[SOLVED] equivalence relation/classes question.

Let A = {-4,-3,-2,-1,0,1,2,3,4,5}. Define the relation R on A as: For all x,y ͼ A, (x,y) ͼ R iff 3 | (x-y).?

Show that R is an equivalence relation.

I need help making sure my answer is correct. Thanks.

R is reflexive iff For all x ͼ A, (x,x) ͼ R.

By definition of R, For all x ͼ A, 3| (x-x).

Since x-x = 0, For all x ͼ A, 3|0. This is true: 3|0 since 0=3(0).

Hence R is reflexive.

R is symmetric iff For all x,y ͼ A, if 3|(x-y) then 3|(y-x).

Suppose m and n are particular by arbitrarily chosen integers such that 3\(x-y). By definition of "divides", since

3|(x-y) then,

x-y = 3k , for some integer k

-(x-y) = -(3k)

y-x = 3(-k), since -k is an integer, this shows that..

3|(y-x) which follows that R is symmetric.

R is transitive iff For x,y,z ͼ A, if 3|(x-y) and 3|(y-z) then 3|(x-z).

Suppose x,y, and z are particular buy arbitrary chosen integers such that 3|(x-y) and 3|(y-z). We must show 3|(x-z) to be true. By definition of "divides", since

3|(x-y) and 3|(y-z) then,

x-y = 3r, for some integer r and,

y-z = 3s, for some integer s, hence

(x-y)+(y-z) = 3r + 3s

x-z = 3(r+s). Since r and s are integers, r+s is an integer and so,

3|(x-z). It follows that R is transitive.

This proof is fitting more for an infinite set but I was wondering if it works in this case as well?

Also can someone show me the equivalence classes so I tell if I am doing them correctly?

For the first one I got..

[0] = {a ͼ A | 3|(a-0) or (0-a)} = {-3,3}

..

[5]

I dunno if I am doing it correctly for this question..