@@xipher5896 that's weird, it is sad that I'm not too familiar with the complex plane yet so I'm not the one to argue in that field but from what I now is that a triangle exists if and only if the sum of any two sides is greater than the third. Is it possible to explain what is this exception you're talking about in a way that I a noob in complex numbers can understand?
@@iMíccoli pfp sauce? [ i=√-1, i²=-1, √[(1)²+(i²)]=√[1-1]=√0=0.What you are agruing about is called the triangle inequality but that only works as long as the sides of the triangle are real. The usual notions of distance changes to Norm when switching to the Argand Plane]
@@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.
Před měsícem+4
@@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?
No, complex numbers are not vectors, strictly speaking. They are something different. They are a basic example of multivectors (a subgroup of a Clifford Algebra). So, it is incorrect to say that they are vectors, even if this is somehow a common mistake, used in applied math!
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.
I agree. Pythagoras theorem does not apply to complex numbers. This question is understandable in the context of the complex plane viewed as vectors in R². Vector analysis provides the (a) solution.
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.
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.
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.
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
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
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) czcams.com/video/9cOgkM0t9NA/video.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.
The Pythagorean Theorem is specific to right triangles in Euclidean space, which have non-negative real lengths. Using your application as illustrated in the figures, a triangle with a=1 & b=2 where a & b are vectors gives you the distance between them as c=1. A proper redefinition would be to state that a & b are magnitude values, while c is a scalar value: |a|² + |b|² = c² However, this is not a proven definition of the theorem. Again, the Pythagorean Theorem applies to length values, and neither -1 nor i are lengths. If the magnitudes are to be implied, it's from a & b describing vectors on a complex plane rather than lengths of the sides of a triangle in Euclidean space, and the notation used in the figure is the notation most commonly used to describe the latter instead of the former. It'd be like writing "6/2+1" and expecting the answer 2 because the slash represents a fraction bar. The notation is unclear, and that's the answer to the question implied by the figure.
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.
i doesn't even make sense to be the side length of a triangle because it isn't a magnitude strictly, it's 1 unit of distance along the "imaginary" direction of the complex plane, making it a magnitude AND a vector. Side length only takes a magnitude, so it's actually of length 1
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.
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.
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.
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.
The notation is ambiguous, if you assume you’re looking at the complex plane you get 2^0.5, but if the labels are for a triangle in a 4 dimensional, non Euclidean geometry where lines can have complex lengths you get 0.
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.
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
Or may we should be acknowledged: 1^2+i^2 is 0, no doubt, but this is not because of Pythagoras therom, just occasionally having the same formula. Pythagoras therom are for length of the edges of triangles, which are scalar values. Generally, only non-negative real numbers are allowed (although negative real numbers work as well). But in concept of geometry, triangles can always be represented by 3 vector values for their vertices if given an original point for reference (virtually all objects can be represented by vertices and edges, yet triangle have only 3 vertices therefore simple enough to have only one combination of edges connecting their vertices). In this case, Pythagoras therom is not practical anymore because we are calculating with vector points, not scalar lengths.
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
It depends on the metric of the plane we are working in. If the metric is positive definite (the usual situation in which we use the Pythagorean thereom), then the length of the segment cannot be i since the distance between two different points must be positive and real and i is neither. If the metric is indefinite, then distance and length don't strictly apply. However a more general concept, the modulus, can be defined. The modulus of that segment can be imaginary, and it works just as described (it squares to -1), and therefore of the hypotenuse, although we can't speak of its length, we can say its modulus is 0. A real physical example of this is a particle moving away from the origin at the speed of light - although it covers non-zero distance and non-zero time (in the origin's frame of reference), it spans a zero interval in spacetime.
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 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!)
If you treat the sides of the triangles as the vectors, then indeed it's sqrt(2). But if you say that the length of the side is i, you can't treat it as 1. That way the hypotenuse is 0. Simply the triangle can't exist as one of its sides has a complex size and the other 0.
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.
