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Math Help - Examples of Relations...

  1. #1
    Junior Member
    Joined
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    Examples of Relations...

    Hello, I am stuck on a problem. The problem is:

    Let A = {1, 2, 3}. Give an example of a relation R on A that is
    a. Transitive, reflexive, but not antisymmetric
    b. Antisymmetric, reflexive, but not transitive
    c. Antisymmetric, transitive, but not reflexive

    Ok, I know what each means,

    Transitive: if (a,b) and (b,c) are in R, then (a,c) is in R.
    Reflexive: if a is in A, then (a,a) is in R
    Antisymmetric: For all a,b in A, if a =/= b and (a,b) are in R, then (b,a) is not in R.

    That being said, I am having difficulty thinking of a relation to satisfy the conditions. Is there a good procedure to try for this type of problem? If anyone can give me a hint or shove in the right direction, that would be helpful. Thank you! I appreciate it.

    Edit, I have given this some thought and have come up with this:

    a. R = {(a,b) \in Z X Z }
    b. R = {(a,b) \in Z X Z, a = b}
    c. R = {(a,b) \in Z X Z, a < b}
    Last edited by Alterah; April 7th 2010 at 10:47 PM.
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  2. #2
    Junior Member
    Joined
    Oct 2006
    Posts
    71
    Quote Originally Posted by Alterah View Post
    Hello, I am stuck on a problem. The problem is:

    Let A = {1, 2, 3}. Give an example of a relation R on A that is
    a. Transitive, reflexive, but not antisymmetric
    b. Antisymmetric, reflexive, but not transitive
    c. Antisymmetric, transitive, but not reflexive

    Ok, I know what each means,

    Transitive: if (a,b) and (b,c) are in R, then (a,c) is in R.
    Reflexive: if a is in A, then (a,a) is in R
    Antisymmetric: For all a,b in A, if a =/= b and (a,b) are in R, then (b,a) is not in R.

    That being said, I am having difficulty thinking of a relation to satisfy the conditions. Is there a good procedure to try for this type of problem? If anyone can give me a hint or shove in the right direction, that would be helpful. Thank you! I appreciate it.

    Edit, I have given this some thought and have come up with this:

    a. R = {(a,b) \in Z X Z }
    b. R = {(a,b) \in Z X Z, a = b}
    c. R = {(a,b) \in Z X Z, a < b}
    What you've come up with looks a bit strange to me.

    Let's just consider a., and leave b. and c. to someone else.

    Think about R = {<1,1>, <2,2>, <3,3>, <1,2>, <2,3>, <2,1>, <1,3>}.
    Convince yourself that R is indeed a preorder that does not have the additional property of antisymmetry.

    In the finite case, there are effective methods for obtaining such results.
    Establishing that might, in itself, be interestng.
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