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quaternions
#3: Post edited
- In the [Wikipedia page](https://en.wikipedia.org/wiki/Quaternion#History), 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 $$ijk=\sqrt{-1}\sqrt{-1}\sqrt{-1}=(-1)^{\dfrac{3}{2}}$$ 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).
- In the [Wikipedia page](https://en.wikipedia.org/wiki/Quaternion#History), 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).
#1: Initial revision
How $ijk=\sqrt{1}$?
In the [Wikipedia page](https://en.wikipedia.org/wiki/Quaternion#History), 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 $$ijk=\sqrt{-1}\sqrt{-1}\sqrt{-1}=(-1)^{\dfrac{3}{2}}$$ 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).