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Thread: Quaternions

  1. #1 Quaternions 
    Forum Professor wallaby's Avatar
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    the identity i<sup>2</sup>=j<sup>2</sup>=k<sup>2</sup>=ijk=-1

    what i don't understand is the ijk part.

    what i understand of it is that since quaternions are not commutative this part of the identity affects the results of quaternion multiplication but...
    since i=j=k= square root (-1)
    then shouldn't ijk=-i=-j=-k?

    i'm obviously missing something here but i can't find what.


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  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
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    i^2, j^2 and k^2 do equal -1, but i doesn't equal j, etc. (They're three different square roots. Another system [octonians, if I spelled it right] has 7.) The ijk=-1 is just part of (one of) the definition. (i*j)*k = k*k = -1. Also, i*(j*k) = i*i = -1.


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  4. #3  
    Forum Professor wallaby's Avatar
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    Quote Originally Posted by MagiMaster
    i^2, j^2 and k^2 do equal -1, but i doesn't equal j, etc. (They're three different square roots. Another system [octonians, if I spelled it right] has 7.) The ijk=-1 is just part of (one of) the definition. (i*j)*k = k*k = -1. Also, i*(j*k) = i*i = -1.
    ahh i see it now.

    thank you.
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