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Equivalence classes

  1. Jun 1, 2012 #1
    Given the set S, where aSb if and only if a - b [itex]\in[/itex] [itex]Z[/itex]

    It is asking for the equivalence class and the answer given is
    S has only one equivalance class for each real number x such that 0 ≤ x < 1. the class [x] is given by {x + k : k [itex]\in[/itex] [itex]Z[/itex]}

    i dun get it, since S is a set of relation where a - b is an element of Z, then there should be a lot of equivalance class for S, however, it states that S has only one equivalence class and its {x + k}
  2. jcsd
  3. Jun 1, 2012 #2


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    Hey look416.

    The equivalence class seems to be defined in terms of the usual way (in terms of a modulus like definition) and since you are dealing with integers this seems appropriate (the final thing is going to be a set of integers).
  4. Jun 1, 2012 #3
    but its stating it owns one equivalence class only, i thought there should be infinite sets of classes?
  5. Jun 1, 2012 #4

    Stephen Tashi

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    The answer said "one equivalence class for each real number....etc. ", not "one equivalence class".
  6. Jun 1, 2012 #5
    ok,but that raise me to another problem
    it says that for each real number x such that 0 ≤ x < 1. , which means that the whole relation only valids for that range?
    requires some help on understanding the relation between the a-b and x
  7. Jun 1, 2012 #6
    Well, there is an equivalence class for 1.5 too (for example), but that is equal to the equivalence class of 0.5 (since 1.5-0.5 is in Z). So every equivalence class can be written as [itex]\{x+k~\vert~k\in \mathbb{Z}\}[/itex] with x in [0,1[.
  8. Jun 1, 2012 #7
    oh, so the variable x is account for the decimals among the integers, i see
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