 Hello and welcome to this session, I discuss the following problem. Let f be a function from w to w, we defined as fn is equal to n minus 1 if n is odd and fn is equal to n plus 1 if n is even, show that f is invertible, find the inverse of f, here w is the set of all whole numbers. Now let us write the solution. Given to us is f is a function from w to w, defined as fx is equal to n minus 1 if n is odd and n plus 1 if n is even, where w is the set of all whole numbers that is 0 comma 1 comma 2 and so on. Now let us first prove that f is invertible, to show that first f is a bidirection, for this first we show that f is 1 1, a comma b belongs to w, then if a and b are even, then f of a is equal to f of b which implies a plus 1 is equal to b plus 1 which implies a is equal to b. Now if a and b are odd, then f of a is equal to f of b which implies a minus 1 is equal to b minus 1 which implies a is equal to b, in both cases is equal to f b which implies a is equal to b. Now if a is odd and b is even, then f of a is equal to a minus 1 is even, f of b is equal to b plus 1 is odd, therefore a is not equal to b which implies f of a is not equal to f of b. Now similarly if a is even and b is odd, then a not equal to b implies f of a is not equal to f of b, thus f is 1 1. Now to prove that, given arbitrary element of w, if a is an odd whole number, then there exists an even whole number minus 1 belongs to w of domain such that f of a minus 1 which is equal to a minus 1 plus 1 which is equal to a. Now if a is an even whole number, then there exists a number plus 1 belongs to w of domain such that f of a plus 1 is equal to a plus 1 minus 1 which is equal to a. We know that f of 1 is equal to 0, thus every element of w in core domain is pre-imagined in domain. But f is 1 1 and on to therefore, this is a bijective function invertible. Now let us find the inverse of f. Now let us find the inverse of f. We observe that n is odd for any y belongs to w, y is equal to n minus 1 by definition is equal to y plus 1, y is even when n is even, y is equal to n plus 1 or n is equal to y minus 1 and y is odd. Thus we define f inverse of y is equal to g of y g such that from w to w such that g of y is equal to y minus 1 if y is odd and y plus 1 if y is even. Inverse of f itself, this is our required inverse. I hope you are interested in this problem. Bye and have a nice day.