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Elements of an Equivalence Class

  1. Nov 30, 2016 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    Question:
    Let ~ be the equivalence relation on the set ℤ of integers defined by a~b if a-b is divisible by 5. Let k ∈ Em belong to the equivalence class of m, and l ∈ En belong to the equivalence class of n. Prove that the product kl belongs to the equivalence class Emn.

    What I've tried:
    Since k ∈ Em, then m - k = 5x, for x ∈ ℤ.
    Also, since l ∈ En, then n - l = 5y, for y ∈ ℤ.
    So, I need to show that m*n - k*l = 5z, for z ∈ ℤ.
    I've tried multiplying (m-k)*(n-l), but to no avail.
    So I tried one case. Let m=20, k=10, n=30, and l=5.
    Well, 20-10=5(2), so 10 ∈ E20, and 30-5=5(5), so 5 ∈ E30
    Also, 600-50=5(110), so 50 ∈ E600, which for this case means kl ∈ Emn.

    Dilemma:
    I'm convinced the product kl belongs to the equivalence class of m*n always, so I'm having trouble actually finding a systematic way of proving it in general, instead of using a specific case.
     
    Last edited: Nov 30, 2016
  2. jcsd
  3. Nov 30, 2016 #2

    fresh_42

    Staff: Mentor

    Why haven't you calculated what you are actually looking for: ##m\cdot n - k \cdot l##?
     
  4. Nov 30, 2016 #3
    I haven't calculated that because I'm not sure what that would represent.
    I know an equivalence class is defined as a relation on a set S with x ∈ S, then the equivalence class of x is
    Ex = {y ∈ S | x ~ y}
     
  5. Nov 30, 2016 #4

    fresh_42

    Staff: Mentor

    Here's the answer:
    And here's the way:
    ##mk-kl = (5x+k)(5y+l)-kl= \dots ##
     
  6. Nov 30, 2016 #5
    I may have made a mistake in my "What I've tried" section.
    I already corrected it in my original post.

    It should have read
    My apologies.
     
  7. Nov 30, 2016 #6

    fresh_42

    Staff: Mentor

    No, my sorry, I had a typo. Of course it's ##mn-kl=(5x-k)(5y-l)-kl= \dots##
    The point is: you already have everything you need in your post. What is ##mn-kl##?
     
  8. Nov 30, 2016 #7
    I'm not sure that I follow the logic.
    mn - kl = (5x+k)(5y+l) - kl

    I expanded this, and trivially, I obtain that mn - kl = mn - kl
     
  9. Nov 30, 2016 #8

    fresh_42

    Staff: Mentor

    You set ##m=5x-k## and ##n=5y-l## which you can do because ##k\in E_m## and ##l\in E_n##.
    This is correct. Also correct is, that ##mn-kl=5z## for some ##z \in \mathbb{Z}## has to be shown, because it would mean ##E_{mn}=E_{kl}## what we need. Now, simply substitute ##m## and ##n## to get
    ##mn-kl=(5x-k)(5y-l)-kl=25xy-5xl-5yk+kl-kl=25xy-5xl-5yk=5z##.
    This line was all what's been missing.
     
  10. Nov 30, 2016 #9
    Ah, I see it now! (Though I think the signs are a bit off, but nonetheless it should be the same)

    Since 25xy + 5xl + 5yk = 5(5xy) + 5(xl) + 5(yk) = 5[5xy + xl +yk], then this is just an 5z, as [5xy + xl + yk] is just some integer.

    Perfect! Thank you so much.
     
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