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In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?

Jyrki Lahtonen
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RubyBlue
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1 Answers1

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With non-real quaternions we can always find two square roots.

You can write any quaternion in the form $$ q=a+b\vec{u}, $$ where $a$ and $b$ are real, and $\vec{u}$ is a unit vector. You probably know that as a quaternion $\vec{u}^2=-1$. Therefore we can treat $\vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.

Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is $$ q^2=(a^2-b^2)+2ab\vec{u}. $$ For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever $q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2\notin\mathbb{R}$, they must both lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $\vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $\mathbb{C}=\mathbb{R}\oplus\mathbb{R}\vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.

Jyrki Lahtonen
  • 133,153
  • Interesting. In the quaternion field $\mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots? – Zhuoran He Oct 30 '17 at 05:20
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    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $\Bbb{H}$: to each unit vector $\vec{u}$ we have $n-1$ non-real $n$th roots of unity in $\Bbb{R}\oplus\Bbb{R}\vec{u}$. Therefore all reals have infinitely many $n$th roots in $\Bbb{H}$. – Jyrki Lahtonen Oct 30 '17 at 07:26
  • I get it. Whenever there's symmetry breaking, e.g. $\sqrt{-1}=\pm i$, $1^{1/3}=1, e^{\pm2\pi i/3}$ in $\mathbb{C}$, quaternions would say that the $\pm i$ is in fact an infinite number of directions $ai+bj+ck\in\mathbb{H}$ with $a^2+b^2+c^2=1$. – Zhuoran He Oct 30 '17 at 17:47
  • I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2\in\mathbb{R}$." I can't make the edit due to the character minimum on edits. – j0equ1nn Nov 25 '17 at 04:02
  • Thanks @j0equ1nn. Such details are anything but clear to me. – Jyrki Lahtonen Nov 25 '17 at 05:47