 Welcome to the session. In this session I will discuss a question which says that let a be a set of first five natural numbers and b be a set of first five prime numbers, let r be a relation defined from a to b as follows. r is equal to a such that x belongs to a, y belongs to b and x is less than equal to y. x plus r and r inverse as such sub-order pairs also determine the range and domain of r and r inverse. Now before starting the solution of this question we should know our result. And that is for any binary relation the inverse of the relation is denoted by r inverse and this r inverse is equal to a such containing order pair y x where the order pair x y belongs to r. So to get the inverse of the relation interchange the first and the second components in every order pair in the given relation. And let us discuss one more thing that if r that is the relation hiding the order pair x y that x belongs to a, y belongs to b and x is less than equal to y, then r inverse the set containing the order pair y x such that y belongs to b, y is greater than equal. Now this result will work out as a key idea but with the solution the set a is given as the set of first five natural numbers the set of first five prime numbers. So given 1, 2, 3 that b is a set prime numbers containing the elements 2, 3, 5, 7 is also given to us. Those ordered pairs x is less than equal to y containing these elements and b is a set containing these elements. Therefore a cross b is a such containing these ordered pairs, these ordered pairs. Now those ordered pairs are 1, 2 as here belongs to a cross b and also here it is 1 is less than 2. Also 1, 3 then after that 1, 5, 1, 7 find this relation less than or equal to the second component which means 1, 3 first result which is given as the key idea. Second components is equal to a set considering 1, 3, 1, that is we are introducing the first and the second components here and by introducing we are getting 2, 1, we are getting and after that 5, 1, 7, 1, 11, 1, 5, 2, 3, 3, 5, 3, 7, 3, 4, 11, 3, 5, 4, 11, 4, 5, 5, 7, 5 and 11, 5. Now we have to find the range and domain of R and R inverse. Now we will let domain is the set of first components of the ordered pairs. So here domain which are these? 1, 2, 3, 4 and 5. Now domain which are these? It is equal to the set containing the element components of the ordered pairs R will be equal to, will be equal to the set the range will be equal to second components of these ordered pairs containing the elements 1, 2, 3, solution of the given question and that's all for this session. Hope you all have enjoyed this session.