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

Because I haven't gotten around to revealing it yet. If I did then all the lottery games will be useless. The oceans would rise - the moon would leave orbit - time would begin flowing backwards... So in a way - I am saving HUMANITY from itself... 🤪

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?

j should satisfy: j x j = -i, just as i x i = -1

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.

a 3D complex number would actually most likely be of the form a+bi+cij, this would allow us to properly represent the number as a vector... I'm pretty sure that I have the solution but there are some gaps that haven't had the time to fill up, as delving into mathematics is just one of the facets of my gnostic path any experienced mathematician is welcome to contact me by replying to this comment, we could finish this paper together may the Monad bless you

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.

Thank you. Appreciated

Umm, what about quaternions? 😂😂😂😂

Chalk Calculus Love it

This was a very clear video. Thank you

What s a 3D number? Never heard of numbers having a dimension

you sir have earned a new subscriber! i love these small presentations of some more involved problems in science, thank you for the great work

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.

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

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.

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.

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.

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

After reading many of the comments I asked chatgbt the following question: Are mathematics with bicomplex numbers possible? Here is the answer! Yes, mathematics with bicomplex numbers is possible and forms a well-established area of study within hypercomplex number systems. Bicomplex numbers are a specific type of hypercomplex number that extend the complex numbers by introducing two imaginary units, typically denoted as \(i\) and \(j\). These numbers can be written in the form \(a + bi + cj + dij\), where \(a, b, c,\) and \(d\) are real numbers, and \(i\) and \(j\) are imaginary units with the properties \(i^2 = j^2 = -1\) and \(ij = ji\). Here's an outline of the mathematical framework and some operations with bicomplex numbers: ### Algebraic Structure 1. **Addition and Subtraction**: - Addition and subtraction of bicomplex numbers are performed component-wise, similar to complex numbers. - If \(z_1 = a + bi + cj + dij\) and \(z_2 = e + fi + gj + hij\), then: \[ z_1 + z_2 = (a + e) + (b + f)i + (c + g)j + (d + h)ij \] \[ z_1 - z_2 = (a - e) + (b - f)i + (c - g)j + (d - h)ij \] 2. **Multiplication**: - Multiplication of bicomplex numbers involves the distributive property and the relations \(i^2 = j^2 = -1\) and \(ij = ji\). - For example, if \(z_1 = a + bi + cj + dij\) and \(z_2 = e + fi + gj + hij\), then: \[ z_1 \cdot z_2 = ae + afi + agj + ahij + bei + bfi^2 + bgij + bhij \cdot i + cej + cfij + cgj^2 + chij \cdot j + deij + dfij \cdot i + dgij \cdot j + dhij \cdot ij \] - Simplify using \(i^2 = j^2 = -1\) and \(ij = ji\). 3. **Conjugation and Norm**: - The conjugate of a bicomplex number \(z = a + bi + cj + dij\) can be defined similarly to complex conjugation but may involve multiple forms depending on the specific algebraic structure. - Norms and magnitudes can be more complex to define, often involving combinations of real and imaginary parts. ### Applications and Further Study Bicomplex numbers and similar hypercomplex systems find applications in various fields: 1. **Mathematical Physics**: - Used to describe phenomena in theoretical physics, particularly in extending theories involving complex numbers. 2. **Signal Processing**: - In certain signal processing problems, hypercomplex numbers can be used to represent multi-dimensional signals. 3. **Fractals and Dynamical Systems**: - Bicomplex numbers can be used to study fractals and other complex structures in higher dimensions. ### Example Calculation Consider two bicomplex numbers \(z_1 = 1 + 2i + 3j + 4ij\) and \(z_2 = 5 + 6i + 7j + 8ij\). Their sum is: \[ z_1 + z_2 = (1 + 5) + (2 + 6)i + (3 + 7)j + (4 + 8)ij = 6 + 8i + 10j + 12ij \] For multiplication: \[ z_1 \cdot z_2 = (1 + 2i + 3j + 4ij) \cdot (5 + 6i + 7j + 8ij) \] Expanding and simplifying using \(i^2 = j^2 = -1\) and \(ij = ji\): \[ = 1 \cdot 5 + 1 \cdot 6i + 1 \cdot 7j + 1 \cdot 8ij + 2i \cdot 5 + 2i \cdot 6i + 2i \cdot 7j + 2i \cdot 8ij + 3j \cdot 5 + 3j \cdot 6i + 3j \cdot 7j + 3j \cdot 8ij + 4ij \cdot 5 + 4ij \cdot 6i + 4ij \cdot 7j + 4ij \cdot 8ij \] \[ = 5 + 6i + 7j + 8ij + 10i + 12(-1) + 14ij + 16i(-1) + 15j + 18ij + 21(-1) + 24ij(-1) + 20ij + 24(-1) + 28(-1) + 32(-1) \] \[ = 5 - 21 + 6i - 16i + 15j + 7j + 8ij + 14ij + 18ij - 24ij - 24 - 28 - 32 \] \[ = -76 - 10i + 22j + 16ij \] In summary, mathematics with bicomplex numbers is not only possible but also provides a rich structure for various advanced mathematical concepts and applications.

j*j=-i

I thought I was drinking too much then...

Sir please continue to make videos.

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.

Title arrogant

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.

i * j = 0

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.

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.

They are out there, just look harder.

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☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂☂

A complex number itself does not exist in real numbers it is just a mask to hide our unability to get √(-1) what is this operation in lR set means ? may be nothing, but we have surpassed this insens by creating an imagiinary number to represent √(-1) we call it i in so way that √(-1)=i but this equation also is not an arithmetic nor an algebraic equattion, we did not find the result of the square root opération of (-1) , we just hid it by a mask i . So i is not a result of √(-1) it is used as a symbol to make calculus easy . Now i did not agree that a 3D complex number does not exist first reason is that you can represent for example z=5+2i-3j in a 3 orthogonal axisis landmak you have the real axis lR , and the imaginary plane ] i lR , j lR [ so you have 5 unit on lR and 2 unit on i lR and (-3) unit on j lR it s like a vectoriel representation so you can represent any bicomplex or 3D complex number by this way. I agree with you that we have a trouble in how to calculate (i times j)=(i*j) notice that the word calculate is not the right word to use here remember we have never calculated √(-1) we have created a symbol wich is a graphical representation or a mask of an uncalculable operation to ease the otherwide calculus so the question is : are we able to find an other mask to make calculus possible in 3D ] lR , i lR , j lR ] it is an other chalenge like the precedent one but not the same Annane Mustapha Blida center Algeria date 2024 07 04 day 5 time 23:47 GMT+1 pc time ☂☂☂☂☂☂☂

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

Thank you for interesting video

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

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.

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.

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?

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

Try j³=-1

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.

'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.

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?

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

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

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. 😊

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

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

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

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