This explanation was my first thought But then I figured this is kinda like saying you have a rope with one end tied to a fence post here and now and the other end tied to a fence post in the future, then asking how long the rope has to be Just because you can write something down doesn't mean it has to mean something But hey, there's probably a formalized answer to this
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
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.
but what if they meant the line had length i this would mean they are going i units up going n units up is adding nxi ixi=-1 so what you have is a line going forward one unit and a line going backwards one unit so this isnt really a triangle anymore but x would then have to be 2
you didn't fix the problem you just changed the definition of i to be 1 i think a more generalized approach is: multiplying the side by i will cause it to rotate an extra 90 degrees, thus the right triangle will become a triangle with 180 degree angle but not sure how to show that this definition is useful
The question itself doesn't even make sense. Picture the real plane being flat (eg. desktop, screen) and the imaginary plane being above/below the real plane, so 1 would be 1 unit to the right of the origin *O,* and 1i would be1 unit above *O,* and -1i would be 1 unit below *O.* The *MAGNITUDE* of any of those would be 1, so "a length of i" makes zero sense. That would imply a vector, in this case a triangle with one side to the right of *O* and the other side above *O,* and you could have a side of -1 or -4 or a complex 8+6i or anything else. Works fine with vectors, but not "lengths of a triangle's sides". Simplify the problem only instead of a side with length i, make it -1. Still comes out to the correct root2, simply because it's all in the real plane, but "length of -1" makes zero sense, as that tries to equate a scalar with a vector. *Lengths must be scalars, not vectors.*
i = anything but the imaginary i that does not exist. x=sqr(1^1+i^2) wherein is just a letter like x not a special mathematical fantasy.And also x>=1 even if i is 0 x is still at least 1 and can never be 0. Your solution is nonsense. |i| is still an imaginary number and when you square it it is supposed to be -1. (Which I do not agree with, but that is another matter.) And if you insist it is only the length and not the i-ness of the number, you are forgetting that you have no way of knowing if one i unit equals 1 x unit. So therefor my solution is correct, i can be anything (except the mathematical "i") or if you wanna nitpick, the relation of i-units to x-units can be anything (even 0) and x is at least 1 and certainly not 0.
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))
"i" can't even be a side lenght. This question makes absolutely no sense when you just show it out of context. If you told me that the top vertex represents i in the complex plane, the right one represents one and the vertex with the right angle is 0, then the lenght is sqrt(2), otherwise the question itself is simply wrong
This is absolutely wrong. The side length is not a vector, it can be seen as the norm of a vector. By definition of norm, it has to be a nonnegative real number. Thus, a triangle having a side of length i simply doesn't make sense, just like a triangle having a side of length -1. Also, complex numbers are not vectors "by definition". You can construct a vector space having ℂ as its underlying set, but ℂ itself is not a vector space and thus its elements are not vectors.
If i is a vector that is perpendicular to 1, then the diagram is a lie and it’s actually a degenerate triangle where the right angle is actually no angle at all, making the length of the “hypotenuse” 0
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 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
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.
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.
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
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.
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.
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
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)
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.
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)
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
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.
x=0 doesn't even make sense anyway.
it does if you think about the formula: x²=1²+i²
since i²=-1 and 1²=1, 1+(-1)=x=0
@@UpdateFreak33x is the side length of the hypotenuse so it must be greater than 1 because it is the greater side of a right triangle.
@@iMíccoli yes this is true, however an exception could occur when using complex numbers.
@@xipher5896 that's weird, it is sad that I'm not too familiar with the complex plane yet so I'm not the one to argue in that field but from what I now is that a triangle exists if and only if the sum of any two sides is greater than the third. Is it possible to explain what is this exception you're talking about in a way that I a noob in complex numbers can understand?
@@iMíccoli pfp sauce? [ i=√-1, i²=-1, √[(1)²+(i²)]=√[1-1]=√0=0.What you are agruing about is called the triangle inequality but that only works as long as the sides of the triangle are real. The usual notions of distance changes to Norm when switching to the Argand Plane]
Nobody said the x is 0! Everybody was saying that x is 0
I see what you did there
Factorial joke
Haha
Factorial lol
You didnt need to explaing guys...
Conclussion: i = 1
i = 1 but verticle(y axis)
The length of i is one
yi = i
y = 1
@@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
Conclusion: Concussion
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.
I'm i meters away from getting a girlfriend
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.
