Let R be a relation on defined by if and only if or . 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 . Notice that . Therefore, , and hence, R is reflexive.
In order to show R is antisymmetric, let and . Assume and . Therefore, or , and or .
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.
Oct 13th 2007, 10:54 AM
Are both of these true?
Oct 13th 2007, 10:58 AM
Yes they are, because...
In the first statement, the first elements in the ordered pairs work. . In the second case, the second elements work. . 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).
Oct 13th 2007, 11:01 AM
Is R antisymmetric?
Oct 13th 2007, 11:15 AM
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 and in .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 ; therefore, (2,1)R(1,2). Since 1 and , . Thus, R is not antisymmetric, and hence, R is not a partial order.
Oct 13th 2007, 11:21 AM
By George, you have.
Oct 13th 2007, 11:22 AM
Ha, thanks a lot.
Disproofs are so much easier than proofs most of the time.