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?

  • 6
    $\begingroup$ Please note, that even for usual complex numbers both $i$ and $-i$ are $\sqrt{-1}$ $\endgroup$
    – lesnik
    Feb 5 '16 at 7:14
  • 4
    $\begingroup$ 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. $\endgroup$ Feb 5 '16 at 7:17

There is a whole sphere of possible answers: $xi+yj+zk$, with $x^2+y^2+z^2=1$. Then $$(xi+yj+zk)^2=x^2i^2+y^2j^2+z^2k^2+xy(ij+ji)+xz(ik+ki)+yz(jk+kj) \\ =-1(x^2+y^2+z^2)=-1$$


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 $|u|^2 = 1$.


In a field, a quadratic has at most 2 solutions, but you're not in a field. Let's unpack that statement.

A field is a thing (set) like the real numbers, the complex numbers, the rational numbers or whatever. It has an element called 0 and an element called 1. You can always calculate $x y$, $x+y$ and $-x$ and you can calculate $y^{-1}$ so long as $y\ne 0$. You expect a lot of things to be true, like $x+y=y+x$, $x+0=x$, $x+(-x)=0$, $x y=y x$, $0 x=0$, $1 x=x$, $x(y+z)=x y+x z$ and $x x^{-1}=1$.

A quadratic is a polynomial of degree 2. In your case, the polynomial is $x^2+1$, and you're looking for the values of $x$ which make $x^2+1=0$. In a field, a polynomial of degree $d$ has at most $d$ solutions. In this case, there are 0 solutions in the rationals or the reals but 2 in the complexes, namely $\pm i$.

Your excellent intuition is now trying to deduce that $x^2+1=0$ has at most two solutions in the quaternions, but the problem is that the quaternions are not a field. They fail the axiom $x y=y x$. They are a thing called a skew field. In a skew field, just like a field, you can always calculate $x y$, $x+y$ and $-x$, and $y^{-1}$ so long as $y\ne 0$. Just like a field, $x+y=y+x$, $x+0=x$, and $x+(-x)=0$. But now things get weaker, and we only impose $0 x=x 0=0$, $1 x=x 1=x$, $x(y+z)=x y+x z$, $(x+y)z=x z+y z$ and $x x^{-1}=x^{-1}x=1$.

It would be nice if polynomials over skew fields had a limited number of solutions, but as Empy2 pointed out, the number of solutions is not even finite. Failing that, it might be nice if some solutions were more natural than others, but in this case even that's not really true. You can't tell one solution to $x^2+1=0$ from another without explicitly mentioning one of $\{i, j, k\}$, which is pretty obviously cheating. It's like the fact that you can't tell $i$ from $-i$ in the complex numbers without mentioning $i$. If you want to go down this rabbit-hole, the key word is automorphism.


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