@@longlivethe9989 the quote I took from a reply this channel made to a comment here, stating "magnitudes are always real numbers"
this.. is the correct counterargument
True.
The hypotenuse being zero is legal on a pseudo-Riemannian manifold. It shows up all the time in special relativity with light cones.
You miss the point, this is the space-time geometry. A photon has energy 1, momentum i and mass √(1²+i²)=0
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?
@@tm30shadowball37 I'm saying that the magnitude is i, not that the magnitude is equal to the modulus of i.
moduli are real numbers, this would mean that i is a real number in the set {0,1,2,3...}
No, complex numbers are not vectors, strictly speaking. They are something different. They are a basic example of multivectors (a subgroup of a Clifford Algebra). So, it is incorrect to say that they are vectors, even if this is somehow a common mistake, used in applied math!
Yes there os mistake in question itself i 1 and 0 are vertices of triangle not sides and mod of i 1 and √2 are sides
What do we mean by "being vectors" exactly? Is it not just C being a vector space over R?
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
@@addwithad and with this logic, what changes when one side is complex? It's not consistent across both examples
@@huhneat1076 |3|^2 + |4|^2 still yields |5|^2
this isn't correct because the length of the triangles side is i, it isn't a vector pointing to i
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
@@addwithad but it's a vector with magnitude i, while in the video you assume it's a vector with the same magnitude as i (so 1)
actually thinking about it as a vector is misleading as the sides of a triangle have no set direction, only magnitude
@@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.
@addwithad So x=(√2)i is a valid answer too, as its "displacement" is √2.
And what if we take (√3)i (i.s.o. i) and 1, what is x? Is it 2, (√2)i, √2?
The real issue here is everyone assuming without it being specified that the triangle is right angled and that pythagoras is applicable
Exactly, it is one of those poorly stated questions.
conclusion: x = √(2)*e^(iπ/4)
I agree. Pythagoras theorem does not apply to complex numbers.
This question is understandable in the context of the complex plane viewed as vectors in R².
Vector analysis provides the (a) solution.
@@coshy2748 it's cuz the notion of distance changes to norm.
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.
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.
underrated comment
(split complex numbers are incredible 😍😍😍)
Yeah, I'm pretty sure this is exactly how distances in Lorentzian space work.
@@Eta_Carinae__ Yep! true as well
by your logic, any complex number with a magnitude of sqrt(2) can be the length
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'm telling them that this is a real triangle that someone has thrown an i on, which is not the same
@@addwithad the dimensions are not meant to be to scale.
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
Coders seeing: " x is not 0! " ->x=0
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
Yes 😂
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)
czcams.com/video/9cOgkM0t9NA/video.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.)
Distance as defined by standard mathematics cannot be imaginary. 1,-1,i,--i all of them are distance 1.
@@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.
The Pythagorean Theorem is specific to right triangles in Euclidean space, which have non-negative real lengths.
Using your application as illustrated in the figures, a triangle with a=1 & b=2 where a & b are vectors gives you the distance between them as c=1. A proper redefinition would be to state that a & b are magnitude values, while c is a scalar value:
|a|² + |b|² = c²
However, this is not a proven definition of the theorem. Again, the Pythagorean Theorem applies to length values, and neither -1 nor i are lengths.
If the magnitudes are to be implied, it's from a & b describing vectors on a complex plane rather than lengths of the sides of a triangle in Euclidean space, and the notation used in the figure is the notation most commonly used to describe the latter instead of the former. It'd be like writing "6/2+1" and expecting the answer 2 because the slash represents a fraction bar. The notation is unclear, and that's the answer to the question implied by the figure.
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.
Okay, by that logic if there was 1 instead of i x would be equal to 0, since distance between 1 vector and itself is 0.
i doesn't even make sense to be the side length of a triangle because it isn't a magnitude strictly, it's 1 unit of distance along the "imaginary" direction of the complex plane, making it a magnitude AND a vector. Side length only takes a magnitude, so it's actually of length 1
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.
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.
Yes, it's a bad question though because if you look at the original, x is pointing the wrong direction to be 1+ 1i, it could also be -1 + 1i
A triangle cannot have an imaginary side 😂
Mathematicians sometimes forget if they should in search of the if they could
But why we think that uint should be in same length?
