Why there are no 3D complex numbers

[@QuicksilverSG]

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

[@reeb3687]

what about 2^log_2(3)?

[@QuicksilverSG]

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

[@Loots1]

@@QuicksilverSG WHAT ABOUT 3!!!!!!!

[@QuicksilverSG]

@@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

[@sciencedaemon]

Irrelevant when geometric algebra exists that is a superset of complex, quaternions, and so forth. It is the proper generalization.

[@u2b83]

@@sciencedaemon While it's true that geometric algebra (GA) provides a more unified framework that includes complex numbers, quaternions, and beyond, it's not necessarily "irrelevant" to study or use these hypercomplex systems like quaternions and octonions. Each has unique properties and advantages for specific applications, especially in physics and computer graphics. GA may offer a more general theory, but hypercomplex numbers still serve as practical tools in various fields, and their unique algebraic structures offer insights that are not always straightforward in the GA context. Both perspectives have value depending on the context and goals.

[@sciencedaemon]

@@u2b83 meh, geometric algebra is a superset of all those things and lacks nothing because of it. They on the other hand can only have very specific applications, cannot be fundamental. That is what makes them irrelevant. There is no real point in using them when you have a better system that does everything they do in a better way. The reason those systems exist is similar to why vector algebra, matrices, and so forth are still used in physics today when GA replaces all of it... personal opinions of those teaching, not sound reasoning leading to the best choice.

[@Outer-Heaven_Supercomputer]

i did read about not being able to compare imaginary numbers (like not being able to say 4i+2>3i-3), i am not an expert on this but i imagine no 3d numbers are there because their dimensions double, is this right?

[@TymexComputing]

Sedenions are two doubled octonions, 16 versors? What are the onions itself? What is a division of zero? can i have two numbers (vectors) that multiplied give zero but zero is not one of them?

[@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

[@t1234-q5z]

School just finished for many people

[@hdthor]

The algorithm.

[@scutyardwilliamgate]

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

[@bumpty9830]

Even more briefly, from a slightly different perspective: Complex numbers are about rotations in two dimensions. There are two real degrees of freedom in complex number because two dimensions allow one rotation axis which needs a basis element (called "i") and one unit is required for multiplication needing another basis element ("1"). The next step up describes rotations in three dimensions*. There are three rotation axes in three dimensions, requiring one element each ("i,j,k" of quaternions) and a unit is still required (quaternion "1"). There are no "three dimensional complex numbers" because there is no spatial dimension with two axes of rotation. * Okay, it describes the double cover of the three-dimensional rotations, but that's a detail that gets missed at this level.

[@sciencedaemon]

Geometric algebra will clear things up for you.

[@bumpty9830]

What about my explanation did you find unclear, @@sciencedaemon?

[@SuicideRedemption100]

When he mentioned why j² should be = to -1, it lead me to think about nonabelian groups, just then i made the connection to SO(3), and your comment explained an aspect that was even more elusive to me

[@nikitademodov3446]

But this does not prove that "3d-numbers" (with some basic properties) can't exist. Just that they wouldn't be describing a rotation.

[@bumpty9830]

Of course you're right, @@nikitademodov3446. But that statement obviously begs the question: "What would it describe, then?" The answer would be a Cayley table, whether written in that form or not, and its non-existence becomes clear in the search.

[@amandeep9930]

We can use the tower theorem. Let F be an extension of R which contains C and is of degree 3. Then, 3= [F,R]=[F,C][C,R]=[F,C]x2. Which is a contradiction.

[@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.

[@sciencedaemon]

It's called geometric algebra.

[@hdthor]

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.

[@avramlevitter6150]

12:35 If we're not sure about assuming commutativity, why are we assuming distributivity?

[@sidharthghoshal]

if you relax enough constraints there exist plenty of 3-dimensional algebras. in some sense satisfying answers relax the fewest constraints

[@ceo1OO]

Michael Penn's explanation was more rigorous... but i think this guy's explanation is more intuitive... He gave a more intuitive demonstration of why there are no 3-D complex numbers... and why the next step has to be 4-D complex numbers instead...📈

[@sciencedaemon]

No, the next number is a multivector of 3 vectors. This is easily grasped in geometric algebra, no need for complex numbers.

