Answer :

It is given that f : W W be defined as

f(n) =


Let f(n) = f(m)


We can see that if n is odd and m is even, then we will have n -1 = m +1.


n – m = 2


this is impossible


Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.


Therefore, both n and m must be either odd or even.


Now, if both n and m are odd, then we get:


f(n) = f(m)


n -1 = m -1


n = m


Again, if both n and m are even, the we get:


f(n) = f(m)


n +1 = m + 1


n = m


f is one – one.


Now, it is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co – domain N is the image of 2r +1 in domain N.


f is onto.


f is an invertible function.


Now, let us define g : W W be defined as



Now, when n is odd:


gof(n) = g(f(n)) = g (n-1) = n -1 +1 = n (when n is odd, then n-1 is even)


And when n is even:


gof(n) = g(f(n)) = g (n+1) = n +1 -1 = n (when n is even, then n+1 is odd)


Similarly, when m is odd:


fog(m) = f(g(m)) = f (m-1) = m -1 +1 = m


And when n is even:


fog(m) = f(g(m)) = f (m+1) = m +1 -1 = m


Therefore, gof = IW and fog = IW


Therefore, f is invertible and the inverse of f is given by f-1 = g, which is the same as f.


Thus, the inverse of f is f itself.


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