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.
@@addwithad great explanation
The confusion arises because most people don't realise that the triangle can only be drawn in the Complex plane, not the Euclidean plane.
well it's not 0! either
1:01 except in this diagram
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.
When the thumbnail is wrong but title is right:
Damn I wish I found that one sooner. I liked it a lot.
The notation is ambiguous, if you assume you’re looking at the complex plane you get 2^0.5, but if the labels are for a triangle in a 4 dimensional, non Euclidean geometry where lines can have complex lengths you get 0.
i read it as factorial at first😂
Its not ordinary geometry. Its hyperbolic one like in Minkowski diagram of special relativity where Pitagoras theorem is
a2-b2=c2. If a=b then c=0.
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.
How length can be negative please answer
Are you still doubting negative lenghts when he calmly adresses IMAGINARY lenghts??
Ask Gojo, he introduced negative lenghts and volume
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
"i" is neither positive nor negative on the "real number" number line. "i" can be thought of as positive on the "imaginary number" number line.
Or may we should be acknowledged:
1^2+i^2 is 0, no doubt, but this is not because of Pythagoras therom, just occasionally having the same formula. Pythagoras therom are for length of the edges of triangles, which are scalar values. Generally, only non-negative real numbers are allowed (although negative real numbers work as well).
But in concept of geometry, triangles can always be represented by 3 vector values for their vertices if given an original point for reference (virtually all objects can be represented by vertices and edges, yet triangle have only 3 vertices therefore simple enough to have only one combination of edges connecting their vertices). In this case, Pythagoras therom is not practical anymore because we are calculating with vector points, not scalar lengths.
Conclusion: x is 0, not 0 factorial.
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
if i is a number:
1^2+i^2=x^2
1+(-1)=x^2
1-1=x^2
0=x^2
x=sqrt(0)=0
if i is a variable:
1^2+i^2=x^2
1+i^2=x^2
x=sqrt(1+i^2)
It depends on the metric of the plane we are working in. If the metric is positive definite (the usual situation in which we use the Pythagorean thereom), then the length of the segment cannot be i since the distance between two different points must be positive and real and i is neither.
If the metric is indefinite, then distance and length don't strictly apply. However a more general concept, the modulus, can be defined. The modulus of that segment can be imaginary, and it works just as described (it squares to -1), and therefore of the hypotenuse, although we can't speak of its length, we can say its modulus is 0.
A real physical example of this is a particle moving away from the origin at the speed of light - although it covers non-zero distance and non-zero time (in the origin's frame of reference), it spans a zero interval in spacetime.
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 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!)
You're absolutely correct.
If you treat the sides of the triangles as the vectors, then indeed it's sqrt(2).
But if you say that the length of the side is i, you can't treat it as 1. That way the hypotenuse is 0. Simply the triangle can't exist as one of its sides has a complex size and the other 0.
Judging by how the vectors added up, wouldn't the real answer then be 1+i? As complex numbers can have a real and imaginary part at the same time
The length of the segment connecting the points (0,0) and (0,i) is 1. That's all!
1^2+1^2=2 x=2^0.5
X is actually the square root of rotating 180° and translating rightward one unit 🤓
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.
This explanation was my first thought
But then I figured this is kinda like saying you have a rope with one end tied to a fence post here and now and the other end tied to a fence post in the future, then asking how long the rope has to be
Just because you can write something down doesn't mean it has to mean something
But hey, there's probably a formalized answer to this
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
Lengths can only be positive integers and irrational numbers as length can never be negative or imaginary as it is a real life magnitude
i² = ±1
a² + b² = c²
1² + i² = c²
1 + ±1 = c²
c² = 0 or c² = 2
c = √2
i = sqrt(-1) so I don't know where you're getting the plus or minus from
@@bruhifysbackup What you have is called a complex.
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.
