Why there are no 3D complex numbers

A simple argument as to why there are no 3D complex numbers. See the following video by Michael Penn for a more elaborate and rigorous proof.
• Why are there no 3 dim...

[@guythat779]

Despite the video being from 7 months ago all the current comments are from within a week Idk what happened byt happy to be part of it

[@alsjeu]

this is called recommendations, pal

[@tristan2338]

School just finished for many people

[@TheLazyVideo]

The algorithm.

[@scutyardwilliamgate]

funny thing is I was thinking of this exact question a couple of weeks ago

[@FallenStarFeatures]

TL;DW: The dimensional progression from scalar, complex, quarternion, to octonion is exponential rather than linear.

[@reeb3687]

what about 2^log_2(3)?

[@FallenStarFeatures]

@@reeb3687 What about it? That's just equal to 3.

[@Loots1]

@@FallenStarFeatures WHAT ABOUT 3!!!!!!!

[@FallenStarFeatures]

@@Loots1 Sorry, 3 is not an integer power of 2.

[@higgledypiggledycubledy8899]

Thank you! ❤️

[@u2b83]

Hypercomplex Numbers Quaternions: Instead of a 3D version of complex numbers, the next step up in dimension leads to quaternions, which are four-dimensional (one real part and three imaginary parts: i, j, k). Quaternions were discovered by William Rowan Hamilton in 1843 and are used in various applications, especially in three-dimensional computer graphics and spatial rotations. Octonions: Further extending, we have octonions, which are eight-dimensional. However, as we increase dimensions, we lose certain desirable properties such as commutativity and associativity. Quaternions are non-commutative, and octonions are non-associative. Summary of Properties Lost Commutativity: Lost with quaternions. Associativity: Lost with octonions. Alternativity: Lost with sedenions. Power-Associativity: Still retained by sedenions but more complex algebras might lose it. Zero Divisors: Introduced with sedenions and present in higher-dimensional hypercomplex numbers. Alternativity: a(ab)=(aa)b (ab)b=a(bb) Power-Associativity a(a^2)=(a^2)a

[@akhileshmachiraju1521]

Two ways to note while observing an imaginary field. 1. It is perpendicular to a field and 2. It is unique to each field. If there are three independent variables, there will three independent imaginary fields corresponding to each independent vector.

[@user-uf7gq8kb7y]

Such a great and easy explanation, Thank you!!

[@boguslawszostak1784]

It depends on what we mean by "3D complex numbers." If we are referring to objects that describe affine transformations in 3D space, then quaternions can be considered as such numbers.

[@realwoopee]

Great explanation!

[@kirillerofeev8758]

So it seems that we can build such a space only if we have a n-element group where each elements has an order less or equal to two?

[@drbonesshow1]

I thought I was drinking too much then...

[@splat752]

I wondered what the starting assumptions would be and I would have liked them to be at the start. For instance linear independence, associativity and distributivity. However as I am interested in geometric algebra I find this a useful stepping stone and it has filled a gap. Nice video.

[@rlf4160]

GA removes the obstacles. Quaternions become a natural consequence.

[@DeeperScience]

I am a physicist, so this is more of a "physicist math" video. If you want the "math math" version, look at the Michael Penn video in the description. Very interesting also, but much more complex, pun intended.

[@azamazo]

There are relevant use cases with more than one complex part (see en.wikipedia.org/wiki/Quaternion for example) that would be relevant to explain in the context of what is described in this video.

[@davidepascu3026]

What would happen if we made also “scalar” multiplication by a real number also not commutative? Is there in that a similar argument? Do we have to be more ingenious? Or is it actually possible in that case?

[@DeeperScience]

If you lose commutativity, you lose a lot of nice properties. It's been a while since I recorded the video, but I think you can make the proof work without commutativity with real numbers. Note that when you get to the next level, which are the quaternions, you lose commutativity between the non-real component (numbers of the form A + B i + C j + D k, capital letters real numbers, i^2=j^2=k^2=-1, you have for example i j = - j i.

[@quandarkumtanglehairs4743]

​@@DeeperScience I see that you have maintained a strict constraint in your starting premises, and retained yourself from arguing algebraic geometry using ij bivector or ijk trivector. I like your maintenance of this constraint, as it shows us exactly WHY we need a non-numeric vector in expressing imaginary values in higher dimensions. I think your presentation here, exactly with this constraint, is perfect. Thank you! I had never considered it, before...

[@Dismythed]

The way to get to the third or any higher dimension with complex numbers is not hard at all. The problem is in the fact that √-1 assumes binary with √1 and a right angle joint with 1 and -1. Therefore, if you want a right angle joint to those two plains, you need a new term. Dimensions are defined by their square roots: √1, √2, √3, √4, √5, etc. Therefore, to get the third dimension, you need a term that gets you to the same value. So then, in order to get you to a value for the third dimension, you need to go up a step. The value of the first and second dimension together is their hypotenuse, which is equal to √2. Therefore, the value of the square root of three dimensions will be √3. But the value of the median angle between the second and third dimension is exactly the same for that between the first dimension. Therefore, we need to reach a value of √2 in the hypotenuse of the second and third dimension. √-1 gives us a second-dimensional angle. Therefore, in order to get a third-dimensional angle, we need something that gets us to the reverse angle of the first dimension if applied three times, just like applying √-1 twice gets us to the reverse angle of the first dimension if applied twice., but it needs to be dependent on √-1 in order to get us to the second dimension. The answer is that we add a √ for each dimension we travel into so that applying √-1 gets us back to the lower dimension from the higher dimension. The answer is that we convert √-1 into its power, namely -1¹’². To add a dimension, we just increase the value of 2 to 3 in the power's denominator. This equates to ³√-1. Therefore, each higher dimension is a greater value of squares. In this case, the cube root. So to get you up and down dimensions, you are adding and subtracting square operators and inverting the value (-1 to 1 or 1 to -1) to get to the other side of that dimension's plane. You're welcome.

[@ianbd77]

Yes I'd wondered about this question.... thanks

[@ostanin_vadym]

Thank you for interesting video

[@TheEternalVortex42]

You kind of miss out on the requirements we are trying to satisfy. It's easy to make a 3-d algebra if we don't care about division: for example take R[x]/(x^3−1). This means you would have numbers of the form a + b ω + c ω^2, where a, b, and c are real and of course ω^3 = 1. The only problem is you get 0 divisors, e.g. (ω - 1)(ω^2 + ω + 1) = 0. Indeed since every cubic has a real root all 3-d algebras will have zero divisors.

[@SiqueScarface]

I am trying to understand this geometrically. If the units not equal to 1 have the property to rotate the vector, it basically means that to keep the space three dimensional, you should not be able to reverse a rotation by i with a rotation by j.

