First we identify the following: {{begin-eqn}} {{eqn | n = 1       | l = \mathbf 1       | r = \left({1, 0}\right)       | c =  }} {{eqn | n = 2       | l = \mathbf i       | r = \left({i, 0}\right)       | c =  }} {{eqn | n = 3       | l = \mathbf j       | r = \left({0, i}\right)       | c =  }} {{eqn | n = 4       | l = \mathbf k       | r = \left({0, 1}\right)       | c =  }} {{end-eqn}} We can see that: {{begin-eqn}} {{eqn | l = \mathbf 1^2       | r = \left({1, 0}\right) \left({1, 0}\right)       | c =  }} {{eqn | r = \left({1 \times 1 - 0 \times 0, 1 \times 0 + 0 \times 1}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({1, 0}\right)       | c =  }} {{eqn | r = \mathbf 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf i^2       | r = \left({i, 0}\right) \left({i, 0}\right)       | c =  }} {{eqn | r = \left({i \times i - 0 \times 0, \left({-i}\right) \times 0 + 0 \times i}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({-1, 0}\right)       | c =  }} {{eqn | r = -\mathbf 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf j^2       | r = \left({0, i}\right) \left({0, i}\right)       | c =  }} {{eqn | r = \left({0 \times 0 - i \times \left({-i}\right), 0 \times i + i \times 0}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({-1, 0}\right)       | c =  }} {{eqn | r = -\mathbf 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf k^2       | r = \left({0, 1}\right) \left({0, 1}\right)       | c =  }} {{eqn | r = \left({0 \times 0 - 1 \times 1, 0 \times 1 + 1 \times 0}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({-1, 0}\right)       | c =  }} {{eqn | r = -\mathbf 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf i \mathbf j       | r = \left({i, 0}\right) \left({0, i}\right)       | c =  }} {{eqn | r = \left({i \times 0 - i \times 0, \left({-i}\right) \times i + 0 \times 0}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({0, 1}\right)       | c =  }} {{eqn | r = \mathbf k       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf j \mathbf k       | r = \left({0, i}\right) \left({0, 1}\right)       | c =  }} {{eqn | r = \left({0 \times 0 - 1 \times \left({-i}\right), 0 \times 1 + i \times 0}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({i, 0}\right)       | c =  }} {{eqn | r = \mathbf i       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \mathbf k \mathbf i       | r = \left({0, 1}\right) \left({i, 0}\right)        | c =  }} {{eqn | r = \left({0 \times i - 0 \times 1, 0 \times 0 + 1 \times i}\right)       | c = from $\left({w, x}\right) \left({y, z}\right) := \left({w y - z \overline x, \overline w z + x y}\right)$ }} {{eqn | r = \left({0, i}\right)       | c =  }} {{eqn | r = \mathbf j       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = w       | r = a_1 \mathbf 1 + b_1 \mathbf i       | c =  }} {{eqn | l = x       | r = d_1 \mathbf k + c_1 \mathbf j       | c =  }} {{eqn | l = y       | r = a_2 \mathbf 1 + b_2 \mathbf i       | c =  }} {{eqn | l = z       | r = d_2 \mathbf k + c_2 \mathbf j       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | ll= \left({w, x}\right) \left({y, z}\right)       | l = =       | o =        | r = (       | c =  }} {{eqn | o =        | r = \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       | c =  }} {{eqn | o = ,       | r = \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       | c =  }} {{eqn | o =        | r = )       | c =  }} {{end-eqn}} which is equivalent to: {{begin-eqn}} {{eqn | ll= \mathbf x_1 \mathbf x_2       | l = =       | o =        | r = \left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) \mathbf 1       | c =  }} {{eqn | o = +       | r = \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) \mathbf i       | c =  }} {{eqn | o = +       | r = \left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) \mathbf j       | c =  }} {{eqn | o = +       | r = \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) \mathbf k       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = w y       | r = \left({a_1 a_2 - b_1 b_2}\right) + \left({a_1 b_2 + b_1 a_2}\right) i       | c =  }} {{eqn | l = z \overline x       | r = \left({d_1 d_2 + c_1 c_2}\right) + \left({d_1 c_2 - c_1 d_2}\right) i }} {{eqn | l = \overline w z       | r = \left({a_1 d_2 + b_1 c_2}\right) + \left({a_1 c_2 - b_1 d_2}\right) i }} {{eqn | l = x y       | r = \left({d_1 a_2 - c_1 b_2}\right) + \left({d_1 b_2 + c_1 a_2}\right) i }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = w y - z \overline x       | r = \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       | c =  }} {{eqn | l = \overline w z + x y       | r = \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 }} {{end-eqn}} as required. {{qed}}