x is not 0! (viral maths meme)

[e-mail] for maths tutoring enquiries

[@iMíccoli]

x=0 doesn't even make sense anyway.

[@UpdateFreak33]

it does if you think about the formula: x²=1²+i² since i²=-1 and 1²=1, 1+(-1)=x=0

[@iMíccoli]

​@@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.

[@xipher5896]

@@iMíccoli yes this is true, however an exception could occur when using complex numbers.

[@iMíccoli]

​@@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?

[@primenumberbuster404]

​@@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]

[@salehsattouf2320]

Nobody said the x is 0! Everybody was saying that x is 0

[@kage2479]

I see what you did there

[@jevilsugoma1743]

Factorial joke

[@alexterra2626]

Haha

[@ARandomGamer-nx1kv]

Factorial lol

[@pedropiata648]

You didnt need to explaing guys...

[@claudi917]

Conclussion: i = 1

[@Garfield_Minecraft]

i = 1 but verticle(y axis)

[@itripleo5780]

The length of i is one

[@addwithad]

yi = i y = 1

[@adammizaushev]

@@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

[@aeugh4200]

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.

[@netanelkomm5636]

I'm i meters away from getting a girlfriend

[@longlivethe9989]

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"

[@Fire_Axus]

this.. is the correct counterargument

[@iMíccoli]

True.

[@huhneat1076]

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.

[@addwithad]

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

[@huhneat1076]

@@addwithad and with this logic, what changes when one side is complex? It's not consistent across both examples

[@keithmokry8066]

The hypotenuse being zero is legal on a pseudo-Riemannian manifold. It shows up all the time in special relativity with light cones.

[@shehannanayakkara4162]

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.

[@tm30shadowball37]

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 :)

[@addwithad]

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.

[@shehannanayakkara4162]

@@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?

[@shehannanayakkara4162]

@@tm30shadowball37 I'm saying that the magnitude is i, not that the magnitude is equal to the modulus of i.

[@addwithad]

moduli are real numbers, this would mean that i is a real number in the set {0,1,2,3...}

[@shophaune2298]

The real issue here is everyone assuming without it being specified that the triangle is right angled and that pythagoras is applicable

[@iMíccoli]

Exactly, it is one of those poorly stated questions.

[@hiccupwarrior89]

this isn't correct because the length of the triangles side is i, it isn't a vector pointing to i

[@addwithad]

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]

@@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)

[@hiccupwarrior89]

actually thinking about it as a vector is misleading as the sides of a triangle have no set direction, only magnitude

[@addwithad]

@@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.

[@faming1144]

@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?

[@pavelgorokhov2976]

You miss the point, this is the space-time geometry. A photon has energy 1, momentum i and mass √(1²+i²)=0

[@danigarcia2294]

conclusion: x = √(2)*e^(iπ/4)

[@coshy2748]

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.

[@primenumberbuster404]

​@@coshy2748 it's cuz the notion of distance changes to norm.

[@AlessioAlessi]

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!

[@muriloporfirio7853]

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.

[@glitchy9613]

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.

[@muriloporfirio7853]

underrated comment (split complex numbers are incredible 😍😍😍)

[@Eta_Carinae__]

Yeah, I'm pretty sure this is exactly how distances in Lorentzian space work.

[@glitchy9613]

@@Eta_Carinae__ Yep! true as well

[@oskarjanson4858]

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.

[@addwithad]

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

[@MrSeezero]

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.

[@AsgharH238]

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) kzbin.info/www/bejne/b5SymJ6DZdlshKM 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]

Distance as defined by standard mathematics cannot be imaginary. 1,-1,i,--i all of them are distance 1.

[@user-rx5dh4le5x]

@@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.

[@watermagle]

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)

[@user-rx5dh4le5x]

@@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.

[@watermagle]

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.

[@Fire_Axus]

by your logic, any complex number with a magnitude of sqrt(2) can be the length

[@vlad3mirx689]

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.