[@afmikasenpai]

Hey guys, I might have missed something, but why does j^2 have to be -1? For quaternions for example we have i^2=j^2=k^2=ijk=-1 and I guess that works because it makes sense when doing calculations but I am thinking if it's possible for other values of j in the 3D attempt, I mean we could have j^3=-1 why not...

[@duffahtolla]

I think it's because the physical representation of multiplying by j is a 90 deg rotation in the (R, j) plane. He explains at 5:12

[@afmikasenpai]

​@@duffahtollaoh thank you, you are right, I don't know why I missed that.

[@TheLazyVideo]

Geometric Algebra has 3D “complex” numbers (similar to quaternions) and higher dimensions as well. The key is that i*j is an irreducible bivector in Geometric Algebra. In Geometric Algebra, a “3D complex number” is not expressed as a + bi + cj, it’s expressed as a + bi + cj + dij. Note the introduction of the bivector ij. And “4D complex number” requires the introduction of the trivector ijk. Also, obligatory, since someone will say quaternions are 4D not 3D: kzbin.infoiUvcYNonkaI Geometric Algebra intro: kzbin.info/www/bejne/bGHdkJumeqanepo

[@schmud68]

Quaternions are a 4-dimensional real algebra, this is precisely why they are written as a +bi +cj +dk for real a,b,c,d... The unit quaternions on the other hand are not an algebra. They are the 3-dimensional Lie group Spin(3) \cong SU(2). In fact, Spin(3) is diffeomorphic to the 3-sphere.

[@quandarkumtanglehairs4743]

​@@schmud68Your face is diffeomorphic in the 3-sphere.

[@schmud68]

@@quandarkumtanglehairs4743 and geometric algebra is a fad

[@quandarkumtanglehairs4743]

@@schmud68 No, it's much more than a fad. All of mathematics comes from geometry, it's in the name, geo-, Earth, and -meter, measure, in Greek. From "taking the measure of the Earth", by which was meant to understand the weights and measures of all things. This comes down to relations and proportions, and we come to concepts like squared numbers and cubed numbers by their geometric origins. Algebraic geometry, and geometric algebra, are continuations past trigonometric identities which fulfill a certain pattern-recognition from the one to the other. I don't think it's merely a fad, at all, but instead is humanity grasping toward a language of a higher intellect, namely, a higher-dimensional comprehension of the Universe alien to our own. What I was noting to our uploader is that his video shows WHY we must bear this other burden of non-numeric values (the bivectors and tricevtors) while attempting to communicate ideas in these other dimensions. It's a necessary constraint, akin to original geometry-proof constraints of only a compass and a straight edge, which I appreciate and helps bridge a gap. I didn't even know about this or consider it before this was recommended to me.

[@schmud68]

​@@quandarkumtanglehairs4743 sure mathematics has historical roots in geometry, and geometry is still a huge part of modern maths, but there are certainly things in maths that are not described geometrically. Maybe because humans are not smart enough, who knows. Geometric algebra is not all of geometry. You can't even do geometric algebra on all Riemannian manifolds (which are the natural higher-dimensional generalisation of surfaces, like a basketball etc). Or if you like physics, you can't always do it on a general spacetime in general relativity. Algebraic geometry, though sounding similar to geometric algebra is a very different thing. I've never used it much so just look at the wiki page to get a taste. Geometric algebra/Clifford algebra, ignoring philosophical viewpoints, is just a mathematical framework to talk about a vector space V with an inner product/dot product. Naturally, when V=R^3 is 3D space, then there are great visualisations which also aid in higher-dimensions when V=R^n. There is also a surprising amount of novel things that come out geometric algebra, like the Pin and Spin groups (I think one calls the Spin group the rotor group in geometric algebra terminology). Like I mentioned above, Spin(3) is essentially the unit quaternions. There is a related fact, in geometric algebra terms, the space of bivectors in 3D is essentially the quaternions themselves. The Pin and Spin groups also have an interesting representation theory, and the geometric algebra, in a way, directly points to so-called pinor and spinor representations. Spinors and pinors are important in quantum physics. The typical example being the use of Dirac spinors in QFT. Pinors and spinors also allow one to get interesting mathematical information about manifolds on which one can do geometric algebra. I am not saying geometric algebra is not useful or insightful, just that it is one part of maths and doesn't really tell you much about a lot of the other parts. I think the fact that you can't always use it on manifolds already says quite a lot. Maybe it is convincing to mention that geometric algebras (Clifford algebras) are completely classified as being isomorphic to matrix algebras Mat_{nxn}(D) or direct sums of matrix algebras Mat_{nxn}(D) + Mat_{nxn}(D) with D the reals, complex numbers or quaternions. Algebraically, this is a very limited selection of possibilities as there are many more types of algebras one could consider.

[@vincentbutton5926]

You started with the axiom/assumption i * i = -1. But what if you don't allow that, and start with the assumption that i * j = -1? It's not going to help solve sqrt(-1) easily, but perhaps it has other properties!

[@NLGeebee]

√-1 doesn’t need to be solved, because there are no solutions.

[@citricdemon]

​@@NLGeebeeMy uncle told me that's not true. he works at Nintendo.

[@ayylmao2410]

@@citricdemonnah ur wrong my dad is roblox

[@NLGeebee]

@@citricdemon in the corporate cafeteria?

[@citricdemon]

@@NLGeebee ask your mother

[@fuseteam]

I just realized that multiplying by i, rotates the vector in the complex plane and a 3D space is formed by 3 planes.....so we need 3 imaginary numbers to do rotations in 3 planes

[@taranmellacheruvu2504]

At the very end, you were only able to use c^2 = -1 because you assumed j commutes. I think this could be avoided by initially just writing the arbitrary number as a+bi+cj, because we know i commutes already.

[@sumitss6lite286]

What if we create new types of numbers assuming 1/0 to exist in different axis?

[@trevoro.9731]

Anything divided by 0 would be not a number, but it can be a mathematical abstraction, convertible to numbers in expressions, also you have to define non-negative 0 to avoid negative/positive option.

[@jeffr.1681]

Anything you create gives you double the axes, not plus one. Start from complex and add a 1/0 term and you get that and that times i, so you have four axes now. Same for an infinitesimal term, which is probably more useful. Add that and you also add i*e. You do both and you get eight axes, and so on.

[@tempname8263]

Your idea reminds me of projective algebra.

[@jeffr.1681]

Thinking more on it, there could be a system without imaginary parts but with an epsilon (infinitessimal) and omega (infinite), with e×e=0, omega × omega=omega, and e×omega=omega×e=1. Not commutative or associative but it does seem to describe a 3d field.

