Split-complex numbers: en.wikipedia.org/wiki/Split-c... Euler’s formula: en.wikipedia.org/wiki/Euler%2... The 3blue1brown video: • Linear transformations... The exp & log graph I mentioned: www.desmos.com/calculator/xm1...
Very cool! I'm not sure how much you know about Representation Theory, but I think it is very related to what you're doing here. In rep theory you do things like mixing groups and vector spaces to create R-algebras (the complex, split complex, and triplex numbers are all examples of R-algebras), and study matrix representations, among many other things. I think a lot of what you did can be rephrased in the language of representation theory, which might be a fruitful way to continue exploring your number system. Super interesting video!
@@brantleyvose2205 I'm a mathematics learner. I'm a fresher in my college studying computer science and engineering. I want to dive much much much deep into mathematics. I need a roadmap and a complete sequence to follow while self studying mathematics. Like the sequence of subjects, and topics to go along. My current level is: i've done calculus 1 2 3. I'm yet to do multivariable calculus and vector calculus, I've done probability and statistics and will dive more deeper, I've done initial discrete mathematics, like set theory, axioms, proofs, propositional logic, relations and functions. I've done initial level of number theory. I've done linear algebra deeply after doing just a little bit of group theory. I learnt about abelian groups and then did linear algebra deeply till inner product space. I'll complete group theory next. And there are other miscellaneous topics that i did, like complex numbers, binomial theorem, inverse trigonometric functions, geometry (coordinate), 3d geometry, parabola, ellipse, circle, hyperbola, sequences and series, especially those arithmetic progression and geometric propression and their properties. I'm yet to do real analysis. So that's pretty much my current level. What should i do to get better? What is the sequence i should follow? Can you suggest me some resources? Also how to learn everything in mathematics ofc not 100% but atleast 95%.
@@sounakroy1933It sounds like you've learned a lot! You know more than I did when I was a freshman. What you learn next really depends on where you want to go. The reason is this: there is no one in the world who knows 95% of mathematics. I would say there isn't anyone who knows 10% either (Some may argue with me. The number depends on how you measure these things.) That's just the amount of mathematics being done. Math is an enormous landscape! Exploring the whole thing is not feasible, but you can chart a course to get anywhere you want to go. With that said, it sounds like you are most interested in pure math, so I think that some real analysis and abstract algebra would open the most doors for you. I started real analysis with Basic Analysis by Jiri Lebl. It covers topics you have already learned from calculus, but with more rigor and detail, and with proofs. It also covers metric spaces which are a great unifying concept in analysis. As for abstract algebra, I started with Introduction to Abstract Algebra by Shapiro, especially the chapters on groups and rings. Lots of more advanced math assumes familiarity with these ideas, so it will be very helpful to have them under your belt. I wish you the best in your mathematical learning!
So, more generally these are Quotient rings of the polynomial algebra. That is, we use polynomial multiplication (which is commutative) and identify some polynomials with 0 - Forming residue classes like when calculating mod N in the whole numbers where for example: 5 ≈ 1 (mod 4) [5] = [1] in Z / (4) = Z/4Z For example, with the complex numbers we identify real polynomials whose difference is a multiple of X²+1: X^3+ 2X^2 + 3 ≈ X^3 + 1 ≈ -X + 1 (mod X^2 +1) Complex numbers = R[X] / (X²+1) = {[p] | p in R[X] } = {a[1] + b[X] | a, b in R}, and i is what we call the residue class of X, i = [X] and 1=[1]. Example: (1+X)X = X+X² ≈ X+-1 (mod X²+1) In other words: [1+X][X] = [X + X²] = [X + 1] =i+1. Split-complex numbers = R[X] / (X²-1), and j = [X] . Tricomplex numbers = R[X] / (X³-1), and i = [X], j = [X²] = i².
You could also think of this structure as an algebra over R (or C for that matter ) with the cyclic group of order 3 as its basis. Of course, these are all equivalent formulations.
Also for Split-complex numbers and Tricomplex, the ideals [x^2-1] and [x^3-1] are not prime, and hence the Split-complex and Tricomplex numbers are not an integral domain (unlike Complex numbers which form field), hence the zero divisors existing as mentioned in the video. This gives some intuition for why the complex numbers are so special and why they are the "largest" algebraically closed field of characteristic zero.
Ok. Just to nitpick here: While we can definitively build the 3-dimensional matrix expansion to get "triplex" numbers in the way you did here, there are actually 2 (3 if you count conjugates as different) to include i and j as perfectly normal complex numbers (of course, all computations which do not involve taking inverses of zero divisors still hold. In fact, all of your zero divisors reduce to 0 in that case). So, here they are: 1) i=j=1. You get back the real numbers (check the definitions: i^2=1=j, j^2=1=i, ij=ji=1) 2) i=(-1+sqrt(3)/2), j=(-1-sqrt(3)/2) (4th roots of unity, you get at least the Eisenstein numbers, if you only take integer combinations) However, what you _want_ to do is take the free algebra generated by your definitions (and realize it by a matrix), so you don't want further restrictions like i=1
I was using the thumb (triplex 'j', pointing at me), index finger (triplex 'i', pointing up), and middle finger (1, orthogonal to the other two) of my left hand to represent the positive sides of 3 axes, and I simulated rotation with your triplex 'i', going from my middle finger to my index finger to my thumb back to my middle finger, and rotation with your triplex 'j', going from my middle finger to my thumb to my index finger back to my middle finger. These rotations match the multiplication rules, especially when mixing your triplex 'i' and 'j'. The thing is that your positive triplex 'i' and 'j' are limited in their reach, as in no amount of triplex 'i' and/or 'j' multiplication can get me to the negative sides, but I can reach the negative sides of my 3 axes by using the negatives of your triplex 'i' and 'j', so I can't say that your positive triplex 'i' and 'j' could represent full circle rotations using my hand demonstration, but the negatives of them could. Maybe we could replace your triplex 'i' and 'j' with their negatives. The minuses in the negative version of your rules cancel out anyway. The triplex '-i' could represent clockwise full circle rotations on each axis and the triplex '-j' could represent counterclockwise full circle rotations on each axis.
you can think of i and j instead as i and i^-1 / i^2, with i defined by i^3=1. you could extend this with a bunch of things, primes being special. like quinplex numbers using 1, i, i^2, i^3, i^4 where i^5 = 1.