[@element4element4]

​@@sciencedaemonYou don't need complex numbers nor geometric algebra to work in any dimension. Including 3. The question is, is there an algebra similar to complex numbers that naturally describe geometry in 3d. Must have a unit and be invertible (associative normed division algebra). If you remove some of those requirements, then you can. Even much better constructions than geometric algebra. But the question is about this, so ignoring complex numbers is changing the question.

[@sciencedaemon]

@@element4element4 no one said you need GA for anything. Stop assuming. The fact is that GA enables a more straightforward approach to math than others. Go ahead, knock yourself out with all the other systems to get to the same result. This a main point of GA, to facilitate a better way of doing the same math. In physics you can certainly use matrices, vector algebra/calculus, linear algebra, and so forth to the same end. GA has it all combined in one (including geometric calculus).

[@element4element4]

@@sciencedaemon It's like someday is asking a question about surfaces in 3d, and I come to point out there is more general topology and and differential geometry. How is that relevant to the question?

[@sciencedaemon]

@@element4element4 it isn't like that at all. You just want to see what you want to.

[@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.

[@markwrede8878]

Aligning a tube at minus one-half on the complex plane would allow a third dimension of spin to the Riemann Zeta function. This allows deduction of the sequential differences among primes as a series of families of primes, i.e. each prime hosts a "gear" notched with the sequential difference from the preceding prime. This gear for twin primes is known as phi. There are solutions for square primes, sexy primes, and so on.

[@AdrienLegendre]

A different approach is to recognize that complex numbers are a type of Clifford algebra and Clifford algebras can be extended to higher dimensions.

[@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

[@girayyillikci3188]

Where does the quartenions stands at?

[@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

[@RohneeshSachan]

Such a great and easy explanation, Thank you!!

[@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

[@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

[@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.

[@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

[@adamel-sawaf4045]

@@SimpleFarm.38 yes but that electricity stuff is unrelated to this, the notation just happens to be similar

[@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.

[@xinpingdonohoe3978]

Just to check, we're taking ω such that ω³=1, but it's separate from the complex values of ³√1 which also work, right? A new element ω which is distinct from 1 and e^(±2πi/3)? I assume so, given the zero divisors line. I guess that would also mean that any odd dimensional algebra would also have zero divisors due to the necessity of a real root. But what about even numbers with odd factors? Degree 6 polynomials don't always have a real root. Or is it because the structure of it would be isomorphic to R[x]/(x³-1) × R[x]/(x³-1) and the structural property of zero divisors would carry over?

[@ashutoshganguly1805]

how about i(dot)j = 0? And also the last part seems a bit fishy, what is ij? what is the operator between them, and how did you assume that these are commuative that you assumed the freedom to multiply i with (ij)?

[@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...

[@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.

[@rclrd1]

It doesn't give anything new. ω^3 = −1 where ω = (1 + i√3)/2

[@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.

[@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

[@spuddy553]

help me to understand, i buy that we cannot express j in terms of i because they are linearly independent, but can't we express R in terms of i? What's happening there?

[@umbraemilitos]

The complex numbers are isomorphic to the even subalgebra of 2D VGA, where the imaginary unit it just the bivector ~ psuedoscalar. The algebra is not closed if you include one more basis vector, and limit to two unique bivectors. This is because (3 choose 2) = 3.

[@sbreheny]

If you define i as causing a rotation in the 1,i plane then wouldn't you define j to cause a rotation in the ij plane? So, j^2 would not equal -1 but -i.

[@dng88]

Can this argument prove i does not exist as well i.e. the “dim” 1 and i multiple etc other will leads to inconsistence i.e. generate a complex number with unreal a and b (just like the 15:00 c^2 = -1) i * 1 = a + bi i * i * 1 = ia - b -1 = ia - b => b=1 and a=0 which is real and consistent with our understanding or it is ok.

[@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.

[@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

[@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 😮

[@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)

[@rivers5320]

What I don't understand is, if using i,j doesn't work, can we express a 3D position with just complex numbers?

[@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.

[@joaodavid2001]

3D complex numbers are quaternions: a scalar dimension and 3 bivector dimensions. 2D complex numbers have 1 bivector dimension. Bivectors describe rotation, vectors indicate translation.

[@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.

[@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.

[@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.

[@Fert456]

Thank you for taking the time to explain that, Boss. It was very clear and helpful

[@Bapuji42]

He's the Boss! Bruce Springsteen!

[@alaksiej5913]

11:49 what if i * j = 0 ?

