Equivalence relation and equivalence class

[foreverdream]

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 appreciate detail explanation and perhaps similar examples or show me how to do this.

 

[HallsofIvy]

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

Since x~ y if and only if 3x^2- y^2 is divisible by 2, y is equivalent to 3 if and only if 3(3^2)- y^2 is divisible by 2. That is, if and only if 27- y^2= 2n for some integer n. That, in turn, gives y^2= 27- 2n. Now, n=1, that is 25 so y= 5 is in that set. 27- 2n= y^2 is the same as 2n= 27- y^2. 27- y^2 is even if and only if y^2 is odd- if and only if y is odd. For example, if y= 7, 27- 49= -22= 2(-11) so 3(3^2)- 7^2= -22 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 3(4)- y^2= 2n so y^2= 12- 2n. What integers, y, satisfy that?

 

[foreverdream]

I understand this but my definition is [3] = {2k+1:k €Z} so what do I do with +1