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Math Help - relations

  1. #1
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    relations

    R1 = {(1,1),(2,2),(3,3),(4,4),(5,5)}
    R2 = {(1,2),(2,3),(3,4),(4,4),(5,1)}

    I'm supposed to determine if these relations are reflexive, irreflexive, symmetric, antisymmetric, and/or transitive.

    Can anyone help me with this? Please, explain your answer to me because I'm lost when it comes to this.

    Thank you!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jzellt View Post
    R1 = {(1,1),(2,2),(3,3),(4,4),(5,5)}
    R2 = {(1,2),(2,3),(3,4),(4,4),(5,1)}

    I'm supposed to determine if these relations are reflexive, irreflexive, symmetric, antisymmetric, and/or transitive.

    Can anyone help me with this? Please, explain your answer to me because I'm lost when it comes to this.

    Thank you!
    For example, reflexive: We must have aRa for all possible values of a. R2 is obviously not reflexive, for example, because 5R1.

    Transitive is probably the one that will take the most work. For transitive we must have that if aRb and bRc then aRc for all a, b, c. Again, R2 is not transitive because 1R2 and 2R3 but we don't have that 1R3.

    I'm not sure what to say about R1 as far as transitivity. I suppose it is transitive because it fits the theorem, but the only possible case in R1 is of the type: if aRa and aRa then aRa, which is rather trivial.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post

    I'm not sure what to say about R1 as far as transitivity. I suppose it is transitive because it fits the theorem, but the only possible case in R1 is of the type: if aRa and aRa then aRa, which is rather trivial.

    -Dan
    Isnt there any requirement on the objects to be distinct?

    I remember a exercise question that asks if reflexivity is an unnecessary condition, since by symmetry and transitivity , aRb and bRa should imply aRa? I thought that the answer to this was that a,b,c in the transitivity condition refer to distinct objects.


    I am not sure though... Want to clarify the idea
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    Isn’t there any requirement on the objects to be distinct?
    First we must assume that these two relations are on the set {1,2,3,4,5}.
    Then the relation R_1 is simply the identity relation and therefore is a equivalence relation.
    As for a requirement on the objects to be distinct, there is no such requirement.
    If we know that S is a transitive relation (a,b) \in S\,\& \,(b,a) \in S \Rightarrow \quad (a,a) \in S
    What you may be recalling is the ‘trick question’ about a transitive and symmetric relation implying reflexive.
    The problem there is one cannot assure all elements are related to themselves.
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  5. #5
    Lord of certain Rings
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    Quote Originally Posted by Plato View Post
    The problem there is one cannot assure all elements are related to themselves.
    Let us assume the set has more than 1 element. Now pick any element 'a', and any other element 'b'.If all elements obey symmetry relation, then for any (a,b) in R, we have (b,a) in R. Now if transitive relation holds for all (a,b) in R and (b,a) in R, we must have (a,a) in R.

    What is wrong with this argument?
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  6. #6
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    Quote Originally Posted by Isomorphism View Post
    Let us assume the set has more than 1 element. Now pick any element 'a', and any other element 'b'.If all elements obey symmetry relation, then for any (a,b) in R, we have (b,a) in R. Now if transitive relation holds for all (a,b) in R and (b,a) in R, we must have (a,a) in R.

    What is wrong with this argument?
    The problem with this argument is the assumption that all elements are related to at least one other. It may be the case that x is not related to any other element, so symmetry and transitivity could still hold for all elements (including x) because those properties only make claims about certain relationships given that those relationships are there to begin with.
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  7. #7
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    Here is a counterexample. R is symmetric & transitive but not reflexive. <br />
A = \left\{ {1,2,3,4,5} \right\}\,\& \,R = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {3,3} \right)} \right\}
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  8. #8
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    Thanks for all the advice...but how about this one:

    R={(1,1),(1,2),(2,1),(3,4),(4,3)}

    Can something be not reflexive and not irreflexive? I ask because it isn't true that aRa for every pair and it isn't true that a(NOT R)a for every pair.

    It is symmetric!

    It isn't transitive because there is (3,4) and (4,3) but no (3,3).

    Does this sound correct?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post

    Can something be not reflexive and not irreflexive?
    yes.

    R = {(1,1), (2,1)} is not reflexive and not irreflexive
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  10. #10
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    So is the last relation that I posted reflexive, irreflexive, or neither? Why?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    So is the last relation that I posted reflexive, irreflexive, or neither? Why?
    you mean R={(1,1),(1,2),(2,1),(3,4),(4,3)} ?

    it is not reflexive, since you don't have (2,2),(3,3) and (4,4)

    it is not irreflexive since you have (1,1)

    reflexive means aRa for ALL a in the set we are considering

    irreflexive means aRa NEVER happens for ALL a in the set we are considering
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