Prove the relation on $\Bbb N \setminus \{0,1\}$ is a partial order I'm a bit new to this material and trying understand some problem I'm solving
Let $R$ be a relation on the set set =  $ \{ N \setminus \{ 0,1\}\} $ that's defined like 
this: $aRb$ if there is an integer x such that: $ b = ax $
Prove that this set is a partial order set, and whether it's weak or strong. Also, check if this is a linear set. 
I started with showing it's reflexive, $aRa \rightarrow a=ax \rightarrow c=1 $
anti symmetric: Let $aRb$ and $bRa$ 
$$b=ax$$
$$a=bx$$
then it's clear that $a=b$  (am I right?)
I couldn't know how to show transitivity, I thought first that
$aRb, bRc$, 
show that  $aRc$
if $b=ax$ and $c=bx$ then $c=ax^2$, is that enough? I know there can be $x^2$ that can satisfy this equation but I don't know if I'm right
Can you please tell me how to check whether it's a weak or strong partial order?
How do I find the minimum and maximum of this set?
Thanks!
 A: for transitivity assume $aRb$ and $bRc$ . So we have $b=ax$ and $c=by$. substitute $b$ for $ax$ to get $c=(ax)y$ and now use associative property to get $c=a(xy)$ so $aRc$.
A: You seem to have a misconception regarding the concept of bounded variable which prompts you to make some mistakes.
Let $A:=\mathbb N\setminus\{0,1\}$.
For all $a,b\in A$, one has $aRb$ if, and only if, $\exists x\in \mathbb N(ax=b)$.
(Note that I changed the set where $x$ lies. If you let $x$ range over the integers, it's not a partial order. For instance you'll get $-2R2\land 2R-2$).
I don't get your proof for reflexivity. If it's even right, it's badly written.
Now let $a,b,c\in A$.
To prove that $R$, as I define it, is reflexive, one proves that $\exists x\in \mathbb N(ax=a)$. The simplest way to prove this is by pinpointing an $x$ that does the trick. Can you find it?
Your try at anti-symmetry is wrong due to the misconception I mentioned above. You want to prove that if $\exists x\in \mathbb N(ax=b)\land \exists y\in \mathbb N(by=a)$, then $a=b$. Assume the antecedent and fix $x,y$ as mentioned. You get $(by)x=ax=b$. What does $xy$ have to be and consequently what do $x$ and $y$ need to be? How does that help you?
Regarding your mistake at proving this, you took $x=y$ which is wrong. And you did the same mistake for transitivity.
For transitivity assume that $\exists x(ax=b)\land \exists y(by=c)$. The goal is to prove that $\exists z\in \mathbb N(az=c)$. Let $x,y$ be as mentioned. Then from $by=c$ and $ax=b$ you get $(ax)y=c$. Can you find an appropriate $z$?
For the rest of the problem, review all the definitions.
