Why is $ijk=i^2$, shouldn't it be equal to $-i$?

Question

Quaternion units are set up as, $i^2=j^2=k^2=ijk$. I am confused as to how this is possible. If $i^2=j^2=k^2=-1$ then by rights, $\sqrt{k^2}=\sqrt{j^2}=\sqrt{-1}=\sqrt{i^2}$ would make $i=j=k$ true, which would mean that $ijk=-i=-j=-k$.

Answer 1

You assume that $x^2=y^2$ implies $x=y$, i.e. that you have a globally defined square-root function. That is not even true for real or complex numbers and it is even more wrong for quaternions.

Where for real numbers you have a canonical square-root function at least on positive inputs ("take the positive solution"), you do not have a canonical square-root function for complex numbers and have to be much more careful with signs here, because now both solutions to $x^2=z$ with given $z\in\mathbb{C}$ are completely equal in their algebraic behaviour. In other words: there is no notion of "positive" in the complex world that can be used to distinguish the two solutions without making arbitrary choices.

In the world of quaternions this is even more wrong, because the equation $x^2=z$ can have infinitely many solutions. For example the solutions to $x^2=-1$ are all the quaternions of the form $ai+bj+ck$ with $a,b,c\in\mathbb{R}$ and $a^2+b^2+c^2=1$, i.e. the space of these solutions is the 2-Sphere ! Now you don't even have a nice way to switch between solutions, because there is a lot more going on here than a simple sign.

Answer 2

What exactly do you mean by $\sqrt\cdot$?

Note that $(1)^2=(-1)^2$ but $1\neq -1$