 Hi and how are you all today? I am Priyanka and let us do the following question together. It says write the following set by using setBuilder method. Now these are the five set which are given to us and all these are in Roaster form. We need to convert them in SetBuilder form. We know that in SetBuilder method all elements have a single common property which is not possessed by any other element outside the set. So we identify the common property of the elements of the set and our setBuilder format is xh to x common property. So the knowledge of this procedure is the key idea we are going to use in order to proceed on with our solution, proceeding on with the solution of the first part. Now we are given in Roaster form 36912. Right? So in set A we see that all elements are multiples of 3 right? And we can write x as 3n. Let this be set A. So we denote symbol x for element of the given set. Single common property will be all elements of all elements multiples of 3. Right? And written as 3n. So the solution will become A is equal to x is to x is equal to 3n and n is greater than equal to 1 less than equal to 4. Now let us verify our property. We have written that x is equal to 3n means each element is equal to 3n and it will change as we are changing the n. x greater than equal to 1 and less than equal to 4. That means n will vary as 1, 2, 3 and 4. And what is our result? We can find that we are getting all the elements that was present in the set. Right? So this means that the answer which we have formed is a correct answer. And this is the answer to the first part. Gone with the second part. Now we are given B is equal to 2, 4, 8, 16, 32. Now we are taking all the elements as x. Now the single common property we have ourselves assumed it to be named as B. So the common property is that x is a square number, isn't it? Denoted by n where n is greater than equal to 1 and also n is less than equal to 5. So let us write it in set builder form and it will be x is 2x is equal to 2 to the power n and 1 is less than equal to n is greater than equal to 1 and it is less than equal to 5. Now this completes our second part and let us verify it also. 2 raised to the power 1 then we have 2 raised to the power 2, 2 raised to the power 3, 2 raised to the power 4 and 2 raised to the power 5. So what is the answer? It will be 2, 4, 8, 16 and 32, isn't it? So that means we have verified our obtained answer also. So this is our required answer of the second part. Proceeding on with the third part let us write it as c is equal to 5, 25, 125 and 625. Similarly as we did above all these are 5 raised to the power n. So our answer can be directly written as the common property first. x is a power of number 5 which is donated as 5 raised to the power n and n is greater than equal to 1 whereas it is less than equal to 4. So let us write it down. Our answer will be b is equal to x is to x is equal to 5 raised to the power n and 1 is less than equal to n whereas it is less than equal to 4. And let us verify it also. 5 to the power 1 gives us 5, 5 to the power 2 gives us 25, 5 to the power 3 gives us 125 and 5 to the power 4 gives us 625. So isn't it all the elements which are present in set c? So it will be c is equal to x is to x is equal to 5n raised to the power n and n is greater than equal to 1 whereas less than equal to 4. So this becomes the answer of the third part. Proceeding on further to the fourth part let us take it as t is equal to 2, 4, 6 and so on. So which is the common property which we are observing? It is that they are all even natural numbers, right? 2, 4, 6, 8 and so on. So the answer will be in set builder form x is to x is an even natural number where we have represented x as the elements of the set and it is never ending so we haven't given any condition to it. This determines our answer of the fourth part. Proceeding on to the fifth part we are given 1, 4, 9 and so on till 100. So which is the common property which we can observe? We can see that x which represent for the elements of the given set is the square of number n where n is greater than equal to 1 and n is also less than equal to 10 because 100 is the square of 10. So our answer will be x is to x is equal to n raised to the past square and 1 is we need to write the property the condition attached to it also. So this becomes the answer of the fifth part. So I hope you understood the question. In this question we will learn to identify the single common property of a given set in roster form and then write it in the set builder form. Right, bye for now.