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Topic: Algebra Family multiplication table for 3D vectors
Replies: 17   Last Post: Mar 13, 2014 7:32 PM

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Robin Chapman

Posts: 301
Registered: 5/29/08
Re: Algebra Family multiplication table for 3D vectors
Posted: Feb 28, 2014 9:24 AM
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In article <280220140632289126%edgar@math.ohio-state.edu.invalid>, annu
Poropudas <hanporop@luukku.com> wrote:

> On Thursday, February 27, 2014 1:00:50 AM UTC+2, Robin wrote:
> > Again you have an algebra in which every element has an inverse,
> > except for the (infinitely many) elements that don't :-(

>
> Yes, The smaller space for this special case of my algebra family
> seems to be the following:
>
> R^3\{plane: z=-y, x=x,y=y}


It's actually R^3 with both the plane y = -z and the line x = y = 0
removed.


>
> This plane must be excluded in this special case of my algebra family.
> This plane goes through origin.
>
> One property which you noticed (in your previous post) seems to be that
> this set is not an integral domain? (if your calculation: e2*e3=e3 ->
> (e2-e1)*e3=0 is correct.For heaven's sake Mr Poropudas, cannot you even check a simple

calculation? Here is the multiplication table, from your previous
posting.

> e1, e2, e3
> e2, -e1+e3, e3
> e3, e3, 2*e2


From it one has e1*e3 = e3 and  e2*e3 = e3, Therefore,
(by distributivity) (e2-e1)*e3 = e2*e3 - e1*e3 = e3 - e3 = 0.

Now, have you decided what properties you want of your algebra?
You are aware that you cannot have an associative three-dimensional
algebra over the reals with a multiplicative identity and
with the property that all nonzero elements are invertible
(there is no three-dimensional division algebra over R).
If you could tell us what you are aiming to achieve, perhaps
we could help you do it.



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