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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:

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

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

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

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).

root 2002-06-10