Empty
  • Empty
How many pairs of numbers are there that are both their additive and multiplicative inverses? (in the quaternions) :P
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Empty
  • Empty
Nahhh no such thing as number theory needed.
freckles
  • freckles
so we want to find a and b such this is true: \[a+b=0 \\ ab=1 \] \[a=-b \\ -bb=1 \\ -b^2=1 \\ b^2=-1 \\ b=\pm \sqrt{-1}=\pm i \\ \]
freckles
  • freckles
\[i+(-i)=0 \\ i(-i)=-i^2=-(-1)=1\]

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freckles
  • freckles
so are we looking for the pair i,-i?
freckles
  • freckles
maybe I should also consider \[(x_1+y_1i)+(x_2+y_2i)=0 \\ (x_1+y_1i)(x_2+y_2i)=1\]
freckles
  • freckles
\[-(x_2+y_2i)^2=1 \\ (x_2+y_2i)^2=-1 \\ x_2+y_2i=\pm i \text{ but this still makes } x_2=0 \text{ and } y_2=\pm 1\]
Empty
  • Empty
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
zzr0ck3r
  • zzr0ck3r
look at a caley table...
Empty
  • Empty
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.
zzr0ck3r
  • zzr0ck3r
the quaternains is a group with one operation, and there are two elements with self inverses by definition
zzr0ck3r
  • zzr0ck3r
Maybe I am confused by the question...
zzr0ck3r
  • zzr0ck3r
how can a pair of numbers have an inverse?
Empty
  • Empty
Another way of asking this question is what are all the quaternion solutions to this system of equations: a+b=0 a*b=1
zzr0ck3r
  • zzr0ck3r
I am still confused. The quaternians are a group. Ill stop cluttering up the question...
Empty
  • Empty
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.
zzr0ck3r
  • zzr0ck3r
that would be an extension. The quaernains are a 8 element group. By definition.
zzr0ck3r
  • zzr0ck3r
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.
zzr0ck3r
  • zzr0ck3r
as another example^^^
Empty
  • Empty
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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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