2. Let \( A=\{x: x \) is a prime number \( <7\}, B=\{5,6\} \) a) Write \( A \) in roster form b) Write \( A x B \) c) Write the number of possible relations from \( A \) to \( B \)

To solve the problem, let's go through each part step by step. ### a) Write \( A \) in roster form The set \( A \) consists of all prime numbers less than 7. The prime numbers less than 7 are 2, 3, and 5. Therefore, we can write: \[ A = \{2, 3, 5\} \] ### b) Write \( A \times B \) The Cartesian product \( A \times B \) consists of all ordered pairs where the first element is from set \( A \) and the second element is from set \( B \). Given that \( B = \{5, 6\} \), we can find \( A \times B \) as follows: \[ A \times B = \{(2, 5), (2, 6), (3, 5), (3, 6), (5, 5), (5, 6)\} \] So, the Cartesian product \( A \times B \) is: \[ A \times B = \{(2, 5), (2, 6), (3, 5), (3, 6), (5, 5), (5, 6)\} \] ### c) Write the number of possible relations from \( A \) to \( B \) A relation from set \( A \) to set \( B \) is a subset of the Cartesian product \( A \times B \). The number of possible relations is equal to the number of subsets of \( A \times B \). The number of elements in \( A \times B \) is 6 (as we found in part b). The number of subsets of a set with \( n \) elements is given by \( 2^n \). Therefore, the number of possible relations from \( A \) to \( B \) is: \[ 2^{|A \times B|} = 2^6 = 64 \] ### Summary of Answers a) \( A = \{2, 3, 5\} \) b) \( A \times B = \{(2, 5), (2, 6), (3, 5), (3, 6), (5, 5), (5, 6)\} \) c) The number of possible relations from \( A \) to \( B \) is \( 64 \).