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Equivalence relation and equivalence class

  1. Feb 2, 2012 #1
    i have two relations given to me which are both defined on the integers Z by

    relation 1: x~y if 3x^2 -y^2 is divisibale by 2

    and relation 2: x~y if 3x^2 -y^2 ≥0

    I have used three properties to figure out that relation 1 is eqivalence relation as it stands for all three properties i.e. reflexive, symmetric and transitive where as relation 2 is not equivalence as its not symmetric

    If this is correct- which I think it is. I have no idea what to do with second part
    which is:
    I have for relation 1: x~y if 3x^2 -y^2 is divisibale by 2 ( which is equivaleance), Show that :
    [3]={2k+1:k # Z} # means belongs to

    i would appriciate detail explaination and perhaps similar examples or show me how to do this.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 2, 2012 #2


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    [3] is the set of all numbers that are equivalent to 3 under this equivalence relation.

    Since x~ y if and only if [itex]3x^2- y^2[/itex] is divisible by 2, y is equivalent to 3 if and only if [itex]3(3^2)- y^2[/itex] is divisible by 2. That is, if and only if [itex]27- y^2= 2n[/itex] for some integer n. That, in turn, gives [itex]y^2= 27- 2n[/itex]. Now, n=1, that is 25 so y= 5 is in that set. [itex]27- 2n= y^2[/itex] is the same as 2n= 27- y^2. 27- y^2 is even if and only if [itex]y^2[/itex] is odd- if and only if y is odd. For example, if y= 7, [itex]27- 49= -22= 2(-11)[/itex] so [itex]3(3^2)- 7^2= -22[/itex] is divisible by 2. Every odd integer is equivalent to 3. Since equivalence classes "partion" the entire set, we then need to look at even numbers if y is equivalent to 3, then [itex]3(4)- y^2= 2n[/itex] so [itex]y^2= 12- 2n[/itex]. What integers, y, satisfy that?
  4. Feb 2, 2012 #3
    I understand this but my definition is [3] = {2k+1:k €Z} so what do I do with +1
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