's angles are right angles. The indicator that is usually used is a small square inside the right angle. Therefore, x could be any of many values.
magnitudes can be imaginary in the split complex numbers (a+bj, j^2 = 1) as abs(a+bj) = sqrt(a^2-b^2), for example abs(j) = i you can imagine this diagram as the split complex number 1+j, which indeed has a magnitude of 0.
This triangle doesn't exist in normal Euclidean space, it probably exists in a weird geometry where points can have complex distances Edit: I realized that complex distances might not be possible because of how distance metrics work, but I heard about geometries with real dimensions and imaginary dimensions from a Wikipedia article and a youtube video (I know they're not the best sources, sorry) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-9cOgkM0t9NA.html en.wikipedia.org/wiki/Complex_polytope (WARNING: This article is a headache, I don't understand 99% of it, I just put it here to show that these weird geometries might exist somehow.)
@@kazedcat exactly, its like saying you own -2 apples, it just doesnt make sense, the same way you cant have negative cardinality you cant have complex distance it just doesnt apply.
Actually you can in Minkowski space and other pseudoeuclidean spaces. That is the case where the meme about zero-length hypotenuse is actually true. The framework of special relativity is built on such spaces and the light in our universe is basically moving in zero-"length" trajectories (to be specific, those "lengths" in Minkowski space are called spacetime intervals)
@@watermagle minkowski space is not built on some sort of imaginary spacetime thats ridiculous, minkowski space is a 4th dimensional real vector space so first get your facts correct then come at me.
The initial formulation of Minkowski space was based on Poincare's observation that time could be understood as fourth spatial dimension, but with an imaginary coefficient. Those formulations are actually isomorphic. The reason why noone uses the "imaginary-time' formulation of special relativity is because it cannot be generalised on curved spaces.
Mathematicians seeing : "x is not 0!" -> x is not 1 -> x < 1 v x > 1 Notice: 0! = 1, v - or, i think everyone know it, but if not, here it is I'm not English speaker, so sry for errors
X is 0 What you forgot in the video is that the second side is perpendicular to the first, so you have to multiply it by i (think of a triangle of sides 1 and 1, and how it would be represented in the complex plane; which also explains the case 1, -1, sqrt2). So you have a side being 1+0i, and other being i*(0+1i)=-1+0i. Adding both vectors get you to the origin with modulus 0, as expected.
I think the real interpretation of i in this case would be rotating the right angle of the triangle by 90 degrees and then getting that the sides overlap making the distance x = 0
Good question, the complex plane is a flat space like a regular graph so the distance function is the same, it is Pythagoras' theorem. If we have a co-ordinate (x,y) on this graph which represents a complex number (x+yi), we find the length of the line between this point and the origin using root(|x|^2+|y|^2). If the number is i, x=0 and y=1 so the expression simplifies to 1 Usually the | | modulus symbols aren't used for Pythagoras' theorem when we learn it in school because we work with real positive numbers not vectors, but in more advanced maths it is important to use them. i should be considered as a unit of direction, not a unit of length. It is like a 90 degree rotation. When we say i, we really mean something more like 1i, trying to separate i from 1 and asking its length is like separating the minus sign from 1 and asking what the length of minus is, it doesn't have one it is just a direction.
I do believe you’re supposed to transform the i vector 90 degrees before the calculation, else the way you’re doing it, a right triangle with both sides 1 will have a hypotenuse 0
I am wondering if we can use physics to examine that example. We need something that has "i" value in one direction and in 90° 1 Value in the other and then check if the power or whatever we use annihilates itself. Would be interesting
It is far easier to explain this by explaining that any unit of measurement is arbitrary. Thus, if the side (1) is measured as one yard, it can also be measured as three feet. Or 36 inches. Or 914.4 millimeters. Etc. Now go ahead and do the calculation on any one of those measurements, and I guarantee the answer will not be zero.
The triangle exists only when the summ of any two sides of it is greater the the third one. With complex numbers comparison doesn't exist → the triangle doesn't exist. Of course if we ignore that the length of the side can be only a real number.
Hi, thanks for your question Displacement is a vector and can be negative depending on where you define the origin point. Length is the size/magnitude of the displacement and doesn't depend on direction. Since complex numbers are defined in the complex vector space we can evaluate the magnitude of these vectors in the complex number space using |z| = √(|x|^2+ |y|^2) where z is a complex number of the form z = x+yi You can also see this property by considering what happens when you square a complex number: |x+yi|^2 = (x+yi)*(x-yi)= x^2 +y^2
The video mentions vectors that have magnitude and direction (eg. degrees) and then abandons the idea. The correct answer is sqrt2 with an angle. The vectors need to be drawn on a vector diagram to define the angle associated with x.
