I've been given the question below and I'm not sure how to show that it is a partial order.
Show that the relation $R =\{(a,b) \mid a \text{ divides } b\}$ over the set $\mathbb{Z}^+$ is a partial order and is not an equivalence reaction.
I understand that it would be called a partial order if is transitive, reflexive and anti-symmetric.
Would you say that is transitive as (1,2) is an element of Z+ and (2,3) is an element of Z+, then (1,3) is an element of Z+?
Would you say it is reflexive as every element is related to itself for being a member of the set Z+ ?
How would you show it is anti-symmetric?
I am new to this topic and so I may sound rather stupid but I genuinely have no idea how to show that is a partial order.
Reflexive is clear since $a$ divides $a$ for all $a \in \mathbb{Z}^+$. Now suppose that $a$ divides $b$ and $a \neq b$ this means that $b$ does not divide $a$ (try this for yourself). Finally suppose $a$ divides $b$ and $b$ divides $c$. We want to show that $a$ divides $c$. We want to show there exists an integer $z$ such that $az = c$. From our hypothesis for some integers $x,y$ we have that \begin{gather} ax = b \hspace{20 pt} by = c. \end{gather} Hence we can substitute to see that \begin{gather} axy = c. \end{gather} Now $xy$ is a natural number, say $z.$ Hence $az = c$ or $a$ divides $c$. Hence this is a partial ordering. Try as an additional exercise to show where the partial ordering breaks down if we take our set to be all integers instead of just positive ones.