Field Contains at least 2 Elements

Theorem
Let $\struct {F, +, \times}$ be a field.

Then $E$ contains at least $2$ elements.

Proof
By definition, $\struct {F, +, \times}$ is an algebraic structure such that:
 * $\struct {F, +}$ is an abelian group whose identity element is $0 \in F$
 * $\struct {F^*, \times}$ is also an abelian group, where $F^* = F \setminus \set 0$.

From Group is not Empty, $\struct {F^*, \times}$ cannot be the empty set.

So there are at least $2$ elements in $F$:
 * the identity element $0$ of $\struct {F, +}$
 * the identity element $1$ of $\struct {F^*, \times}$.