You made a wrong assumption. The side is already perpendicular, so since i is a vector, it too would be perpendicular but from the given side. Therefore, assuming we stay within the 2D space of the screen, i of the perpendicular line would no longer be perpendicular, but at the other end making x zero. If i were negative (-i), the rotation would be the opposite direction, like you mentioned it was a vector. Therefore, the formula is actually true, even for complex numbers. The issue is that you assumed the perpendicular line was made perpendicular by i. In reality, it does not make much sense having a sidelength of i just as having a geometric length of -1 wouldn’t make much sense either.
Hi I'm trying to understand your comment a little better so that I can respond, are you saying that the 2D space of the triangle is a real projection of the complex plane? I'm suggesting that the shape itself must exist in a 2D complex plane z=x+yi so that it can be aligned with the triangle and the lengths solved.
@addwithad the triangle is just in a regular 2D space as most geometry works. Thus a sidelength of i would be a rotation of the given side. Now, if you assume that the number i is clarifying the direction of the side in complex space, then what you did was right, but that depends on if you assume that this triangle is drawn in complex space and its not a triangle of side lengths of 1 and i. So basically, I guess both answers can be correct depending on the assumptions made of the nature of the triangle. I assumed it to be a normal geometric triangle with an abnormal vector side length. You assumed the vector side length to define the rightness of the triangle.
@addwithad though, if you wanted to go a step further with my assumption, if i is perpendicular to the given side, we could go to 3D space and it would have an entire circle of solutions ranging from 0
@@markstavros7505 I spent a painful amount of time trying to comprehend you; but tell me, did you mean that the triangle is not actually a right triangle and so could have any solutions ranging from 0 to 2. I think you meant that if the triangle is in the complex plane, then í is perpendicular so it's a right triangle, but if i is in the real plane, then it's value is equal to one and it's not given that the triangle is a right. I don't know trigonometry so I couldn't understand the last statement.
@c.jishnu378 if we assume that we can go into 3D space after the projection of the I vector, then yes, since technically no coordinates are given. The direction of the sidelength 1 is unknown
I remember hearing this idea that complex numbers are vectors, but I couldn't buy it. It's like saying that C = R^2. Learning about the history of vector analysis, the vectors we're familiar with came _after_ when C and H were defined, and were constructed specifically to avoid the isomorphism with the special unitary group. If anything, complex numbers and quaternions are probably best thought of as an object that encodes in multiplication, what vectors encode in _both_ dots and crosses. But just because you can ultimately do the same work with them, doesn't make them the same thing.
This problem is wrong since the beginning because it is a geometry problem, and the size of the side of a polygon cannot be an imaginary number. Even if we imagine that it could be, in a hypotetical world, then the correct answer will be zero.
When I saw the video I thought I would be dissatisfied, if the answer would be sqrt 2. So, I am. The answer is 0 if the catheti are just perpendicular lines, not just vectors on the complex plane. We just need to extend the classic geometry to allow something like this.
It's very interesting point of view... Your way for me looks very close to möbius strip topology. Triangle is the basis with "i"-number to build möbius strip... Is It posible to visualise "your way"in 3-d form?.. with spherical diagonal line as "x"π and ortogonal oriented triangle sides "i" and "1" ...
I learned from a video that imaginary numbers came about because there was no way to describe negative area. As in, if a square's area is -1, it's side length would be i. All I'm saying, is that all this magnitude talk, to me sounds like undermining a legitimite abstract idea to potentially be learned more about.
I don’t agree, on the grounds that: If the side that is Measured as “i” was actually imaginary, it would be at a right angle to the plane if the screen, and therefore wouldn’t be visible to us, as I side effect the hypotenuse would also be hidden, and would be essentially 0 to us. Though an argument could be made that the measurement of the hypotenuse could also be -1+i
Undefined is not equal to zero. Yes if you interpret the problem geometrically side length i do not exist but that means x= undefined. x will not be zero it will be undefined.
You’re interpreting and answering the question as if the triangle is on the complex plane, but I don’t think that’s what the question is asking. Even though the question doesn’t make sense on a geometrical plane (because there can’t be a length of i), you can’t just convert the question to something else because it makes more sense.
The question did not specify that it is in a geometric plane. You are free to interpret it however you like. Even if it is in a geometric plane distance i do not exist so x= undefined therefore x is not 0.
@@kazedcat Yes, what I meant was that I think the intent of the question was to refer to a right triangle oriented in any direction with leg “lengths” i and 1, and not a right triangle with a specific orientation in the complex plane. I agree that the answer isn’t 0 in either case (unless it’s possible somehow for a length to be imaginary)
Why is x necessaring the square root of two? Now that we are using complex numbers, what's to stop it from being a complex number with a magnitude sqrt2?