[@addwithad]

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

[@eyalbryan9708]

Coders seeing: " x is not 0! " ->x=0

[@enumberfan]

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

[@eyalbryan9708]

Yes 😂

[@KrasBadan]

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.

[@jasonnong3305]

i read it as factorial at first😂

[@random_Person347]

The confusion arises because most people don't realise that the triangle can only be drawn in the Complex plane, not the Euclidean plane.

[@GameJam230]

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

[@Eta_Carinae__]

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.

[@fotatata6964]

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

[@DewageDonBodhiUthpalaAlwis]

But why we think that uint should be in same length?

[@addwithad]

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.

[@kornelviktor6985]

​@@addwithad great explanation

[@ricijs-tavsvietejaisgejars2364]

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

[@SennGorSan]

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.

[@Gk2003m]

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.

[@1_1bman]

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.

[@addwithad]

I'm telling them that this is a real triangle that someone has thrown an i on, which is not the same

[@1_1bman]

@@addwithad the dimensions are not meant to be to scale.

[@KazmirRunik]

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.

[@aisawaloki1571]

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.

[@markstavros7505]

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.

[@addwithad]

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.

[@markstavros7505]

@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.

[@markstavros7505]

@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

[@c.jishnu378]

​@@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.

[@markstavros7505]

@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

[@colon-Thorn]

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

[@davidbrisbane7206]

The length of the long side of the triangle in the Argand diagram is |1 + i| = √(1² + 1²) = √2

[@mathslove51]

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.

[@SURok695]

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.

[@Alex-5d-space]

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" ...

[@TheMathManProfundities]

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.

[@MrJuliancarroll]

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.

[@guyejumz6936]

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.

[@acompletelyawesomenameyay2587]

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

[@kazedcat]

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.

[@davidsousaRJ]

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.

[@c.jishnu378]

When the thumbnail is wrong but title is right:

[@solemnwaltz]

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

[@bat127]

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.

[@alex_ramjiawan]

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.

[@user-wx6oi4fm7z]

How length can be negative please answer

[@synd9554]

Are you still doubting negative lenghts when he calmly adresses IMAGINARY lenghts??

[@HarisRehmanGG]

Ask Gojo, he introduced negative lenghts and volume

[@addwithad]

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

[@jamesharmon4994]

"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.

[@axospyeyes281]

the thing is just that this isn't a right triangle

[@Lokalgott]

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

[@bable6314]

Photons are like this, more or less.

[@razi_man]

Of course X is not Zero. They're Maverick Hunters.

[@Aristotle000001]

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.

[@sperner9069]

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.

[@kazedcat]

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.

[@sperner9069]

@@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)

[@kazedcat]

@@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.

[@hihihihilll1227]

how are you sure it is a right angled triangle

[@shruggzdastr8-facedclown]

Technically, isn't the hypotenuse of a unit triangle of positive slope on the complex plane equal to: 1+i -- not sqrt2?

[@Garfield_Minecraft]

when i get real be like

[@matthewgilbie4087]

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!)

[@bable6314]

You're absolutely correct.

[@DavyCDiamondback]

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?

[@addwithad]

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)

[@DavyCDiamondback]

@@addwithad your confusing the complex plane with two orthogonal vectors of length 1 and i

[@Himasthla]

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

[@svcjunior5526]

Is it correct to use imaginary numbers to determine the length? I don't think so

[@activetutorial]

Complex numbers are not vectors.

[@DanDart]

hmm... |x| is sqrt2 anyway because we're plotting on the complex plane anyway?

[@supernoob7422]

Does an imaginary length even make sense?

[@A.V.F.P]

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

[@crochou8173]

Yeah...tell me about su2

[@_John_Sean_Walker]

i² = ±1 a² + b² = c² 1² + i² = c² 1 + ±1 = c² c² = 0 or c² = 2 c = √2

[@bruhifysbackup]

i = sqrt(-1) so I don't know where you're getting the plus or minus from

[@_John_Sean_Walker]

@@bruhifysbackup What you have is called a complex.

