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Thread: Why does ijk=-1

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    Bohandas's Avatar

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    Default Why does ijk=-1

    I'm trying to understand hypercomplex numbers by extrapolating from 3blue1brown's explanation of regular complex numbers but it's not working.

    It seems like it should just be equal to k because the number line would just be rotated into each plane sequentially. Am I understanding it wrong, or does 3blue1brown's explanation only apply to the ordinary complex numbers?

    ALSO, on an unrelated complex number issue, would a transfinite imaginary or complex number necessarily be a special kind of transfinite number? It seems to me that complex numbers don;t meet the definition of either cardinal or ordinal
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    Default Re: Why does ijk=-1

    Quote Originally Posted by Bohandas View Post
    I'm trying to understand hypercomplex numbers by extrapolating from 3blue1brown's explanation of regular complex numbers but it's not working.

    It seems like it should just be equal to k because the number line would just be rotated into each plane sequentially. Am I understanding it wrong, or does 3blue1brown's explanation only apply to the ordinary complex numbers?

    ALSO, on an unrelated complex number issue, would a transfinite imaginary or complex number necessarily be a special kind of transfinite number? It seems to me that complex numbers don;t meet the definition of either cardinal or ordinal
    To some extent because it was picked to be that way.
    It's easy to see it from the quaternion multiplication table i*i=-1 j*k=i therefore i*(j*k)=-1 and reassuringly (i*j)*k=-1 as well

    So looking at it visually
    1*i -> i
    i*i -> -1
    So on the zero j,k plane 1 is changed (rotated) to i
    And similarly for the unit&j unit&k planes

    Notice of course that multiplication by 1 doesn't rotate.
    Now lets look at the other two options.
    j*i -> -k
    k*i -> j
    These are clearly not left unaffected, if it were a simple pure rotation to the i axis then of course they would be (I'm sure it's been considered).
    However if you look at it again, it does look a bit rotationy. You could imagine looking down the i axis and seeing it rotate.

    So to consider a three dimensional anology, it's like a plane yawing 90 degrees from north to east, and in the same motion rolling 90 degrees so up is now [south]. (again this has of course been considered but makes horrible maths, for some reason). However in three dimensions the roll and yaw share an axis, this is not the case in quaternion (I'm not sure separating it into two independent rotations works).

    So going back to (i*j)*k
    We;re multiplying by j so we rotate the (1,j) axes but our point is at zero with respect to these so nothing happens, we rotate the (i,k) axis so our point is now at k
    [We should be doing this at the same time, it's one rotation, we get away with it because one of them has no effect, but we would have to be careful]
    So now our initial vector is in the k axis.
    And we're multiplying by k so we rotate the (1,k) axes.

    [TLDR it is a rotation, but you don't just rotate as though watching from above]
    Last edited by jayem; 2021-03-06 at 05:47 AM.

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    Default Re: Why does ijk=-1

    The simplest way to understand complex numbers is that i is a rotation from the real axis 90 degrees counterclockwise toward the imaginary axis. Extrapolating to 4 dimensions is going to make my head hurt, but let's try. I am going to assume that i,j,k, each behave as rotations in their own plane. But rotating i by j doesn't start on the j-plane, so rotating about an axis orthogonal to the j-plane can't get you onto the j-plane. Likewise, it can't get a real number or the i-plane and j-plane would be congruent, so it has to end up on the k-plane. It works out to be a bit cross-product like (albeit possibly with commutation), so ij=k, jk=i, and ki=j.
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    Default Re: Why does ijk=-1

    Quote Originally Posted by Rockphed View Post
    The simplest way to understand complex numbers is that i is a rotation from the real axis 90 degrees counterclockwise toward the imaginary axis. Extrapolating to 4 dimensions is going to make my head hurt, but let's try. I am going to assume that i,j,k, each behave as rotations in their own plane. But rotating i by j doesn't start on the j-plane, so rotating about an axis orthogonal to the j-plane can't get you onto the j-plane. Likewise, it can't get a real number or the i-plane and j-plane would be congruent, so it has to end up on the k-plane. It works out to be a bit cross-product like (albeit possibly with commutation), so ij=k, jk=i, and ki=j.
    Which isn't a surprise as basically he wanted vectors and cross products 40 years too early.

    I looked it up, and I was right historically i*j*k=-1 is axiomatic and written on Brougham bridge. Although 3B1B's presentation of complex numbers is based on a more modern thinking of what they mean, so he'd probably start by constructing the rotational analogue and then derive the historical axioms as the interesting parallel rather than the other way round. A third option would be to start with the multiplication timeline.

    Again as Rockphed implies, many of the other options have the wrong sort of consequences.
    If i,j,k aren't symettric then that's a bit messy (though you do have Tessarines, in this case you have values such that a*b=0).
    If the orthoganal 'vectors' were unaffected by the rotation then the resultant 'direction' is basically dominated by the last two numbers (I think), etc... (which I'm sure you could do something with but isn't a good 4D complex number)

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    Default Re: Why does ijk=-1

    Quote Originally Posted by Bohandas View Post
    ALSO, on an unrelated complex number issue, would a transfinite imaginary or complex number necessarily be a special kind of transfinite number? It seems to me that complex numbers don;t meet the definition of either cardinal or ordinal
    -1 isn't even a cardinal or ordinal, so it would be pretty silly to add a square root of -1. If you extend the ordinals to include negatives, subtraction, division, square roots of positive numbers, etc., you end up with the surreal numbers, although the operations end up being somewhat different from cardinal or ordinal arithmetic. You can then add a square root of -1 to get the surcomplex numbers.
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    Default Re: Why does ijk=-1

    One "mathematician's answer" is that that is what is required to make the quaternions a "normed division algebra."

    That is to say, they follow most of the nice algebraic rules that real numbers do, except they are missing a couple features to fit the extra dimensions in. And it has been proven that the only such algebras are the one dimensional real numbers, the two dimensional complex numbers, the four dimensional quaternions, and the eight dimensional octonions. At each step, one has to give up another property that the real numbers have.

    Also, the discoverer of the quaternions was William Rowan Hamilton. I don't see him mentioned directly by name above, so I'll put that here.
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    Default Re: Why does ijk=-1

    Quote Originally Posted by gomipile View Post
    One "mathematician's answer" is that that is what is required to make the quaternions a "normed division algebra."

    That is to say, they follow most of the nice algebraic rules that real numbers do, except they are missing a couple features to fit the extra dimensions in. And it has been proven that the only such algebras are the one dimensional real numbers, the two dimensional complex numbers, the four dimensional quaternions, and the eight dimensional octonions. At each step, one has to give up another property that the real numbers have.

    Also, the discoverer of the quaternions was William Rowan Hamilton. I don't see him mentioned directly by name above, so I'll put that here.
    The giving up a property bit might be a useful thing here for illustration by what they don't give up. They maintain associativity.

    so ijk=i(jk)=(ij)k,

    jk=i

    ijk=ii=-1

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