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Homework Help: Prove that a relation is an equivalence relation

  1. Oct 14, 2008 #1
    Please be nice to me, I'm new here. Anyway, help to solve this maths problem would be much appreciated:
    1. The problem statement, all variables and given/known data
    Work out a detailed proof (below) that the relation on the integers defined by p~q if and only if 7|p-q is an equivalence relation:
    a) the relation is reflexive
    b) the relation is symmetric
    c) the relation is transitive

    2. Relevant equations
    p~q if and only if 7|p-q

    3. The attempt at a solution
    a) (I'm pretty sure this is done right)
    If relation is reflexive then:
    x[tex]\in[/tex]S[tex]\rightarrow[/tex] (x,x) [tex]\in[/tex]R
    Therefore x~x
    7|x-x since x-x=0 and 7|0
    Therefore relation is reflexive.

    That's the easy bit. Now:
    b)If relation is symmetric then:
    x~y [tex]\leftrightarrow[/tex] y~x

    And I don't know how to go on from there. Please help me!!
    Last edited: Oct 14, 2008
  2. jcsd
  3. Oct 14, 2008 #2


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    Homework Helper


    [tex] x \sim y [/tex]

    so that

    7 \mid x - y

    To show that

    [tex] y \sim x [/tex]

    you need to show that

    7 \mid y - x

    How can you do that?

    For the transitive part, begin by assuming

    x \sim y & \text{ so } 7 \mid x - y \\
    y \sim z & \text{ so } 7 \mid y - z

    Write out what these two statements mean, and you should see why it follows that

    x \sim z
  4. Oct 14, 2008 #3


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    Science Advisor

    You mean "reflexive".

    'quote]That's the easy bit. Now:
    b)If relation is symmetric then:
    x~y [tex]\leftrightarrow[/tex] y~x

    And I don't know how to go on from there. Please help me!![/QUOTE]
    x~y means 7 divides x-y which means x-y= 7n for some integer n.

    y~ x means 7 divides y- x which means y- x= 7m for some m. Knowing that x- y= 7n, y- x= 7 times what?

    "Transitive": if x~y and y~z then x~z.

    Okay, you know x~y so x- y= 7n for some integer n.
    You know y~ z so y- z= 7m for some integer m.
    Therefore x- z= 7*what?
    (hint: what is (x- y)+ (y- z)?)
  5. Oct 14, 2008 #4
    Ok, focusing on the symmetric bit for now (sorry about that major typo, HallsofIvy):


    I can see that this is leading to some sort of a proof, but I don't really know what to write. Is something like the following enough for proof?:
    m and n have a common factor of 7, so x-y and y-x are always divisible by 7. Therefore x~y[tex]\leftrightarrow[/tex]y~x.
  6. Oct 14, 2008 #5
    I've proved the transitivity now, thanks for the help you two. Unfortunately,I've just realised that there's more:
    Fill in the blanks:
    "The equivalence class containing 5 is given by
    [5] = {n[tex]\in Z[/tex]|n has remainder _ when divided by _}"
    Am I supposed to put in 0 and 7? If it is, that seems like a bit of a random question. If it isn't, then I have no idea what's going on!
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