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Relations- reflexive, symmetric, anit-symmetric, transitive

  1. Oct 22, 2014 #1
    • Note that forum guidelines require the homework template is filled out
    Suppose that

    R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
    R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
    R3={(2,4),(4,2)} ,
    R4={(1,2),(2,3),(3,4)},
    R5={(1,1),(2,2),(3,3),(4,4)},
    R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

    Determine which of these statements are correct.
    Check ALL correct answers below.


    A. R3 is symmetric
    B. R5 is transitive
    C. R6 is symmetric
    D. R2 is reflexive
    E. R1 is not symmetric
    F. R4 is symmetric
    G. R4 is transitive
    H. R4 is antisymmetric
    I. R3 is reflexive
    J. R1 is reflexive
    K. R3 is transitive
    L. R5 is not reflexive
    M. R2 is not transitive


    A. R3 is symmetric as (x,y) implies (y,x). So (4,2),(2,4)
    B. R5 is transitive as there is no (x,y) (y,z)
    C. R6 isn't symmetric as there is (1,4) but no (4,1)
    D. R2 ia reflexive as (1,1),(2,2),(3,3),(4,4) are elements
    E. R1 isnt symmetric as (4,4) isnt an element
    F. R4 isnt symmetric as there is no (2,1) for (1,2)
    G. R4 is transitive as there is no (x,y) (y,z) relation
    H. R4 isn't antisymmetric as its not symmetric
    I. R3 isnt reflexive as there is no (2,2) for (2,4)
    J. R1 is reflexive as there is as all (x,x) are satisfied
    K. R3 isnt transitve as there is no (2,2) for (2,4),(4,2)
    L. R5 is reflexive as all (x,x) satisfied
    M. R2 is transitive as (1,1) for (1,2),(2,1)

    Hi, I got a mark of 0% for this answer, and have no idea where I am going wrong. Can anybody help? Thanks Kevin
     
  2. jcsd
  3. Oct 22, 2014 #2
    Determine which of these relations are reflexive. The variables x, y, x', y' represent integers.

    A. x∼y if and only if xy is positive.
    B. x∼y if and only if x+y is even.
    C. x∼y if and only if x−y is a multiple of 10.
    D. x∼y if and only if x−y is positive.
    E. x∼y if and only if x+y is odd.

    A. xy is positive, so xy>0. So if x >=1, then (x,x) is satisfied. True
    B. If x+y is even, then x and y are both even or both odd. So odd(1,1) is even and even(2,2) is even. Therefore true
    C. x-y is a multiple of 10, but x-x is always 0, so its false.
    D. x-y is positve, but if x(1) and y(0) then (x,x) is 0, not positive, therefore false.
    E. x+y is odd therefore x is odd and y is even, or x is even and y is odd. But even + even, and odd + odd is always even. Therefore false.

    Again..I got a mark of 0% for this answer, and have no idea where I am going wrong. Can anybody help?
     
  4. Nov 1, 2014 #3
    x∼y if and only if x−y is a multiple of 10 is true and so is x∼y if and only if x+y is even :)
     
  5. Nov 2, 2014 #4
    If it makes you feel any better I'm stuck on question 2.
     
  6. Nov 2, 2014 #5
    I got these sorted. What part of q2 are you stuck with?
     
  7. Nov 2, 2014 #6
    I got Q2 eventually; I also got Q5 but Q's 3 and 4 are beyond me.
    Question 3 goes right over my head,
    I keep getting Question 4 wrong and no clue why.

    Question 4:

    (1 pt) Determine which of these relations are reflexive. The variables x, y, x', y' represent integers.

    A.
    x∼y if and only if x+y is positive.
    B. x∼y if and only if xy≥0 \).
    C. x∼y if and only if x+y is odd.
    D. x∼y if and only if xy is positive.
    E. x∼y if and only if x−y is a multiple of 10.


    I think that B and D are reflexive, not sure about E. I get it wrong whether or not I tick E though.
     
  8. Nov 2, 2014 #7
    D is not reflexive as 0(0) for (x,x) is not positive.
    E is reflexive as x -x will always be 0. 0 is a multiple of EVERY number.
    yes, b is reflexive as (0)0 is always >= 0.


    And how about q3?
     
  9. Nov 2, 2014 #8
    Suppose that

    R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
    R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
    R3={(2,4),(4,2)} ,
    R4={(1,2),(2,3),(3,4)},
    R5={(1,1),(2,2),(3,3),(4,4)},
    R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

    Determine which of these statements are correct.
    Check ALL correct answers below.


    A. R2 is not transitive
    B. R4 is antisymmetric
    C. R4 is symmetric
    D. R3 is transitive
    E. R3 is reflexive
    F. R4 is transitive
    G. R1 is not symmetric
    H. R1 is reflexive
    I. R2 is reflexive
    J. R6 is symmetric
    K. R5 is transitive
    L. R3 is symmetric
    M. R5 is not reflexive

    ----------------------------------------

    Don't even understand the notation in this one, despite reading the notes twice.
     
  10. Nov 2, 2014 #9
    i can't see your answers, so can't tell where you are going wrong? Can you put up a screeshot/grab

    oh sorry, they are your answers, right?
     
  11. Nov 2, 2014 #10
    Sorry, I probably should have just marked the ones I ticked.
     

    Attached Files:

  12. Nov 2, 2014 #11
    A) r2 is transitive as for (x,y)[1,2](y,z)[2,1] there is a (x,z)[1,1]
    k) r5 is transiitve as there is no terms For (x,y)(yz) implying (x,z), therefore the theory is true because we can't prove it false(maths logic)

    everything else looks right
     
  13. Nov 2, 2014 #12
    A) Even though it doesn't apply to the whole set?
    K) >Not proven to be false. > Therefore true.

    You have to be joking me with the logic for those two! Thanks so much for the help, as you may have noticed I would have been screwed if you haven't have helped me with those. Dave C., Coffee on me if you ever want one.
     
  14. Nov 2, 2014 #13
    A) But it does apply for the whole set. There is no other terms fitting the criteria of (x,y)(y,z) only (1,2)(2,1) as you can see y is 2 in both. Can you show me another (x,y)(y,z) in the set?
    k) we came accross that logic in truth tables if you remember, we had lets say x implies y. If x is false, that means y is automatically true, as we cant prove otherwise.
     
  15. Nov 2, 2014 #14
    A) So as long as any elements {x,y,z} in the set follows the pattern (x,y) , (y,z), (x, z) the set is transitive?
    The reason I thought it wasn't transitive is that there's nothing linking (2,2), (3,3), or (4,4) to each other.
     
  16. Nov 2, 2014 #15
    ...No, its not {x,y,z} in the set(as that would imply we could take any 3 numbers of our choice!), its the elements linked by "y" with (x,y) (y,z)........y just has to be in the 2nd position in the first set, and 1st position in the second set.
    There is actually another example in that set. (1,1)(1,2). as you can see y is 1 here, so do we have (x,z)[1,2]? Yes
     
  17. Nov 2, 2014 #16
    "y just has to be in the 2nd position in the first set, and 1st position in the second set"

    and this has to be true for EVERY set in the function; or ANY set in the function?
     
  18. Nov 2, 2014 #17
    yes, it must be true for EVERY 2 pairs to which it applies to
     
  19. Mar 2, 2016 #18
    i think R4 is not Transitif because it has (1,2), (2,3) then (1,3) is not in the R4. so does (2,3), (3,4) then (2,4) and its not in R4 either. it has (x,y), (y,z) in the relation but it doesnt has (x,z)
     
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