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Homework Help: Relation which is reflexive only and not transitive or symmetric.

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Relation which is reflexive only and not transitive or symmetric?

    2. Relevant equations
    No equations just definitions.

    3. The attempt at a solution
    I can find a relation for the other combinations of these 3 however, I cannot find one for this particular combination. It is possible that none exist but I cannot find would like confirmation of this.
  2. jcsd
  3. Aug 12, 2011 #2
    It might have been good if you included the necessary definitions. That is a good way to begin this sort of problem. Let the set have three elements, x, y, and z. In order to be NOT symmetric, what MUST you have? In order to be NOT transitive, again, what MUST you have? And, to be reflexive, you MUST have what? Try it, and even if you don't get it, tell us what you tried.
  4. Aug 12, 2011 #3
    Reflexive relation means a is related to a.
    Example of reflexive only: ......
    Example of reflexive: Parralel
    Example of non reflexlive: Is greater than

    Symmetrix means that if A is related to B than B is related to A:
    Example of symmetric: Perpendicular
    Example of only symmetric: Has opposite parity to.

    Transitive: If A is related to B and B is related to C then A is related to C.
    Example of transitive: is greater than
    Example of non transitive: perpindicular

    I understand the three though i should probably have put this under relevant equations so sorry about that, I cannot in spite of understanding the different types of relation think of a relation which is reflexive but not transitive or symmetric
  5. Aug 12, 2011 #4
    Okay, break it down. If the set is {x,y,z} then what pairs must you have to meet the definition of reflexive?

    Now, forget that and make a new thought. If it is NOT transitive, what pairs MUST you have to make a NOT transitive relation?

    Now, again EXASPERATED thought. lol! I put seperate and spell checked messed it up... Anyway, SEPARATE thought. What ordered pairs MUST you have to have NOT symmetric relation?

    Can you answer those three questions independently of each other? Don't worry yet about one relation that meets all three at the same time.
  6. Aug 12, 2011 #5
    To make a relation reflexive you must have ({x,x},{y,y}{z,z})

    I don't think there is any required pairs for not having transitivity.

    Again i don't think there are any pairs required for not having symmetric relation.

    Sorry I am clearly missing the point you are driving towards, it seems impossible to have the relationship in the question but I cannot clearly express why.
  7. Aug 12, 2011 #6
    It's okay, now I know why you aren't getting it so I can help you.

    A relation is transitive IFF: if (x,y) and (y,z), then (x,z).

    But, you see, if you have (x,y) but you do NOT have (y,z), then the relation IS transitive, because the antecedent is false. Do you see now how to approach this? you must include (x,y) AND (y,z) and NOT (x,z) to make the relation NOT transitive.

    So, after you are sure you understand that, apply it to the symmetric definition. You will again need to have a certain ordered pair present in order to make the relation NOT symmetric.
  8. Aug 12, 2011 #7
    Okay thanks i had not realised that so that means for a relationship to not be symmetric you must have: (x,y) (y,z) and (x,z)?

    and for transitivity (x,y) and (y,z) and not (x,z)

    So since (x,z) must be in the set for symmetric yet not in the set for transitivity it is not possible.

    Thanks very much i had not realised that if the primary conditions of symmetry/ transitivity were not there then it made the sets symmetric/ transitive although this makes perfect sense.
    Thank you very much.
  9. Aug 12, 2011 #8
    With all of that being said, ask yourself is your relation: {(x,x),(y,y),(z,z)}, symmetric and transitive? If it is symmetric and transitive what element(s) can you add so it isn't?
  10. Aug 12, 2011 #9
    No. To be not symmetric you must have (x,y) but NOT (y,x)

    that would be NOT transitive

    Also, no.
    This set IS possible. :(
  11. Aug 13, 2011 #10
    For a set to be not to be symmetric we must have (x,y) but not (y,x)

    For a set to be not to be transitive we must have we must have (x,y) and (y,z) but not (x,z).

    So far we have a set which requires ({x,x},{y,y}{z,z},{x,y}{y,z}) but must not have {(y,x),(x,z)}
    Am i right in saying that (z,y) is also not allowed as we have (y,z) and symmetry is not allowed?
    This leave only one element left undiscussed in the set which is the element {z,x}
  12. Aug 13, 2011 #11
    To answer JonF:
    ({x,x},{y,y}{z,z}) is an equivalence relation.
    As stated above:
    For a set to be not to be symmetric we must have (x,y) but not (y,x)
    For a set to be not to be transitive we must have we must have (x,y) and (y,z) but not (x,z).

    so my set must have [{x,x},{x,y}{y,y}{y,z}{z,z}]
  13. Aug 13, 2011 #12
    This is not necessarily true.

    A relation is symmetric iff: for all a and b in the set, a R b => b R a. That means if we have a R b, then we must have b R a. If we have just one case where a R b, but not b R a, then the relation is not symmetric.

    In your example, you have {x, y} (also written x R y), but you don't have {y, x}, so the relation is not symmetric. The relation will continue to be not symmetric even if you have {y, z} and {z, y} and {x, z} and {z, x}. The symmetry was lost when we declared x R y but not y R x.
  14. Aug 13, 2011 #13
    thanks for the clarification
  15. Aug 13, 2011 #14
    Yay you got it! :)
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