Integers are Euclidean Domain

Theorem
The integers $\Z$ with the mapping $\nu: \Z \to \Z$ defined as:
 * $\forall x \in \Z: \map \nu x = \size x$

form a Euclidean domain.

Proof
From Integers form Ordered Integral Domain we have that $\struct {\Z, +, \times, \le}$ forms an ordered integral domain.

For all $a \in \Z$, the absolute value of $a$ is defined as:


 * $\size a = \begin{cases}

a & : 0 \le a \\ -a & : a < 0 \end{cases}$

By Product of Absolute Values on Ordered Integral Domain we have:
 * $\forall a, b \in \Z: \size a \cdot \size b = \size {a b}$

Since $\size b > 0$ it follows from Relation Induced by Strict Positivity Property is Compatible with Multiplication that $\size a \cdot \size b \ge \size a$.

The second criterion:


 * For any $a, b \in R$, $b \ne 0$, there exist $q, r \in R$ with $\map \nu r < \map \nu b$, or $r = 0$ such that:
 * $ a = q \circ b + r$

follows from the Division Theorem.