These rings are actually relatively easy to describe: adjoining an element z to ℝ satisfying z^n=1 for some positive integer n yields a product of k = ⎣k/2⎦ + 1 rings: ℝ[z]/(z^n-1) ≅ ℝ × ℂ^(k-1) if n is odd (n = 2 k - 1), ℝ[z]/(z^n-1) ≅ ℝ × ℂ^(k-2) × ℝ if n is even (n = 2 k - 2), with z mapping to the element ( 1, ζ, ζ^2, …, ζ^(k-1) ) where ζ=exp(2 π i / n) is a primitive n-th root of unity. This isomorphism is also known as the Real Discrete Fourier Transform: the left side is the domain of periodic real-valued discrete-time signals with period n, generated as ℝ-algebra by the time-shift operator z (where z^n=1 forces periodicity with period n) and the right side is the corresponding frequency spectrum.
This was so cool! It's basically like watching new math get invented on the spot and then discovering its properties. As Vi Hart once said, "Mathematics is about making stuff up and seeing what happens." Something I found interesting is that the exponential function has separate functions for its even and odd parts, which are cosh and sinh respectively (cos and sin are effectively the imaginary versions). What was used here could really only be called the "modulo 3 = 0" part, the "modulo 3 = 1" part, and the "modulo 3 = 2" part, which were A(x), B(x), and C(x) respectively, which didn't even have names before.
I know I'm very late here, but ... The functions A(x), B(x), C(x) are also a very nice basis for the solutions of the differential equation y''' = y. This corresponds nicely with the transformations in the split-complex numbers, which have the solutions of y'' = y, and in the complex numbers, which have (two of) the solutions of y(4) = y.
I think a nicer way to say for the functions corresponding to complex numbers (sin, cos) is that they are the solution of y'' = -y, as it is sort of symmetric with the split-complex number where y'' = y, while not producing extra solutions. Maybe y(4) = y could correspond to something like a "split-quadruple number"
This is very related to Johannes' comment about polynomial quotients. That is, it's no coincidence that exp(t[X]) in R[X] / solves P(D)y = 0. Indeed, using the standard ansatz y = exp(tx), we easily see that x satisfies P(x) = 0.
The matrix representation is very neat, and it helped me understand the split complex numbers a lot better. Just like i rotates the plane by 90 degrees, j swaps the two axes, reflecting them across x=y. This is why j^2=1, but j != +-1, because a reflection across x=y is its own inverse but is not the same thing as a dual reflection across the two axes or the identity transformation. It reminds me of the relationship between * and 0 in game theory. They have the same magnitude, but different values.
@@themathhatter5290 That's an interesting thought. I think the number i just swaps/rotates the axes in a "counterclockwise" order (1->i i->j j->1), and the number j swaps/rotates the axes in a "clockwise" order (1->j j->i i->1). So ij=1 just means that the axes rotated forward and then back equals to no change at all. i^2=j and j^2=i just mean that two forward axis swaps equal 1 backward swap, and vice versa
If you exponentiate an imaginary number, the result walks along the circumference of the unit circle, and gives a complex number that performs that rotation. If you exponentiate a split-complex number (without scalar part), the result walks along the unit _hyperbola,_ and gives a split-complex number that performs that _hyperbolic_ rotation. This is why the split-complex numbers are sometimes called the "hyperbolic numbers," and why I sometimes call the complex numbers "elliptic numbers." They are intimately linked with hyperbolic and elliptic/spherical geometry respectively. There's another cousin called the dual numbers, which when exponentiated, the result walks along a straight line, and gives a dual number that performs that _translation._ They're the flat/euclidean geometry counterpart. They also have the most hilariously trivial exponential ever: exp(εx) = 1 + εx.
You could make this algebra even more interesting by considering the coeffients to be complex numbers, i.e. a+bi+cj with a,b,c in C. This is similar to how the quaternions are constructed from the complex numbers.
Doing this adds at least one really neat property to the triplex numbers -- you can say that every triplex number has not one, but two conjugates. For the triplex number z = a + bi + cj, these are z† = a + ωbi + ω²cj and z‡ = a + ω²bi + ωcj, where ω is the complex number exp(2πi/3). The conjugate z' in the video is just the product of these, so the norm is the product of the original number and both of its conjugates: N(z) = z z† z‡. And unlike with z', we also have N(z†) = N(z‡) = N(z). Also interestingly, (z†)† = z‡, and correspondingly (z‡)‡ = z†.
@@rossjennings4755 That would be a different algebra altogether, though: the complex-triplex numbers, or hyper-triplex numbers. The idea behind the triplex numbers is that they form a ring with no irreducibles, where the three algebras over R^2 are subalgebras and subrings of the triplex numbers, and the triplex numbers are the smallest ring/algebra with this property. They are also nice in that they are to modular arithmetic mod 3 what the complex numbers are to modular arithmetic mod 2. So yes, you are right that the complex-triplex numbers are interesting to consider, but I would not say they are necessarily better than the triplex numbers. It is application dependent.
