Use set-roster notation to list the distinct equivalence classes of R.
Let A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7} and define a relation R on A as follows:
For all m, n ∈ A, m R n ⇔ 3|(m2 − n2).
It is a fact that R is an equivalence relation on A. Use set-roster notation to list the distinct equivalence classes of R. (Enter your answer as a comma-separated list of sets.)
We are looking to partition A into subsets that are mutually exclusive and whose union is equal to A.
Translation: We want to split A into groups, where members in each group are equivalent to each in the R sense described in the problem statement.
So to check if -3 is equivalent to -2, i.e. if -3 R -2, we need to check if 3|(-3)2-(-2)2, or equivalently if 3|5. 3 doesn't divide 5, so -3 is not equivalent to -2. They will be in separate groups.
Next we might check if -3 R -1. Does 3|(-3)2-(-1)2? That is, does 3|8? No. So -1 will be in a different equivalence class from -3.
We might check if -1 is equivalent to -2. Does 3|(-1)2-(-2)2? Yes, 3|-3.
To summarize, we know -3 is by itself, so far. {-3}. We also know that -1 and -2 are equivalent, i.e. -1 R -2. So, we have a list of subsets going, { {-3, and maybe more to come}, {-1,-2, and maybe more to come}}.
Continuing in this way, we may find more equivalencies. We needn't check all of them. Let's find some more to see why this is.
Having accounted so far for -3 with -2 and -1, let's check if -3 R 0. Does 3|(-3)2-(0)2? Yes, 3|9. So it is the case that -3 R 0. We can update our answer {{-3,0,and maybe more that equal these}, {-1,-2, and maybe more that equal these},and maybe more equivalence classes other than these}
We needn't check if 0 R -1 or 0 R -2. Why? Because of transitivity. If 0 R -1 then by transitivity we would have to have -3 R -1, because we already have -3 R 0. But this isn't so (we checked that -3 R -1 is false), and we can be sure that it's not the case that 0 R -1. Ditto for 0 R -2.
Continuing we might try -3 R 1. By now we might notice this is the same as -3 R -1 [ (1)2 = (-1)2 ], which we already said (at least twice before) is false. Also we know that 3|0=(-1)2-(1)2, that is, we know that -1 R 1. And so another reason -3 shouldn't equal 1. Similarly m R -m for any m. And so we can update our partition: { { -3, 0, 3, maybe more}, {-1, -2, 1, 2, maybe more}, maybe more}.
Check 0 R 4. Does 3|02-42? No, 3|-16 is not true. So 4 isn't with 0. What about 4 R 1? Does 3|16-1? Yes, 3|15.
5 R 0? Does 3|25? No. 5 R 1? Does 3|25-1=24? Yes. So both 4 and 5 are with 1.
6 R 0? Does 3|36? Yes. 6 is with 0.
7 R 0? Does 3|49? No. 7 R 1? Does 3|48? Yes, so 7 is with 1. Turns out we only had two groups:
{ {-3,0,3,6}, {-2,-1,1,2,4,5,7} }.
Here there were two R-equivalence classes. Elements inside each class are R-equivalent to each other. And members from different classes are not R-equivalent. Each element of A falls into an equivalence class, and the classes don't overlap.
Hope that helps.
