For an example of a noncommutative division ring, let
\begin{equation*}
1 =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\quad
{\mathbf i}
=
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix},
\quad
{\mathbf j} =
\begin{pmatrix}
0 & i \\
i & 0
\end{pmatrix},
\quad
{\mathbf k} =
\begin{pmatrix}
i & 0 \\
0 & -i
\end{pmatrix}\text{,}
\end{equation*}
where \(i^2 = -1\text{.}\) These elements satisfy the following relations:
\begin{align*}
{\mathbf i}^2 = {\mathbf j}^2 & = {\mathbf k}^2 = -1\\
{\mathbf i} {\mathbf j} & = {\mathbf k}\\
{\mathbf j} {\mathbf k} & = {\mathbf i}\\
{\mathbf k} {\mathbf i} & = {\mathbf j}\\
{\mathbf j} {\mathbf i} & = - {\mathbf k}\\
{\mathbf k} {\mathbf j} & = - {\mathbf i}\\
{\mathbf i} {\mathbf k} & = - {\mathbf j}\text{.}
\end{align*}
Let \({\mathbb H}\) consist of elements of the form \(a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k}\text{,}\) where \(a, b , c, d\) are real numbers. Equivalently, \({\mathbb H}\) can be considered to be the set of all \(2 \times 2\) matrices of the form
\begin{equation*}
\begin{pmatrix}
\alpha & \beta \\
-\overline{\beta} & \overline{\alpha }
\end{pmatrix}\text{,}
\end{equation*}
where \(\alpha = a + di\) and \(\beta = b + ci\) are complex numbers. We can define addition and multiplication on \({\mathbb H}\) either by the usual matrix operations or in terms of the generators \(1\text{,}\) \({\mathbf i}\text{,}\) \({\mathbf j}\text{,}\) and \({\mathbf k}\text{:}\)
\begin{gather*}
(a_1 + b_1 {\mathbf i} + c_1 {\mathbf j} +d_1 {\mathbf k} ) + ( a_2 + b_2 {\mathbf i} + c_2 {\mathbf j} +d_2 {\mathbf k} )\\
= (a_1 + a_2) + ( b_1 + b_2) {\mathbf i} + ( c_1 + c_2) \mathbf j + (d_1 + d_2) \mathbf k
\end{gather*}
and
\begin{equation*}
(a_1 + b_1 {\mathbf i} + c_1 {\mathbf j} +d_1 {\mathbf k} ) ( a_2 + b_2 {\mathbf i} + c_2 {\mathbf j} +d_2 {\mathbf k} ) = \alpha + \beta {\mathbf i} + \gamma {\mathbf j} + \delta {\mathbf k}\text{,}
\end{equation*}
where
\begin{align*}
\alpha & = a_1 a_2 - b_1 b_2 - c_1 c_2 -d_1 d_2\\
\beta & = a_1 b_2 + a_2 b_1 + c_1 d_2 - d_1 c_2\\
\gamma & = a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2\\
\delta & = a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2\text{.}
\end{align*}
Though multiplication looks complicated, it is actually a straightforward computation if we remember that we just add and multiply elements in \({\mathbb H}\) like polynomials and keep in mind the relationships between the generators \({\mathbf i}\text{,}\) \({\mathbf j}\text{,}\) and \({\mathbf k}\text{.}\) The ring \({\mathbb H}\) is called the ring of quaternions.
To show that the quaternions are a division ring, we must be able to find an inverse for each nonzero element. Notice that
\begin{equation*}
( a + b {\mathbf i} + c {\mathbf j} + d {\mathbf k} )( a - b {\mathbf i} - c {\mathbf j} - d {\mathbf k} ) = a^2 + b^2 + c^2 + d^2\text{.}
\end{equation*}
This element can be zero only if \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are all zero. So if \(a + b {\mathbf i} + c {\mathbf j} +d {\mathbf k} \neq 0\text{,}\)
\begin{equation*}
(a + b {\mathbf i} + c {\mathbf j} + d {\mathbf k})\left( \frac{a - b {\mathbf i} - c {\mathbf j} - d {\mathbf k} }{a^2 + b^2 + c^2 + d^2} \right) = 1\text{.}
\end{equation*}