[@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

[@untio]

and are you sure that no matter how you define j, it is not possible? I mean you tried with squart of -1 for j, but, are you sure there could not be any other value?

[@marasmusine]

Thanks for the video, I hadn't thought about that before. Hello maths and physics people, I have a question. I recall that imaginary numbers come up in physics sometimes, such as quantum physics. Do quaternions come up in physics anywhere?

[@gaborszucs2788]

would be fun to see what if we don't restrict a, b, c to R but rather to C. E.g. -1 = c^2 would have a solution

[@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?

[@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.

[@timothygolden5321]

Another route that preserves the usual algebraic properties exists: polysign numbers. Consider that the reals have two possible signs and that those signs are modulo two behaved under product. So it is easily stated that a three-signed system will be modulo three behaved: --=+.-+=*,-*=-,++=-,+*=+,**=*, and so a three-signed value -a1+a2*a3 multiplied by another value -b1+b2*b3 will be none terms in general, consolidating again to a three-signed value: -a1b3-a2b2-a3b1+a2b3+a32+a1b1*a3b3*a1b2*a2b1, which is just the foil of the three termed expressions. What makes polysign thrive though is that the old law of balance -a+a=0 (which is the P2 or real version) becomes in P3: -a+a*a=0, and so the generally three termed expressions above can generally be reduced to two as for instance -1+2*3=+1*2, just as a general P2 value -a1+a2 will be reducible to one familiar term known as the reduced value. The geometry of P3 is implied by the balance no different than the opposed rays of P2 form the line. P3 are three equally opposed rays forming the plane, and the rotational nature of the product exposed their similarity to C, yet P3 are built upon the same laws as P2, and so mathematically speaking polysign can take a superior position. Indeed P3 are C, though they are in a new format. Pn are developable by these same laws and will be n-1 dimensional. They do commute and associate just as the familiar algebra of R(or P2) and C(or P3) so that we can speak of expression like z1(z2@z3)=z1z2@z1z3, where '@' is a universal zero sign, which becomes necessary because the plus sign of old has now taken the notation of sign two. It is no longer a universal identity sign, and the mnemonic of the number of strokes to draw the sign being its numerical counterpart is sensible so that @,-,+,*,# can get us ordinary superposed expressions up to P5, with the zero sign @ being the same as the high sign. This is no different really than the recognition that an unsigned value implies the identity sign in usual notation. So some of ordinary notation is preserved, but any time a '+' sign is used for superpostion, or a '-' sign is used for inversion there is a notational conflict. Possibly these notational difficulties have been enough confusion to prevent the greats from exposing polysign numbers. Historically it was difficult enough to get negative numbers accepted as 'real' and this is exposed in reading Descartes where these values are literally called 'false' roots, and likewise the even more difficult 'imaginary' roots are so named as even less approachable. The higher sign systems do come with some interesting consequences as a result of the clean algebraic behavior. For instance in P4 we have: ( - 1 + 1 )( + 1 # 1 ) = * 1 - 1 # 1 + 1 = 0 and it will be known already to mathematicians steeped in associative algebra that P4 must take on a form consistent with RxC, and P5 that of CxC, and so forth. Yet with polysign we never even needed the Cartesian product to develop these higher spaces, and that they contain images of the lower spaces is fine, but under product there are dimensional collapses that are possible. As well improper transforms are readily exposed as natural to polysign, as say for instance in doing a product survey of P4 upon a sphere one will find that sphere turning into a pancake and then turning inside out, eventually taking the shape of a rod as well. It should be mentioned too that there is a lesser sibling P1 whose geometry and qualities in fact have correspondence with time, and should we consider the family of polysign numbers P1 P2 P3 | P4 P5 P6 ... where that bar indicates the unusual product behavior, and so correspondence to spacetime can be gleaned. In effect, though no physics have been done here, we have a candidate for emergent spacetime, and this then suggest that polysign will one day play a strong role in physics. Pretty clearly electromagnetism will be coming along on the ride, as we are now engaging in a structured form of spacetime rather than the usual claims of isotropic space. This said, relative reference frames do recover some sense of relativity within such structured forms, and arguably the polysign numbers with their own sense of balance have somewhat captured relativity within their numerical form, where no actual need for working in the reduced form exists for the simple arithmetic exposed thus far. Indeed it is the act of rendering a value which exactly imposes that balance. Such a geometical and algebraic principle is new to humans. If you are uncomfortable with that interpretation you could simply insist upon working in the reduced form unconditionally, but the necessity of that is not actual. The very term 'two dimensional' explicitly ties back to P2, as in P2xP2, yet you can see that through the eyes of polysign this is a misnomer. It is possible that the Cartesian product itself is such a misnomer as a constructor of higher dimensional space. We do indeed top out physically in three dimensions, and the sensibilities of the Cartesian product as borrowed from physical sensibility will need to be investigated further. Claims for instance in group theory to construct the lowly sum as a function mapping SxS onto S are entirely unbelievable from this vantage. Why would you care to make such a complicated construction? If a is in S, and b is in S then could their sum simply be in S? Each and every usage of the Cartesian product deserves scrutiny. The fact is that R as P2 are no longer fundamental. This casts aside the footing of mathematics as it has been built out to date. That this endeavor leads to the development of a new complex number P3 built upon the same laws that build P2, and that a strange underling P1 carries the paradox of time within its zero dimensional yet unidirectional geometry: polysign really are quite different. To admit that the ray is more fundamental than the line could be a fine starting point.

