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Math Help - Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

  1. #1
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    Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    The first part of my assignment was creating 9 relations based on the set M := {A,B,C} itself with ascending cardinality

    I ended up having this, I was assured that this is correct:

    R1= {(A,A)}
    R2= {(A,A),(A,B)}
    R3= {(A,A),(A,B),(A,C)}
    R4= {(A,A),(A,B),(A,C),(B,A)}
    R5= {(A,A),(A,B),(A,C),(B,A),(B,B)}
    R6= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C)}
    R7= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A)}
    R8= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B)}
    R9= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B),( C,C)}

    Ok now I need to check R1, R2, R4 for properties.

    R1= {(A,A)}
    R1 is not reflexive, because it does not contain (B,B) and (C,C)
    R1 not sure about transitivity. Can I use (A,A) for the formal definiton aRb ^ bRc => aRc? I dont know how far I can reach out with the definition.
    Do I need 3 different pairs or is one pair also enough to prove it? Like: I have (A,A) and (A,A). Therefore I also have (A,A)?
    R1 is symmetrical (A,A) is the same as (A,A)
    R1 is also anti symmetrical (even though I struggle with the definition, but I know that in this case it is because A and A are identical.)

    R2= {(A,A),(A,B)}
    R2 is not reflexive, because it does not contain (B,B) and (C,C)
    R2 is transitive? Almost same case like above. Can I prove like that? (A,A) and (A,B), therefore (A,B)
    R2 is not symmetrical?
    R2 is anti symmetrical because (A,A) is also a part of (A,B)

    R4= {(A,A),(A,B),(A,C),(B,A)}
    R4 is not reflexive, because it does not contain (B,B) and (C,C)
    R4 totally no clue at all if it is transitive.
    R4 is not symmetrical ?
    R4 is anti symmetrical because (A,A) is also in (A,B),(A,C),(B,A)?

    This took some time to wrote down, I hope most of it is correct.
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  2. #2
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    You are mostly correct.
    Any one element relation is transitive by default. So R_1 is.

    R_4 is not anti symmetrical, (A,B) \wedge (B,A) \not{\Rightarrow~} A = B
    Thanks from Cyganek
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Thank you very much!

    Could you please be so kind and explain anti-symmetry for a dumbo like me? Is this arguement for example "(A,A) is part of (B,A)" correct and a solid proof?
    And why is R2/R4 not symmetrical? I know it makes sense that it is not, but how would I argue?
    Is R4 transitive or not?
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Quote Originally Posted by Cyganek View Post
    Could you please be so kind and explain anti-symmetry for a dumbo like me? Is this arguement for example "(A,A) is part of (B,A)" correct and a solid proof?
    And why is R2/R4 not symmetrical? Is R4 transitive or not?
    There is no point in re-inventing the wheel.

    Have a look at this page and at this one.

    BTW R_4 is transitive.
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    transitive means, as you say "if aRb and bRc then aRc". Here, where you have "ARB" symbolized by (A, B) that is "if (X, Y) and (Y, Z) then (X, Z)".
    For 1) {(A, A)} that is only "if (A, A) and (A, A) then (A, A)"
    Of course "if (X, X) and (X, X) then (X, X)" is always true as long as (X, X) is in the relation so we don't often worry about that and can require that "X" is not the same as "Y" and that "Y" is not the same as "Z". But in a case where there are no such pairs, we can use the basic logical fact that "if X is false then the statement 'if X then Y' is TRUE for all Y.
    Either way we get that the relation is transitive.

    I hate to contradict Plato but R4= {(A,A),(A,B),(A,C),(B,A)} is not transitive because
    We have (B, A) and (A, C) but do NOT have (B, C).
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Quote Originally Posted by Plato View Post
    BTW R_4 is transitive.
    I disagree. (B,A),(A,B) \in R_4 but (B,B) \notin R_4
    Thanks from Cyganek
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Thank you both!

    One (maybe stupid question) about the last thing you wrote:

    In R2 I said (A,A) ^ (A,B) => (A,B)

    Cant I say the same here?

    (A,A) ^ (A,B) => (A,B)
    and
    (A,A) ^ (A,C) => (A,C)
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    To show transitivity, you must show all pairs that satisfy the pattern yield the correct outcome. So, it is not a question of "can you say the same". You must say them to prove it is transitive.
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Quote Originally Posted by HallsofIvy;804737 I hate to contradict Plato but R4= {(A,A),(A,B),(A,C),(B,A)} [b
    is not[/b] transitive because
    We have (B, A) and (A, C) but do NOT have (B, C).
    Quote Originally Posted by SlipEternal View Post
    I disagree. (B,A),(A,B) \in R_4 but (B,B) \notin R_4
    You are both correct. Actually I don't know which one I looked at. That what I get not replying with quote.
    Last edited by Plato; November 17th 2013 at 01:10 PM.
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    Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

    Okay thank you very much guys! To make it clear and sum it up:

    R1= {(A,A)}
    R1 is not reflexive
    R1 is transitive
    R1 is symmetrical
    R1 is anti symmetrical

    R2= {(A,A),(A,B)}
    R2 is not reflexive
    R2 is transitive
    R2 is not symmetrical
    R2 is anti symmetrical

    R4= {(A,A),(A,B),(A,C),(B,A)}
    R4 is not reflexive
    R4 is not transitive
    R4 is not symmetrical
    R4 is not anti symmetrical

    Is this correct?
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