 Hello and welcome to the session. In this session, we will solve the following question and the question says that A is equal to the set containing X such that X is a factor of 9, B is equal to the set containing X such that X is a factor of 18, then prove that A intersection B is equal to B intersection A. Let's start the solution now. First, we will find the elements of the sets A and B. Now, we are given that the set A contains all the factors of 9. So, set A is equal to the set containing the elements 1, 3, 9 since 1, 3 and 9 are the only factors of 9. Also, we are given that the set B contains all the factors of 18. So, set B is equal to the set containing the elements 1, 2, 3, 6, 9 and 18. As we know that all these numbers are the factors of 18. Now, let us find out what is A intersection B. This is equal to the set containing the elements 1, 3, 9 intersection the set containing the elements 1, 2, 3, 6, 9, 18. We know that A intersection B contains the elements which are common between the sets A and B. We can see that the common elements between these two sets are 1, 3 and 9. So, we get A intersection B is equal to the set containing the elements 1, 3, 9. Now, we will find B intersection A. This is equal to the set containing the elements 1, 2, 3, 6, 9 and 18 intersection the set containing the elements 1, 3, 9. Again, we can see that the common elements between these two sets are 1, 3 and 9. This implies B intersection A is equal to the set containing the elements 1, 3, 9. Therefore, A intersection B is equal to B intersection A, hence proved. With this we end our session. Hope you enjoyed the session.