 Hello and welcome to the session. Let us understand the following question. Check the injectivity and subjectivity of the function. Given to us, f is a function from n to n given by fx is equal to x square. Now before writing the solution, let us see the key idea to the problem. Let us understand what is injectivity and subjectivity. A function from x to y is defined to be one-one detective. If images of distinct elements are distinct, every x1, x2 belongs to x of x1 is equal to f of x2 implies x1 is equal to x2 and a function from x2 y is said to be onto a subjective. If every element of y or every y belongs to y, there exists is equal to y. Now let us write the solution. Let us check for one-one. x1, x2 belongs to n such that f of x1 is equal to f of x2 then it implies x1 square is equal to x2 square by definition which implies x1 is equal to plus minus x2. Now since does not contain negative numbers, so x1 is equal to minus x2 is not possible. Therefore, x1 is equal to x2. Check for one-two. For each y belongs to n in co-domain x belongs to n of domain x is equal to y which implies x square is equal to y which implies x is equal to plus minus under root y that is f of plus minus root y is equal to plus minus root y the whole square which is equal to y. As we know set of natural numbers n does not contain square roots thus we can say that for every y belongs to co-domain does not exist x is equal to plus minus root y which belongs to domain f of x is equal to y from n to n is not onto the function f is one-one onto therefore function is injective but not subjective. This should be problem bye and have a nice day.