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Quotient set of an equivalence relation

  1. Oct 12, 2013 #1
    On the set of Z of integers define a relation by writing m [itex]\triangleright[/itex] n for m, n [itex]\in[/itex] Z.

    m[itex]\triangleright[/itex]n if m-n is divisble by k, where k is a fixed integer.

    Show that the quotient set under this equivalence relation is:

    Z/[itex]\triangleright[/itex] = {[0], [1], ... [k-1]}

    I'm a bit new the subject of Set Theory so I'm a bit unsure as to how to go about solving this.
    Last edited: Oct 12, 2013
  2. jcsd
  3. Oct 12, 2013 #2


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    [itex]\triangleright[/itex] seems to be undefined. You say it is a relationship without saying what it is.
  4. Oct 12, 2013 #3
    sorry it wasn't clear from my post, i've rewritten the post to be a bit more clear.

    The relationship is:
    m[itex]\triangleright[/itex]n, if m-n is divisble by k, where k is a fixed integer.
  5. Oct 12, 2013 #4


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    But you also haven't said what "[1]", "[2]", ... "[k]" are. Of course, they are the equivalence classes but unless your question is "why are there k classes" rather than "why are the classes what they are", you need to specify "what they are"! Essentially what you wrote was that you want to prove that the "quotient set" is the "set of equivalence classes" which is basically the definition of "quotient set"!
    (Similar to writing "Luna= Luna"!)

    Start with k= 1. All numbers in the same equivalence class with it ([1]) are numbers n such that n- 1 is divisible by k. That is the same as saying "n- 1= mk" for some m or n= mk+ 1. That is, all numbers that are 1 more than a multiple of k: [1]= {..., -2k+ 1, -k+ 1, 1, k+ 1, 2k+ 1, ...}. Similarly, [2]= {..., -2k+ 2, -k+ 2, 2, k+ 2, 2k+ 2, ...}. You can do that until you get to [k]= {..., -2k, -k, 0, k, 2k, ...} which is the same as [0]. If we try to do the same thing with k+ 1, we get [k+1]= {..., -2k+ (k+1), -k+ (k+1), k+ 1, k+ (k+ 1), 2k+ (k+ 1), ...}= {..., -k+ 1, 1, k+ 1, 2k+ 1, 3k+ 1, ...} which is exactly the same as [1].
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