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Math Help - Sets and Relations

  1. #1
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    Question Sets and Relations

    Sorry not much of a maths wiz so I don't even know if I've posted this in the correct section. Apologies if not.

    Just doing some revision for a first year maths exam and there is a question on sets and relations asking to determine if something is reflective, symmetric and transitive.

    For example

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

    Is symmetric and transitive but not reflective. I don't understand how this is worked out. Could some one please explain this to me?
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  2. #2
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    Quote Originally Posted by eddielee View Post
    Sorry not much of a maths wiz so I don't even know if I've posted this in the correct section. Apologies if not.

    Just doing some revision for a first year maths exam and there is a question on sets and relations asking to determine if something is reflective, symmetric and transitive.

    For example

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

    Is symmetric and transitive but not reflective. I don't understand how this is worked out. Could some one please explain this to me?
    I believe you meant reflexive not reflective.

    A relation R on a set A is called reflexive if (a,a) \epsilon R for every element a \epsilon A.

    R does not contain (3,3) so it is not reflexive.

    A relation R on a set A is called symmetric if (b,a) \epsilon R whenever (a,b) \epsilon R, for all a,b \epsilon A. Since (1,2) \epsilon R and (2,1) \epsilon R, R is symmetric.

    I'll leave transitive to you.
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  3. #3
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    Thank you so much.

    Even though 3 isnt in R is still symmetric? And sorry to sound stupid but, I presume that this symbol means = or something along those lines but what is <br />
\epsilon<br />
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  4. #4
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    Sorry I understand the symmetric thing now. It just clicked. Out of the groups in R they all have to either be like (1,1) so the same number of if there like (1,2) there needs to be a (2,1).
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  5. #5
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by eddielee View Post
    Thank you so much.

    Even though 3 isnt in R is still symmetric? And sorry to sound stupid but, I presume that this symbol means = or something along those lines but what is <br />
\epsilon<br />

    You're welcome. \epsilon means is a member or element of.
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  6. #6
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    Quote Originally Posted by oldguynewstudent View Post
    You're welcome. \epsilon means is a member or element of.
    [tex]\in[/tex] gives the element symbol \in.
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  7. #7
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    Right okay, So for example a \in X means a is an eliminate of X.

    With transitive if it was {(1,2),(3,4),(4,5)} it would be transitive and {(1,2),(3,4),(5,6)} wouldn't be?
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    Or do they have to go round in a loop? So it would have to be {(1,2),(3,4),(4,1)} to be transitive?
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  9. #9
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    Quote Originally Posted by eddielee View Post
    With transitive if it was {(1,2),(3,4),(4,5)} it would be transitive and {(1,2),(3,4),(5,6)} wouldn't be?
    {(1, 2), (3, 4), (4, 5)} is NOT a transitive relation, since (3, 5) is not a member.

    {(1, 2), (3, 4), (5, 6)} IS a transitive relation.

    I suggest you review the definition of 'transitive relation'.

    R = {(1, 2), (3, 4), (4, 5)} has in its field the "linking" member 4, but R LACKS the "linked" ordered pair (3, 5).

    T = {(1, 2), (3, 4), (5, 6)} has no "linking" members in its field, so it can't fail to be transitive, since it can't miss having any "linked" ordered pair, since, as mentioned, there are no "linking" members in the field.
    Last edited by mr fantastic; May 24th 2010 at 12:36 PM. Reason: Notation.
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  10. #10
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    Quote Originally Posted by eddielee View Post
    Or do they have to go round in a loop? So it would have to be {(1,2),(3,4),(4,1)} to be transitive?
    No, "looping" is not required for transitivity. Also, the relation you just mentioned is not even transitive, since it lacks (3, 1); also it lacks (4, 2), and having (4, 2) would require also having (3, 2) .

    You need to review the exact definition of 'transitive relation'.
    Last edited by mr fantastic; May 24th 2010 at 12:33 PM. Reason: Notation.
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  11. #11
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    Quote Originally Posted by eddielee View Post

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

    Is symmetric and transitive but not reflective.
    No, R is reflexive; however R is not reflexive on S. That is, R is reflexive on the field of R, but R is not reflexive on S, since R lacks (3, 3) as a member.
    Last edited by mr fantastic; May 24th 2010 at 12:34 PM. Reason: Notation.
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