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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!

George N. Missailidis
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isaacbernstein
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  • $ij=k$ 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. – Arthur Aug 21 '17 at 06:14
  • Thank you, but why are we allowed to assume that ijk = -1? – isaacbernstein Aug 21 '17 at 06:14
  • You have to assume something in addition to what you are assuming. There is more than one choice of assumption that gets you where you want to be. – Gerry Myerson Aug 21 '17 at 06:17
  • We are allowed to assume whatever we want, as long as we accept the consequences. $ijk=1$ is equivalent to $ij=k$, given that $k^2=-1$ is already established, and it is what is needed to arrive at the quaternions. If we had some other definition, like $ij=i+j$ then two things happen: 1) we still have no idea what, for instance, $ik$ is, so we would need to add that as an axiom too 2) it wouldn't give us a division ring. Note that $i^2=j^2=k^2$ says nothing about what you get when you multiply different letters together, only when you multiply a letter with itself. – Arthur Aug 21 '17 at 06:17
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    From $i^2=j^2=k^2=-1$ alone, you cannot even prove $i\ne k$ or $j\ne k$. – Hagen von Eitzen Aug 21 '17 at 06:33
  • In $\Bbb{H}$ any element $u=ai+bj+ck$ such that $a^2+b^2+c^2=1$ satisfies $u^2=-1$. This implies that you can find infinitely many distinct quaternions with squares equal to $-1$. So to prove the isomorphism you absolutely must have something extra relating the $i,j,k$ to each other (not just saying that their squares are $=-1$). My main point being that there exists division algebras generated by three quantities $i,j,k$, all with squares $-1$, but where the relation $ij=k$ does not hold. Simply pick a collection of three unit vectors that don't form a right-handed orthonormal system. – Jyrki Lahtonen Aug 21 '17 at 06:37
  • To give a simple example. Let $i,j,k$ be the usual quaternions. Let $i'=i$, $j'=j$, $k'=-k$. Then $$i'^2=j'^2=k'^2=-1,$$ but $$i'j'=-k'.$$ Therefore the equation $i'j'=k'$ cannot be a consequence of $i'^2=j'^2=k'^2=-1$. – Jyrki Lahtonen Aug 21 '17 at 06:41

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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 $ij=k$.

skyking
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