Proof for 3d contradiction (Quaternions)

Prove by Contradiction. Why 1 real and 2 complex numbers won't work.

Suppose we only have 2 complex numbers. $i, j \in C$ . Assuming closed multiplication, the multiplication between the two numbers would result in another value consisting of the two numbers.
Where $a, b, c \in R$

i j = a + i b + j c

Multiply the left hand side by $i$ .

- j = - b + i a + i j c

Substituting $i j$ into the above expression.

\begin{aligned} - j & = - b + i a + (a + i b + j c) c \\ 0 + i 0 + j 0 & = (a - b) + i (a + b) + j (c^{2} + 1) \end{aligned}

Thus this contradicts the statement that $a, b, c \in R$ .

c^{2} = - 1