 Hi and welcome to the session. My name is Reshi. Let us do one question. Question is, let star with a binary operation on n where n is the set of all natural numbers given by a star b is equal to Lcm of a and b. Find 5 star 7 comma 20 star 16. Second part is a star commutative. Third part is a star associative. Fourth part is, find the identity of star in set n. Fifth part is, which elements of n where n is the set of natural numbers are invertible for operation star. First of all, let us understand the key idea to solve the first part of the question. A binary operation star on a set a is a function star from a cross a to a. We denote star a b by a star b where a and b are the elements of set a. Let us now start the solution. We know binary operation star is defined on n by a star b is equal to Lcm of a and b. This is given in the question. Now, we have to find in the first part 5 star 7 and 20 star 16. So, 5 star 7 we can write is equal to Lcm of 5 and 7, which is equal to 35. So, we get 5 star 7 is equal to 35. Now, similarly we will find 20 star 16, which is equal to Lcm of 20 and 16 and Lcm of 20 and 16 is equal to 80. So, we get 20 star 16 is equal to 80. This completes the first part of the question. Now, in second part we have to find a star commutative below a star b is equal to Lcm of a and b. b star a is equal to again Lcm of a and b. This implies a star b is equal to b star a, right? Since a star b is equal to b star a for every a b belonging to n, where n is the set of all natural numbers, star is commutative. So, this completes our second part. Now, the third part of the question is a star associative. Let us start the solution of third part. We know a star bracket b star c is equal to Lcm of a b and c. Similarly bracket a star b star c is equal to Lcm of a b and c star c is equal to a star bracket b star c. So, we can write since a star b star c is equal to a star bracket b star c for every a b c belonging to n, where n is the set of all natural numbers, star is associative. The fourth part of the question is, find the identity of star in n at a without required identity element. Is equal to a is equal to e star a, right? R a is equal to a for every a belonging to set of natural numbers. Now, a star a is equal to Lcm of a and e. Lcm of a and e is equal to a. This is only possible when e is equal to 1. So, this implies e is equal to 1. So, 1 is the identity element in n is the set of all natural numbers. Now, the question is, which elements of n are invertible for operation star? Let us start with the solution of fifth part now. Let us assume a be an invertible element in n. There exists b belonging to n such that a star b must be equal to 1. We know a star b must be equal to identity element and the required identity element we had already obtained is 1. So, we put a star b equal to 1. Now, we know a star b is further equal to Lcm of a and b. So, we get Lcm of a and b is equal to 1. But Lcm of a and b is equal to 1. This is only possible when a and b both have the values equal to 1. So, we get a is equal to b is equal to 1. We know if a star b is equal to the identity element, then b is the inverse of a. So, we get a is equal to b is equal to 1. So, a was our required invertible element. So, we get 1 as an required invertible element, the element invertible operation star. The required answer for the first part is 5 star 7 is equal to 35 and 20 star 16 is equal to 80. Required answer for second part is yes. Required answer for third part is yes. Required answer for fourth part is 1 and the required answer for fifth part is also 1. So, this completes the session. Hope you understood the session well. Take care and goodbye.