1. ## Partial Order Relations

Let R be a relation on $\displaystyle \bold{N}^2$ defined by $\displaystyle (x,y)R(u,v)$ if and only if $\displaystyle x \leq u$ or $\displaystyle y \leq v$. Is R a partial order? Prove your answer.

This is what I came up with so far.

So, just to be courteous to the people who read this, a relation on a set is called a partial ordering if the relation is reflexive, antisymmetric, and transitive.

Therefore, I tried to prove that R is reflexive, antisymmetric, and transitive.

Here's what I did.

Claim: R is a partial order.

Proof: For R to be a partial order, it must be reflexive, antisymmetric, and transitive.
In order to show R is reflexive, let $\displaystyle (x,y) \in \bold{N}^2$. Notice that $\displaystyle x \leq x$. Therefore, $\displaystyle (x,y)R(x,y)$, and hence, R is reflexive.
In order to show R is antisymmetric, let $\displaystyle (x,y) \in \bold{N}^2$ and $\displaystyle (u,v) \in \bold{N}^2$. Assume $\displaystyle (x,y)R(u,v)$ and $\displaystyle (u,v)R(x,y)$. Therefore, $\displaystyle x \leq u$ or $\displaystyle y \leq v$, and $\displaystyle u \leq x$ or $\displaystyle v \leq y$.

After this point, I am not sure how to continue. I am thinking that I would have to break this into cases, but I'm not sure how to break it into cases because I have two different statements that both involve "or". It's looking like the transitive proof will also have cases, but I'm not sure how to attempt those either. If I can clear up the antisymmetric one, the transitive one can probably follow pretty easily. I will post a continuation after I get some hints on how to go about continuing. Thanks.

2. Are both of these true?
$\displaystyle \left( {1,2} \right)R\left( {2,1} \right)\quad \& \quad \left( {2,1} \right)R\left( {1,2} \right)$

3. Yes they are, because...

In the first statement, the first elements in the ordered pairs work. $\displaystyle 1 \leq 2$. In the second case, the second elements work. $\displaystyle 1 \leq 2$. I guess I'm still not seeing how that's going to help me prove this :-(. (we just started relations, so i'm having trouble getting used to the idea right now).

4. Is R antisymmetric?

5. Oh, let's see.

I'll show you what i'm thinking. Critique it and/or confirm it.

Here you go. (i'll put comments in blue.)

So, your example was. Consider the ordered pairs $\displaystyle (x,y) = (1,2)$ and $\displaystyle (u,v) = (2,1)$ in $\displaystyle \bold{N}^2$.So for R to be antisymmetric, if x ~ y and y ~ x, then x = y. So, for a counterexample, we find x and y such that the hypothesis is true and the conclusion is false. Notice that x = 1 < 2 = u; therefore, (1,2)R(2,1). Also, notice that $\displaystyle v = 1 \leq 2 = y$; therefore, (2,1)R(1,2). Since 1 $\displaystyle \not= 2$ and $\displaystyle 2 \not= 1$, $\displaystyle (1,2) \not= (2, 1)$. Thus, R is not antisymmetric, and hence, R is not a partial order.

6. By George, you have.

7. Ha, thanks a lot.

Disproofs are so much easier than proofs most of the time.

Thanks a lot man. I'll click the thanks button.