Комментарии •

  • @protocol6
    @protocol6 6 лет назад +31

    I love this pair of videos plotting Complex and Split-Complex (Hyperbolic) numbers. I'm sad that you didn't complete the set with Duals.

    • @xyz.ijk.
      @xyz.ijk. Год назад +1

      He was concerned they would fight 😅 and it would all amount to naught.
      But he also didn't include the x^2=1 element ( I forget the symbol for this one ... the epsilon is for the duals ... what's thus one?).😊

    • @protocol6
      @protocol6 Год назад

      @@xyz.ijk. i²=-1, ε²=0, and j²=1 is the usual convention for mathematicians. For EEs, j²=-1 (because of the symbol for current) which is annoying. I usually just refer to the broader set as \kappa since 𝜿² ∈ ℝ covers all of the conic rotations except the parabola.

    • @xyz.ijk.
      @xyz.ijk. Год назад

      @@protocol6 I appreciate what you wrote ... sorry that my attempt at humor on the dual numbers fell flat. I'm familiar with them and have an engineering background -- so I'm very familiar with the frustrations of j²=-1 ! But in addition to ε²=0 (I couldn't get the εpsilon on my cell earlier), there is another system where [symbol]²=+1. I don't remember the symbol, or whether it was Michael Penn or it was someone else, but they were building a system and it was considered to "complete" the squares for the three items: -1, 0, 1.

    • @protocol6
      @protocol6 Год назад

      @@xyz.ijk. I caught the joke. It was very punny.
      I've not seen another symbol for √1 but I'd be curious to see it if you can figure out what it was. The only thing else I can think of might be something to do with 𝔖 (\mathfrak{S}) which is sometimes used for unity.
      Perhaps the lowercase? 𝔰, ß or ſ

    • @protocol6
      @protocol6 Год назад +1

      @kwccoin3115 Complex are only the -1 version. With 0, Rotations are linear and they are properly called Dual numbers. They model Gallilean relativity (with infinite speed of light) and also are useful for automatic differentiation. With 1, they are called Split-Complex and model a flat Minkowski space-time with rectangular hyperbola rotations. Elliptical rotation is anything 0 other than 1 are the non-rectangular hyperbolas. I don't think you can get a true parabola without perspective but that's something you can do when you move to 4 dimensions. Just swap out the second coefficient of all three with a 3-vector (and use the dot product where necessary in multipliction) for the 4 dimensional quaternion versions of all three.

  • @plasmaballin
    @plasmaballin 5 лет назад +16

    The split complex plane is a 2-D Minkowski spacetime, and the magnitude is the spacetime interval. This could be a very useful tool for understanding special relativity.

    • @johnsherfey3675
      @johnsherfey3675 5 лет назад +5

      To quote Wikipedia :
      "Since the early twentieth century, the split-complex multiplication has commonly been seen as a Lorentz boost of a spacetime plane.[2][3][4][5][6][7] In that model, the number z = x + y j represents an event in a spacio-temporal plane, where x is measured in nanoseconds and y in Mermin’s feet. The future corresponds to the quadrant of events {z : |y| < x }, which has the split-complex polar decomposition z = p(e^(aj)). The model says that z can be reached from the origin by entering a frame of reference of rapidity a and waiting ρ nanoseconds. The split-complex equation
      e^(aj) * e^(bj) = e^((a + b)j)
      expressing products on the unit hyperbola illustrates the additivity of rapidities for collinear velocities. Simultaneity of events depends on rapidity a;
      ....
      en.wikipedia.org/wiki/Split-complex_number
      You can go down to the links at the bottom to see where it's use and refrenced.

  • @user-hh5bx8xe5o
    @user-hh5bx8xe5o 6 лет назад +12

    Though this system may seem weird at first, once you understand it's the algebra of the "swapping" or transposition of vectors if you think in a combinatorial way or reflections in a geometric way, it's much more easier to grasp.
    Basically, these numbers describe the functions f: (v,w) -> (w,v) (and (w,v) -> (v,w) or any reflection through a line so that f°f=Id) in the plane just like the complex numbers arise from plane rotations and scaling (or in functional terms f: (v,w) -> (w,v) and (w,v) -> (-v,-w) so that f°f=-Id)).
    The paths along hyperbolas are "continuous" reflections where the tangent vector to the curve is the reflection of the position vector just like on a circle the tangent vector is orthogonal to the position vector.
    In summary, where complex numbers described the algebra of continuous rotations and dilatations (spirals or also called vortexes), the split complex do the same with continuous reflections and dilatations (sometimes called anti-vortex).

