 Hello and welcome to the session, let us discuss the following problem today. Let every function from n to n be defined by f of n is equal to n plus 1 by 2 if n is odd and n by 2 if n is even for all n belongs to n, state whether the function f is bijective. Just define your answer. Now let us write the solution. Now let us first check for 1 1. We take that f of x is equal to f of y which implies x is equal to y or not. Now consider 1 comma 2 belongs to n. Now f of 1 is equal to 1 plus 1 by 2 which is equal to 1 and f of 2 is equal to 2 by 2 which is equal to 1. Thus f of 1 is equal to f of 2 but 1 is not equal to 2. Therefore f is not 1 1. Now let us check for on 2. Let n be any arbitrary element of n. We have to prove that for every n belongs to n whether even or odd there exist its pre-imagined n. This is an odd natural number. 1 is also an odd natural number minus 1 which is equal to putting to n minus 1 in our given function we get 2 n minus 1 plus 1 by 2 which is equal to n. Now if n is an even natural number then n is also an even natural number. f of 2 n is equal to 2 n by 2 which is equal to n. It belongs to n whether even or odd there is a pre-imagined n but since n is not 1 1 therefore the function is not bijective. You understood the problem. Bye and have a nice day.