# Complex Numbers cannot be Extended to Algebra in Three Dimensions with Real Scalars

## Theorem

It is not possible to extend the complex numbers to an algebra of $3$ dimensions with real scalars.

## Proof

Aiming for a contradiction, suppose that $\set {1, i, j}$ forms a basis for an algebra of $3$ dimensions with real scalars.

Let $1$ and $i$ have their usual properties as they do as complex numbers:

$\forall a: 1 a = a 1 = a$
$i \cdot i = -1$

Then:

$i j = a_1 + a_2 i + a_3 j$

for some $a_1, a_2, a_3 \in \R$.

Multiplying through by $i$:

$(1): \quad i \paren {i j} = \paren {i i} j = -j$

and:

 $\ds i \paren {a_1 + a_2 i + a_3 j}$ $=$ $\ds a_1 i - a_2 + a_3 i j$ $\ds$ $=$ $\ds a_1 i - a_2 + a_3 \paren {a_1 + a_2 i + a_3 j}$ $\ds$ $=$ $\ds a_1 i - a_2 + a_1 a_3 + a_2 a_3 i + {a_3}^2 j$ $\ds \leadsto \ \$ $\ds 0$ $=$ $\ds \paren {a_1 a_3 - a_2} + \paren {a_1 + a_2 a_3} i + \paren { {a_3}^2 + 1} j$ from $(1)$

But this implies that ${a_3}^2 = -1$, which contradicts our supposition that $a_3 \in \R$.

Hence the result by Proof by Contradiction.

$\blacksquare$