I think I found an error in the presentation. At 13:32, you provide the three matrices as: a = |1 0 0| |0 1 0| |0 0 1| b = |0 0 1| |1 0 0| |0 1 0| c = |0 1 0| |0 0 1| |1 0 0| But at 16:35, you write the matrix as: |a c b| |b a c| |c a b| That bottom row isn't right. It should be: |a c b| |b a c| |c b a| The determinant that you present is the correct one for the matrix that I've shown, not the one you wrote down. Not a big deal...graphical equivalent of a typo.
I also was looking to see if anybody commented this already. Actually, the upper matrix at 13:32 also has the transposition error. Indeed, the same error appears to crop up each time he shows that matrix, likely because he is re-using the same graphic.
@@jursamaj Thanks for pointing this out. I was watching this and was very very confused. It's been a while since I used matrices, but this was not making sense to me.
This is one nice chunk of mathematics and entertainment. I thoroughly enjoyed myself, and this comment is to show appreciation, and to nudge the algorithm about the quality of this here video. Cheers, I hope you have a good week.
If you allow for matrices with complex entries, you can actually represent triplex i&j with 2x2 matrices triplex i = [1/4 + i sqrt(3)/4, 3/4 - i sqrt(3)/4] [3/4 - i sqrt(3)/4, 1/4 + i sqrt(3)/4] triplex j is the same with complex-conjugated entries
hey! i examined the functions A, B, C before when i was looking for functions who were equal to their own triple-derivative and constructed a power series from them. theyre like a mix of an exponential function and a (co)sine) function. try plotting them all in desmos, theyre super cool. because of this property, they form a cycle like this: dC(x)/dx = B(x), dB(x)/dx = A(x), dA(x)/dx = C(x) note that although they have a simple series representation, their plot is quite complex, and you can subtract (1/3)e^(-x) from A(x), B(x) or C(x) to get a visually-nicer function which is also simpler to express in terms of known functions, which is to say: A(x) - (1/3)e^(-x) = (2/3)e^(-x/2) cos(sqrt(3)x/2) = (2/3)r(x) (note that tau=2pi) dr(x)/dx = d( e^(-x/2) cos( sqrt(3)x/2 ) )/dx = e^(-x/2) cos( sqrt(3)x/2 + tau/3 ) = g(x) dg(x)/dx = d( e^(-x/2) cos( sqrt(3)x/2 + tau/3 ) ) = e^(-x/2) cos( sqrt(3)x/2 - tau/3 ) = b(x) db(x)/dx = d( e^(-x/2) cos( sqrt(3)x/2 + tau/3 ) ) = e^(-x/2) cos( sqrt(3)x/2 ) = r(x) this means these 'simplified' functions (which have more complex taylor series representations) also have the property that they are equal to their own triple-derivative, and further, notice how they naturally conceptually correspond to 3 equidistant rotations along a circle which seems to be a common theme of triplex numbers when i was looking into these a little i wanted to see if there were any equivalents of "trigonometrric identities" for r(x), g(x), b(x) (which i named such after the red-green-blue) and found the following: r(x) + g(x) + b(x) = 0 (this implies A(x) + B(x) + C(x) = e^x which should be obvious given their taylor series) 2r(x)^2 = r(2x) + e^(-x) 2g(x)^2 = b(2x) + e^(-x) 2b(x)^2 = g(2x) + e^(-x) hope this is interesting! maybe it can give you some ideas
I like how when there are two elementary objects, they are always labeled as + and - or similar, and when there are three then the go-to labels are R G and B
This is a very good video. To be honest this is the best one I have seen in years and it contains only one mistake. The mistake is about the conjugate. Most people who read this know that on the complex plane or in the space of quaternions the conjugate is given via a flip in the real axis. But if you do this for these 'triplex' numbers it goes wrong so what is going on? Well actually a better definition of the conjugate is just the first row of the matrix representation. For complex numbers and quaternions this gives the same conjugate as flipping in the real axis. If we write a triplex as Z = x + iy + jz, the conjugate is given as Z' = x + iz + jy. If you now multiply Z against Z' and equal this to 1 you get ZZ' = 1, that gives two equations: One equation of a sphere and the other is a cone. The intersection of this sphere and cone is the solution to ZZ' = 1 and that is what I call an exponential circle. The most simple exponential circle is e^it = cos t + i sin t, most of you know that beast. Well if you solve ZZ' = 1 you get the same in three dimensions. By sheer coincidence last week I wrote a post on this detail, I refer to triplex numbers as circular numbers because the matrix representations are so called circular matrices. Here is a link to my post from last week: 3dcomplexnumbers.net/2022/02/10/once-more-the-sphere-cone-equation/ I hope it is acceptable that I promote my own post, I have no commercial interest with that whatsoever so lets hope it does not get removed...
Beautiful work, you’ve been thorough, certainly bridged some gaps. At 38:12 in the natural logarithm expression, the coefficients for i and equally for j can be written also as ⅓ ln √ { (a+b+c)³ / (a³+b³+c³ - 3abc) } ± ⅓ √3 tan⁻¹ {√3 (b - c)/(2a - b - c)} where ± indicate in the case of the plus sign for the coefficient of i and the minus sign for the coefficient of j . Restrictions a + b + c ≠ 0 a³ + b³+ c³ - 3abc ≠ 0 2a - b - c ≠ 0
There is a more generalized way to find the conjugate given the matrix representation. We know that for any matrix A, A * adj(A) gives a diagonal matrix with entries equal to the determinant of A, and the determinant is equal to the norm of A. So, calculating adj(A) for a generalized element of the ring A gives you the conjugate essentially for free.
