Show that for $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1$. 
Show that  $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1.$

I am pretty sure I can easily google the multiplicative inverse in $\mathbb{H}$, but can you give me a hint on how to determine the inverse von an arbitrary $a \in\mathbb{h}$ myself? This task comes directly after showing that the conjugation is a ring antihomomorphism. ($a\ \bar+ \ b\ = \bar a+\bar b, a\ \bar  *\ b = \bar b*\bar a$ and $ 1 = \bar1$)
 A: This is a somewhat nonstandard approach, but I think it illustrates a nice model for $\mathbb H$.
Note that $\mathbb H$ is isomorphic to the algebra of matrices over $\mathbb C$ of the form
$$\begin{pmatrix}
a+bi & c+di\\
-c+di & a-bi
\end{pmatrix}$$
which is easy to verify: you just need to check that it contains a copy of $\mathbb R$ (namely the matrices with $b=c=d=0$) and that it contains $i,j,k$ such that $i^2=j^2=k^2=ijk=-1$, which come from setting everything but $b,c$ and $d$ respectively equal to $0$.
Now the inverse is just the inverse of a $2\times 2$ matrix, which is easy to compute and can easily be shown to have the same form.
A: You should just solve for ($(x  + y i + z j + w k)(a+b i + c j + d k) = 1.$ Notice that you will get a system of linear equations in $x, y, z, w.$
A: 
Show that  $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1.$
This task comes directly after showing that the conjugation is a ring antihomomorphism.

Inverses are easily found inside the quaternions with conjugation, multiplication and division.
Really, the only computational fact you need is that $a\overline{a}\in \Bbb R$. By looking at what conjugation does to quaternions, you can easily see that $\overline{q}=q\iff q\in \Bbb R$.
But then since conjugation is an antihomomorphism, $\overline{q\overline{q}}=\overline{\bar q}\bar q=q\bar q$, so $q\bar q$ is real.
But then look: $q(\frac{\bar q}{q\bar q})=1$ finds a right inverse for $q$. Since this is true for all nonzero quaternions,  $\frac{\bar q}{q\bar q}$ has a right inverse also, which is necessarily $q$. Thus $\frac{\bar q}{q\bar q}=q^{-1}$.
