The set $\mathbb{Z}$ is totally ordered

Question

Having the following definition of the $\leq$-Relation in $\mathbb{Z}$:

For $a, b\in \mathbb{Z}$ we define $$ a \leq b : \iff b-a \in \mathbb{N} $$ Show that $(\mathbb{Z}, \leq)$ is totally ordered.

I managed to show that $\leq$ as defined above is reflexive, antisymmetric and transitive. I am clueless on how to go about showing that the set is totally ordered.

I remember how it's being done via induction for the $\leq$-Relation in $\mathbb{N}$ but I doubt that I can apply this here. For natural numbers introduce the set: $$ M:= \lbrace n \in \mathbb{N} \mid n \leq m \text{ or } m \leq n \rbrace \text{ for fixated }m \in \mathbb{N} $$ And show that $M = \mathbb{N}$ via induction.

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I also thought about applying some of the definitions of the relation to introduce $\mathbb{Z}$, namely $$\text{for $c,d,e,f \in \mathbb{N}$}, \ (c,d) \ \sim (e,f) : \iff c+f = d + e $$ which defines an equivalence relation with class $[c,d]$ $$ \mathbb{Z}:= \lbrace [c,d] \mid (c,d) \in \mathbb{N}^2 \rbrace $$

Exhausting google also didn't help me, although I have worked with the terms given in http://en.wikipedia.org/wiki/Total_order

Answer 1

To show that it is totally ordered it just remains to show that for all $a,b$ you have $a \le b$ or $b \le a$. That is, you have to show $a-b $ or $b-a$ is a natural number. This follows from the fact that for each integer $c$ one has $c$ or $-c$ is a natural number. I am not sure if you can use that, but even if not it should not be that hard to establish. How, exactly would depend on how you defined the integers in relation to the natural numbers.

Answer 2

You can use trichotomy property of $\mathbb{Z}$ which says every Integer (it is actually for Reals,so for integers too) number is either positive, negative or zero. By that if you pick $a$ and $b$ $\in \mathbb{Z}$ then $a-b\in \mathbb{Z}$ and thus $a-b \ge0 $ or $\le 0$, $\implies $ $a \ge b$ or $b \ge a$ resp.