A = {1, 2, 3, 4} andR = {(1, 1), (1, 2), (2, 1), (2, 2), (3,4), (4, 3), (3, 3), (4, 4)}.Shows that R is an equivalence relation on A hence find partition of A inducedby R.24. let A = {1, 2, 3,4}. Let R & S be an equivalence relations on A given asR = {(1, 1), (1, 2), (2, 1), (2, 2), (3,4), (4, 3), (3, 3), (4, 4)}S= {(1, 1), (2, 2), (3,1), (1, 3), (3, 3), (4, 4)}find partition of A induced by 𝑅−1 ∩ 𝑆−1 , 𝑅−1 , 𝑅 ∩ 𝑆

To show that R is an equivalence relation on A, we need to verify three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity: For every element a in A, (a, a) must be in R. In this case, we can see that (1, 1), (2, 2), (3, 3), and (4, 4) are all in R, satisfying the reflexivity property.

2. Symmetry: If (a, b) is in R, then (b, a) must also be in R. Looking at the pairs in R, we can see that (1, 2) and (2, 1) are both present, as well as (3, 4) and (4, 3). Therefore, R satisfies the symmetry property.

3. Transitivity: If (a, b) and (b, c) are in R, then (a, c) must also be in R. Checking the pairs in R, we can see that (1, 2) and (2, 1) are both present, as well as (3, 4) and (4, 3). Therefore, R satisfies the transitivity property.

Since R satisfies all three properties, it is an equivalence relation on A.

To find the partition of A induced by R, we need to group the elements of A based on their equivalence classes. Each equivalence class consists of elements that are related to each other under R.

From the given pairs in R, we can see that the equivalence classes are:

• [1, 2]
• [3, 4]

Therefore, the partition of A induced by R is {{1, 2}, {3, 4}}.

Now, let's move on to finding the partition of A induced by R^(-1) ∩ S^(-1), R^(-1), and R ∩ S.

To find the partition induced by R^(-1) ∩ S^(-1), we need to find the pairs that are common in the inverse of R and the inverse of S. In this case, the inverse of R is the same as R, and the inverse of S is {(1, 1), (2, 2), (1, 3), (3, 3), (4, 4)}. The common pairs are {(1, 1), (2, 2), (3, 3), (4, 4)}. Therefore, the partition induced by R^(-1) ∩ S^(-1) is {{1}, {2}, {3}, {4}}.

To find the partition induced by R^(-1), we need to find the equivalence classes based on the inverse of R. Since the inverse of R is the same as R, the partition induced by R^(-1) is the same as the partition induced by R, which is {{1, 2}, {3, 4}}.

To find the partition induced by R ∩ S, we need to find the pairs that are common in R and S. In this case, the common pairs are {(1, 1), (3, 3), (4, 4)}. Therefore, the partition induced by R ∩ S is {{1}, {3, 4}}.

I hope this helps! Let me know if you have any further questions.