Technically, isn't the hypotenuse of a unit triangle of positive slope on the complex plane equal to: 1+i -- not sqrt2?
i is not a logical side length, and neither should it be interpreted as a vector. Essentially, the basis of the question is wrong.
actually, it IS 0!
hmm... |x| is sqrt2 anyway because we're plotting on the complex plane anyway?
but what if they meant the line had length i
this would mean they are going i units up going n units up is adding nxi ixi=-1 so what you have is a line going forward one unit and a line going backwards one unit so this isnt really a triangle anymore but x would then have to be 2
when i get real be like
you didn't fix the problem you just changed the definition of i to be 1
i think a more generalized approach is: multiplying the side by i will cause it to rotate an extra 90 degrees, thus the right triangle will become a triangle with 180 degree angle
but not sure how to show that this definition is useful
It could happen on a flat earth.
The question itself doesn't even make sense. Picture the real plane being flat (eg. desktop, screen) and the imaginary plane being above/below the real plane, so 1 would be 1 unit to the right of the origin *O,* and 1i would be1 unit above *O,* and -1i would be 1 unit below *O.*
The *MAGNITUDE* of any of those would be 1, so "a length of i" makes zero sense. That would imply a vector, in this case a triangle with one side to the right of *O* and the other side above *O,* and you could have a side of -1 or -4 or a complex 8+6i or anything else. Works fine with vectors, but not "lengths of a triangle's sides".
Simplify the problem only instead of a side with length i, make it -1. Still comes out to the correct root2, simply because it's all in the real plane, but "length of -1" makes zero sense, as that tries to equate a scalar with a vector.
*Lengths must be scalars, not vectors.*
Vectors says it’s 1+1i right?
i = anything but the imaginary i that does not exist. x=sqr(1^1+i^2) wherein is just a letter like x not a special mathematical fantasy.And also x>=1 even if i is 0 x is still at least 1 and can never be 0.
Your solution is nonsense. |i| is still an imaginary number and when you square it it is supposed to be -1. (Which I do not agree with, but that is another matter.) And if you insist it is only the length and not the i-ness of the number, you are forgetting that you have no way of knowing if one i unit equals 1 x unit. So therefor my solution is correct, i can be anything (except the mathematical "i") or if you wanna nitpick, the relation of i-units to x-units can be anything (even 0) and x is at least 1 and certainly not 0.
complex unit "i" is a vector, it cannot be a physical value sadly
Right,x is not 0!
It’s actually 0
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))
"i" can't even be a side lenght. This question makes absolutely no sense when you just show it out of context. If you told me that the top vertex represents i in the complex plane, the right one represents one and the vertex with the right angle is 0, then the lenght is sqrt(2), otherwise the question itself is simply wrong
The length of the long side of the triangle in the Argand diagram is |1 + i| = √(1² + 1²) = √2
Does an imaginary length even make sense?
This is absolutely wrong. The side length is not a vector, it can be seen as the norm of a vector. By definition of norm, it has to be a nonnegative real number. Thus, a triangle having a side of length i simply doesn't make sense, just like a triangle having a side of length -1.
Also, complex numbers are not vectors "by definition". You can construct a vector space having ℂ as its underlying set, but ℂ itself is not a vector space and thus its elements are not vectors.
If i is a vector that is perpendicular to 1, then the diagram is a lie and it’s actually a degenerate triangle where the right angle is actually no angle at all, making the length of the “hypotenuse” 0
Of course X is not Zero.
They're Maverick Hunters.
No, the vektor is x= -1+i 😂
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" ...
no, it's not zero factorial
Does complex length make any sense? Even negative length makes no sense!
Is it correct to use imaginary numbers to determine the length? I don't think so
Complex numbers are not vectors.
the thing is just that this isn't a right triangle
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
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.
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.
Wrong. Complex numbers are not vectors, they are different animals altogether.
ok but how tf do u have a length of i 💀
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
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.
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.
If x=0! or x=0, you are claiming that i = 1 which is obv not true
Question is wrong it shouldn't be sides they are values of its vertices i 1 and 0 and modulus of 1 i and ✓2 are sides
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
If x = 0 i = 1 and therefore x also = √2
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)
@@sperner9069 Even if it is possible for length to be imaginary it will violate the axiom distance(a,b)=0 if and only if a=b.
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.
Sure but the original problem has x pointing to the left, this is the wrong direction for 1 + i yet the side is labelled 1 not -1
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)
@@addwithad your confusing the complex plane with two orthogonal vectors of length 1 and i
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
Photons are like this, more or less.
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.