[@christianorlandosilvaforer3451]

​@@jeffr.1681actually those are call hyperreal numbers

[@elinope4745]

I thought it was complex numbers that determine the difference between emanating fields and spacial turns (rotations).

[@KurdaHussein]

I think we can also answer as: since there were a problem in R ( it was √(-1) ) that led to birth of "i" (complex numbers) , so what problem is there ( in R or C ) which leads to birth of "j" ??? so just putting a random letter which based on no problem makes no sense at all.

[@DeeperScience]

There are at least two motivations for the video. First, there is something called the quaternions, that are 4-D. One may wonder if there is something between the complex numbers and the quaternions. Second, as mentioned in the videos, complex numbers are very useful for making 2-D calculations. One may want to do the same in 3D, since our physical space is 3D.

[@SimpleFarm.38]

i is current in electricity so they shifted the i to be j so electrical equations would make sense. I’m still sorting it out though

[@MichaelJamesActually]

Why can't you have j in terms of i? You have i*i = -1, so those axes are not truly orthogonal

[@vikraal6974]

If j is dependant on i then j becomes redundant. There is no need for j and our system reduces to 2D complex numbers.

[@LydellAaron]

Can we put two complex numbers together, circles, to make a sphere? That's what I've been doing.

[@baselinesweb]

Thank you. I'll be back for more.

[@balthazarbeutelwolf9097]

one slight objection: in your derivation you used commutativity of multiplication. That does not hold for quaternions, so it is a big step to assume it for 3D numbers.

[@tomgavelin2879]

Where did he assume the commutativity?

[@balthazarbeutelwolf9097]

@@tomgavelin2879 when multiplying an equation with j he freely distributed the j around. To begin with, when multiplication is not (necessarily) commutative you have to clarify whether you multiply on the left or on the right.

[@tomgavelin2879]

I am aware of commutativity and distributivity. I just do not see anywhere in this video where commutativity was assumed where it didn't apply. Do you have a time stamp of what you're talking about? I am curious because I feel like I'm missing something obvious.

[@jeremypaton4300]

It looks like the only commutative assumption in the last step is the commutativity of scalar multiplication (ie a * z = z*a if a is a real number and x is a 3-d complex number). Any multiplication of i and j are not assumed to be commutative here in any step, as far as I can see.

[@tomgavelin2879]

​@@jeremypaton4300ok awesome, that's what I thought. Thought I was taking crazy pills for a second hahaha. Thanks for the check

[@ArchAwesomist]

You can describe 2 D space from 3d, but 3d is described from 4d space

[@NLGeebee]

What about z* = (a, b) + ci. With (a, b) being a point on the horizontal 2D plane and c on the vertical Im-axes? Then rotate it by multiplying with (0,i) or (i,0) or (i,i). Just a wild idea, I leave the maths to (dis)prove it to you. 😊

[@nicolasbesson4392]

The greatest teacher of all time damn

[@twixerclawford]

What would happen if c (in a+bi+cj) was not a real number, but rather a complex number?

[@Adam-rt2ir]

You'd get quaternions: a+bi+cj+dk = a+bi+(c+id)j for real numbers a, b, c, d

[@twixerclawford]

@@Adam-rt2ir in that case, quaternions could be thought of as a+bj where a and b are complex numbers?

[@Adam-rt2ir]

@@twixerclawford the Cayley-Dickson construction tells us the precise way in which the quaternions are numbers of the form a+bj where a, b are complex numbers. Same with octonions being pairs of quaternions, and so on.

[@DF-ss5ep]

​@@Adam-rt2irWoah

[@H_fromDiscord_real]

what if the 3d complex number is cbrt(-1) instead of sqrt(-1)? someone please explain how this would work out

[@jamesweatherley7764]

cbrt(-1) is just -1, so it won't help.

[@H_fromDiscord_real]

@@jamesweatherley7764 ah

[@jeremyalm9006]

The sound of the chalk on the board is awful. I wonder if it’s just the way the audio was recorded or if he’s using some nasty chalk like Crayola.

[@tempname8263]

@@H_fromDiscord_real Your idea is still valid though. In your algebra your unit cubed would equal -1. Which means that in your algebra cbrt(-1) equals to either your unit or -1. Multivalued functions are nothing new. Many equations can have multiple equally valid solutions (until new constraints are given).

[@H_fromDiscord_real]

@@tempname8263 wow i never thought of it that way 😮

[@Tabu11211]

by 5:58 I think I already get it. You would have two numbers to go to -1 but that leaves us with an incomplete 3d. So Im assuming that if j was complex that would fix it bringing us to 4d numbers.

[@Tabu11211]

rip xD

[@Unknown-vj4yb]

What if we take negative roots of complex numbers Like (--(a + ib))^1/2 =?

[@ManuelFortin]

Not sure what you mean exactly, but this will be a complex number (distribute the ^1/2, and the -1 ^1/2 becomes i.

[@splat752]

​@@ebog4841What about quarternions?

[@JH-le4sd]

@@splat752 They are 4-d

[@e.s.r5809]

As written, yes! That "implied" -1^(1/2) out the front would be ±i. Complex numbers do indeed also have a positive and negative root. Consider that (-i)^2 = (-1)(-1)(i × i) = -1. Very technically, i is not defined as the square root of -1. i is defined as the number such that (±i)^2 = -1.

[@eddie5484]

Well, obviously i * j doesn't equal 1. That's not how rotations in 3-d works. We can still have 3-d complex numbers if we keep it clear that multiplication means to rotate into a direction. For example; start with a unit of the real direction - 1. Multiplying ti by i means rotating that vector into the i direction. Then rotate the resulting vector into the j direction. It's still a vector n 3-d space. It's just not the same as doing the multiply by j first and then by i. If you insist on doing it the wrong way, requiring i * j = 1, you're effectively saying rotations in 3-dimensions are impossible.

[@StephenBoothUK]

If i^2 is -1 does it not logically follow that j^2=-i? j*i is therefore ij and multiplying a real by j rotates it in the super-imaginary plane.

[@Purified-Bananas]

I guess the closest thing to a 3D complex number is if you take a 3D vector as a rotation of |V| around the axis V / |V|.

[@Purified-Bananas]

Better, take |A| = 1 to mean a rotation of 90 degrees.

[@flavioxy]

can you make it 3D by using quaternions and leaving one of a,b,c,d as 0?

[@euanthomas3423]

Yes. You then get standard 3D vector analysis which is quaternions with the real parts always equal to zero. There was a big argument in the 19th century about which was better, vectors or quaternions. Most physicists, particularly Heaviside and Gibbs, thought quaternions cumbersome and decided to split off the vector part and just worked with that.

