Isn't it cheating to consider an $ \mathbb{R}^3 $ vector as a "pure quaternion"? In Jack Kuipers' book he says (p 114):

How can a quaternion, which lives in $\mathbb{R}^4$ operate on a vector, which lives in $\mathbb{R}^3$ ? 

His answer:

A vector v $\in$ $\mathbb{R}^3$ can simply be treated as though it were a quaternion q $\in$ $\mathbb{R}^4$ whose real part is zero.

Then later we define that special property:
If p is a pure quaternion (ie $\Re({p}) =0 $):

$$
qpq* = w
$$
Where $\Re(w)=0$ as well.
Ok, so how does he get away with skipping over the fact that an $ \mathbb{R}^3 $ vector is real and a pure quaternion is purely imaginary?
 A: All that you require is that there is a one-to-one mapping (an isomorphism) between $\mathbb{R}^3$ and the space $\mathfrak{I}(\mathbb{H})$ of purely imaginary quaternions. Since they are both three-dimensional vector spaces, such an isomorphism trivially exists.
More concretely, you can say that the pure imaginary quaternion $x{\bf i} + y{\bf j} + z{\bf k}$ maps to the vector $(x,y,z)$ with the obvious mapping in the other direction.
He is saying that since the space of pure imaginary quaternions and the space of vectors in three dimensions both have three degrees of freedom (and are both linear) you may as well consider them to be the same space. Since the mapping $p \mapsto qpq^*$ preserves the property of being pure imaginary, we have an induced action of quaternions on the space of real three-dimensional vectors.
A: The quaternions $\mathbb H$ have a very rich algebraic structure . They are a real division algebra endowed with an anti-involution $q\mapsto q^*$ [anti- alludes to the reversal $ (pq)^*=q^*p^*$ ].  
Inside $\mathbb H$ you have a completely canonical three dimensional subspace $\mathbb H_0$, defined  algebraically by the condition $q^*=-q$, and called "pure quaternions" or "imaginary quaternions" (the adjectives pure and imaginary have historical roots but no mathematical content).
It is then easy to see that your map $\mathbb H \to \mathbb H:p\mapsto qpq^*$
leaves $\mathbb H_0$ invariant.
Indeed if $ p\in \mathbb H_0$, then by definition of $\mathbb H_0$ we have $p^*=-p$ and so $r^*=(qpq^*)^*=q^{**}p^*q^*=
q(-p)q^*=-(qpq^*)=-r$, so that as announced $r=qpq^*$ is also in $\mathbb H_0$.