[@EastBurningRed]

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.

[@addwithad]

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

[@doobu8249]

no, it's not zero factorial

[@SkylineGmd]

Right,x is not 0! It’s actually 0

[@CyCloNeReactorCore]

actually, it IS 0!

[@uplink-on-yt]

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.

[@wildfire_]

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.

[@addwithad]

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

[@ilplolthereturn7525]

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):

[@addwithad]

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

[@ilplolthereturn7525]

@@addwithad i know that now

[@WookieRookie]

Does complex length make any sense? Even negative length makes no sense!

[@z4zuse]

1:01 except in this diagram

[@ValidatingUsername]

Vectors says it’s 1+1i right?

[@benjfr5723]

|i|^2 + |1|^2 = |-i*sqrt(2)|^2. Therefore x = -i*sqrt(2)

[@addwithad]

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

[@second_sheep690]

do you have discord server

[@addwithad]

no, sorry

[@kaynight64]

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))

[@rockstarpunkthegamer2631]

Sqrt(2)

[@pro_faitex___5153]

Если вектор тянется от 0 до i то длинна вектора равна 1 так что решение не верно вы не правильно поняли условие

[@ultrio325]

Or, yknow, write the question clearly next time

[@52soccerstar]

Eulers theorem could help

[@addwithad]

re^(θi) = r[cos(θ) +i*sin(θ)] Let: rcos(θ) = 1 r*i*sin(θ) = i θ = cos^-1(1/r) θ = sin^-1(1/r) cos^-1(1/r) = sin^-1(1/r) Solution r=root(2)

[@Bruh-bk6yo]

Get it? Cuz the Pythagorean theorem states that ||a||²+||b||²=||c||²?

[@addwithad]

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

[@Bruh-bk6yo]

@@addwithad ||x|| means a norm.

[@user-yx4jh6gi5n]

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!

[@rockstarpunkthegamer2631]

Hahahahhaahhahauahaua

[@Atrix256]

x = ε

[@TheSheep1]

0!=1

[@FishSticker]

Why would you say the answer is root 2 and not -1+i

[@addwithad]

The side is labelled 1 not -i so it's problematic

[@iminediamonds]

obviously x isn't 0! 0! = 1

[@Lex_rGd_128]

So, what makes us think that x will be a real number? 🤔

[@c.jishnu378]

The length of x is a real n.o, length is always a real n.o.

[@c.jishnu378]

Teh direction is complex.

[@Lex_rGd_128]

@@c.jishnu378 At the same time, the length of one of the legs of our triangle is equal to i... Doesn't one contradict the another?

[@c.jishnu378]

@@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.

[@c.jishnu378]

@@Lex_rGd_128 because it "contradicts", we take the absolute value to make it real.

[@bobbyheffley4955]

X=sqrt 2

[@aepokkvulpex]

X is actually the square root of rotating 180° and translating rightward one unit 🤓

[@adammizaushev]

If one of the sides is i, why is x=sqrt(2) rather than 1+i? It's either (1, i, 1+i) or, if you take the magnitudes, (1, 1, sqrt(2))

[@addwithad]

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.

[@adammizaushev]

@@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))

[@addwithad]

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.

[@adammizaushev]

@@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)

[@kazedcat]

​@@adammizaushevimaginary lengths are inconsistent with the 3 axioms of distance.

[@A.V.F.P]

By your wrong calculation the answer should have being 1+1i

[@alex_ramjiawan]

Nope.

[@lucascaldasdecarvalhoferre5757]

Paguei 600 R$ numa calculadora HP Prime e ela me deu o resultado 0 … quero meu dinheiro de volta

[@VuachoicacGaming-RongSamVangTV]

|i| = 1

[@valentinodrachuk5692]

That is wrong

[@Masonova1]

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.