Show that the divisibility relation $\text{ | }$ is an ordering relation in $\mathbb{N}$

Question

Show that the divisibility relation $\text{ | }$ is an ordering relation in $\mathbb{N} \setminus\left\{0\right\}$ and state if this ordering is total or partial.

I'm not sure how we do this but I think we have to show 3 things?

1. reflexive
2. anti-symmetric
3. transitive

If so then for

1. We have $a \text{ | }b \Rightarrow \exists k \in \mathbb{N}$ such that $b=ka$

2. Every number divides itself because $a \text{ | }a$ is possible because $a=1 \cdot a$

If $(a \text{ | }b) \wedge (b\text{ | }a)$ then $ka=b$ and $a=jb$

If $\exists j,k \in \mathbb{N}$ then $kjb=b$ and so $kj=1 \Rightarrow k=j=1 \Rightarrow a = b$

3. If $a \text{ | } b \text{ there }\exists k$ so $b=ka$ and if $b \text{ | }c$ there $\exists j$ such that $c=jb$ so $c=j(ka) \Rightarrow c=(jk)a$ so $c$ is a multiple of $a$

Is it alright till here? I have no idea how to say if this is total or partial? But I think it's partial because all these steps will work for $N$ and $Z$..?

Answer 1

Everything's good.

An order relation $R$ on $A$ is total if, for every $a,b\in A$, we have $a\mathrel{R}b$ or $b\mathrel{R}a$. Otherwise it is partial. Can you decide it for divisibility?