Nice vid man, i can feel your pain going through all those derivation, but seeing that taylor series in terms of exp,sin,cos was worth it, hope u have a nice day : D
I came here looking to learn what Split complex numbers were. Stayed because it was interesting. Nice video, hope you get your work in a wikipedia entry
Very nice presentation. Let me add some thoughts. First, the "diagonal" basis e1, e2, e3 shows, that triplex numbers are isomorphic to RxC with multiplication rule (a,z)(b,w)=(ab, zw). Second, representing an algebra as a subalgebra of a matrix algebra of much higher dimension (nine in this case) is a mere fun fact imho. Last, the formulae for A(x), B(X), C(X) can be easily derived from the fact that they are obviously solutions to y+y'+y''=exp(x). Here we have exp(x)/3 as one solution and to this we add any solution of the homogeneous equation y+y'+y''=0 with the standard approach y=exp(cx) leading to 1+c+c^2=0. Then you solve for the initial values of y, y', y''.
At 31:38 I so have been off and on wondering about A(x), B(x) and C(x) while thinking about how sin/cos came out of the power series and always wondered about other numbers like every 3rd or 5th. This scratches the itch for the 3rd, at least for now until I want to delve further.
sinh/cosh act as the even and odd parts of the exponential function. (sin/cos are just for the imaginary variant where the sign flips.) A, B, and C I would then probably call the "modulo 3 parts." Following the same logic, it should absolutely be possible to extend this and find the "modulo n" parts of the exponential function (or an arbitrary function). For the modulo 5 parts, you'd probably have something like 1 + xi + (x²/2) j + (x³/3!)k + (x⁴/4!) l + x⁵/5! +..., Yielding functions A(x) = 1 + x⁵/5! + ..., B(x) = x + x⁶/6! + ..., C(x) = x²/2 + x⁷/7! + ..., D(x) = x³/3! + x⁸/8! + ..., and E(x) = x⁴/4! + x⁹/9! + ... I'm not sure what these would look like in terms of normal sin(h), cos(h), log, exp, etc.
@@angeldude101 the general formula seems to be that Σ{n=0...∞}(x^(Nn+K)/(Nn+K)!) = (Σ{n=1...N}(exp(xcos(nτ/N))cos(xsin(nτ/N)-Knτ/N))/N, though I've only confirmed that algebraically for N = 4 and 5
Not necessarily. He addresses why this construction is not the same as in the video, but if you're interested in a more formal description I would start by investigating quotient rings. By taking quotients of the polynomial ring R[x] you can describe these algebraic structures explicitly. For these triplex numbers it might seem appropriate to take a quotient of R[X, Y] and follow through by brute force, but actually since we have X=Y^2 we can see that they are just the quotient ring R[X]/. And all the maths in the second half of the video come from this! Just goes to show how much structure is hidden implicitly in a construction so regular as the ring of real polynomials.
@@xatnu If the question is about a group with one binary operation then it is about a an Abelian group with three elements. If the question is about a ring of certain characteristic then there more than one answer.
@@CasualGraph Hmmm, can different triplex systems, with different definitions for the imaginary parts, be derived that contain the real, complex and dual, or real, dual, and split complex, or complex, split complex and dual, or even all of them?
Good catch! I was thinking the same, because what happens when you represent a purely real triplex? You have a=1, b=c=0, but that doesn't give you the identity matrix, which is the unique analog of 1 in 3x3 matrices. Putting 'a' along the diagonal does, and that mostly fixes the position of b and c, up to symmetry between b and c.
Personally, the *proof* of j not being either +/- 1 at around 7:00 is confusing or at least kinda wonky. We are defining it in terms of j^2 = 1. So the act of squaring is baked into the very foundation of what j is. Therefore, working with j only, for me at least, means implicitly working with a square root without it having to be signaled. When the operation returns +/-1 , it's not like some new form of ambiguity, but rather simply an effect of taking the square root prior. Sqrt (1) fulfills the (1+j) * (1-j) multiplication requirement. At least that's what common sense seems to imply.
Yeah, I think you’re right that mapping j -> 1 or j -> -1 will not necessarily “break” anything and will result in true equations. (Specifically, this is true because it’s equivalent to taking R[X]/[X^2-1]) But that’s a little boring and we can have this j be this funky, not 1 or -1 object and get a more interesting structure out. Also, looking at the matrix representation, j corresponds to reflection over y=x, which is neither the identity nor the negative identity, so that’s cool.
