Solve number 12-15

Step 1: Solve Question 12. Given the sets: A = \1, 2\ B = \2, 3\ C = \3, 4\ a) Compute A (B C) First, calculate the intersection of B and C: B C = \2, 3\ \3, 4\ = \3\ Next, calculate the union of A with the result: A (B C) = \1, 2\ \3\ A (B C) = \1, 2, 3\ b) Compute (A B) (A C) First, calculate the union of A and B: A B = \1, 2\ \2, 3\ = \1, 2, 3\ Next, calculate the union of A and C: A C = \1, 2\ \3, 4\ = \1, 2, 3, 4\ Finally, calculate the intersection of these two results: (A B) (A C) = \1, 2, 3\ \1, 2, 3, 4\ (A B) (A C) = \1, 2, 3\ The answers for Question 12 are: *a) \1, 2, 3\ *b) \1, 2, 3\ Step 2: Solve Question 13. Given the sets: A = \1, 2\ B = \a, b\ Find the Cartesian product A × B. The Cartesian product A × B is the set of all ordered pairs (x, y) where x A and y B. A × B = \(1, a), (1, b), (2, a), (2, b)\ The answer for Question 13 is: \(1, a), (1, b), (2, a), (2, b)\ Step 3: Solve Question 14. Given the sets: A = \1, 2, 3\ B = \a, b, c\ A function f: A B is defined such that f(1) = a, f(2) = b, and f(3) = c. Determine if this function is injective. A function is injective (one-to-one) if every distinct element in the domain maps to a distinct element in the codomain. In other words, if x_1 ≠ x_2, then f(x_1) ≠ f(x_2). In this case: f(1) = a f(2) = b f(3) = c Each element in A (1, 2, 3) maps to a unique element in B (a, b, c). No two distinct elements in A map to the same element in B. Therefore, the function is injective. The answer for Question 14 is: Yes, the function is injective. Step 4: Solve Question 15. Given the sets: A = \1, 2\ B = \2, 3\ Show that A B = B A. First, calculate A B: A B = \1, 2\ \2, 3\ A B = \1, 2, 3\ Next, calculate B A: B A = \2, 3\ \1, 2\ B A = \1, 2, 3\ Since both A B and B A result in the set \1, 2, 3\, we have shown that A B = B A. The answer for Question 15 is: A B = \1, 2, 3\ and B A = \1, 2, 3\, therefore A B = B A.