# Quaternions Defined by Ordered Pairs

## Theorem

Consider the quaternions $\Bbb H$ as numbers in the form:

$a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$

where:

$a, b, c, d$ are real numbers;
$\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are entities related to each other in the following way:
 $\displaystyle \mathbf i \mathbf j = - \mathbf j \mathbf i$ $=$ $\displaystyle \mathbf k$ $\displaystyle \mathbf j \mathbf k = - \mathbf k \mathbf j$ $=$ $\displaystyle \mathbf i$ $\displaystyle \mathbf k \mathbf i = - \mathbf i \mathbf k$ $=$ $\displaystyle \mathbf j$ $\displaystyle \mathbf i^2 = \mathbf j^2 = \mathbf k^2 = \mathbf i \mathbf j \mathbf k$ $=$ $\displaystyle - \mathbf 1$

Now consider the quaternions $\Bbb H$ defined as ordered pairs $\left({x, y}\right)$ where $x, y \in \C$ are complex numbers, on which the operation of multiplication is defined as follows:

Let $w = a_1 + b_1 i, x = c_1 + d_1 i, y = a_2 + b_2 i, z = c_2 + d_2 i$ be complex numbers.

Then $\left({w, x}\right) \left({y, z}\right)$ is defined as:

$\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$

where $\overline w$ and $\overline x$ are the complex conjugates of $w$ and $x$ respectively.

These two definitions are equivalent.

## Proof

First we identify the following:

 $\text {(1)}: \quad$ $\displaystyle \mathbf 1$ $=$ $\displaystyle \left({1, 0}\right)$ $\text {(2)}: \quad$ $\displaystyle \mathbf i$ $=$ $\displaystyle \left({i, 0}\right)$ $\text {(3)}: \quad$ $\displaystyle \mathbf j$ $=$ $\displaystyle \left({0, i}\right)$ $\text {(4)}: \quad$ $\displaystyle \mathbf k$ $=$ $\displaystyle \left({0, 1}\right)$

We can see that:

 $\displaystyle \mathbf 1^2$ $=$ $\displaystyle \left({1, 0}\right) \left({1, 0}\right)$ $\displaystyle$ $=$ $\displaystyle \left({1 \times 1 - 0 \times 0, 1 \times 0 + 0 \times 1}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({1, 0}\right)$ $\displaystyle$ $=$ $\displaystyle \mathbf 1$

 $\displaystyle \mathbf i^2$ $=$ $\displaystyle \left({i, 0}\right) \left({i, 0}\right)$ $\displaystyle$ $=$ $\displaystyle \left({i \times i - 0 \times 0, \left({-i}\right) \times 0 + 0 \times i}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({-1, 0}\right)$ $\displaystyle$ $=$ $\displaystyle -\mathbf 1$
 $\displaystyle \mathbf j^2$ $=$ $\displaystyle \left({0, i}\right) \left({0, i}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0 \times 0 - i \times \left({-i}\right), 0 \times i + i \times 0}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({-1, 0}\right)$ $\displaystyle$ $=$ $\displaystyle -\mathbf 1$
 $\displaystyle \mathbf k^2$ $=$ $\displaystyle \left({0, 1}\right) \left({0, 1}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0 \times 0 - 1 \times 1, 0 \times 1 + 1 \times 0}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({-1, 0}\right)$ $\displaystyle$ $=$ $\displaystyle -\mathbf 1$
 $\displaystyle \mathbf i \mathbf j$ $=$ $\displaystyle \left({i, 0}\right) \left({0, i}\right)$ $\displaystyle$ $=$ $\displaystyle \left({i \times 0 - i \times 0, \left({-i}\right) \times i + 0 \times 0}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0, 1}\right)$ $\displaystyle$ $=$ $\displaystyle \mathbf k$
 $\displaystyle \mathbf j \mathbf k$ $=$ $\displaystyle \left({0, i}\right) \left({0, 1}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0 \times 0 - 1 \times \left({-i}\right), 0 \times 1 + i \times 0}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({i, 0}\right)$ $\displaystyle$ $=$ $\displaystyle \mathbf i$
 $\displaystyle \mathbf k \mathbf i$ $=$ $\displaystyle \left({0, 1}\right) \left({i, 0}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0 \times i - 0 \times 1, 0 \times 0 + 1 \times i}\right)$ from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ $\displaystyle$ $=$ $\displaystyle \left({0, i}\right)$ $\displaystyle$ $=$ $\displaystyle \mathbf j$

Let:

$w = a_1 + b_1 i$
$x = d_1 + c_1 i$
$y = a_2 + b_2 i$
$z = d_2 + c_2 i$

and so:

$\overline w = a_1 - b_1 i$
$\overline x = d_1 - c_1 i$

Then substituting for $(1)$ to $(4)$ above, we have:

 $\displaystyle w$ $=$ $\displaystyle a_1 \mathbf 1 + b_1 \mathbf i$ $\displaystyle x$ $=$ $\displaystyle d_1 \mathbf k + c_1 \mathbf j$ $\displaystyle y$ $=$ $\displaystyle a_2 \mathbf 1 + b_2 \mathbf i$ $\displaystyle z$ $=$ $\displaystyle d_2 \mathbf k + c_2 \mathbf j$

Notice the way $\mathbf j$ and $\mathbf k$ are configured. See that they are what appears to be in the wrong order.

We can then demonstrate the equivalence by showing that:

 $\displaystyle \left({w, x}\right) \left({y, z}\right) \ \$ $\displaystyle =$  $\displaystyle ($ $\displaystyle$  $\displaystyle \left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) + \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i$ $\displaystyle$ $,$ $\displaystyle \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) + \left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) i$ $\displaystyle$  $\displaystyle )$

which is equivalent to:

 $\displaystyle \mathbf x_1 \mathbf x_2 \ \$ $\displaystyle =$  $\displaystyle \left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) \mathbf 1$ $\displaystyle$ $+$ $\displaystyle \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) \mathbf i$ $\displaystyle$ $+$ $\displaystyle \left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) \mathbf j$ $\displaystyle$ $+$ $\displaystyle \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) \mathbf k$

where:

$x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
$x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$

in accordance with Quaternion Multiplication.

Thus:

 $\displaystyle w y$ $=$ $\displaystyle \left({a_1 a_2 - b_1 b_2}\right) + \left({a_1 b_2 + b_1 a_2}\right) i$ $\displaystyle z \overline x$ $=$ $\displaystyle \left({d_1 d_2 + c_1 c_2}\right) + \left({d_1 c_2 - c_1 d_2}\right) i$ $\displaystyle \overline w z$ $=$ $\displaystyle \left({a_1 d_2 + b_1 c_2}\right) + \left({a_1 c_2 - b_1 d_2}\right) i$ $\displaystyle x y$ $=$ $\displaystyle \left({d_1 a_2 - c_1 b_2}\right) + \left({d_1 b_2 + c_1 a_2}\right) i$

So:

 $\displaystyle w y - z \overline x$ $=$ $\displaystyle \left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) + \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i$ $\displaystyle \overline w z + x y$ $=$ $\displaystyle \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) + \left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) i$

as required.

$\blacksquare$