If x is the complex number 1 + i, it's pointing in the wrong direction on the question. Sure, it could be one of several complex numbers with magnitude sqrt(2)
Ridiculous question, i is not a length so we cannot have a triangle with a side of i. What you have worked out is based on a side of |i| instead which is quite a different thing.
came here expecting a nice little essay about what it might mean for a distance to be a complex number. instead i got a blatant misunderstanding of the complex plane. the geometry of the complex plane is not in any way linked with the geometry of the plane the right triangle sits in. the complex plane is just a visualization tool. people are telling you that the quantity i represents the distance of that side of the triangle. you are telling them that that is invalid. this is true, but the entire premise of the hypothetical is to ask, "what if it were valid? what kind of results would we get?" and you are refusing to entertain the notion.
I have to disagree. Puting aside other issues, for your solution to make sense, one must interpret the “1” and the “i” assigned to the sides of the triangle as vectors in the complex plane, but interpret “x” as a length. Otherwise, the answer would be 1+i. Sqrt(2) is just a vector in the same direction as 1. If we agree that the problem is poorly posed, then any solution will be problematic.
This is kind of odd because the original question states i as a distance. This is wrong in a practical sense, but also in a mathematical sense. What you’ve done here is placed a line from 0 to 1, then from 1 to 1 + i. A line going from 0 to 1 is just that, it’s a line of distance 1. A line going from 1 to 1 + i is also a line of distance 1. Effectively, you’ve simply drawn a 1 by 1 triangle in the *complex* plane. This is not what the original question was implying. The triangle in the complex plane is theoretical. The question states a ‘real’ triangle of side length i. Indeed, it is true that you can’t apply pythagoras theorem to it, because a square with side length i does not have a positive area, but you also can’t chart it on the complex plane. The point of the question is how imaginary numbers can break otherwise practical concepts.
There are a lot of problems with the original question for example that it isn't a triangle if x = 0, another is that the line x is drawn leaning left like the vector -1 + i, but the side is labelled i not -1
wouldn’t x just be the vector 1+i if you’re going to interpret it that way? you can say the magnitude of x is sqrt(2) but the question is asking for x and not the magnitude.
So... Now we measure length (as scalar) with vectors? The triangle sides already have a direction of their own. The conclusion I'm coming to (as a youtube commenter, which is all the qualification I base this on) is that you can't use complex numbers as lengths, making the original problem invalid. Yet I'm sure somebody, somewhere, actually makes use of imaginary numbers as lengths to perform engineering somewhere.
This is Brent's Triangle. It's from a viewer of a maths channel, I believe its Blackpenredpen. It's a meme, but also a very cool complex shape. This guy here is misinterpreting the joke. The triangle lengths are i, 1 and 0. These are not vectors.
This doesn’t seem like a satisfying answer. If you had a right triangle with sides 3, 4, and x, x does not equal 1, even though laying out 3 and 4 on the number line, that is the distance between them. Similarly, if you had a triangle with sides i and 1, and an angle between them of 20, your solution here does not apply. In fact, i is not a length or a vector. It’s imaginary. And this is an imaginary triangle. Because i’s only function is to extend the natural numbers to include sqrt(-1), it stands to reason that all functions should work the exact same with i as with any other number It seems like you’ve essentially just invented a new triangle with sides 1 and 1 instead of engaging with the i-ness of the problem (Don’t mean to sound rude here, sorry if I did!)
I'm probably wrong, but I don't agree with this. i is not a vector, it's a complex number. You can represent a complex number on a plane with two axes, but that's just a representation. It's a bit like saying the number 6 looks like an upside down 9. Pythagoras's theorem says you square the number and the square of i is -1 by definition. The entire thing is a bit of a nonsense. A real triangle can no more have a side of length i than it can have a length of -1 but if you want to follow Pythagoras and the definition of complex numbers then I say the "answer" is 0. Take your answer and work back. Your triangle has hypotenuse of sqrt(2) and one side has length 1. Let's say the other side has length x. So x^2 +1^2 = 2. Therefore x^2 =2-1 = 1. i is not a solution for x so there is no triangle with sides of length 1, i, and sqrt(2) as you're saying.
