Is a quaternion a way to divide vectors?

Question

This is a naive question but I read popular science articles where it is stated that quaternions define vector division, without further explanations

Answer 1

No. Quaternions can be regarded a specific example of vectors, since the set of quaternions forms a vector space that is isomorphic to $\mathbb{R}^4$ under quaternion addition and scalar multiplication, but a "vector", broadly defined, is any object that belongs to some vector space. The general axioms for an abstract vector space do not allow division and so some vector spaces (e.g. the real or complex numbers) might have a natural division operator that can be defined, while other vector spaces (e.g. the set of 3x5 real matrices) may not. If you are concerned with three-dimensional vectors as used in mechanics, these do not have a natural division operation.

The most general construction that allows division (to my knowledge--but I'm not an algebraist) is a division ring. The most commonly used construction that allows division is a field. The quaternions themselves are often worked with as a division algebra over the field of real numbers and are important because there are only a finite number of such algebras (up to isomorphism) with several nice properties (see the Frobenius theorem). You might also be interested in looking at the Cayley-Dickson construction and Clifford Algebras.

Answer 2

"Yes", but in a very qualified sense. (That is, what @SZN wrote is true, but I'd put a different emphasis on things.)

First, we could ask about "division" of two-dimensional (real) vectors. The complex numbers (which form a field) give a way to make sense of this.

In three dimensions? It turns out that there is no real-three-dimensional field, so, "no".

In four dimensions? Skipping over the three dimensional case was part of Hamilton's inspiration: yes, the Hamiltonian quaternions for a "non-commutative field", also called "skew field" or "division ring" or "division algebra".

But there are no more (real) division algebras than these (by Frobenius' theorem or others). There are further complicated objects (e.g., "octonians"), but these (by the theorem) cannot quite be division algebras.

So, _in_general_, no, there's no reasonable definition of division for vectors, but in some special cases there is.