[@thomasolson7447]

This one has me confused. I'm trying to visualize the planetary motion as a wave. Quaternions pose a problem. [t, f(t)] where f(t) has a real, and three imaginary parts. So, it's five dimensional. I guess that might explain electro-magnetism.

[@mihirbpi]

Is there a way to make it work if we try something other than j^2 = -1?

[@DeeperScience]

Good question. The first issue is that you then no longer have the relationship "multiplying by j is equivalent to rotating 90 degree", which makes the complex number useful in calculations. Also, this will not work. You can look at the video linked in the description for a deeper understanding of the subject. If you want, you can try to come up with a simple proof using a contradiction similar to the one in my video yourself. Suppose j^2=a+b i + c j, i j = d +e i + f j, and do manipulations until you get to a contradiction. Since you have 6 parameters a to f to play with, the algebra may be a bit ugly.

[@mihirbpi]

@@DeeperScience Thanks! I will look into it

[@tempname8263]

@@DeeperScience This is still useful

[@tempname8263]

This is exactly what geometric algebra does. Depending on the algebra, you can have geometric units which square to, for example, -1, 0 or +1. As well as some exotic ones. -1 (usually we use biunits pair for that) encodes rotation, 0 translation and +1 hyperbolic rotation (repelling 'force')

[@prbprb2]

I think you should try to start with i, j and then show that no, matter what the rules are for multiplication using i and j, that not all numbers can have inverses. That may have a bigger scope (lack of a division algebra), but I don't quite understand clearly the assumptions you have made here. Sincerely .....

[@shyamfrancis9350]

Please don't stop making videos

[@DeeperScience]

I know the channel has been inactive for a while. I have a nice one in preparation about the movement of spinning tops and similar objects. I just have to find some time to record and edit it.

[@humanrightsadvocate]

What if we define j such that: j² = -1 ji = i ji² = -j

[@ahmedbensedik4186]

The problem is such choice of j: why j must such that j^2=-1???

[@canonicalgio7837]

surely it would be i^2 = -1 and j^2 = -í?

[@sanjinred]

But they exist in 3D, e.g electromagnetic field, but they are only represented mathematically in 2 dimensions. Yet what I have learned is that imaginary numbers exist ortogonal to one of the x or y axis. if you look at the xy axis the imaginary numbers lie inwards 180 degrees from the xy point of view. But they are mathematically represented with only one of the real axis with an argument (angle) relative to one of the R number axis. They could be represented in 3D with the xy axis but I see no practical use for that.

[@arkadiuszkoszewski8380]

What if j^2=j? The only problem will be with negative j.

[@sergehog]

The mistake in your logic is that you assume that i*j must be a number. But it's not. It's so-called bivector, which is squares to -1 by itself. The whole algebra you've created is the most basic one, containing quaternions. So, the object w + x*i + y*j + z*i*j directly corresponds to classic quaternion. After all, it is 3D complex numbers

[@schmud68]

i think that assuming ij is a number is the point of the whole video... He shows that a collection of numbers of the form a +bi +cj with i^2=-1, j^2=-1 cannot express ij in terms of such a number without reaching a contradiction or degenerating back to the complex numbers. Hence, you are forced to let ij be its own element. If you make no more assumptions, you still have to deal with iji, jij, (ij)^n and (ji)^n as independent elements (funnily enough (iji)^2 = (jij)^2 = -1). So it leads to an infinite-dimensional real algebra with generators i,j. Of course quaternions arise if you instead require ij=-ji which implies also (ij)^2=(ji)^2=-1. Instead you could require ij=ji which implies (ij)^2 = (ji)^2 =1 and actually corresponds to the commutative real algebra C\otimes_R C. Indeed, just map i \mapsto i \otimes 1 j \mapsto 1 \otimes i then ij = ji \mapsto i \otimes i. An interesting point is that C \otimes_R C is actually not a Clifford algebra/GA, precisely because i and j do not anticommute. Funnily enough the real algebra C \otimes_R C is isomorphic to the real algebra C \oplus_R C and this fact is useful in the classification of Clifford algebras. To me it is more a statement about group presentations, you really have a group G generated by (1,-1,i,j) subject to the relations that (-1)^2 = 1 -1 i = i (-1) -1 j = j (-1) i^2 = -1 j^2 = -1 you cannot state a relation like ij = 1,-1,i,j without introducing dependence between your generators as i=\pm j or contradicting the already imposed relations. As in the real algebra case, if you impose no extra relations iji, jij, (ij)^n, (ji)^n, then they are independent elements of the group. To return to the real algebra case you just take the real group algebra R[G] of G and quotient by the ideal generated by e_{-1} + e_{1} = 0, where e_{g} are basis vectors of R[G].

[@Darisiabgal7573]

The problem is the if we look down the real axis the rotation of the axes j relative to I is not normal, worse, it’s undefined. For example any equation in which 0i exists, j is indistinguishable from I. As a result it cannot be a bivector because with a bivector one axis needs to be relatable to the second, otherwise the bivector is undefined.

[@darioabbece3948]

I guess that if we set i*j=0 everything should work, provided the fact that this operation is undefined as someone said. It can look paradoxical but having something in i*j is paradoxical in itself from a geometric interpretation

[@fuseteam]

So you're telling me that......ij = k :D

[@Cyberautist]

I guess I am too stupid to understand the math. But didn’t a guy named Hamilton came up with a formula for complex numbers in 3D? en.m.wikipedia.org/wiki/Quaternion

[@ricardgavalda6135]

quaternions are not 3d but 4d. The simple proof by @carl3260 a bit higher up kind of suggests that if you try to add a 3rd dimension, you are cornered into adding a 4th one. That's what Hamilton worked out, how "4d complex numbers" are defined, and showed that they are consistent.

[@an1rb]

Why can't (1, w, w^2) be used? w = complex cube root of -1

[@sillygoofygoofball]

the cube root of -1 is represented in the two dimensional complex numbers. It is a particular linear combination of 1 and i. Look up “roots of unity” on Wikipedia for more info. You can use them to get the n-th roots of any complex number

[@camberbanbury3691]

They are out there, just look harder.

[@agnarrenolen1336]

If R^2 are points in 2D Euclidean space, what would points in C^2 be?

[@mismis3153]

Quaternions ?

[@user-qy8mv7wm5s]

C is isomorphic as a set to R² so points in C are like points in R². Similarly C² is isomorphic to R⁴ so points in C² would be equivalent to points in 4-dimensional real euclidean space. Iirc for all n in N we have that C^n is isomoprhic to R^(n/2) so you can make connections between complex spaces and real spaces for any dimension. TL;DR: points in C² are like quaternions

[@AM-bw3ze]

​@@user-qy8mv7wm5sDoes that mean there are no 3d complex numbers because the complex numbers would need dimension 3/2?

