Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A little help with symmetric, reflexive and transitive

  1. Sep 11, 2008 #1

    It's been a while since I've had to determine whether a relation is reflexive, symmetric or transitive so I would appreciate a bit of guidance. I now that being reflexive means the relation xRx for all x and being symmetric implies xRy and yRx for all x, y and transitive implies that if (xRy and yRz) then xRz.

    But here are my dilemmas:

    1. isFatherOf on the set of people: Based on my thinking this is not reflexive. If I have Bob, and put him in the relation Bob isFatherOf Bob, someone can't be the father of themselves, so that fails. At least this is how I'm reasoning this and for some reason I can't see any other way to reason it. Secondly, it's not symmetric. If again I use Bob and Ted and put them in the relation, Bob isFatherOf Ted, Ted cannot be father of Bob so this relation also fails. Lastly, for being transitive that also fails. Using Bob and Ted, I'll add Henry and put them into the relation as follows: Bob isFatherOf Ted and Ted isFatherOf Henry therefore Bob cannot be father of Henry since Ted is.

    So my final answer is it is neither reflexive, symmetric, antisymmetric or transitive.

    Again, these are my ways of reasoning it out. I hope I'm on the right path with this.

    2. The relation R = {(x,y)} | x^2 + y^2 = 1} for x and y real numbers.
    Here do I need to make sure the numbers I use to test the relation for x and y when put in the equation x^2 + y^2 = 1 to make the equation equal 1? I mean if that's the case then x and y can only be (0,1) and (1,0) to have the equation hold. Which then would state that the relation is not reflexive, antisymmetric, and transitive but is symmetric.

    Lastly, 3. The relation R = {(x,y) | x mod y = 0} for x, y in the set {1, 2, 3, 4}.
    Again for this one, there is an equation involved. So do I take all combination's in the set in which x mod y will equal 0? Like (1,1), (2,2), (3,3), (4,4), (2,1), (3,1), (4,1) all of those (x,y) will produce x mod y = 0. So base on this my answer is as follows:

    The relation is reflexive since xRx for all x.
    The relation is not symmetric since 2R1 (2,1) is present but 1R2 (1,2) is not.
    The relation is transitive since 2R2 (2,2) and 2R1 (2,1) implies 2R1 (2,1).
    The relation is antisymmetric since (1,2), (1,3) and (1,4) are not present.

    So my final answer for this one is the relation is reflexive and transitive but not symmetric.

    Thanks in advance for any help provided, it is greatly appreciated.
    Last edited: Sep 11, 2008
  2. jcsd
  3. Sep 11, 2008 #2
    Opps I also left out antisymmetric. I don't understand how if aRb and bRa then a=b. I could use a little guidance on this one too. Thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook