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Comments on How $ijk=\sqrt{1}$?

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How $ijk=\sqrt{1}$?

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In the Wikipedia page, I can clearly see that $$i^2=j^2=k^2=ijk=-1$$

But if we consider them separately

$$i=\sqrt{-1}$$ $$j=\sqrt{-1}$$ $$k=\sqrt{-1}$$

So $$ijk=\sqrt{-1}\sqrt{-1}\sqrt{-1}=(-1)^{\dfrac{3}{2}}=\sqrt{-1}$$ It totally doesn't satisfy what I was looking for. In section, "Multiplication of basis elements" they assume that $ij=k$ and $ji=-k$. Even I can't find out why they are non-commutative (I know that matrices is non-commutative but can't find relation between matrices and complex number).

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2 comment threads

As to why they're not commutative, think of them in terms of their application to describe rotation. ... (1 comment)
Quaternions are not complex numbers (1 comment)
As to why they're not commutative, think of them in terms of their application to describe rotation. ...
Fred Wamsley‭ wrote over 2 years ago

As to why they're not commutative, think of them in terms of their application to describe rotation. Grab something from a set of dice, a cube with unique markers on each face. Make a note of which side is facing you, then rotate it around the x axis and then around the y axis. Make a note of the side facing you now. Put it back the way it was, now rotate it around the y axis first and then the x axis. The side facing you is not the same one you got the first time. 3-D rotations don't commute, so things that can represent them don't commute.