# Math Help - [SOLVED] equivalence relation/classes question.

1. ## [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..

2. Originally Posted by Kitizhi
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.

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.
It looks good to me.
Here is one equivalence class: $\{-4,-1,2,5\}$.
What are the other two?

3. I got {-4,-1,2,5}, {-3,0,3} and {-2,1,4}.

Thats all of them correct?

4. YES.