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I dont think I really need to elaborate, do I? If you know what quaternions are then you know there are several imaginary-value options to choose from, or axes, along which the $\sqrt{-1}$ may exist.

By definition $i^2 = j^2 = k^2 = -1$ but no indication is given of going the other way.

I suppose we could just say that it simply does not have an inverse, but I feel like that is a cheap, un-thought-out cop-out. Does anyone know differently or can it be proven?

I wonder since $\sqrt{-1}$ is $i$ and $i^2 = -1$ its clear that the inverse operation does exist with complex numbers, and complex numbers are a subset of the quaterions.

In fact, each of the axes $i,j,k$ by themselves are considered imaginary. And indeed, any plane in quaternion space that comprise one imaginary axis and the real axis can be interpreted as one of three distinct complex planes.

So it begs the question, to me, why must the $\sqrt{-1}$ be $i$ and not $j$ or $k$? Or some combination of values whose magnitude is $1$ and whose square is $-1$. In the grander scope of quaternion space where three distinct imaginary axes exist, what is the $\sqrt{-1}$? Or perhaps have we decided the operation ought not be defined due to its ambiguity?

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    $\begingroup$ Please note, that even for usual complex numbers both $i$ and $-i$ are $\sqrt{-1}$ $\endgroup$ – lesnik Feb 5 '16 at 7:14
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    $\begingroup$ And even for real numbers, both $-2$ and $2$ are square roots of $4$, and it is only by a common convention that we choose to let $\sqrt{4}$ refer to the positive square root. $\endgroup$ – Eric Wofsey Feb 5 '16 at 7:17
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There is a whole sphere of possible answers: $xi+yj+zk$, with $x^2+y^2+z^2=1$. Then $$(xi+yj+zk)^2=x^2i^2+y^2j^2+z^2k^2+xy(ij+ji)+xz(ik+ki)+yz(jk+kj) \\ =-1(x^2+y^2+z^2)=-1$$

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To do this you decompose it into real and vector path. That is rewrite it as:

$$ c +xi + yj + zk = (c, u)$$

where $u = (x,y,z)$. Now we have that quaternion multiplication can be written as:

$$(a, u)(b, v) = (ab - u\cdot v, av + bu + u\times v)$$

which means the square becomes:

$$w^2 = (a, u)^2 = (a^2 - |u|^2, 2au + u\times u) = (a^2 - |u|^2, 2au)$$

For this to become $-1$ you would have $2au=0$ which means that either $a$ or $u$ must be null. But if $u$ is null then $a^2-|u|^2 = a^2 \ge 0$, so we can conclude that $u$ is not null and therefore $a$ is. So we have apart from $a=0$ that $a^2 - |u|^2 = -|u|^2 = -1$ that is $|u|^2 = 1$.

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