If I can prove that $ij = k$, given that $i^2 = j^2 = k^2 = -1$, then it will be easy to prove the other quaternion formulas. However, I'm having a lot of trouble getting past this step. I started by setting up the following equation, such that $(a, b, c, d) \in \mathbb{R}$

$ij = a + bi + cj + dk$

Obviously, I need to prove that $a = b = c = 0 \land d= 1$. I've done all sorts of wacky things with this equation and I haven't gotten any closer to proving that $ij = k$. I would greatly appreciate it if somebody could point me in the right direction. Thanks in advance!

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    $\begingroup$ I don't think you can show this based on what you have. Aren't you missing the condition that $ijk=-1$? $\endgroup$ – Morgan Rodgers Aug 21 '17 at 6:12
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    $\begingroup$ That is, this is a condition you need, you can't prove it from the other relations. $\endgroup$ – Morgan Rodgers Aug 21 '17 at 6:12
  • $\begingroup$ $ij=k$ is one of the defining axioms of the quaternions. It can't be proven. In fact, you can use the four elements $1,i,j$ and $k'=(1+i+j+k)$ to generate your quaternions, and in that case, $ij=k'-i-j-1$, and the us no way to tell which you're using unless you add $ij=k$ to the definition. $\endgroup$ – Arthur Aug 21 '17 at 6:14
  • $\begingroup$ Thank you, but why are we allowed to assume that ijk = -1? $\endgroup$ – isaacbernstein Aug 21 '17 at 6:14
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    $\begingroup$ From $i^2=j^2=k^2=-1$ alone, you cannot even prove $i\ne k$ or $j\ne k$. $\endgroup$ – Hagen von Eitzen Aug 21 '17 at 6:33

You can't do that with that alone. You could for example still have $i=j=k$ which would make $ij=i^2 = -1$.

The traditional way is to set $i^2=j^2=k^2=ijk = -1$, but even that is not enough. You have to rely on some other properties which may be taken for granted. Then the proof is $ij = -ijk^2 = -(ijk)k = -(-1)k = k$, but then you've used some other rules as well. You've used the associative law (without which by the way $ijk$ is ambiguous), but also the fact that $(-1)^2 = 1$ and that $-1$ commutes with anything.

You can also use the property that $kx=0\leftrightarrow x=0$ and that $x-y=0\leftrightarrow x=y$ together with the right distributive law. Then we have $(ij-k)k = ijk-k^2 = (-1)-(-1) = 0$, therefore $ij-k=0$ so $ij=k$.


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