If you choose the following N-function, lets call it M, as M(x) = (N(x))^(2/3), then you get all the nice properties like involution of x' and stuff. Because the new conjucate, lets call it x*, would have to have the property xx* = M(x) = N(x)^(2/3), so multiplying by N(x)^(1/3) on both sides, gives x(N(x)^(1/3) x*) = N(x), so since xx' = N(x), we get x* = 1/N(x)^(1/3) * x' = 1/sqrt(M(x)) * x'. Because N(ax) = a^3 N(x) for real a, for M it is M(ax) = a^2 M(x). So, (x*)* = M(x*)/x* = M(1/sqrt(M(x)) * x') / x* = (1/sqrt(M(x)))^2 * M(x') / x* = 1/M(x) * M(x') / x* = 1/M(x) * (N(x'))^(2/3) / (1/(N(x))^(1/3) * x') = 1/M(x) * (N(x))^(4/3) * N(x)^(1/3) / x' = 1/N(x)^(2/3) * N(x)^(5/3) / x' = N(x) / x' = x nice
Pretty interesting video. I was thinking about how to represent these triplex numbers graphically: Think about complex numbers. Multiplying by i rotates a complex number by 90° (or 2π/4, or τ/4 radians), because i² = -1 (analogous to how two 90° rotations in a plane lead to a 180° rotation which is the same as multiplying by -1). Thus, i⁴ = 1: you need 4 90° rotations to complete a full turn, coming back to 1. In these triplex numbers, we have instead i³ = 1 and j³ = 1, which could be thought of as 120° rotations (or 2π/3, or τ/3 radians) because you need 3 of them to complete a full turn and come back to 1. We could then define i to be a counterclockwise rotation by τ/3 radians in a 2D plane, and j to be the opposite: a clockwise rotation by τ/3 radians (or viceversa, that works fine too). This would be completely consistent with the following definitions: - i² = j, because turning τ/3 radians counterclockwise twice (2τ/3) is the same as turning τ/3 clockwise once (-τ/3) - j² = i, for similar reasons - ij = ji = 1, because i and j are opposite rotations, so i undoes j and j undoes i and we always end at 1 In complex numbers, we can define in a 2D plane the imaginary axis as being ortogonal to the real axis (i.e. forming a 90° angle). In triplex numbers we could define in a 2D plane the "i" axis as the one which forms a counterclockwise 120° angle (or +τ/3 radians) with the "1" axis... ...and the "j" axis as the one which forms a clockwise 120° angle (or -τ/3 radians) with the "1" axis. If we define triplex numbers this way, we get i + j = -1, and that 1 + i + j = 0 because the three numbers 1, i and j visually complete an equilateral triangle when being added. Thus, it would make total sense that e_1 = (1+i+j)/3 has the property that (e_1)² = e_1, because it's just 0² = 0. Similarly, e_2 = (2-i-j)/3 ends up being 3/3 = 1, so it also makes sense that (e_2)² = e_2, because it's just 1² = 1. And well, (e_1)(e_2) = 0. Then we have e_3 = (i-j)/sqrt(3). i - j happens to sit at a 90° counterclockwise angle with respect to 1, but its distance from the origin is exactly sqrt(3), so when we divide by sqrt(3) to get e_3 we get exactly a 90° rotation, pretty much like the "vanilla i" in complex numbers. Thus (e_3)² = -1, and so if e_2 = 1, it makes sense that (e_3)² = -(e_2). Similarly (e_2)(e_3) is just 1*e_3 = e_3, and (e_1)(e_3) = 0*e_3 = 0. So e_1 would be just 0, e_2 is just 1, and e_3 is just a 90° counterclockwise rotation, which means (e_3)² = -1. As for the exponential function, yeah, it's harder to define exp(xi) or exp(xj) in this way, but we could define: - exp(xe_1) = exp(x*0) = 1 - exp(x*e_2) = exp(x*1) = exp(x) - exp(x*e_3) = exp(x*"vanilla i") = cos(x) + "vanilla i"*sin(x) = cos(x) + (e_3)sin(x) which are consistent with the results in the video. Seeing that i is a 120° rotation, which could be expressed as -1/2 + (sqrt(3)/2) * "vanilla i", or -1/2 + (sqrt(3)/2) * e_3, we could try and directly define: exp(xi) = exp(-x/2) * exp(xe_3 * sqrt(3)/2) = exp(-x/2) * (cos(x*sqrt(3)/2) + (e_3)sin(x*sqrt(3)/2)) thus exp(xi) = exp(-x/2) * (cos(x*sqrt(3)/2) + (i-j)/sqrt(3) * sin(x*sqrt(3)/2)) Similarly: exp(xj) = exp(-x/2) * (cos(x*sqrt(3)/2) - (i-j)/sqrt(3) * sin(x*sqrt(3)/2)) What do you think about this? I'd love to read your thoughts.
i havent gone thru the entirety of your derivations, but just going intuitively, this feels like a problem that cannot be solved in the 2 dimensions. If it were solvable, the representations of i and j in terms of "vanilla i" should have had no mathematical issues. he says in the video that there would be some issues, based on whether random hidden divisions by zero are popping up when assuming typical multiplication and division behaviour. i would invite you to review these assumptions in your derivations. i donot want to leave u with incomplete answers, but i have limited time, so forgive me. but i would invite you to look at your visualisation in terms of spheres, 3d planes and steradians. the 2d plane visualisation you came up with might just be a planar cross section of the 3d case. wishing you a great day, whenever you do read this.
@@Nubbdy Thanks for your answer, I forgot to take into account those issues, a 3D representation would be better but I still couldn't develop a good one yet, because of limited time too. I was thinking about 1, i and j in 3 separate ortogonal axes, so multiplying by i would (maybe) rotate the three of them around the (1+i+j) axis such that 1 ends at i, i ends at j and j ends at 1. And multiplying by j would be a similar rotation, but in the opposite direction. I still haven't handled how to multiply general triplex numbers though. Maybe the process isn't exactly a rotation, but something similar. I also thought about multiplying by -1, which in a 3D space has a completely different effect from multiplying by -1 in a 2D plane. The latter does not reverse orientation of objects in the 2D plane, so it's the same as a 180° rotation. The former does reverse orientation in 3D space and cannot be expressed as rotations because they cannot do that (afaik), so multiplying by -1 in triplex numbers would be a completely different operation from multiplying by i or j. It's interesting though, maybe in the future I can think about this a bit more.
@@nomechomath5644 yup, i think you are on the right track. i would also recommend two things. firstly, for the confusion about rotation / not, please review the rotation / translation / other transformation matrices used in coordinate geometry / graphics. first results on google images, easy to find. it might be that the axis for rotation across the bases change based on the basis. you may be right, it might be a new type of transformation as well, maybe call it twist? secondly, the e1, e2, e3 bases might be the easier route to visualisation, since the math is simpler for them. lastly, this is hard and going to take time. this is basically inventing (/ completing the invention) of an entirely new "sphere" of math. dont underestimate its importance / difficulty, but do value your personal time first. it took complex numbers ages to find real world relevance, but quantum physics wouldnt be as easy without it. wishing you luck!
the matrix representation for i was super surprising for a second because i found it on my own in an unrelated context. i was trying to solve a linear recurrence relation by forming an infinite vector with the terms of the recurrence as components, and i ended up needing a matrix representation for a transformation that brings each basis vector to the next-indexed one (e1 -> e2, e2 -> e3, and etc.), and got sidetracked and starting looking at finite versions of this, which includes the matrix representation for i (it brings e1 to e2, e2 to e3, and e3 to e1), which makes perfect sense in the context of these numbers now
Hi Torin. If you want to have a little more challenge, please consider ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-9pR3-CqnvV8.html and other videos in the olympiad playlist.
