Element of Integral Domain is Divisor of Itself

Theorem

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.

Then every element of $D$ is a divisor of itself:

$\forall x \in D: x \divides x$

Proof

Follows directly from the definition of divisor:

$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$

$\blacksquare$