question_answer 58)

Let f :  be defined as f(n) = n ? 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.  

Answer:

Let x1 and x2 are two distinct elements of W. Injective : Case ? I : When both x­1 and x2 are even :             As       Case ? II : When both x1 and x2 are odd :    f(x1) = x1 ? 1 and f(x2) = x­2­ ? 1 As         f(x1) Case-II : When x1 is even and x2is odd :  f(x1) = x1 + 1 and f(x2) = x2 ? 1            As Similarly f(x1)  for odd x1 and even x2 Therefore, in all cases                   Hence f is one-one function. Surjective : Let y be any element of W then f(y ? 1) = y, if is odd and f(y + 1) = y, if y is even Therefore, corresponding to every element y of       W, there exists elements y ? 1 (or y + 1) of W and f :         Hence f is onto function. Thus f is both one-one and onto function.  is invertible. To find inverse of f : As  f(n  - 1) = n, if n is odd And f(n + 1) = n, if n is even So, f?1(n) =       = f(n)