Quaternions

[@angelmendez-rivera351]

I think people watching the video are getting confused on how it is possible that ijk = -1. ii = -1, and jj = -1. It should be noted that ij = k is essentially a definition: k is just a name we use for the product ij. Now, notice that i(ij)j = (ii)(jj) = (-1)(-1) = 1, yet (ij)(ij) = i(ji)j = -1. Therefore, ij = -ji. This is important to realize. Now, from (ij)(ij) = -1 and ij = k, it follows that ijk = kk = (ij)(ij) = -1, hence ijk = -1. Notice that this is not in contradiction with ii = jj = kk = -1, because (ijk)(ijk) = (-1)(-1) = 1, but (ijk)(ijk) = -(ijk)(ikj) = (ijk)(kij) = -(ijk)(kji) = -i(j(kk)j)i = ijji = -ii = 1, so there is no contradiction. The confusion lies with the fact that viewers are mistakenly saying ij = ji, when in fact, ij = -ji.

[@pinklady7184]

Thank you for another great video on mathematical notations. Very helpful and many thanks.

[@mathflipped]

Here H stands for Hamiltonian quaternions because there are also so-called generalized quaternions.

[@michaeloconnor8071]

Great videos btw ! Been watching your Real Analysis series since I'm taking it this coming semester.

[@ThunfischXXx]

If it makes sense for the specific symbol, could you include more visualizations?

[@tayebtchikou1646]

Thak you very much for this series, it was very helpful to me. I hope to make other videos about the hyper complex numbers set, gamma matrices and other symbols that are most used in Physics.

[@brightsideofmaths]

Great suggestion!

[@HungDuong-dt3lg]

How is ijk=-1 when i^2 =j^2=k^2 = -1 ? Like if you square both sides of the equation you get negative 1 on the left and positive one one the right. I'm confused.

[@camicus-3249]

If I understand you correctly, you're saying: (-1)^2 = 1 (ijk)^2 = i^2 * j^2 * k^2 = -1 * -1 * -1 = -1 If so, the key thing to realise is that when squaring ijk, (ijk) * (ijk) does not equal i^2 * j^2 * k^2. If they were commutative you could happily simplify ijkijk by shifting the letters around, but for quaternions you *can not* shift them around. It can be simplified, though. (Just with extra care)

[@brightsideofmaths]

@@camicus-3249 Thanks for the correct explanation! The noncommutativity is something one can easily forget.

[@stashdsouza4602]

If the complex numbers lead to the algebraic completion of the reals, then what do the Quaternions lead to?

[@brightsideofmaths]

Please note that the quaternions don't form a field anymore.

[@LLL_elder]

As noted, they no longer form a field and are constructed for geometric reasons. Complex number multiplication encodes 2D dilation and rotation. In that same way, quaternions perform 3D dilation and rotation

[@stashdsouza4602]

@@brightsideofmaths Yes. Indeed, commutativity of the multiplication has to be forgone.

[@elreturner1227]

I’m about to sound so dumb right now I’m to curious to just give up now if j and k squared both equal -1 and i squared equals -1 then what’s the difference do they just have different properties or is it something else?

[@brightsideofmaths]

Compare it to the complex numbers: i and (-i) both square to -1, but they are different numbers.

[@mtssmreddy5172]

Each I, j & k represent different dimensions like i is a orientation that will land you in certain dimension so does j & k in their respective dimensions

[@elreturner1227]

@@mtssmreddy5172 I get that but I can be expressed as sqrt(-1) what can j and k be expressed as

[@mtssmreddy5172]

@@elreturner1227 i = j = k = _/ -1 ( square root of -1 ) but still I, j & k are different so we can't use normal multiplication between i or j or k because it convey's i, j & k are equal each landing on same imaginory line but in reality it isn't so we need to invent new specific multiplication system for them i.e i * j = k, here we don't take normal multiplication rule, we invent specific multiplication rule for the new multiplication The use of _/ -1 is to land real numbers

[@mtssmreddy5172]

@@elreturner1227 if we define specifically i is _/ -1 which lands a real number on particular imaginary line i j is _/ -1 which lands a real number on particular imaginary line called j but not i j is _/ -1 which lands a real number on particular imaginary line called k but not i or j

[@michaeloconnor8071]

I'm confused how ijk=-1 when i,j,k=sqrt(-1). Wouldn't that mean ijk=-sqrt(-1)?

[@malawigw]

never think of them as sqrt(-1), instead think of them as basis elements with a certain multiplication table en.wikipedia.org/wiki/Quaternion#Definition The definition of i,j & k are such that they ahve anticommuting multiplication, ij = k and ji = - k and kk = -1 thus you have ijk = kk = -1

[@brightsideofmaths]

For having sqrt(-1), you need a definition of sqrt first. We have a common one in the real numbers, but you probably don't know the one in the complex numbers or even for the quaternions. Therefore, best avoid using sqrt(-1) altogether :)

[@michaeloconnor8071]

@@brightsideofmaths Okay, thankyou ! The other explanation clarified things too. Very hard to work with numbers and variables that look like them but dont act like them haha.

[@brightsideofmaths]

@@michaeloconnor8071 Yes, if you are used to work with matrices, you should look at a matrix representation of the quaternions. Then everything is clear immediately because for matrices we do not expect commutativity or a unique definition of sqrt( ).

[@Florian.Dalwigk]

Finally :)

[@brightsideofmaths]

I see, you were waiting for these numbers :D

[@ekisvioleolivaradamos6701]

I really learned a lot from this channel even if I'm bad at this... Hehehe