    • @dirrelito
      @dirrelito 2 года назад

      Curious. In the sense as a lie algebra relating to a lie group?

  • @DDranks
    @DDranks 8 лет назад +20

    Thanks for the video! I knew about the hyperbolic trig functions, but split complex numbers were new to me, so I got some mind=blown moments. Very enlightening.

    • @Math_oma
      @Math_oma 8 лет назад +9

      +Pyry Kontio
      You should check out the video on the Lorentz transformation. It's quite remarkable that the split-complex numbers can be used in special relativity to express hyperbolic rotations in spacetime.

  • @jacobeek65
    @jacobeek65 6 лет назад +21

    Fascinating stuff! Would it be possible to define theorems in a Split-Complex analysis analog to theorems in Complex Analysis? For example: Cauchys integral formula, the Residue Theorem etc?
    Thanks for the video.

  • @adumont
    @adumont 2 года назад +8

    There's also their cousins, the Dual Numbers where instead of I and J, we use epsilon such that epsilon^2=0. It has very interesting properties related to differentiation.

    • @xyz.ijk.
      @xyz.ijk. Год назад +2

      Im glad you wrote this. I have a question for you. I'm familiar with Dual Numbers epsilon^2 = 0 system ... and I know there is also a system where "something"^2=1, similar to the Dual Numbers in its context (separate from the quaternions, and split complex numbers ... and even Graphman's triplex numbers (really excellent)), but I can't find the "something"^2=1 where it is treated like the Dual numbers. Do you know this system? I saw a video on it -- perhaps during the Summer of Math marathon?

  • @daboffey
    @daboffey Год назад +1

    One consequence of this is that the split complex numbers are no more a metric, as |z| = 0 does not imply z = 0, furthermore, |z| < 0 for some z.

  • @ericbischoff9444
    @ericbischoff9444 7 лет назад +6

    Thank you for the great video! It's a fascinating topic, and the explanations make it really simple.

  • @61rmd1
    @61rmd1 4 года назад +3

    Very nice video and topic; I remember something similar when I studied relativity many years ago...I have two questions: 1) I don't see any link to something like the "argument" of the split complex number, while for the classical complex ones it plays a crucial role; 2) could we imagine something analogous to the "split-quaternions", as a generalisation of split-complex? I'm thinking on a kind of hypersplit-complex...I don't know much about it, I've never worked with split complex numbers...Thanks for sharing this beautiful stuff

    • @Tokhaar
      @Tokhaar 3 года назад

      The argument is x...

    • @SimonClarkstone
      @SimonClarkstone Год назад

      I'd guess that since the argument of a complex number Arg(x+yi) = atan2(y,x), the argument of a split-complex number is the hyperbolic equivalent: Arg(a+bj) = atanh2(b,a) which will often (always?) = atanh(b/a)
      Apparently that's ½ln((a+b) / (a-b))
      I think that's also the thing that adds normally when combining velocities?

  • @plushloler
    @plushloler Год назад +1

    At 27:02 you say that numbers lying on the hyperbola passing through 2 and -2 will have a magnitude of 2, but that’s not correct. They’ll actually have a magnitude of 4. If you graph x^2 - y^2 = 2, the hyperbola will pass through the square root of 2.
    So, numbers lying on a hyperbola x^2 - y^2 = z, will have the magnitude z^2.

    • @plushloler
      @plushloler Год назад

      No wait I’m wrong, you’re right. I’ll delete my comment :/

  • @angelmendez-rivera351
    @angelmendez-rivera351 5 лет назад

    Another way to explain the distinction between complex number multiplication and split complex number multiplication is by explaining the “imaginary” component as the determinant of some matrix and the scalar or real component as an inner product of two vectors and with the distinction between ac - bd and ac + bd coming from the inner product metric. The complex number metric has (1, -1), whereas the split complex metric has (1, 1), and the difference between 1 and -1 aligns with + bd and - bd. The dual numbers can be explained in exactly the same way, except that the metric would be (1, 0). Strictly speaking, these metrics would not represent actual inner products, since the manifold is not Riemannian, but it is a generalization of the inner product.

