Let A = {−6, −5, −4, −3, −2, −1, 0, 1, 2} and define a relation R on A as follows: For all m, n ∈ A, m R n ⇔ 5|(m2 − n2).

Question

Let A = {−6, −5, −4, −3, −2, −1, 0, 1, 2} and define a relation R on A as follows:

For all m, n ∈ A, m R n ⇔ 5|(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.)

For this problem what I did was (-6)^2 - (-6)^2 =5

(-6)^2 - (-5)^2 = 5

and so on until (2)^2 = 5

I repeated the same step for all (-5)^2 - (-6)^2 =5

(-5)^2 - (-5)^2 = 5

(-5)^2 - (2)^2 =5

For the respected number I was multiplying with I put all values that were divisible by 5 in one brace and the ones that were not in another.

Example for -6 I did, divisible by 5 {-6,-4,-1,1,4,6} , not divisible by 5 { -5,-3,-2,0,2,3}

for -5 { -5,0} , { -5,-3,-2,0,2,3}

Is this the correct way to approach this problem?

Would the positive numbers in A have the same set as the numbers?

Answer 1

To find the equivalence classes, we can start from an arbitrary element of our set A, let's say 2, and find all other elements m such that 2 R m. This way we find the equivalence class of 2, denoted by [2]. Then we can continue with remaining elements until we find all classes.

Now to find the equivalence class of 2, we have [2] = {m in A | m^2 - 2^2 is divisible by 5}

We can try all elements of A and realize that the only values of m that satisfies this are m=-3 and m=-2 because 9-4=5 and 4-4=0.

Therefore, the equivalence class of 2 is {2,-3,-2}

Now we can remove {-3,-2,2} from A and continue with the remaining elements. Here is the set of equivalence classes that we get at the end:

{-3,-2,2} , {-6,-4,-1,1} , {-5,0}

Alternative method: another way to find the equivalence classes is to realize that 5 | (m^2-n^2) if and only if m^2 and n^2 have the same remainder when divided by 5. Hence, we could find the remainder of m^2 for all elements of A and that elements with the same result would end up in the same equivalence class:

m : -6 , -5 , -4 , -3 , -2 , -1 , 0 , 1 , 2

m^2%5 : 1 , 0 , 1 , 4 , 4 , 1 , 0 , 1 , 4

Hence {-6,-4,-1,1} are in the same class, {-5,0} are in the other class, and {-3,-2,2} form the last class.