Partial order relations and posets Suppose you are given to prove this question:
Let $P=\{2,3,4,\ldots,10\}$.
Define $\le$ by $a\le b$ if and only if $a\mid b$, i.e, $a$ divides $b$. Prove that it is a partial order on $P$.
I think you should solve it in this way---
1) $2$ divides $2$ itself so reflexive property holds. And so on. 
I want to tell that relation is a subset of $P\times P$. But when we take the pair $(2,2)$ or $(2,4)$, where from they are taken? Directly considering any $2$ elements of $P$? or considering $(2,2)$ as a subset of $P\times P$?
if the answer is 1, how can you tell it a partial order relation?
 A: We have the set $P = \{2,3,4,5,6,7,8,9,10\}$, and we define a relation $\leq$ on $P$ by $a \leq b$ iff $a | b$. This means that we define a relation $\leq \subseteq P \times P$ such that $(a,b) \in \leq$ iff $a | b$, in other words, the relation is a subset of $P \times P$. We sometimes (often) write $a \leq b$ instead of $(a,b) \in \leq$.
Notice that the numbers $2,3,4,\dots,10$ are members of $P$, whereas when we consider $a \leq b$, this really means that $(a,b) \in \leq$, so in this context they are members of $\leq$, which is a subset of $P \times P$.
Given this relation, we need to prove that it is reflexive, antisymmetric, and transitive in order for it to be a partial order.


*

*Reflexivity: Any non-zero number divides itself, so $a | a$ for any $a \in P$ which means that $(a,a) \in \leq$.

*Antisymmetry: If $a \leq b$ and $b \leq a$, then this means that $a | b$ and $b | a$. Since all elements of $P$ are positive, this means that $a = b$.

*Transitivity: If $a \leq b$ and $b \leq c$, then $a | b$ and $b | c$ and this implies that $a | c$, so $a \leq c$.


Hence $\leq$ is a partial order on $P$.
