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Math Help - relations

  1. #1
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    relations

    I was wondering if someone could quickly check to see if these properties are right or wrong. I drew the directed graph, but I'm finding understanding the properties somewhat confusing.

    I have this relation:
    {(3, 3), (3,4), (3,2), (1,3), (1,1), (1,4), (1,2), (4,3), (4,4), (4,2), (2,2)}
    I think this relation is:
    reflexive, NOT symmetric, transitive, NOT irreflexive, NOT antisymmetric, NOT strictly antisymmetric, dichotomous, trichotomous.

    I think I have the most trouble with transitive , dichotomous, and trichomotous. Could someone please check to see if my answers are right?

    Thank you.
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  2. #2
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    Quote Originally Posted by PvtBillPilgrim View Post
    I was wondering if someone could quickly check to see if these properties are right or wrong. I drew the directed graph, but I'm finding understanding the properties somewhat confusing.

    I have this relation:
    {(3, 3), (3,4), (3,2), (1,3), (1,1), (1,4), (1,2), (4,3), (4,4), (4,2), (2,2)}
    I think this relation is:
    reflexive, NOT symmetric, transitive, NOT irreflexive, NOT antisymmetric, NOT strictly antisymmetric, dichotomous, trichotomous.

    I think I have the most trouble with transitive , dichotomous, and trichomotous. Could someone please check to see if my answers are right?

    Thank you.
    It is transitive and dichotomous, but not trichotomous. For example we have that 3R4 and 4R3. Both cannot be allowed in a trichotomous relation.

    -Dan
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  3. #3
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    Doesn't it have to be trichotomous to be dichotomous?

    Thanks for the help.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    Doesn't it have to be trichotomous to be dichotomous?

    Thanks for the help.
    Dichotomous means aRb and/or bRa exist.

    Trichotomous means that only one of aRb, bRa, or a=b are true.

    So it can be dichotomous with 3R4 and 4R3 both existing, but this precludes trichotomy since we can't have both 3R4 and 4R3.

    -Dan
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  5. #5
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    Well, it seems later in my assignment I need to prove that a relation is dichotomous if and only if it is trichotomous and reflexive. This relation contradicts that, does it not?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    Well, it seems later in my assignment I need to prove that a relation is dichotomous if and only if it is trichotomous and reflexive. This relation contradicts that, does it not?
    No because only one of aRb, bRa need to exist to make it dichotomous.

    -Dan
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  7. #7
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    Right, but it can be dichotomous without being trichotomous?

    How does that make any sense? In this example, you're saying it's dichotomous, but not trichotomous. Wouldn't the fact that it's not trichotomous make it not dichotomous by this theorem I have to prove?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    Right, but it can be dichotomous without being trichotomous?

    How does that make any sense? In this example, you're saying it's dichotomous, but not trichotomous. Wouldn't the fact that it's not trichotomous make it not dichotomous by this theorem I have to prove?
    (shrugs) All I can say is that these are the pages I got the definitions from. Here and here.

    I agree that it would seem to make sense that the two terms be linked in the sense that you would expect if it were trichotomous that it would also be dichotomous, based on the prefixes, but apparently this is not the case.

    -Dan
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  9. #9
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    Thanks for the help. I looked at those pages too and it would seem this specific relation is dichotomous, but not trichotomous.

    Anyway, thanks.
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