Clarification of definition of "inverse" with quaternions
In mathematics in general, if we have some object $\alpha$, we usually define its inverse $\alpha^{-1}$ to be the object that "undoes" $\alpha$. That is to say, $\alpha^{-1}$ returns the identity element when it is combined with $\alpha$.
$$\alpha \alpha^{-1} = identity$$
This identity element is sort of a "do nothing" object, and the form it takes depends on the objects we're talking about and on the operations used to combine them.
So in a sense, $\alpha^{-1}$ does the opposite of whatever $\alpha$ does and gets us back to where we started.
That might sound abstract, so here are some concrete examples:
the additive identity for the real numbers is $0$ because when we add it to some number $r$, we just get $r$ back again: $r + 0 = r$. The additive inverse for a real number $r$ is the number that produces $0$ after an addition. Obviously this number is $(-r)$ since $r + (-r) = 0$
the multiplicative identity for the real numbers is $1$ because $r \cdot 1 = r$. The multiplicative inverse for a real number $r$ is the number that produces $1$ after multiplication, that is $\frac{1}{r}$ because $r \cdot \frac{1}{r} =1$. Note that $0$ has no multiplicative inverse.
the multiplicative identity for $n \times n$ matrices is the $n \times n$ identity matrix $I_n$ with $1$'s on the diagonal and $0$ elsewhere. The multiplicative inverse of a matrix $A$ is a matrix $A^{-1}$ such that $AA^{-1} = I_n$
the multiplicative identity for a complex number $z = a+ib$ is $1+i0$ since $z \cdot 1 = z$. The multiplicative inverse is some complex number $w$ such that $zw = 1$. You can show that $w = \frac{1}{z} = \frac{z^*}{zz^*} = \frac{z^*}{\|z^{}\|^{2}}$
Finally, quaternions...
- the multiplicative identity for a quaternion $q = a+ib+jc+kd$ is $1+i0+j0+k0$ since $q1 = q$. The multiplicative inverse is some quarternion $x$ such that $qx = 1$.
Writing a formula for this number $x$ might not be immediately obvious, but you can check that $x = \frac{1}{q} = \frac{q^*}{qq^*} = \frac{q^*}{\|q\|^{2}}$, with $q^* = a-ib-jc-kd$ does satisfy the formula.
If $\alpha = a+bi+cj+dk$ is a quaternion, $\alpha^{-1} = \frac{\overline{\alpha}}{||\alpha||}$ where $\overline{\alpha} = a-bi-cj-dk$ and $||\alpha|| = a^2+b^2+c^2+d^2$.
This makes intuitive sense when you understand some basic properties of the norm, as $\alpha \overline{\alpha} = ||\alpha||$ (this is easily, if tediously, confirmable by calculation) and so $\alpha \cdot \left( \frac{\overline{\alpha}}{||\alpha||} \right)= 1$.