# Guessing the quotient of

1. Feb 8, 2004

### Caldus

If I have an equivalence relation acting on all integers (Z): a ~ b if any only if 3a + b is a multiple of 4, then here is what I think the quotient set is:

The equivalence class of 0 = {x belongs to Z | x ~ 0} = {x | 3x = 4n for some integer n}. (The set would look like {0, 4, 8, 12, 16...}.)

The equivalence class of 1 = {x belongs to Z | x ~ 1} = {x | 3x + 1 = 4n for some integer n}. (The set would look like {1, 5, 9, 13, 17...}.)

The equivalence class of 2 = {x belongs to Z | x ~ 2} = {x | 3x + 2 = 4n for some integer n}. (The set would look like {2, 6, 10, 14, 18...}.)

The equivalence class of 3 = {x belongs to Z | x ~ 3} = {x | 3x + 3 = 4n for some integer n}. (The set would look like {3, 7, 11, 15, 19...}.)

So based on that, I conclude that there are 4 elements in the quotient set. Each element contains one of the sets above.

Am I accurate here? Thanks.

2. Feb 8, 2004

### HallsofIvy

Yes, you are correct. And did you notice that the equivalence classes are precisely the equivalence classes "mod 4"? Since 3 and 4 are relatively prime, If 3x is divisible by 4, then x is divisible by 4: if 3a and b are congruent mod 4, then so are a and b.

3. Feb 8, 2004

### Caldus

I just realized something. Don't I need to include negative numbers as well?

4. Feb 8, 2004

### Caldus

Nevermind, I already took care of them didn't I?

5. Feb 8, 2004

### Caldus

Also, for another problem: a^2 - b^2 acting on Z (a ~ b)...

Is the partition for this problem going to be split into two parts:
1. Numbers that are multiples of 3 (0, 3, 6, 9, 12...)
2. Numbers that are not multiples of 3 (1, 2, 4, 5, 7, 8...)

(Also, is -3 considered a multiple of 3? I'm getting myself confused here...lol...)

Am I accurate again here?