# Unwind quaternion multiplication

I am trying to understand quaterions division.

Imagine I have the following equation, where every member is a quaternion:

$$Q = (qq_1)(qq_2)...(qq_n)$$

I suppose that, if I maintain the order of division exactly reverse to the order of multiplication, i get (by division I mean multiplication by a reciprocal of a given $q_n$, this notion is taken from wikipedia):

$$q^n = \frac{Q}{(q_nq_{n-1}...q_1)}$$

$$q = \sqrt[n]{\frac{Q}{(q_nq_{n-1}...q_1)}}$$

Do these relations hold for quaternions? Will this give me back the original value of $q$ as one of the roots?

UPDATE

As Jason mentioned in the comments, there're two ways to divide: left and right. This is an important detail. Will this obscurity prevent me from receiving the original quaternion, or is it possible to pick the correct (left or right) division to actually get the correct result?

• Since quaternionic multipication is not commutative, I don't think your formula for $q^n$ holds in general. Such a formula would require you to commute $q$ past $q_i$ for all $i$. In fact, it has a further issue. There is no such thing as quaternionic division, only left or right division. So, the notation $\frac{Q}{q_n \cdots q_1 }$ is meaningless. – Jason DeVito Oct 9 '15 at 14:13
• @JasonDeVito: yes, this kind of question bothers me. However, according to the Wikipedia, this is possible (see my edit). – noncom Oct 9 '15 at 14:17
• Maybe I'm missing what you're referring to, but wikipeida does have things like $\frac{Q}{\| q_n \cdots q_1\|}$, but this makes sense as the denominator is a real number and these commute with everything. So, left and right division, in this case, are the same, so you may as well denote it by regular division. – Jason DeVito Oct 9 '15 at 14:49
• Also, there is an explicit example where you can't recover $q$ from knowledge of $Q$ and $q_1,... ,q_n$. Thus, there can be no `"nice" formulaic way to solve for $q$. Suppose $n = 2$, $q_1 = q_2 = 1$, and $Q = -1$. Tthen you get are trying to solve $-1 = q \cdot 1 \cdot q\ \cdot 1 = q^2$. This has uncountably many solutions: each purely imaginary unit quaternion. – Jason DeVito Oct 9 '15 at 14:51
• Yes, there is $q = \pm i$, and $\pm j$ and $\pm k$, but, for example, $\cos(\theta)i + \sin(\theta) j$ is a solution for any $\theta$. More generally, if $q = ai + bj + ck$ with $a^2 + b^2 + c^2 = 1$, then $q$ is a solution, and all solutions have this form. – Jason DeVito Oct 9 '15 at 15:02