[@timothygolden5321]

In relation to the video here polysign P4 are the 3D form of the complex numbers. P4 in polysign take their geometry as rays emanating from the center of a simplex outward to its vertices. These naked rays take the geometric balance: -1+1*1#1=0. While they are nonorthogonally constructed their algebra is extremely simple. Their arithmetic product preserves associativity and commutativity though it does come at some other costs. For instance the behavior |z1z2|=|z1||z2| does not hold generally in P4 and up. Sadly, if you are getting excited about the P4 Mandelbrot set: it is just an extruded form of the P3 version (C). Associative algebra predicts this. Polysign demonstrates it independently.

[@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

[@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.

[@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

[@kappasphere]

Going with what you said, 1+xi+yj+wij still just forms a 4D vector space, so it doesn't make much sense to me that you'd call it a 3D version.

[@cristianm7097]

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

[@Eianex]

what if we allow a,b,c € C ? or a,b,c € '3DC' ?

[@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.

[@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.

[@jaafars.mahdawi6911]

This is complex made simple. Good job. One objection though. You first said we're not gonna assume commutativity because it's not guaranteed (as is the case with quaternions), then on the last board, while disproving i.j=arbitrary 3D complex nb, commutativity was taken for granted. i hence wonder if again we're more careful not to presume this property whether things will change. Thanks anyway.

[@chsgs9009]

makes me wonder if there is anything interesting that results from allowing a,n and c to be in the Complex 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.

[@elinope4745]

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

[@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

[@sabriath]

well you are assuming the same properties to extrude the third dimension equating the second....but the second was built on the idea of a specific set, so if you have the same set, then you are in that same plane, not a third. In order to get the third, you need an entirely new set, which is the division of zero, and that set is actually pretty amazing. and before anyone says "you can't divide by zero"....tangent would disagree as a division of zero is 90 degrees through the tangent functionality. since two 90 degree turns is 180, you have the process by which to produce the fold across the third dimension. I'll leave you humans to figure out the rest, have fun.

[@cristianoo2]

What about Quaternions? They are used to compute pretty much every rotation in computer animations nowadays. Q = w + ai + bj + ck, (i, j, k) represent a 3D space !

[@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

from Morocco thank you for this clear explanation

[@PrimordialOracleOfManyWorlds]

is this true in polar coordinates?

[@rclrd1]

Relation beteen Cartesian coordinates and polar cordinates, expressed in terms of complex numbers, is simply x + iy = r (cosθ + isinθ).

[@PrimordialOracleOfManyWorlds]

@@rclrd1 tyvm.

[@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')

[@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. 😊

[@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

[@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

[@Fetrovsky]

j²=-1 is obviously a non-starter because then j == i. What you need is a new value from an operation that is undefined in the real or imaginary axis, like 0/0, 0^0, or something like that.

[@davidgoffredo1738]

Both -2 and 2 square to 4, but -2 does not equal 2.

[@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

[@agnarrenolen1336]

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

[@mismis3153]

Quaternions ?

[@Totu-t7j]

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]

​@@Totu-t7jDoes that mean there are no 3d complex numbers because the complex numbers would need dimension 3/2?

