Integers are Euclidean Domain

Theorem
The integers $\Z$ with the mapping $\nu: \Z \to \Z$ defined as:
 * $\forall x \in \Z: \nu \left({x}\right) = \left \vert {x} \right \vert$

form a Euclidean domain.

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

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


 * $\left\vert{a}\right\vert = \begin{cases}

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

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

Since $\left\vert{b}\right\vert > 0$ it follows from Relation Induced by Positivity Property is Compatible with Multiplication that $\left\vert{a}\right\vert \cdot \left\vert{b}\right\vert \ge \left\vert{a}\right\vert$.

The second criterion:


 * For any $a, b \in R$, $b \ne 0$, there exist $q, r \in R$ with $\nu \left({r}\right) < \nu \left({b}\right)$, or $r=0$ such that:
 * $ a = q\circ b + r$

follows from the Division Theorem.