Prove the relation on $\Bbb N \setminus \{0,1\}$ is a partial order

Question

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!

Answer 1

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.

Answer 2

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 $aRc$.