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I've been looking up Quaternion multiplication and many resources have stated that
$j*k=i$
and also in other sources I've found
$k*j=-i$


But I have not found any sources stating
$j*k=-i$
and I also have not found any sources stating
$k*j=i$


I know this sounds like a dumb question, but does the commutative property of multiplication not apply when dealing with higher dimensional complex number systems?
$j*k≠k*j$       ?

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    $\begingroup$ Quaternion multiplication (what sources did you end up looking up...?) $\endgroup$
    – user296602
    Commented Jan 27, 2016 at 5:05
  • $\begingroup$ @T.Bongers So even in the article you linked it says jk=i, kj=-i :o Is this really the case? jk≠kj ? $\endgroup$
    – Albert Renshaw
    Commented Jan 27, 2016 at 5:07
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    $\begingroup$ Yes, and the next boldface-titled section is "Noncommutativity of multiplication." $\endgroup$
    – user296602
    Commented Jan 27, 2016 at 5:08
  • $\begingroup$ @T.Bongers Fascinating, I'm surprised I was never taught this before haha! $\endgroup$
    – Albert Renshaw
    Commented Jan 27, 2016 at 5:09
  • $\begingroup$ @user26857 There was a downvote, it has since then been removed. I'm not sure why it's pathetic, I simply wanted to know what I had done wrong so that I could avoid getting more downvotes in the future? I'm sorry that I've offended you sir $\endgroup$
    – Albert Renshaw
    Commented Jan 27, 2016 at 7:02

1 Answer 1

Reset to default
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Yes, quaternions multiplication is not commutative.

https://en.wikipedia.org/wiki/Quaternion

$ i \times j = k $

$ j \times i = -k $

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  • $\begingroup$ Thank you! That's fascinating. I'd like to understand why this is the case, in an intuitive sense. $\endgroup$
    – Albert Renshaw
    Commented Jan 27, 2016 at 5:08
  • $\begingroup$ Quaternion is used in computer graphics and robotics to interpolate 3D rotations through conjugate multiplication en.wikipedia.org/wiki/Quaternions_and_spatial_rotation 3D rotations are not commutative, for example, rotate left and then rotate up is in general not the same as rotate up than rotate left, so in order for quaternion to be useful, it cannot be commutative, in some sense, it is designed that way. $\endgroup$
    – Andrew Au
    Commented Jan 27, 2016 at 5:12
  • $\begingroup$ That makes perfect sense! Thank you Andrew, I'll accept this answer once StackExchange's time limit hits 0 $\endgroup$
    – Albert Renshaw
    Commented Jan 27, 2016 at 5:14

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