Results 1 to 3 of 3

Math Help - Equivalence Relations

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    6

    Equivalence Relations

    I'm having trouble comprehending the concept. For instance, I have this problem, asking whether or not this is an equivalence relation:

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

    on the set (1, 2, 3}

    I understand that for this to be an equivalence relation it must be reflexive, symmetric, and transitive, but I am confused as to how to check for these properties.

    For instance, I see reflexive properties in (1,1), (2,2), and (3,3) but not in the others. Is that a problem? And how do I check for transitive and symmetric?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    As long as the set has, in this case, (1,1), (2,2), and (3,3), then its reflexive - the fact that 1 is ALSO, for example, equivalent to 2 has no bearing on the reflexivity. 1 is related to 1, 2 to 2, 3 to 3, and that's what you need.

    For symmetry, you need to show that if a is related to b then b is related to a for ALL a and b.

    Here, you don't need to worry about (1,1), (2,2), and (3,3) - clearly they're symmetric. Consider (1,2) - this means 1 is related to 2. For R to be symmetric, 2 must also be related to 1, i.e., (2,1) must be in your set (which it is). That takes care of everything - you've checked and "paired up" every element in R - note that (1,1), (2,2), and (3,3) are their own "pairs."

    For transitivity, you need to show that if (a,b) is in R and (b,c) is in R, then (a,c) is in R. (In other words, if a is related to b and b is related to c, then a is related to c.)

    Consider (1,2) in combination with (2,1) for example. We have:

    1 is related to 2.
    2 is related to 1.

    The question for transitivity, then, is: "Is 1 related to 1"? It is, since (1,1) is also in the set.

    Make sense?

    Hope this helps!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    6
    Makes so much more sense than the book! Thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 19th 2011, 01:09 PM
  2. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: April 29th 2010, 04:30 PM
  3. Replies: 10
    Last Post: January 14th 2010, 12:28 PM
  4. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 14
    Last Post: October 1st 2009, 03:03 PM
  5. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 16th 2008, 01:08 PM

Search Tags


/mathhelpforum @mathhelpforum