  • @OmegaCraftable
    @OmegaCraftable 8 лет назад +3

    this is really fascinating, I had never thought about generating a form of complex numbers around the hyperbolics rather than the trigonometrics!

    • @Math_oma
      @Math_oma 8 лет назад +1

      +OmegaCraftable Yeah, apparently it also has some role to play in the math behind special relativity, although I'm not too knowledgeable about that topic. But, as you can see, this number system is no more difficult to learn about than the standard complex numbers.

    • @dantebroggi3734
      @dantebroggi3734 8 лет назад

      Split-complex numbers have bearing in relativity because, in time-space, the distance between an event an an arbitrary origin is:
      d^2 = x^2 + y^2 + z^2 - t^2
      The last 2 terms of which is the distance for Split-complex numbers.
      For more info, via Sci-Fi, I recommend Greg Egan's Orthogonal Series at:
      www.gregegan.net/ORTHOGONAL/ORTHOGONAL.html

    • @46pi26
      @46pi26 6 лет назад

      Mathoma Yeah, one of the easiest places to see naturally hyperbolas is actually in QM for me, with the wavefunction, e^i(k*x-omega*t), where the "t" term is negative and the "x" term is positive. As such, hyperbolic magnitudes are easy to spot.

    • @Igdrazil
      @Igdrazil Год назад

      There is FOUR types of complex numbers. One circular euclidian one generated by the usual i who satisfies i^2=-1. Two hyperbolic types generated by j and k who satisfies j^2=k^2=+1. And a fourth projective one € who satisfies €^2=0

  • @good4usoul
    @good4usoul 3 года назад +1

    11:12
    0. Define magnitude for complex numbers.
    a: Multiply the two complex numbers and put in a form of real + imaginary
    b. Perform the *complex magnitude* squared operation, by taking the square of the real part plus the square of the imaginary part.
    c. Expand out the multiplication, and cancel out the abcd terms
    d. Fully factor out the result
    e. Convert the factors into squares of *complex magnitudes.*
    For split complex numbers, replace step 0 with the definition of split complex numbers.

    • @xyz.ijk.
      @xyz.ijk. 2 года назад

      This was an extremely useful comment. Thank you.

  • @Kosekans
    @Kosekans 7 лет назад +4

    Is it just me, or do you have a very nice to look at, almost beautiful, kind of handwriting? You should sell T-shirts with that.
    Well, let me know, if this is weird ... xD

    • @46pi26
      @46pi26 6 лет назад

      Yeah it's not just you; I'd buy a t-shirt with his handwriting in a heartbeat.

  • @Tabu11211
    @Tabu11211 2 года назад +3

    does this have something to do with photons?

  • @thedarkspeedninjashadittsux
    @thedarkspeedninjashadittsux 6 лет назад +5

    8:24 Does that mean |j| = i? Since |j| = √(0²-1²)

    • @EtherDais
      @EtherDais 4 года назад

      On this note, what is the square root of j? I have a solution which requires i

    • @arvindganesh7539
      @arvindganesh7539 2 года назад

      Bingo, that is the reason why j =/= 1 and j =/= -1, which are potentially true from the fact that j^2 = 1. The absolute value works as follows: |1| = 1 and |-1| = 1 under the definition of absolute value used for real number field. Whereas | j | = i, under the definition used for split-complex numbers (and the real number definition is not applicable here). So j does not equal 1 or -1

    • @Igdrazil
      @Igdrazil Год назад

      No. The moduli in « i » world and the one in « j » world are distinct. Its square, the QUADRANCE is defined for each in a distinct way :
      In « i » world where i^2=-1 : Q(z)=Q(a+bi)=|a+bi|^2=(a+bi)(a-bi)=a^2+b^2 , so Q(i)=1 and |i|=1 (or -1 if the moduli was not defined as a real positive number)
      In « j » world where j^2=+1 : Q’(z) =Q’(a+bj)=|a+bj|^2=(a+bj)(a-bj)=a^2-b^2 , so Q’(j)=-1 and |j|=i or -i, which shows the ambiguity and troublesome of the « moduli » concept. Reason why it’s better to stay with the QUADRANCE concept. Negative quadrance correspond to « space like » vectors (of « imaginary length » whatever that « means »…).
      In fact, the i and j can be clearly seen in 2*2 real matrix representation as : i=[0 1 / -1 0] and j=[0 1 / 1 0]. It’s as simple as that. No need for « esoteric mysteries ». And there is even a third type of complex numbers, also hyperbolic as j, generated by k=[1 0 / 0 -1] who satisfies obviously k^2=j^2=-i^2=+1. They form a « twin » non commutative algebra, alternative to Hamilton Quaternions which have same i but different j and k, all three squaring to -1 and satisfying ijk=-1. Quaternions are less simple since j and k CANOT be represented by real 2*2 matrices. Only by complex 2*2 matrices or real 4*4 matrices. Quaternions are suited for 3D euclidian rotations. Whereas those Dihedrons are suited for hyperbolic rotations, as in Relativity.

