Ordering

The total ordering defined in $\mathbb{N}$ can be extended into a total ordering in $\mathbb{Z}$.

Definition 6.2.13   Let $a$ and $b$ be two integers. Then we denote $a \leq b$ if, and only if, there exists a non negative integer $k$ such that $b=a+k$.

The proof that $\leq$ is an ordering is similar to the proof of theorem 1.28; therefore wo do not develop it here. On $\mathbb{Z}^+$, this ordering coincides with the ordering studied above, in subsection 1.3. Moreover we have:

$$\forall x \in \mathbb{Z}^-, \forall y \in \mathbb{Z}^+, \; x \leq y.$$

Denote $x=\overline{(0,b)}$ and $y=\overline{(a,0)}$ for some natural numbers $a$ and $b$; then the following hold:

$$y=\overline{(a,0)}=\overline{(0,b)}+\overline{(a,-b)} =\overline{(0,b)}+\overline{(a+b,0)}=x+\overline{(a+b,0)}.$$

As $\overline{(a+b,0)} \in \mathbb{Z}^+$, we are done. The ordering in $\mathbb{Z}$ is compatible with the addition and with the multiplication. The proofs are similar to the corresponding proofs in $\mathbb{N}$ (v.s. Propositions 1.29 and 1.30).