Let f : W → W 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.
It is given that f : W → W be defined as
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.