38:49 Imo this is very nice. Provided you define the function: triexp(x) = (1 + 2cos(2pi/3 *x) / 3 You get: i^x = triexp(x) + triexp(x-1)i + triexp(x + 1)j These functions not being predefined has more to do with these numbers being a novel concept.
One thing I found is that you could use a vector representation for those numbers ! You constantly get signals that i might be either 1 or e^(2iπ/3) ; but then, you can try to map it on the vector (1 ; e^(2iπ/3) ). If you give such vectors term-by-term addition and multiplication, you get : 1 (1 ; 1) j (1 ; e^(-2iπ/3) ) e1 (1 ; 0) e2 (0 ; 1) e3 (0 ; i) Exponentiation and logarithm then happen to behave term-by-term as well. I'm a physicist, so I find the idea of i being a litteral superposition of 1 and e^(2iπ/3) very compelling.
@@Anuclano It's the main weakness of this approach : the inverse transformation is a bit troublesome. If we map a number z on the vector (a ; b+i×c ) Then 3z = (a+2b) + (a-b + c√3)×i + (a-b - c√3)×j (Or something like that, my calculations may include mistakes) It's fine because it all behaves linearly, but still, it doesn't work as nicely as when using it with the e1 ; e2 ; e3 decompisition.
@@akiel941 I have coded triplex numbers into Mathematica code so that they can be manipulated as anything else. The code: $Pre = (# /. ({j -> {1, E^(2 I \[Pi]/3)}, k -> {1, E^(-2 I \[Pi]/3)}}) /. {x_, y_} -> FullSimplify[(x/3 + Im[y]/Sqrt[3] - Re[y]/3) j + (x/3 - Im[y]/Sqrt[3] - Re[y]/3) k + 1/3 (x + y + Conjugate[y])] // FullSimplify // Expand) &;
My mind is *evaporating* , but I _understood!_ Great video! Also it seems that at the time of me writing this comment, this video is getting a spike in views. Maybe the algorithm has finally blessed this channel?
Hi Adam. If you want to have a little more challenge, please consider ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-9pR3-CqnvV8.html and other videos in the olympiad playlist.
Okay I was playing around a little bit with trying to make a representation that was constructed from the idea that there exists a basis of the cyclic group of order 3 over the complex numbers such that it can be expressed as a direct sum of three one-dimensional representations when I passed over something that got missed when you talked about idempotents. mainly your second idempotent can be expressed as the sum of two other idempotents in the field of complex numbers (1+iz+jz^2)/3+(1+iz^2+jz)/3 where z is the complex third root of unity and this matters in particular for forming an actually diagonal basis (the diagonal basis isn't just a term to throw around, but is a basis that makes a matrix diagonalize) for an equivalent representation with both i and j (also 1, but that's a given) being a diagonal matrix with i being [[1, 0, 0] [0, z, 0] [0, 0, z^2]] and j being [[1, 0, 0] [0, z^2, 0] [0, 0, z]] This representation is exactly the one you'd get from finding all the irreducible sub-modules and using them to compose a larger one
Very cool! Nice to see other people that are as obsessed with weird abstract number systems as me. ^^ Have these triplex numbers been studied by other people before, or did you discover them yourself?
They have been studied before. This topic is a pretty well-researched topic in mathematics. This is the study of algebras over finite dimensional vector spaces, and it culminates in the theory of Clifford algebras.
Triplex numbers seem to only be documented in this video, but I think that the concept of extension of the reals and number systems in general is a well-researched thing.
I personally would like more if we just defined a multiplication on ordered triplets of numbers. Say, [a₀, a₁, a₂] ⋂ [b₀, b₁, b₂] = [a₀b₀+a₁b₂+a₂b₁, a₀b₁+a₁b₀+a₂b₂, a₀b₂+a₁b₁+a₂b₀]. Then we can call j₃ = [0, 1, 0] and use it to write [a₀, a₁, a₂] = a₀ + a₁j₃ + a₂j₃²
The impossibility, the voice with now music, the typeface... it's just maths but it felt like a slowly haunting horror movie, I was waiting for a jumpscare 😰 Amazing video though, thank you!! :-D
I was working on a similar system couple of months ago. It had these identities i³=ij=ji=-1, j³=1, i²=j and j²=-i. I think it is a mix between complex numbers and split complex numbers. Let's call them "Mixed-Triplex Numbers". My N function for a mixed triplex written like a+bi+cj was a³-b³+c³-3abc. But I haven't worked on diagonal basis, exponential of a mixed triplex and natural logarithm of mixed triplex. Maybe I should work on those one day. And lastly, the matrix forms were i=[[0 0 -1]; [1 0 0]; [0 1 0]] and j=[[0 -1 0]; [0 0 -1]; [1 0 0]]
Is this basically just a bachelor's thesis you felt like making into a video? Good job on the effort. Seriously, there are PhD works with less work out into them.