0:31 i was told that i=(-1) so I would aproach it like this a^2 + b^2 = c^2 1^2 + i^2 = c^2. Put in the nums 1 + (-1)^2 = x^2. i=(-1) 1 + 1 = x^2 x = root of 2 I'm no math guy so if I did something wrong explain in simple terms
Not bad, but i is the square root of -1 which means i × i = -1 1^2 + i^2 = c^2 1 + (-1) = c^2 0 = c^2 0 = c But the other thing which is mentioned in the video is that when calculating distance is that a, b and c are distances which means they all need to be either 0 or positive numbers like on a ruler, so we use the lengths of each part. The length of i is 1
All I saw in those not well spent 2 minutes is: - I don't understand this, I'll just solve something else and say I'm right Bro, the *length* of the side is i not the position of the point
The moduli are implied since a,b,c are real numbers in the original theorem when it was used for applications like building structures. If you see the link below, you don't even need to put i into the theorem, only the number multiplying it, which is just 1: en.wikipedia.org/wiki/Pythagorean_theorem#Complex_numbers
Distances are always positive real numbers. End of story. :D Therefore the original right triangle is simply mislabeled. i cannot be a length, but its modulus 1 can.
I must disagree with both answers. The video is correct - i is not a length. But if i and 1 are vectors, not lengths, than x must also be considered as a vector, not a length. So adding up the perimeter vectors clockwise, you get i + x - 1 = 0 (back at the starting point). Hence x = 1 - i (which has length of sqrt(2), but is not sqrt(2))
Me reading the comments pretending to understand (I only clicked on this video because I want to study maths but am still in 9th grade so I didn't even know what i was until fairly recently):
It's all about the "imaginary" number i, numbers which have i in them are known as complex numbers, maybe you can understand some basics from the wiki page :) en.wikipedia.org/wiki/Complex_number
Whoever said x=0 is like dude how'd you think that way? I knew the length just by looking at it💀 Since i= square root of -1 So x= square root of [(-1)² + 1²] ; (-1)² can be simplified as +1 ; 1² is also +1 So x= root of (1+1) which is the equal to square root of 2
Это ложь, которая стоит на подмене понятий. Sqrt(2) это расстояние между точками на комплексной плоскости а не длина гипотенузы. Ты единое пространство разбил на два одномерных. Позор!
Good question. The short answer is that it is a poorly defined question and I decided to find the lengths of the triangle's sides since the triangle is in different orientations in different memes. If you look at the title page, it would actually be i - 1 in that orientation but we don't know which ways the vectors are pointing, only that there is a triangle with the same lengths.
@@addwithadirregardless of the orientation of the triangle, you have given a complex representation (i) for the one side and a length representation for the other side (sqrt(2)) simultaneously, i.e. mixing the units of measurement. Either (1, i, 1+i) or (1, 1, sqrt(2))
I would say that the units are mixed in the ambiguous question, I've given them side lengths (|i|, |1|, |1+i|) = (1, 1, sqrt(2)) to make sense of the question.
@@addwithad by adding the '|' symbols, you discard the ability of the sides to be complex-valued, which is the sense of the meme. (|i| = 1 is a real number) The fun was in imagining lengths themselves having complex values (like, there is something S, whose length is complex |S| = i)
They're equal algebraically, I think this is about how I've written the distance function just as being Pythagoras' theorem since there is only one distance. The complex distance is defined better as |z|= sqrt( x^2 + y^2) where z is a complex number z = x + yi, we have x = 1, y = 1 since z = 1+i
@@Lex_rGd_128 í is not a n.o, it is a direction. í is defined to be the midway of + and -, it is not like an x-y plane where we are considering different "variables" with their each "units", the complex plane is basically an expanded; just as real as a number line. So every length in a complex plane is a real n.o, unlike an x-y plane which can have both x and y or only x or y in any length.
But .... Pythagoras doesn't say that |h|^2= |a|^2 +|b|^2. At least, we were never taught it with those absolute values. Are you saying that there's no length we can associate with i? Why not? Why are lengths immune from imaginary values? Where's that rule of nature? You can have imaginary time. Imaginary numbers of sheep. Imaginary numbers of RU-vid videos. But we're not allowed to contemplate triangles with imaginary lengths?
I think the triangle is malformed. The depiction suggests we are to treat the imaginary unit as a distance, which doesn't make sense as distances are Euclidean idealizations.
I dont understand why people find it unreasonable that the length is 0, but are okay with the fact that the other length is i. In the problem one side is with length 1 and the other with length i, which is not defined because length can only be positive and real. That's when the solution should stop. There is no way to find the length of a side of a triangle that's undefined.
You're right, but that isn't what he said. Measuring a length L from a point A to a point B is the same than measuring from B to A. Hence L = |A - B| = |B - A| f : z --> |z| goes from C to R+ which means a length of i would be undefined. That way, you rather should say that |2i| > |i| which, this time is true.
@@andy_lamaxyou can definitely compare the magnitude of complex numbers. Look at the first reply C to R+ means the function takes in a complex number and gives out a positive real number
