answer:First, let's find the symmetric difference of A and B, which is denoted by A △ B. The symmetric difference is the set of elements that are in either A or B, but not in both. A △ B = (A ∪ B) - (A ∩ B) = {1, 2, 3, 4} - {2, 3} = {1, 4} Now, let's find the symmetric difference of B and A, which is denoted by B △ A. B △ A = (B ∪ A) - (B ∩ A) = {2, 3, 4, 1} - {2, 3} = {1, 4} As we can see, A △ B = B △ A = {1, 4}. The symmetric difference is commutative. Now let's find the Cartesian product of (A △ B) and (B △ A): (A △ B) × (B △ A) = {1, 4} × {1, 4} = {(1, 1), (1, 4), (4, 1), (4, 4)} Now let's find the Cartesian product of (B △ A) and (A △ B): (B △ A) × (A △ B) = {1, 4} × {1, 4} = {(1, 1), (1, 4), (4, 1), (4, 4)} As we can see, (A △ B) × (B △ A) = (B △ A) × (A △ B). Therefore, the symmetric difference of A and B is commutative with respect to the Cartesian product.

answer:First, let's find the symmetric difference of A and B, denoted as A ∆ B. The symmetric difference is the set of elements that are in either A or B, but not in both. A ∆ B = {1, 4} Now let's find the symmetric difference of B and A, denoted as B ∆ A. B ∆ A = {1, 4} Notice that A ∆ B = B ∆ A, which means the symmetric difference is the same in both cases. Now let's find the Cartesian product of (A ∆ B) and (B ∆ A). The Cartesian product of two sets is the set of all ordered pairs (a, b) where a is in the first set and b is in the second set. (A ∆ B) x (B ∆ A) = {(1, 1), (1, 4), (4, 1), (4, 4)} Since the Cartesian product of (A ∆ B) and (B ∆ A) is the same, the result is the same.

answer:The symmetric difference of two sets A and B, denoted as A Δ B, is the set of elements that are in either A or B, but not in both. In other words, it is the union of the two sets minus their intersection: A Δ B = (A ∪ B) - (A ∩ B). Now, let's find the symmetric difference of A and B: A = {1, 3, 5} B = {2, 4} A ∪ B = {1, 2, 3, 4, 5} A ∩ B = {} (since A and B have no common elements) A Δ B = {1, 2, 3, 4, 5} - {} = {1, 2, 3, 4, 5} Now, let's find the Cartesian product of A Δ B with itself: (A Δ B) × (A Δ B) = {1, 2, 3, 4, 5} × {1, 2, 3, 4, 5} The Cartesian product of two sets A and B, denoted as A × B, is the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B. In this case: (A Δ B) × (A Δ B) = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5)} Now, let's address the commutativity of the symmetric difference operation. The symmetric difference operation is actually commutative, meaning that A Δ B = B Δ A. This is because the union and intersection operations are both commutative, and the symmetric difference is defined in terms of these operations: A Δ B = (A ∪ B) - (A ∩ B) = (B ∪ A) - (B ∩ A) = B Δ A In our example: A Δ B = {1, 2, 3, 4, 5} B Δ A = {2, 4, 1, 3, 5} = {1, 2, 3, 4, 5} As we can see, A Δ B = B Δ A, which shows that the symmetric difference operation is commutative.

answer:First, we need to find the symmetric difference (△) between sets A and B. The symmetric difference is the set of elements which are in either of the sets, but not in their intersection. A △ B = (A ∪ B) - (A ∩ B) A ∪ B = {1, 2, 3, 4, 5, 6} A ∩ B = {3, 4} A △ B = {1, 2, 5, 6} Now, we need to find the Cartesian product (x) of (A △ B) and C. (A △ B) x C = {(1, 7), (1, 8), (1, 9), (2, 7), (2, 8), (2, 9), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9)} Now, let's find the symmetric difference between sets B and A. B △ A = (B ∪ A) - (B ∩ A) B ∪ A = {1, 2, 3, 4, 5, 6} B ∩ A = {3, 4} B △ A = {1, 2, 5, 6} Notice that A △ B = B △ A, which is {1, 2, 5, 6}. Now, we need to find the Cartesian product (x) of (B △ A) and C. (B △ A) x C = {(1, 7), (1, 8), (1, 9), (2, 7), (2, 8), (2, 9), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9)} Comparing the two Cartesian products, we can see that (A △ B) x C = (B △ A) x C.