Thread: Equivalence Classes

How do you find the distinct equivalence classes of R?

A = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. R is defined on A as follows:
For all x, y elements of A, x R y <=> 3|(x-y).

I have the answer, but I do not understand the work leading up to the answer. I appreciate any help.

I am having the same issue with this one:
A = {-4, -3, -2, -1, 0, 1, 2, 3, 4}. R is defined on A as follows:
For all (m,n) elements of A, m R n <=> 5|(m^2 - n^2).

Originally Posted by lovesmath

How do you find the distinct equivalence classes of R?

A = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. R is defined on A as follows:
For all x, y elements of A, x R y <=> 3|(x-y).

I have the answer, but I do not understand the work leading up to the answer. I appreciate any help.

I am having the same issue with this one:
A = {-4, -3, -2, -1, 0, 1, 2, 3, 4}. R is defined on A as follows:
For all (m,n) elements of A, m R n <=> 5|(m^2 - n^2).

In the first case you want to a multiple of 3.
So are in the same class because a multiple of 3.

You do the next one: multiples of 5.

So for 3|(x - y), do you find the combinations of numbers that give you remainders 0, 1 and 2, and each of those combinations will be its own equivalence class?

Originally Posted by lovesmath

So for 3|(x - y), do you find the combinations of numbers that give you remainders 0, 1 and 2, and each of those combinations will be its own equivalence class?

All you do its look at each pair of numbers.
Is their difference a multiple of 3?
If yes, those two numbers belong to the same class.
If no, they are not in the same class.

So is one class.

For the second relation, 5|(m^2 - n^2), I got the following equivalence classes:
{0}, {-4, -1, 1, 4}, {-3, -2, 2, 3}

Are these correct?

Originally Posted by lovesmath

For the second relation, 5|(m^2 - n^2), I got the following equivalence classes:
{0}, {-4, -1, 1, 4}, {-3, -2, 2, 3} Are these correct?

Yes they are correct. Good for you.