  • @diarmuidkeane1
    @diarmuidkeane1 Месяц назад

    should the definiton be stricter ie : j^2 =1 _and_ j not equal to +/-1 . ?

  • @johnhare8208
    @johnhare8208 4 года назад

    The numbers on the x=y & x=y- asymptote could be x=0 in the euclidian plane if you were to superimposed the asymptotes with a 90 degree bend then all the hyperbolic curves would be straight lines like the asymptotes

  • @farklegriffen2624
    @farklegriffen2624 Год назад +1

    Okay but why didn't you explain how |0+1j| = i?? What does it even mean for something to have a magnitude of i?

    • @Igdrazil
      @Igdrazil Год назад

      It doesn’t ! The moduli concept is ill defined in hyperbolic « j » world. And if one wants to use the notation, it doesn’t mean anymore a « length » or a « magnitude ». Best is to stay at the level of the QUADRANCE and avoid this « moduli » ill defined concept.

    • @thes7274473
      @thes7274473 6 месяцев назад

      It doesn't mean anything to have a magnitude i. Magnitude is defined by a function called a norm, and two of the requirements for a norm are that magnitudes can only be nonnegative real numbers and that 0 is the only number allowed to have a magnitude 0. The modulus on split-complex numbers breaks both of these requirements, and is therefore not a norm.

  • @sdmartens22
    @sdmartens22 7 лет назад

    Very good video, can't wait to watch the rest!

  • @usdescartes
    @usdescartes Год назад

    I mean, I get how j makes things easier, but couldn't we just do the same by keeping track of the cosh part and sinh part of a number during our calculation? We could color the sinh part of a number red and just do the same thing. At ANY time during our calculation, we could take j away from both sides and get a true statement, yes?

  • @samuelgage3598
    @samuelgage3598 Год назад +2

    Why is j imaginary?

  • @diarmuidkeane1
    @diarmuidkeane1 Месяц назад

    first thing i thought of was spacetime diagram when seeing the transformation at the beginning - also reading here the connection with minkowski spacetime - so its a bit surprising how split-complex number isn't more prevalent in the "everyday"
    - i think the name doesn't help - how about (circular) complex numbers and hyperbolic complex numbers

  • @henrmota
    @henrmota 3 года назад +1

    Isn't this related to conjugate complex? At the time I am at 10:09

  • @JeanDAVID
    @JeanDAVID 2 года назад

    Can we assume for space time geometry a split complex computation compatible as t has a negative sign wrt espace coordinates

  • @AJ-et3vf
    @AJ-et3vf Год назад

    Awesome video! Thank you!

  • @collegemathematics6698
    @collegemathematics6698 2 года назад +1

    Something doesnt quite right!! The invention of compex numbers come from the need of solving x^2=-1 and other equations like it.
    When we put j^2=1 there no problem here. its just +1 or -1. So why there is j in the first place?!

  • @wherestheshroomsyo
    @wherestheshroomsyo 6 лет назад

    It looks all of the split complex number identities hold when I substitute j = 1 or j = -1. Including the e^(jx) = cosh(x) + jsinh(x).

  • @atlkurtulmus6993
    @atlkurtulmus6993 Год назад

    @30:45 shouldnt AB be located to the north of A?

  • @craigwall9536
    @craigwall9536 6 лет назад

    I like this video. I'll be back for more.

