How many pairs of numbers are there that are both their additive and multiplicative inverses? (in the quaternions) :P

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How many pairs of numbers are there that are both their additive and multiplicative inverses? (in the quaternions) :P

Mathematics
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Nahhh no such thing as number theory needed.
so we want to find a and b such this is true: \[a+b=0 \\ ab=1 \] \[a=-b \\ -bb=1 \\ -b^2=1 \\ b^2=-1 \\ b=\pm \sqrt{-1}=\pm i \\ \]
\[i+(-i)=0 \\ i(-i)=-i^2=-(-1)=1\]

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so are we looking for the pair i,-i?
maybe I should also consider \[(x_1+y_1i)+(x_2+y_2i)=0 \\ (x_1+y_1i)(x_2+y_2i)=1\]
\[-(x_2+y_2i)^2=1 \\ (x_2+y_2i)^2=-1 \\ x_2+y_2i=\pm i \text{ but this still makes } x_2=0 \text{ and } y_2=\pm 1\]
Perhaps, I haven't completely figured it out yet, but I know there is more than just the pair (i, -i) since quaternions allow us to use (j,-j) and (k,-k). I wonder if there is more
look at a caley table...
Like some combination of these values? It wouldn't show up on the cayley table. I'm not convinced at least, maybe I don't understand what you mean.
the quaternains is a group with one operation, and there are two elements with self inverses by definition
Maybe I am confused by the question...
how can a pair of numbers have an inverse?
Another way of asking this question is what are all the quaternion solutions to this system of equations: a+b=0 a*b=1
I am still confused. The quaternians are a group. Ill stop cluttering up the question...
Haha it's fine, it's just they're more than just a group since they extend the complex numbers. For instance I suspect that the pair \(\frac{\sqrt{2}}{2}i+\frac{\sqrt{2}}{2}j\) and its negative might be both additive and multiplicative inverses.
that would be an extension. The quaernains are a 8 element group. By definition.
we extend from the reals to the complex with i. but the group generated by i under multiplication has only 4 elements. So R[i]=C but {i|i^n} does not.
as another example^^^
Well I am talking about quaternions in the sense that they're normally used in my experience. Yeah ok so here is part of expanding the solution to infinitely many pairs, but there may be more since I'm not using all the parts of the quaternion at once: \(\hat i \cos \theta + \hat j \sin \theta\) When added to its negative is clearly 0, so additive inverses, now let's multiply them: \((\hat i \cos \theta + \hat j \sin \theta\ )(-\hat i \cos \theta - \hat j \sin \theta)\) \(\cos^2 \theta + \sin^2 \theta - \sin \theta \cos \theta (\hat i \hat j + \hat j \hat i )\) Since \(\hat i \hat j = -\hat j \hat i\) that second term goes away and we have the pythagorean identity on the left, which is just 1.
the quaternions are an associative algebra actually, which has far more structure than a simple group (which would only be with respect to a single operation). anyways, consider \(q=a+\vec v\) and \(-q=-a-\vec v\) so $$\begin{align*}-qq&=(-a-\vec v)(a+\vec v)\\&=(-a^2+(-v)\cdot v)+(-a\vec v-a\vec v+(-\vec v)\times\vec v)\end{align*}$$since \((-\vec v)\times\vec v=-(\vec v\times\vec v)=0\) and \((-v)\cdot v=-(v\cdot v)=|v|^2\) we can write: $$-qq=(-a^2+|v|^2)+(-2a\vec v)$$and we want \(-a^2-|v|^2=1\) and \(-2a\vec v=0\) so it follows either \(a=0\) or \(\vec v=0\); in the first case, we get that \(|v|^2=1\) works so all pure imaginary unit quaternions work. in the second case we get that \(\vec v=0\) so \(-a^2=1\) which has no real solutions
so in other words \(q=bi+cj+dk\) for \(|q|=b^2+c^2+k^2=1\) works for its own additive inverse to be its multiplicative inverse

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