[@volodyanarchist]

Why not define j completely differently. Like sqrt(j)=-1

[@LydellAaron]

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

[@element1192]

There absolutely can be 3-dimensional complex numbers, but no system of 3d numbers is useful.

[@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...

[@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.

[@humanrightsadvocate]

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

[@ianbd77]

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

[@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

[@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.

[@irvingrabin]

At the very beginning the lecturer postulated that if we had a third dimension, then J ^ 2 must be -1. There were no explanation why. And the rest of the presentation is based on that assumption, which needs to be justified.

[@Frisbieinstein]

You can have complex 4d, 6d, etc.. Or use geometric algebra in any dimensions. I don't like quaternions and so forth. Too weird.

[@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 .....

[@realwoopee]

Great explanation!

[@avramlevitter6150]

The assumption of associativity is also important because then you'd have to figure out why there's associativity in the real dimension but not in the hyperplane perpendicular to the real dimension

[@ahmedbensedik4186]

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

[@cantkeepitin]

Because of rotation in Re,j plane, as analogy for Re,i complex plane

[@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.

[@sciencedaemon]

The reason they do not exist is simply because complex numbers are a limitation, a subset of more "complex" numbers found in geometric algebra (of which quaternions, etc. are a subset), that generalizes to all dimensions and different geometries.

[@element4element4]

@@sciencedaemon Complex numbers and quaternions are special cases of lots of other things as well, geometric algebra is not unique. The questions, is there an associative division algebra in 3d. And no, geometric algebra is not such a thing. Only when you specialize it to complex numbers or quaternions it gains those properties.

[@sciencedaemon]

@@element4element4 that is because there only 4 algebras: reals, complex numbers, quaternions, octonions with division algebras. It is not some fault of GA when it is the superset of those numbers. Also, trying to refer to GA as (not) unique is irrelevant. It is a broader, more encompassing system, not a specific math only useful in one way or another.

[@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.

[@philippfeiffer1452]

Wouldn't a 3d complex be (z1,z2,z3) where zi are each complex. That is how reals are extended to 3d.

[@philippfeiffer1452]

Then you define the addition, inner product, etc the same way as the reals. Using the complex addition and multiplication rules. So, 3d complex can be defined. Check out quantum mechanics.

[@philippfeiffer1452]

The only difference in a complex vector space you also have a duel space that is isomorphic.

[@canonicalgio7837]

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

[@arkadiuszkoszewski8380]

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

[@peta1001]

I did pass all the math university exams while studying electronics engineering. However, there were a couple of theories that always looked like convenient tricks to many of us. One of them is everything about negative numbers and therefore complex numbers as well. Today you can here more and more often that math brought Physicists and Astrophysicists to a stalemate in most of the modern theories (the theory of everything, the string theory, quantum dynamics theories and assumptions...). One can only wonder if the reason for the stalemate may be all these convenient assumptions (including the square root of minus one, zero factorial etc.). However, many with PHD titles are utterly religious nevertheless. They may not believe in God, but they certainly (without questioning) believe in everything the well-known mathematicians from the past concluded... then it may be that the earth is flat as well.

[@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.

[@freshrockpapa-e7799]

Why not i*j=0?

[@coffeyjjj]

because, i*j=0 i=0/j i=0. same reasoning leads to j=0. so, i*j=0 means i=0, or j=0, or i=j=0. so, i*j=0 is not a terribly useful path.

[@freshrockpapa-e7799]

@@coffeyjjj But you would be able to operate with i and j separately, it would be useful, it seems the only way to do it. You can have i*j=0 without necessarily having either i or j = 0 depending on the properties you want i and j to have.

[@coffeyjjj]

@@freshrockpapa-e7799 - i already explained, with crystaline clarity, precisely why your idea is invalid. it would have been encouraging had you expended the very small effort to actually understand what i wrote---which you clearly didn't. however, since you claim _"you can have i*j=0 without either i or j = 0"_ then demonstrate it. if you can't demonstrate this claim, you are clearly engaged in something unrelated to math.

[@freshrockpapa-e7799]

@@coffeyjjj I told you, there are many ways to have i*j not equal 0 without either being 0 depending on how exactly you define i and j and what properties you want them to have. Why are you being so pedantic anyway?

[@coffeyjjj]

@@freshrockpapa-e7799 - "...many ways..."? ok then, again, simply demonstrate ONE way. just one. if u can.

[@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.

[@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.

[@ahmedbelloufa2503]

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