  • @gguevaramu
    @gguevaramu 5 лет назад +1

    Mathoma. I saw a few months ago your series of complex, split-complex and dual numbers. I also saw all about quaternions. So, quaternions are the generalization of complex space rotation.
    If you make an analogy with split-complex numbers, they permit you to do the unit hyperbolic rotation only in the plane. now that I am studying special relativity, I wonder if it is possible to have something equivalent to quaternions for hyperbolic rotations in the 4-Dim space-time? A formulation like this could be very helpful for special relativity.
    On the other hand, did you make a visualization of dual numbers like you did for circular numbers and hyperbolic numbers?
    In advance. THANKS A LOT FOR YOUR JOB

    • @Math_oma
      @Math_oma 5 лет назад +1

      Yeah, check out the geometric algebra material (and the texts I recommend) because it's easy to formulate all this in that system.

  • @danielprovder
    @danielprovder 6 лет назад +2

    where is the dual number one?

  • @Rocknrolldaddy81-xy8ur
    @Rocknrolldaddy81-xy8ur 16 дней назад

    Asymptotes are meridians?

  • @platficker
    @platficker 7 лет назад +1

    How about split complex numbers with complex coefficients?

    • @Math_oma
      @Math_oma 7 лет назад

      +platficker
      Explore it.

    • @EtherDais
      @EtherDais 4 года назад

      Tesserines

    • @百合仙子
      @百合仙子 2 года назад

      A 3-D Minkowski spacetime?

    • @Igdrazil
      @Igdrazil Год назад

      These are the well known Hamilton Quaternions.

  • @imnimbusy2885
    @imnimbusy2885 11 месяцев назад

    I obtained the same matrix form for split complex numbers by defining a Real Number as r= p + n where p is positive real (in set R+) and n is negative real (in R-):
    1. n1, n2 in R-; n1*n2 in R+
    2. p1, p2 in R+; p1*p2 in R+
    3. p1 in R+\{0}, n1 in R-; p1*n1 in R+
    4. 0 is in R+
    5. r1 in R; r1*0 = 0
    6. p1 in R+, n1 in R-; p1*0 = 0 = n1*0
    for p, q in R, p = a+b, q= x+y, or p = , q=
    a, x in R+,
    b, y in R-,
    (a + b)(x + y) = … =
    Expand by using distributive and commutative properties and 1-6.
    The expansion represents the split-complex matrix [a b, b a]

  • @maxwellsequation4887
    @maxwellsequation4887 3 года назад

    Awesome!!!!

  • @rasitcakir9680
    @rasitcakir9680 Год назад

    Amazing

  • @michaelkasprzak6064
    @michaelkasprzak6064 7 лет назад

    Is this the same j as in quaternions

    • @Math_oma
      @Math_oma 7 лет назад +2

      +Michael Kasprzak
      No, because it squares to +1, not -1. The letter 'j' is merely a letter, it's the algebraic property that's important.

    • @Igdrazil
      @Igdrazil Год назад

      No this j is the DIHEDON one with 2*2 real matrix representation j=[0 1 /1 0]. The Hamilton Quaternions do not have 2*2 real matrix representation, except for the « i » sub algebra part, which is common to Dihedrons and Quaternions. But appart from that, for Quaternions representations, 2*2 (classical) complex numbers matrices is needed, or 4*4 real matrices. Which means that there is a second hyperbolic type of complex numbers, generated by k with 2*2 real matrix representation k=[1 0 / 0 -1], which generates, with 1, i and j, the Dihedrons algebra, twin to the Quaternions, but much simpler though partly hyperbolic, thus much richer.

  • @chamelius
    @chamelius 2 года назад

    OF COURSE, the point I picked to track st the beginning was The Origin...

  • @BRYDN_NATHAN
    @BRYDN_NATHAN 2 года назад

    thank you.
    #(ei)x=sinx+icos

  • @8443rslmm
    @8443rslmm 6 лет назад +1

    Also the zero-complex numbers exists. h^2=0 is new imaginary number. blog.hani.co.kr/reslieu/58147

    • @JCCyC
      @JCCyC Год назад

      h is the infinitesinal! These numbers are properly ordered like the reals, and the h coordinate is the "tiebreaker." Oh, you provide a link. Probably all that is there. Gonna read... Page is offline. :(

  • @CasualGraph
    @CasualGraph 8 лет назад +1

    This relation to hyperbolics was perhaps the only thing that could have made me care about split complex numbers, I just wish the video was a bit more accessible.

    • @Math_oma
      @Math_oma 8 лет назад

      Perhaps you could let me know which parts were inaccessible, just to keep it in mind for future videos.

