How do I prove that $ij = k?$

Solution 1:

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 $ij=k$.