[@michaeldeoz]

why you defined "j" as the same as "i" ? What if "j" is another "impossible" solution? i is √-1 what if j is -1! or some else, but not the same as √-1 ?

[@sillygoofygoofball]

well you want j to not be a real number, and you want it to have simple algebraic properties. You can’t have j=infinity, for example, because you would have no useful multiplication rule for it

[@randyzeitman1354]

Why must j^2 = -1? Wouldn’t it have a component that’s both real and I?

[@cfc6214]

He want the same 90 degree rotation properties as when u multiply by i

[@cepson]

It seems to me (not a mathematician) that the problem is that the complex plane doesn't really describe a two-dimensional system in the same way that the x-y plane does. The complex plane doesn't represent dimensions, it just represents complex numbers.

[@susmitislam1910]

No, that's not the problem. You could do all of plane euclidean geometry with complex numbers - that is part of the reason why they're so ubiquitously useful. 3D complex numbers don't exist because it's just not possible to define an algebra comprised of 3 linearly independent basis vectors that satisfy the properties mentioned in the video. There does exist a 4D extension of complex numbers called the quaternions, and they can, similar to complex numbers in 2D, be used to study 3D geometry.

[@Apostate1970]

A way for nonmathematicians to think about this is to reason by analogy as follows: In the same way that negative times negative just returns to positive, and in the same way that cube roots don't require imaginary numbers (since a cube root of a negative real will just return a negative real), so too do you not need, and indeed you can not have, a distinct new type of imaginary with only three dimensions. More explicitly, imaginaries were only introduced because they had to be ... because there had to be *something* that corresponded to second roots of negative reals. These obviously couldn't be either negative nor positive reals, so they had to be something else. The argument that such things must exist relies on the idea of something called "algebraic closure", but without going into that you should be able to grasp the intuition. Basically when we take second roots it works nicely for everything except negatives. The intuition is that there's a missing piece of the puzzle, and that if and only if imaginaries exist can we can fill in that missing piece and make all of algebra work nicely over all of the reals. Since the reals all fit on a single line, the imaginaries can't fit on it, at least not in any direct and natural way. The natural way to conceive of them then is to put them on their own line, orthogonal to the reals, making a second dimension. The analogy above, about -*-, and cube roots, is relevant because it's no coincidence that this second dimension occurs with the second root. The question then is if, and when, new and higher kinds of imaginaries, beyond the regular one of i, can or need to be introduced. The answer is that they can be (whether they "need to" or not). But, again going back to the analogy above, and reasoning about the complex numbers, they only work out at "evenly even" roots and powers. That is, they only work out at powers of 2. So dimension 1 = 2^0 = reals. Dimension 2 = 2^1 = complex = 1 real and 1 imaginary. Dimension 4 = 2^2 = quaternion = 1 real and 3 imaginary. Dimension 8 = 2^3 = octonion = 1 real and 7 imaginary. Dimension 16 = 2^4 = sedonion = 1 real and 15 imaginary. And it keeps going up like that, by powers of 2. Again, if you try to make it work for any other dimensions the proposed extra imaginary numbers simply collapse back either to the reals or to one of the lower imaginaries. We call i the first imaginary, and we typically call the next group i, j, k. (h is the real part, so the dimensions would be h, i, j, k.) I'm not sure what the naming convention is for the octonions and beyond. Just like the imaginaries have real world, practical significance despite being "imaginary" (for example they naturally express relationships between electrical and magnetic fields, can be used to simplify computer information storage and retrieval problems, are required to parsimoniously represent questions in both quantum mechanics and relativity, and lots more stuff like that), so too do the quaternions, h, i, j, k. Some people even think the octonions have real world significance, but I think most regard them (and all the higher number systems) as mostly an idle curiosity.

[@tempname8263]

Yeah, you're sort of right. It doesn't represent physical space, but components of a rotor. It's not an x-y plane, it's scalar-xy mapping. In geometric algebra higher dimensions have more complex structures (pun intended), for which this visualisation won't work. And tbh, we don't need it really.

[@CliffSedge-nu5fv]

The complex plane is two-dimensional. A dimension need not be a physical spatial direction.

[@tempname8263]

@@CliffSedge-nu5fv People should be calling it really two-numerical. But things got tricky once some sillyheads decided to classify complex numbers as numbers...

[@FractalMannequin]

And if we drop associativity?

[@harrisonbennett7122]

Are you familiar with the Cayley-Dickson construction? I'm not sure if you're aware but when moving to the octonions, we then drop asscociativity.

[@henkhu100]

The whole explanation is based on the fact that there is an assumption that j squared is -1. But that immediately means that j equals i or -i. Extending the rel numbers with i means that we have a new value that can not be written as combination of real numbers. So a real 3D extension of the complex numbers means that we introduce a value( and with that a new dimension for the third axis) that is not a a combination of complex numbers. And that is not the case for j if you introduce that j squared is -1. Then j is just a complex number and the situation for i is different as an extension of the real numbers: i is an extension of the real numbers and j is not an extension of the complex numbers.

[@nihilsson]

Have you heard of quaternions? In those i²=j²=k²=-1, but i, j, and k are all different. According to your argument quaternions do not exist.

[@matta5463]

That was a very lengthy, very wrong comment.

[@ciple8330]

@@nihilsson For quaternions, *ijk = -1* as well, so j and k can't both be -i or i. However, if you ONLY define *i²=j²=k²=-1* then they could all just be complex numbers. So henkhu100's comment holds for quaternions.

[@henkhu100]

@@matta5463 Did you see his title: there are no "3D complex" numbers. Keeping that in kind his text is wrong. Because we have 1D real numbers, 2D complex numbers, 4D quaternions etc (8D, 16D, ...) But we have not a 3D system that fits. But that is something else then writing about 3D complex numbers. Just like 4D numbers are no complex numbers.

[@henkhu100]

@@nihilsson Did you see his title: there are no "3D complex" numbers. Keeping that in kind his text is wrong. Because we have 1D real numbers, 2D complex numbers, 4D quaternions etc (8D, 16D, ...) But we have not a 3D system that fits. But that is something else then writing about 3D complex numbers. Just like 4D numbers are no complex numbers.

[@VinniusHKruger]

Maybe 3D vector space + a time complex axis ?

[@liobello3141]

I think you are confusing maths and physics.

[@hornkneeeee]

that would still have 4 components like a quarternion

[@cristianm7097]

I think e^pi*i = -1 is the closure of C for all equations with coefficients in C ?

