Results 1 to 5 of 5

Math Help - Equivalence Relation

  1. #1
    Member
    Joined
    Jan 2009
    From
    Kingston, PA
    Posts
    96

    Equivalence Relation

    in RxR, let (x,y)R(u,v) if a(x^2)+b(y^2)=a(u^2)+b(v^2) , where a,b >0. Determine if the relation R is an equivalence relation.

    I know it has to satify the Reflexive, Symmetry, & Transitive properties but i only did relations with xRy... any hints please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Are you able to write down what those three properties would translate to for this specific relation?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by nikie1o2 View Post
    in RxR, let (x,y)R(u,v) if a(x^2)+b(y^2)=a(u^2)+b(v^2) , where a,b >0. Determine if the relation R is an equivalence relation.

    I know it has to satify the Reflexive, Symmetry, & Transitive properties but i only did relations with xRy... any hints please
    This is the exact same thing. (x,y)\sim (x,y) because 1x^2+1y^2=1x^2+1y^2...etc.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,971
    Thanks
    1635
    Drexel28 is doing the case a= b= 1. I read the problem as saying that a and b can be any positive numbers.

    To have reflexivity, we must have (x, y)R(x, y) for any numbers x and y: we must have ax^2+ by^2= ax^2+ by^2. Well, that's certainly true so this is reflexive. Notice that this is true be the two sides of that equation are true. This relation is reflexive because "=" is reflexive.

    To have symmetry, we must have that if (x, y)R(u, v) then (u, v)R(x, y).
    The first says that ax^2+ by^2= au^2+ by^2. Because "=" is symmetric, that is the same as au^2+ bv^2= ax^2+ by^2.

    To have transitivity, we must have that if (x, y)R(u, v) and (u, v)R(p, q), then (x, y)R(p, q). I'll leave it to you to write down the equations and use the fact that "=" is transitive to show this is true.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    From
    Kingston, PA
    Posts
    96
    Thank you all very much..i was just confused on if i had to show (x,y)R(x,y) for reflexive and if i had to show (u,v)R(u,v) and so on... Have a greay day!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 6th 2011, 11:46 PM
  2. equivalence relation and equivalence classes
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 7th 2010, 06:36 PM
  3. Equivalence relation and order of each equivalence class
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2009, 09:03 AM
  4. equivalence relation
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: January 12th 2009, 05:33 PM
  5. Equivalence relation and Equivalence classes?
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 7th 2009, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum