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

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## 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:

\(\displaystyle i \paren {a_1 + a_2 i + a_3 j}\) | \(=\) | \(\displaystyle a_1 i - a_2 + a_3 i j\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_1 i - a_2 + a_3 \paren {a_1 + a_2 i + a_3 j}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_1 i - a_2 + a_1 a_3 + a_2 a_3 i + {a_3}^2 j\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \paren {a_1 a_3 - a_2} + \paren {a_1 + a_2 a_3} i + \paren { {a_3}^2 + 1} j\) | 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.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem