Norm of Unit of Normed Division Algebra

Theorem
Let $\left({A_F, \oplus}\right)$ be a normed division algebra.

Let the unit of $\left({A_F, \oplus}\right)$ be $1_A$.

Then:
 * $\left \Vert {1_A} \right \Vert = 1$

where $\left \Vert{1_A}\right \Vert$ denotes the norm of $1_A$.

Proof
By definition:
 * $\forall a, b \in A_F: \left \Vert{a \oplus b}\right \Vert = \left \Vert{a}\right \Vert \left \Vert{b}\right \Vert$

So:


 * $\left \Vert{1_A}\right \Vert = \left \Vert{1_A \oplus 1_A}\right \Vert = \left \Vert{1_A}\right \Vert \left \Vert{1_A}\right \Vert$

So $\left \Vert{1_A}\right \Vert \in \R$ is idempotent under real multiplication.

From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill.

But $\left \Vert{1_A}\right \Vert$ can not be $0$ as that would make:
 * $\forall a \in A_F: \left \Vert{a}\right \Vert = \left \Vert{1_A \oplus a}\right \Vert = \left \Vert{1_A}\right \Vert \left \Vert{a}\right \Vert = 0 \left \Vert{a}\right \Vert = 0$

which contradicts the definition of the norm.

Hence the result.