# Why quaternions flopped

After last week's post about the relationship between quaternions and the dot and cross products of vectors, I was asked why I thought quaternions didn't end up being very useful. I'd guess that the short answer is that they really don't model the real world very well.

It definitely isn't because multiplication of quaternions isn't commutative. Matrix multiplication also isn't commutative, yet matrices have ended up being very useful.

Instead, I'd guess that the problem with quaternions is that calculus doesn't work very well with them. With the complex numbers, there's a nice, clean way to characterize analytic functions: *df*/*dz** = 0. But that doesn't seem to extend to the quaternions very well.

If we have a quaternion

q=q_{0}+q_{1}i+q_{2}j+q_{3}k

and its conjugate

q* =q_{0}-q_{1}i–q_{2}j-q_{3}k

then we have

q_{0}^{2}+q_{1}^{2}+q_{2}^{2}+q_{3}^{2}

which looks about right.

But we also find that we can write

q* = -(q+i q i+j q j+k q k) / 2

And because *q* and *q** are connected in this way, we can never find a nice condition like *df*/*dq** = 0 to characterize functions of a quaternion variable like we can for analytic functions. So I'd guess that's also part of why they haven't ended up being very useful.

## Peter L. Griffiths

Hamilton in his letter of 17 October 1843 to John Graves seems very unsure about the relationship between +1, -1, i and j. He asks what are we to do with ij when i and j are the unequal roots of a common square. In fact there is no law of arithmetic which makes ij equal to anything but +1. It is these doubts of Hamilton which are the source of his fallacious theory of the non-commutative properties of the multiplication of imaginary numbers. All multiplication whether of real or imaginary numbers is commutative.

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## Peter L.Griffiths

Further to my comment of 2 September 2011, on 16 October 1843 Hamilton alleged that i^2=j^2=k^2=ijk=-1. This equation is incorrect because every real, imaginary and complex number has only two square roots, so k^2=-1 is wrong unless k can equal either i or j. It can likewise be demonstrated that every real, imaginary and complex number has only three cube roots, four fourth roots, five fifth roots etc.

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## Peter L. Griffiths

Further to my comment of 27 February 2012, -1 can only have two square roots, nevertheless -1 can have three cube roots which are cos60+isin60, cos180+isin180 which equals -1, and cos300+isin300. Hamilton seemed to have no understanding of these matters.

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## Peter L. Griffiths

Why do the roots of +1,-1,+i and -i always add up to zero? The answer is that computing all these roots is a matter of going round the angles of a circle and finishing at the 360 degrees (=0 degrees) starting point. Incidentally, if you have computed the roots of +i then the roots of -i are easily obtained by substituting -i for +i.

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## Peter L. Griffiths

Further to my comment of 27 February 2012 in which I say that every real, imaginary and complex number has only three cube roots, four fourth roots, five fifth roots etc, it is possible that this principle applies to equations of 3 degrees, 4 degrees, 5 degrees etc, so that an equation of 5 degrees may have 5 possible results, whether real, imaginary or complex. I propose this as a theory rather than an accepted principle. For a start Cardan’s suppressed 3 degree reversion should be investigated with this in mind.

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