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Math Help - [SOLVED] equivalence relation/classes question.

  1. #1
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    Question [SOLVED] equivalence relation/classes question.

    Let A = {-4,-3,-2,-1,0,1,2,3,4,5}. Define the relation R on A as: For all x,y ͼ A, (x,y) ͼ R iff 3 | (x-y).?
    Show that R is an equivalence relation.

    I need help making sure my answer is correct. Thanks.

    R is reflexive iff For all x ͼ A, (x,x) ͼ R.
    By definition of R, For all x ͼ A, 3| (x-x).
    Since x-x = 0, For all x ͼ A, 3|0. This is true: 3|0 since 0=3(0).
    Hence R is reflexive.

    R is symmetric iff For all x,y ͼ A, if 3|(x-y) then 3|(y-x).
    Suppose m and n are particular by arbitrarily chosen integers such that 3\(x-y). By definition of "divides", since
    3|(x-y) then,
    x-y = 3k , for some integer k
    -(x-y) = -(3k)
    y-x = 3(-k), since -k is an integer, this shows that..
    3|(y-x) which follows that R is symmetric.

    R is transitive iff For x,y,z ͼ A, if 3|(x-y) and 3|(y-z) then 3|(x-z).
    Suppose x,y, and z are particular buy arbitrary chosen integers such that 3|(x-y) and 3|(y-z). We must show 3|(x-z) to be true. By definition of "divides", since
    3|(x-y) and 3|(y-z) then,
    x-y = 3r, for some integer r and,
    y-z = 3s, for some integer s, hence
    (x-y)+(y-z) = 3r + 3s
    x-z = 3(r+s). Since r and s are integers, r+s is an integer and so,
    3|(x-z). It follows that R is transitive.

    This proof is fitting more for an infinite set but I was wondering if it works in this case as well?

    Also can someone show me the equivalence classes so I tell if I am doing them correctly?

    For the first one I got..
    [0] = {a ͼ A | 3|(a-0) or (0-a)} = {-3,3}
    ..
    [5]

    I dunno if I am doing it correctly for this question..
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  2. #2
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    Quote Originally Posted by Kitizhi View Post
    Let A = {-4,-3,-2,-1,0,1,2,3,4,5}. Define the relation R on A as: For all x,y ͼ A, (x,y) ͼ R iff 3 | (x-y).?
    Show that R is an equivalence relation.

    R is reflexive iff For all x ͼ A, (x,x) ͼ R.
    By definition of R, For all x ͼ A, 3| (x-x).
    Since x-x = 0, For all x ͼ A, 3|0. This is true: 3|0 since 0=3(0).
    Hence R is reflexive.

    R is symmetric iff For all x,y ͼ A, if 3|(x-y) then 3|(y-x).
    Suppose m and n are particular by arbitrarily chosen integers such that 3\(x-y). By definition of "divides", since
    3|(x-y) then,
    x-y = 3k , for some integer k
    -(x-y) = -(3k)
    y-x = 3(-k), since -k is an integer, this shows that..
    3|(y-x) which follows that R is symmetric.

    R is transitive iff For x,y,z ͼ A, if 3|(x-y) and 3|(y-z) then 3|(x-z).
    Suppose x,y, and z are particular buy arbitrary chosen integers such that 3|(x-y) and 3|(y-z). We must show 3|(x-z) to be true. By definition of "divides", since
    3|(x-y) and 3|(y-z) then,
    x-y = 3r, for some integer r and,
    y-z = 3s, for some integer s, hence
    (x-y)+(y-z) = 3r + 3s
    x-z = 3(r+s). Since r and s are integers, r+s is an integer and so,
    3|(x-z). It follows that R is transitive.
    It looks good to me.
    Here is one equivalence class: \{-4,-1,2,5\}.
    What are the other two?
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  3. #3
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    I got {-4,-1,2,5}, {-3,0,3} and {-2,1,4}.

    Thats all of them correct?
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  4. #4
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    YES.
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