[@HerbertLandei]

I tried to have symmetrical components x,y,z, with x²=x, y²=y, z²=z, xy=z, xz=y, yz=x. which works fine except for associativity, so for three numbers in this system a(bc) != (ab)c, which makes them useless. There seems to be something deeper that prevents useful numbers with three components to exist, even if you don't limit yourself to "complex-like" numbers.

[@theupson]

i feel like 3 component vectors have some application in certain small domains. can you amplify on "useful numbers with three components"?

[@HerbertLandei]

@@theupson Numbers that follow the usual laws of reflexivity, associativity etc. Of course Vectors work, but the multiplication is either trivial (scalar product, the three dimensions don't interact) or anti-commutative (cross product), which isn't really nice, e.g. there are two ways to define it (left- and right-handed)

[@ahmedbelloufa2503]

Equivalence in over 7'39": not clear !! Geometrically only (perhaps)

[@filippocontiberas]

Somebody claims he has invented that 3d complex numbers you refers on the video, please check it out on: Italian journal of pure and applied mathematics (IJPAM) n. 29 year 2012, pages from 187 to 300. I'm not expert in math so I can only suggest you to see it personnally.

[@donwald3436]

11:02 if multiplication is not commutative then i(-1) is not -i?

[@vikraal6974]

Scalar multiplication is still commutative.

[@donwald3436]

@@vikraal6974 Why is that a special case why is that not stated?

[@donwald3436]

@@vikraal6974 Is it just so it works lol!

[@donwald3436]

@@vikraal6974 omg math clique.

[@donwald3436]

@@vikraal6974 can you stop deleting my comments

[@agustincabrera3551]

Why i.j = 1? if i^2 = -1 and j^2 also = -1... If you do i.j I assume it should be -1. In that case -j = -i

[@akademefoundation2596]

It's a tessarine. He isn't looking at how the operators actually work or following the vector analysis or anything. Give him credit for trying. He erased my comment where I pointed him to how j works, or the documentation showing it's been known and used that way for over a century by the guy who invented vector calculus. Using it right solves his problem. Using the imaginary numbers wrong... clearly does not.

[@SotirakisPeklivanas]

Correct me if I am wrong, a complex number can be represented by a one dimensional plane which would just be the x line (ax+ibx). By simple reasoning the 3D plane would be represented by ax+bix, cy+djy, ez+fkz. Just like real numbers can be drawn in a 3D coordinated graph, the complex numbers can be represented by a 6D coordinated graph. When you calculate the argument and angle you will always obtain results without the i , j and k. You cannot represent a 6D graph on paper, but you can represent a+bi , c+dj , e+fk on a 3D plane for calculation purposes to find the angle and argument within the 6D plane. Therefore 3D complex numbers do exist, as well as the 2D complex plane. Off course you will have to first find the three arguments, then work out the resultant argument. Complex numbers such as these could possibly work in a linear direction such as 3 phase electricity, but this idea is beyond my pay scale.

[@cfc6214]

Something seems wrong… a and b are in ortogonal axes… on the other hand if u add all 3 numbers u get a 4D complex number since a+c+e are all real

[@SotirakisPeklivanas]

A complex number does exist on the real line. We just don't know where. So we draw a perpendicular line where the a sits. i represents a unique number that generates contradictions when you resolve it. Our math only copes with a whole answer, no two halves (-1×1). You cannot add the dimensions together as the narrator has done in the video. It creates an absurdity. You can only multiply. If you want an area then l×h not l+h, a volume is l×h×w not l+h+w. If you follow the rules of engagement, then you will realise that the higher complex dimensions do exist.

[@CliffSedge-nu5fv]

You are wrong (and probably crazy). Nothing you said makes sense.

[@cfc6214]

@@CliffSedge-nu5fv that was a direct answer, LOL

[@MR-ie7lh]

​​@@CliffSedge-nu5fv thank you for saying that. I almost wasted time with his comments about "one dimensional PLANES" lol

[@rcglinski]

I think this is like another question below. It seems like having a complex number and taking its first derivative with respect to time would accomblish making a 3-d number?

[@Tletna]

The video itself is fine but I disagree with the assumptions made. Also, we could have ai + bj + ck = T and then define a braid like relationship between the three. Or, we could look for solutions to T^3 = -1 other than -1.

[@BangkokBubonaglia]

Thank you for this. This is so simple that even my 14 year old children can understand. I love Michael Penn's channel, but he is often rigorous to the exclusion of making it easy.

[@DeeperScience]

Funny you mentioned your 14-year-old. I explained this to my 13-year-old daughter the other day.

[@PeintreGaucher]

I believe the assumption that j.j=-1 is not the correct one. What if j.j.j=-1 and we consider j^3+1=0 has 3 roots that are not real, nor complex but are ‘bi-complex’ (in the form of a+bi+cj, a b and c real). I’m not enough good in math to go that far, but I don’t like the simple rotation that is done by stating j^2=-1.

[@PeintreGaucher]

And also, what is ‘bi-complex’ numbers are in the form of a+bi+cj+dij? Again, I’m not an expert in math, but reasons here are debatable

[@cfc6214]

I believe it doesn’t matter how u define the product, at the end… u wouldn’t get anything working with 3 and u need quaternions.

[@eugenekochnieff7055]

What about i x j =0

[@noone7692]

Then you will have to define the undefined numbers ie divided by zero numbers. You will be forced to define x/0 where x belongs to both complex and real numbers.

[@TheOneAndOnlyNeuromod]

Why even allow i and j to mix AT ALL? i doesn’t mix with the reals as is. So, j shouldn’t be allowed to mix with i except as another complex term. That would work! The whole thing wouldn’t end up becoming a pure real unless BOTH i and j were multiplied by themselves and even number of times become either -1 or 1 real to remove its imaginary contribution.

[@davidhb6147]

quaternions?

[@CliffSedge-nu5fv]

Four-dimensional numbers

[@localhero1978]

Just let i×j = i× j like we do with3×i = 3i

[@Patrik6920]

.. Leon would turn in his grave .. just ridiculous...

[@Crash-yp7ll]

i, j, & k would be orthogonal vectors, just as x, y, & z; values, and therefore would have to be handled by vector algebra (?) - The possibility of this approach seems not to be addressed in this video.

[@mattias2576]

I think (havent actually looke into it though) is that you can avoid the problems faced in the video for trying to define multiplication. This is done by essentially "escaping" the new dimension, so for i,j and k if we were to define i*j it was already showed in three dimensions that none of the options with 1,i and j would work so we choose i*j=k and same for the others i*k=j and so on. We essentially just have room using the extra dimension to define our multiplication in a nice way. Again, i did not actualky do any calculations so the above just gives the jist of why i think it might work in 4 but not 3

[@akademesanctuary1361]

Nice try. First, x+iy alone goes into three dimensions by rotating into z. Second, for your thinking, you have to use Heaviside's selective j where ij=-1, not the tessarine ij=+1. Then (jx+iy)^2=-(x^2+y^2) and you can then step out of this arithmetic box and define the other imaginary numbers to expand from two to three dimensions. Mind you by 3-D I am referring to coordinate axes. The actual dimensions as manifolds to construct that are four.

[@connorfrankston5548]

To a certain extent this is not true, you can consider a 3-dimensional geometric algebra to be the generalization of the complex numbers, and voila.

[@arturogonzalez6232]

Sorry for the misunderstanding but for 3d geometric algebra do you mean 3d in space being equal to an extension of imaginary coordinates?

[@carl3260]

Slightly briefer (sim to the last section): Assm: ij = a + bi + cj a, b, c in R (1) x i on left: ij = b/c - a/ci -1/cj (2) Compare coeffs of j in (1), (2) c = -1/c => c^2=-1. Contradicts c in R So ij is not a lin comb with real coeffs. So ij is lin indep, hence j doesn’t add 1 new dimension. Similar consideration for ji in absence of commutativity.

[@nagendramr]

Fundamental confusion with the explanation is assuming complex plane and 2d x, y plane is same. But it is not.. To understand complex number properties we need to understand diferrence between 2 axioms.. If we rotate complex axis 90degree we will get real in negative but 2d we cannot rotate and even rotate xy coordinates will stay same it can never be converted between x and y For example a+ib if we rotate then we get ia-b Straight line x+y if we rotate -x+y Therefore both are not equal.. It just coincide betwen our thinking.. Therefore complex number 3d is qual to 1st or 2nd dimension.. 11:36

[@mpcformation9646]

Your video is superficial because you miss the starting crucial point of giving a clear definition or even try to address the question : « what is a number ». So your « proof » is mainly empty because it stands on the sand ground of an undefined definition. And indeed, « i » for instance is actually less a « number » than an « oriented area », despites its arithmetic and algebraic apparent behavior « as a (almost) usual number ». And moreover, by the way, there is not only one « complex number structure » in 2D, but THREE, directly related to the three subalgebras of 2D real matrix algebra. And thus in this deeper regard, there is somehow no « (complex) number » in 2D, but instead, more correctly : ONE SCALAR, TWO VECTORS, and ONE PSEUDO-SCALAR (UNIT ORIENTED AREA). Which leads to three different versions, depending on the precise algebraic structure chosen. That’s the actual unmasked picture of such « three-fold » ALGEBRAICO-GEOMETRIC STRUCTURE. The « purely number » question is thus too naive and narrow minded. It misses the key point of the fundamental inseparability of Algebra and Geometry, despites the usual mainstream misleading perroting blabla. And moreover, in the same line of thoughts, there IS actually « generalised complex numbers » in 3D : ONE SCALAR, THREE VECTORS, THREE BIVECTORS (ORIENTED AREA), and ONE PSEUDO SCALAR (ORIENTED VOLUME = DETERMINANT FORM). You’re thus just proving that there is no coherent structure of the (obvious naive and incomplete) form that you « historically » (pre Clifford and Grassman) suppose. But by no way you’re proving what you pretend to prove : that there is no coherent algebraic complex number structure in 3D. On the contrary, THERE IS! You are making a similar mistakes than the usual superficial blabla of false beliefs about the uselessness of divergent series. They are often on the contrary more rich and useful than boring convergent ones. In brief here, you urgently need to update two centuries of intensive developpements on Clifford Algebras.

[@briangwaze9530]

i * j = 0

[@Pepesmall]

Why is it an issue to express j in terms of I? If you switch to matrices, shouldn't it be expected that one axis would be the complex conjugate of the imaginary axis, and the other the conjugate transpose, because everything needs to be in terms of i to be complex. It's like treating i as the Planck length of the system where every unit has to be a quantity in terms of that unit. You need 4 points to create a 3 dimensional object or space anyway, so you need a 3x4 or 4x3 matrix for any expression within that space at least. And really to define the bounds of the space it would need to be 3x8 for the 8 corners of a cartesian cube. But instead of measuring everything as an arbitrary position on a plane couldn't you just make the center of the object the origin and thereby you could measure it's 3 dimensional space as quantities in relation to each other to create the object or space within it that you need?

[@walterbrown8694]

Complex numbers only deal with 2 orthogonal axes and are easily handled with planar math. In my 89 years (over 30 as an electrical engineer) I have never encountered a need for 3D complex number representation or notation. If somebody comes up with a 3 dimensional Smith Chart (Impedance Globe ?) maybe I'll change my mind.

[@landsgevaer]

There is more to science than electronics, would you agree?

[@doctorrare1259]

In computer graphics, 3D rotations are made with quaternions. In electrical engineering maybe you can remake the Maxwell equations in quaternion form, after all, cross product is the quaternion product with real part cero. However, practical results would be the same, due to that, i think no one has interest in using quaternions in your field.

[@sanjinred]

I guess there is no use for 3d i numbers as the electrical and magnetic forces are orthogonal to eachother

[@Moriarty1982]

What would happen when we try say: I*J*I=I As you'd describe it, it sounds like the transformation of a Curvature Tensor

[@tomgreen4409]

Do this with: u = ai + bj + ck and v = di + ej + fk. With: ij = k, jk = i, ki = j.

[@cfc6214]

U r mixing concepts here. Where is the real axe? U only define 3 imagin.according to u ij should be real and from there u would encounter incongruence

[@tomgreen4409]

@@cfc6214 There are no such thing as imaginary axes. Do the algebra in 2D with [x, y], in 3D with [x, y, z], with x, y, z all reals.

[@cfc6214]

@@tomgreen4409 then we have no complex analysis just geometry. U can make an isomorphism between complex numbers and vector space. That is why complex axes have meaning. It is just a transformation

[@SilverFerrum]

j*j=-i

[@ifroad33]

Quaternions are 3D, no?

[@user-od5ri1fj8m]

4D (1, i, j, k)

[@ifroad33]

@@user-od5ri1fj8m Ah, I see

[@FabienAurejac]

@@user-od5ri1fj8m yes but they serve perfectly 3D since the real part can represent a potential, and the complex parts the 3D axis.

[@ScottSpadea]

I don't believe complex numbers are ever necessary. You can always replace them with sine and cosine.

[@mattias2576]

Complex numbers are often used specifically (solving wace equation for example) to not have to do cumbersome sine and cosine stuff, sure it could be done that way, its just much easier not too.

[@ScottSpadea]

@@mattias2576 As a game programmer, I prefer working with sine and cosine trig, and if a language features is optional, I try to avoid it for consistency and simplicity. I feel like imaginary complex numbers, breaks the rules of algebra too much to accept, redefining multiplication is a serious thing, that should be justified by it doing something that can't be done otherwise. it takes syntax that math already used, and redefined it, so now nobody can use i as a variable or it gets confusing. I am glad every use of complex numbers can be turned into sine and cosine. i j and k should be used for iterators.

[@mattias2576]

@@ScottSpadea i cannot speak to it in terms of programmikg but as a physicist, complex numbers are a very good tool for waves and similar things and also are completely necessary for quantum mechanics. Complex numbers are not always just equivalent to a different way of writing sine and cosine, QM is the best example i have for when complex numbers are fundamentally required, the schroedinger equation always gives complex solutions due to its structure. Also how does it redefine multiplication? It is associative, commutative, distributive with respect to addition etc. I can see why they might be overkill in some places but that does not mean they are not practical in other areas, electrical engineering use them a lot for describing electricity for example

[@ScottSpadea]

@@mattias2576 You claimed "complex numbers are completely necessary for quantum mechanics." but they aren't. you can use cos and sin to solve the same problems, every time. The only reason complex numbers are ever used, are dealing with circles on the complex plane, and you can always just use sin and cos. And if you say it takes more writing, I don't care. All computer code can be translated to bytes, if you can solve the equations on a computer, you can solve it with nothing but an array of bytes and a small list of op codes, for add, get, set, jump, jump if zero, bitwise not, shift left and shift right. Simulating spherical waves that interfere just doesn't require any weird math, you can do it with scaled and shifted sine waves to get the same interference patterns. scale sine waves by their EMWaveLength, Shift them by the measured points relative distance from each source, add the waves, and you have interfering spherical waves. If you complain that doesn't take into account the phase difference, like Gemini AI keeps arguing, it does if you wobble the wave source locations with sine waves, toward and away from the measurement point. Those sine waves are offset by the difference in time each wave peaks.

[@ScottSpadea]

@mattias2576 Simulating spherical waves that interfere doesn't require complex math, you can do it with scaled and shifted sine waves to get the same interference patterns. scale sine waves by their EM Wave Length, Shift them by the measured point's relative distance from each source, add the waves, and you have interfering spherical waves. If you want a phase difference, you can wobble the wave source locations with sine waves, toward and away from the measurement point. Those sine waves are offset by the difference in time each wave peaks. complex numbers are never actually needed, they don't provide more accurate answers, and if a computer can solve the equations, so can an array of bytes and basic op codes.

[@SotirakisPeklivanas]

This argument is flawed from the beginning. The equation is not 3D. When you think of I and j then you think of x,y. You can have x=y as an equation, you can have x-y=0 as an equality. You cannot have a+ib+jc. It does not represent cartesian or polar coordination.

[@u2b83]

The 3rd dimension is the real component aR + bi + cj, so the 3D coordinates are (a, b, c) It's true that traditional Cartesian coordinates deal with real numbers in all dimensions. However, when we extend into hypercomplex numbers, like in the case of bicomplex numbers or quaternions, we introduce additional imaginary coordinates. For instance, in a bicomplex number a+bi+cj, a is real, and bi and cj are orthogonal imaginary components, forming a kind of 3D space in this hypercomplex system.

[@annanemustaph]

I suggest to doing reseachs on what I call a multicomplex number Z=x + i₁ y₁ + i₂ y₂ + ....in un where x is the real part of Z and Σik yk are the imaginary parts of Z on ] lR, i₁ lR , ....., in lR [ (n+1) dimentional set☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂

[@AgnaktoreX]

Quaternions are the complex numbers that belong to the 3D space so I am wondering why you choose such a misleading title.

[@GhostyOcean]

Most people have heard of complex numbers but rarely ever hear about quaternions. The naive would assume you could just "add j" and make it work, I certainly did when I thought about extending complex numbers to 3D.

[@AgnaktoreX]

@@GhostyOcean well you add j and k and then you make it work so its not that far fetched

[@IMAHMS]

When you put Pi as a dot in the line of real numbers.. are you sure?

[@cfc6214]

Man… it exists, dot must be somewhere, buy very sharp

[@IMAHMS]

@cfc6214 no , it is not fully in there, it is tangential, like a perfect curve cannot be drawn with dots; a true circle cannot be pixilated, just like infinity is not a dot on that line

[@IMAHMS]

@cfc6214 you cannot draw a line and claim it has all numbers; a line is based on assumptions which cannot include some values, namely that it is made of equal and consecutive units.

[@IMAHMS]

@cfc6214 Pi and infinity ars real values that exist in the world, what doesn't exist in the physical world is those "numbers": you can never measure anythig so accurately as to say this is definitely 1 nor find two exactly equal things in the world to say 1+1=2, it's all mental.

[@cfc6214]

U r been too philosophical, Lol. I am engineer. I see the drawing as an abstraction…

[@davidwright8432]

'We introduce 'some number' i such that i^2 = -1'. But what the hell does this even mean, having said no real can have this property? Even if you simply stipulate a symbol you can give this meaning, what confidence have you it can be multiplied by any number - as you do when setting up what I'll call an 'i-axis' marked 0i, 1i, etc., all the coefficients of i being a real? You need some deeper motivation, such as an explanation that the idea of number has been extended before, from integer to negative integer and so on. Blithely saying 'our number is the solution to X^2 = -1' flat out contradicts the earlier assertion that there is no such number!!!! This is a difficulty you have no way round, or don't see. Of course there are complex numbers! But I'm criticizing your explanation, not their existence.

[@santividal9387]

But he said 'there are no real numbers that satisfy this equation', thusly saying that this number is not 'real'. This doesn't mean that it doesn't exist, it means it's another type of number... Perhaps I'm misunderstanding your question but it doesn't make a lot of sense when I read it

[@CliffSedge-nu5fv]

@davidwright8432 You sound confused.

[@mikeblackburn5966]

Try j³=-1

[@CliffSedge-nu5fv]

Then j = -1 and two complex numbers.

[@leeharrison2722]

Uh ... I do not know what you mean by "3D complex numbers" ... but there are quaternions that handle 3D rotations beautifully (better than Euler matricies), and then octonions ... and that is the end of the division algebras.

[@derendohoda3891]

quaternions are 4d

[@marfmarfalot5193]

Title arrogant

[@CliffSedge-nu5fv]

What?

[@hamidb.3402]

Because complexe set is just an other way to talk about 2 dim vectorial space.