L Let A = {1,2,3,4,5,6,7,8} and define an equivalence relation by a-b if and only if a-b is divisible by 3. Verify that this is an equivalence relation. What are the equivalence classes? What is the quotient space? Attach File Browse Local Files Browse Content Collection Moving to another question will save this response.
To verify that the relation a-b is divisible by 3 is an equivalence relation, we need to show that it satisfies the three properties of reflexivity, symmetry, and transitivity.
1. Reflexivity: For any number a, a-a is always divisible by 3 since a - a = 0 which is divisible by any number including 3. Therefore, the relation is reflexive.
2. Symmetry: If a-b is divisible by 3, then b-a will also be divisible by 3 since the difference between a and b doesn't change when the order is reversed. Therefore, the relation is symmetric.
3. Transitivity: If a-b is divisible by 3 and b-c is divisible by 3, then the sum of these differences (a-b)+(b-c) = a-c will also be divisible by 3 because the sum of two numbers divisible by 3 is also divisible by 3. Therefore, the relation is transitive.
Since the relation satisfies all three properties, we can conclude that it is an equivalence relation.
Now, let's find the equivalence classes:
1 is in the same equivalence class as 4 and 7 because 1 - 4 = -3 and 1 - 7 = -6 are divisible by 3.
2 is in the same equivalence class as 5 and 8 because 2 - 5 = -3 and 2 - 8 = -6 are divisible by 3.
3 is in the same equivalence class as 6 because 3 - 6 = -3 is divisible by 3.
To summarize, the equivalence classes are:
[1] = {1, 4, 7}
[2] = {2, 5, 8}
[3] = {3, 6}
The quotient space is the set of all equivalence classes, which in this case is {[1], [2], [3]}.
To verify that the relation a-b is an equivalence relation, we need to show that it satisfies the three properties of reflexiveness, symmetry, and transitivity.
1. Reflexivity: For any element a in the set A, a-a = 0, which is divisible by 3. Therefore, the relation satisfies reflexivity.
2. Symmetry: If a-b is divisible by 3, then b-a is also divisible by 3. This is because if a-b = 3k, then b-a = -(a-b) = -3k = 3(-k), which is divisible by 3. Thus, the relation satisfies symmetry.
3. Transitivity: If a-b is divisible by 3 and b-c is divisible by 3, then a-c is also divisible by 3. This is because if a-b = 3k and b-c = 3m, then a-c = (a-b) + (b-c) = 3k + 3m = 3(k+m), which is divisible by 3. Hence, the relation satisfies transitivity.
Therefore, the relation a-b is an equivalence relation.
To determine the equivalence classes, we need to find all elements that are related to each other.
Let's consider the elements in the set A and their relations:
1 is related to 4 and 7.
2 is related to 5 and 8.
3 is related to 6.
Since these are the only distinct equivalence classes, the equivalence classes in this case are:
[1] = {1, 4, 7}
[2] = {2, 5, 8}
[3] = {3, 6}
The quotient space is the set of all equivalence classes, denoted by A/~. In this case, the quotient space is:
A/~ = {[1], [2], [3]} = {{1, 4, 7}, {2, 5, 8}, {3, 6}}.