    • @CasualGraph
      @CasualGraph 8 лет назад +1

      Well, honestly it's a video production quality thing. The microphone makes you sound constantly muffled, yet it still picks up breaths. Your overall vocal performance is fine when you're actually talking, but it's really slow. I assume you're going unscripted, and the overall effect of that is that you have to pause in the video to get your words together. While your "um"s and "so"s are kept down, there's also the occasional stutter that I imagine comes from the lack of a script. None of that compares to the amount of silence generated whenever you're writing something out in the video. You might've noticed that Numberphile will tend to edit out the act of the writing since viewers can read faster than hands can write, and they don't want a seven minute video becoming twenty five. Your timing issues make it so that the best way to watch this 40 minute video involves periodically jumping forward to skip the writing pauses and find out where you were going with whatever you were saying. The fact that the overall video is rather difficult to watch is the largest thing getting between your presentation of this information and the people who would want to see it. Hope I could help.

    • @Math_oma
      @Math_oma 8 лет назад +2

      I appreciate your criticism and am working to gradually get better at the cosmetic issues. However, I respectfully reject most of your stylistic advice. My aim in my videos is to go through technical details behind the topic of the video and that takes time. I do not do Numberphile-style videos (pop math) where where almost no substance is covered and the viewer likely cannot do any calculations on his own after viewing a video. As a result, my viewers and subscribers tend to be of a higher caliber than Numberphile's (compare the two comment sections) even though I likely won't make any videos as popular as theirs.

    • @CasualGraph
      @CasualGraph 8 лет назад +1

      Okay, now I'm a bit confused. I can appreciate your desire to focus more on depth than pop appeal, but I don't seem to recall ever having mentioned any of that. Not to sound too defensive, but my response was more about cutting down silent time, not how to alter the content for a broader audience. I recommended cutting out an unnecessary amount of silence, not halving the content and making titles more catchy. I guess I'm a bit bothered that you're rejecting "stylistic advice" on grounds that I don't think apply to them.

    • @Math_oma
      @Math_oma 8 лет назад

      Well, you brought up Numberphile and you and I both know that Numberphile is pop math. In particular, its splicing out of proof/justification (an important aspect of math) suits the genre. I can understand your point about cutting down on silent time if I'm doing something tedious like copying down an equation and I'll try to minimize it. I could always stick some Justin Bieber in the background if it gets boring.

  • @sgurdmeal662
    @sgurdmeal662 4 года назад +1

    Alolan complex numbers

  • @kaleido9631
    @kaleido9631 Год назад

    ........ but 1 squared already equals 1. It's not an imaginary number. Not saying it's not valid, i just... I AM CONFUSION.

  • @BboyKeny
    @BboyKeny Год назад

    It seems similar to Geometric Algebra

    • @CouchTomato87
      @CouchTomato87 Год назад

      That's because it's included in GA. You can have elements that square to -1, 0, or 1.

  • @fuseteam
    @fuseteam 6 лет назад +3

    i'm inclined to define k²=i

    • @thedarkspeedninjashadittsux
      @thedarkspeedninjashadittsux 6 лет назад +1

      Fuseteam Then k = (1+i)/√2

    • @sergioh5515
      @sergioh5515 6 лет назад +3

      The Dark_Speed_Ninja/ShadikkuX that's if k is a number that you solve for but he obviously defined k in such a way that it is it's own respective number

    • @sergioh5515
      @sergioh5515 6 лет назад +2

      The Dark_Speed_Ninja/ShadikkuX it's like saying j^2=1 but j IS not necessarily 1!

    • @thedarkspeedninjashadittsux
      @thedarkspeedninjashadittsux 6 лет назад +1

      Sergio H Ok, that sounds like this number k would have interesting properties.

    • @johnsherfey3675
      @johnsherfey3675 5 лет назад

      I did i^2 = 0 which is interesting I want to do pi, e and phi next.

  • @HoSza1
    @HoSza1 Год назад

    That other guy who also do backflips in his math videos, he already told us about split complex and double complex numbers so yeah...

  • @namename8986
    @namename8986 2 года назад

    video too quiet

  • @46pi26
    @46pi26 6 лет назад +20

    0/10 not enough vector cooking

  • @Hhhhhhhhhh767
    @Hhhhhhhhhh767 7 лет назад

    This hearts my head coz im dumb lol