The earlier parts of this video are a testament to how poorly mathematics are taught in general, and to how poorly complex numbers are understood by laypeople who are at least acquainted with them, but are no means well-versed on the subject. The host in the video mentioned how the equations j^2 = 1 and i^2 = -1 are quadratic equations, and that quadratic equations do nothing, but beg to be solved. This is a misconception. The reality is that quadratic equations are just that: equations. Solving them is certainly an important ability to have, but it is not as though there exists some metaphysical obligation to solve these equations. The equation i^2 = -1 does not imply i = sqrt(-1). In fact, the concept of the square root of a number is technically ill-defined in most contexts. Also, the equations i^2 = -1 and j^2 = 1 are not equations to solve for i and j. No, these equations are simply relationships that express the multiplication table for a number system. In the same way that a multiplication table for integers exists, which everyone learns in grade school at some point in their lives, other number systems have multiplication tables too. The multiplication table for complex numbers is the (2, 2) table where the first column has the equations 1·1 = 1, 1·i = i, and the second column has the equations i·1 = i, i·i = -1. All other complex multiplications can be derived from this table, by using the distributive and associative property of multiplication over addition, and remembering that every complex number can be written as a·1 + b·i, where a and b are arbitrary real numbers. For split-complex numbers, the situation is very similar, but the right-bottom quadrant of the multiplication table simply reads j·j = 1 instead. So what distinguishes the split-complex numbers from the complex numbers is their multiplication table. For dual numbers, the multiplication table is given ε·ε = 0. These are special cases of an algebra. An algebra is a vector space with a multiplication that has the distributive property and is compatible with scalar multiplication. The Euclidean space R^2 is a vector space over the scalar field R. If we introduce some such multiplication between vectors u and v in R^2, then we have an algebra. There are two basis vectors in R^2, {1, w}. The three possible algebras are the algebras where w^2 = -1, w^2 = 0, and w^2 = 1, and in all three, the basis vector 1 is the identity of the algebra multiplication. The triplex numbers are very similar. The primary distinction is that the triplex numbers form an algebra over the vector space R^3 instead, where the basis vectors are labeled {1, i, j}. The multiplication table is given by (1, 1) = (1, i) = (i, 1) = (1, j) = (j, 1) = 1, (i, i) = j, (j, j) = i, and (i, j) = (j, i) = 1. I am using (•,•) to denote the algebra multiplication between triplex numbers, to distinguish it from the scalar multiplication between real numbers or real numbers and triplex numbers. The above equations complete the multiplication table for the triplex numbers. Since triplex multiplication is associative, and since 1 is the identity element of this multiplication, this means the triplex numbers form a ring, and a commutative ring, at that. But unlike the complex numbers, they do not form a field, because they have zero divisors, as talked about in the video. Characterizing the zero divisors is quite tricky too, and factoring simple polynomials such as x^2 - 1 suddenly becomes difficult. If we have that x^2 = 1, then (a + b·i + c·j)^2 = a^2 + 2·a·(b·i + c·j) + b^2·j + 2·b·c + c^2·i = (a^2 + 2·b·c) + (2·a·b + c^2)·i + (2·a·c + b^2)·j = 1, implying a^2 + 2·b·c = 1, 2·a·b + c^2 = 0, 2·a·c + b^2 = 0. We can still study the triplex numbers to quite a lot of depth, though. For example, the video notices that i^3 = i·i^2 = i·j = 1. So the powers of i have a modulo 3 rotation, with 1, i, j. In other words, i^(3·n) = 1, i^(3·n + 1) = i, and i^(3·n + 2) = j. Meanwhile, j^0 = 1, as usual, and j^1 = j, but also, j^2 = i, as given by the table, so j^3 = j·i = 1. So j^(3·n) = 1, j^(3·n + 1) = j, and j^(3·n + 2) = i. This produces interesting results when considering polynomials. Also, it means that j·j^(3·n + 2) = 1, so j^(3·n + 2) = j^(-1) = i, and similarly, i^(-1) = j. So the basis vectors are invertible. These are the type of things we can study. Any function that can be defined on an arbitrary ring structure can be defined on the triplex numbers, and using the order topoology of the real numbers, the corresponding topology on the triplex numbers can also be done, and so functions such as the exponential function can be defined, and this also means we can do calculus, though only to a limited extent. Also, the video points out how the complex numbers and the split-complex numbers are subrings of the triplex numbers, but it turns out that the dual numbers also form a subring. So all the algebras of R^2 are subalgebras of R^3, and the triplex numbers are the smallest algebra for which this is true. This makes the triplex numbers important for mathematical study, and in modular arithmetic as well. Another interesting property is that every element that is not a zero divisor has a multiplicative inverse. So there is no such a thing as an irreducible element or a composite element: you only have units, or zero divisors. This is another important property of the triplex numbers. As you can see, there is more to algebraic structures than just being able to solve equations. Algebraic structures are about modelling situations and creating a rigorous foundation for different kinds of arithmetic. The fact that the complex numbers happen to "create" solutions for the equation x^2 + 1 = 0 is just a consequence of the fact that the complex numbers are the algebraic closure of the real numbers.
Despite your valiant effort, I literally fell asleep. But never mind, that has it's value too, I'm saving this video for sleepless nights ;). (Sorry, I really fell asleep, just woke up and couldn't resist. It's a great video, and an interesting concept, but it does need a fresh mind.)
18:56 It's not difficult to prove. If x is your number, represented as a matrix, then its conjugate will be x'=det(x) x^-1. The determinant of that new matrix depends on the dimension d of your representation. You get that, N(x') = det(x)^d det(x)^-1 = N(x)^(d-1) So, complex numbers can be represented with d=2, and their simplest norm will have the property N(x)=N(x'), but for tri-complex numbers, you need d=3, so there's an extra power of 2.
