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  • one year ago

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

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  1. Empty
    • one year ago
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    Nahhh no such thing as number theory needed.

  2. freckles
    • one year ago
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    so we want to find a and b such this is true: \[a+b=0 \\ ab=1 \] \[a=-b \\ -bb=1 \\ -b^2=1 \\ b^2=-1 \\ b=\pm \sqrt{-1}=\pm i \\ \]

  3. freckles
    • one year ago
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    \[i+(-i)=0 \\ i(-i)=-i^2=-(-1)=1\]

  4. freckles
    • one year ago
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    so are we looking for the pair i,-i?

  5. freckles
    • one year ago
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    maybe I should also consider \[(x_1+y_1i)+(x_2+y_2i)=0 \\ (x_1+y_1i)(x_2+y_2i)=1\]

  6. freckles
    • one year ago
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    \[-(x_2+y_2i)^2=1 \\ (x_2+y_2i)^2=-1 \\ x_2+y_2i=\pm i \text{ but this still makes } x_2=0 \text{ and } y_2=\pm 1\]

  7. Empty
    • one year ago
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    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

  8. zzr0ck3r
    • one year ago
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    look at a caley table...

  9. Empty
    • one year ago
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    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.

  10. zzr0ck3r
    • one year ago
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    the quaternains is a group with one operation, and there are two elements with self inverses by definition

  11. zzr0ck3r
    • one year ago
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    Maybe I am confused by the question...

  12. zzr0ck3r
    • one year ago
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    how can a pair of numbers have an inverse?

  13. Empty
    • one year ago
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    Another way of asking this question is what are all the quaternion solutions to this system of equations: a+b=0 a*b=1

  14. zzr0ck3r
    • one year ago
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    I am still confused. The quaternians are a group. Ill stop cluttering up the question...

  15. Empty
    • one year ago
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    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.

  16. zzr0ck3r
    • one year ago
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    that would be an extension. The quaernains are a 8 element group. By definition.

  17. zzr0ck3r
    • one year ago
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    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.

  18. zzr0ck3r
    • one year ago
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    as another example^^^

  19. Empty
    • one year ago
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    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.

  20. anonymous
    • one year ago
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    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

  21. anonymous
    • one year ago
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    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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