At 18:50 you say N(x')=N(x)^2, I was thinking about this statement, so, I start with xx'=N(x), now apply N to both sides to get N(xx')=N(N(x)), now the RHS is just N(x)^3, because in the matrix representation, you just have the determinant of a 3x3 matrix with only N(x) in the diagonal and zero everywhere else, so you have N(x)N(x')=N^3(x) so you get N(x')=N(x)^2. I think you should have included this in the video, would have given the viewer some idea about the use of determinants here for the "norm" here.
A few months ago I was kind of attempting to kind of explore the possibility of a kind of number system with infinite sums & probability distributions.. e.g. Sum(1 - 1 + 1 - 1 + 1 ...) = Dist(0: .5, 1: .5)... because the sum doesn't converge, but it spends 1/2 of its time at value 0 and 1/2 at value 1... so then if you pull the first term of the sum out, that implies that Dist(0: .5, 1: .5) = 1 + Dist(0: .5, -1: .5) and ends up getting into some interesting identities that are fun to explore. I'm nowhere near good enough at math to do anything about it, but it's fun to play around with, and I was wondering if there's anyone doing anything like that you know of that I can read up on? Curious to see what it would look like to describe the gaussian distribution with those represenations... I guess maybe it would have to be an infinite sum of infinite sums for the distributions to be continuous... or I guess a series with somehow infinte cycles?
Nice! Although it is not new to me. Dennis Morris has also worked out some of this. Moreover, he has added something that is not mentioned here. You can define other trig functions which have no ordinary angles, but have angles that are 2 dimensional vectors. He has also shown that you canh construct multi-complex numbers of any number of dimensions. In his system, the complex numbers and the split complex numbers are just two-dimensional numbers. The triplex numbers are two types of two dimensional numbers, four types of three-dimensional numbers, and eight types of four dimensional numbers, one of which is the quaternions, and another is the anti-quaternions, and there are six others.
The existance of a root of 1 different to 1 and -1 could potentially be explained by the fact that squaring a number is not an injective operation in the reals, and we have no reason to assume it would be defined as injective when applied to numbers beyond the set of the reals.
I want to make a new imaginary component, m, in which I say that m = 1/0. So it resolves (-n)! where n is a real integer, and 0/0 = 0m = 0. Also there are some things to look out for, like not pairing m with a real number. Here is why: Suppose you could do that safely. If so, what is 0*m? Well it could be 0m, which is 0, or it could be, by definition, (1/0)*0 = 1. We get 2 different answers. So you have to pair up m with a real number at all times except for the definition itself. Edit: So this already exists as an extended real number line, as pointed out by Blinded.
This is outstanding! It would have been nice to explore to what extent the triplex numbers form a 3-dimensional real algebra (i.e. demonstrating the commutative and associative laws for multiplication, and the distributivity of multiplication over addition, etc.) - the matrix representation would make the computations for this a lot easier I think. People often say that there is no 3-dimensional analogue of the complex numbers (including Hamilton himself, who tried for years to create a 3-dimensional analogue of C until he realised he needed 4 dimensions), but you went and did it anyway! :D
I believe that Hamilton had a more restricted icriteria for a 3-dimensional analogue for C, which this doesn't fulfill. Notably, he may have looked for a ring where the conjugate had identical properties to that in C, as this ring doesn't have symmetry of the conjugate. It's certainly interesting to see how the nonequivalent analogues work here, though.
Just like how e^x which is equal to the power series of x^n/n! Is the solution of y’=y and how cosh and sinh which are the respective power series of only odd or even parts of the above series are the solutions of y”=y, I think that A(x), B(x) and C(x) are the solutions of y’’’=y and they can be related to exponentials and cosines and sines, pretty cool
This gives me group ring vibes. In particular, the triplex numbers feel like the group ring of the cyclic group of order 3 over the real numbers; similarly, the split-complex numbers feel like the group ring of the cyclic group of order 2 over the real numbers. Huh.
Great video! You did not actually find the diagonal basis. In the diagonal basis the triplex number should be representable as the identity matrix: the rules of e2*e3 show the non-identity matrix of complex numbers. Therefore a similar diagonal representation should be derived from your e2 and e3. The same would occur if you diagonalize and find the eigenvectors of the triplex numbers, e.g. the normalized eigenvectors for {{a,c,b},{b,a,c},{c,b,a}}, where e1 stays the same, but e2 and e3 look a bit more strange, but complementary.
*before watching the video* So, you want to find numbers that have these properties? Yeah, that's easy, using the typical definition of i: i^2 = -1 let's call new numbers n and m; i know that: n^2 = m m^2 = n nm = 1 well, you can easily see one way to solve it: n = 1 + 0i; m = 1 + 0i but that would be to easy, so the next thing: -1 as a solution for both of them nearly works but n^2=m and m^2=n are impossible in this context If I assume that m != n (!= means doesn't equal to) then both have to be imaginary numbers, and both have to cancel each others imaginary part out, so I know it is written as n = c + di m = c - di So substitute in nm = 1 (c + di)(c - di) = c^2 - (di)^2 = c^2 + d^2 = 1 c^2 + d^2 = 1 But wait, there is one more thing n^2 = m nm = 1 n^3 = 1 So we really need to find all solutions for n^3 = 1 And using our friend's, Euler's formula: e^(i*a) = cos(a) + isin(a) We can assume that multiplying imaginary numbers sums their angles. Why? Because: e^(i*a) * e^(i*b) = e^(i*a + i*b) = e^i(a+b) And we know that n^3 = 1 So for angle a, 3a = 2π So a = 2/3 π Now we calculate w = cos(2/3 π) + isin(2/3π) We get w = -1/2 ± √3/2 i So, let's say n = -1/2 + √3/2 i m = -1/2 - √3/2 i If you don't believe me, you can check the check the dependencies yourself
