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

Solution:

f : W → W is defined as f (n) = {n - 1, if n is odd and f (n) = n + 1, if n is even}

For one-one:

f (n) = f (m)

If n is odd and m is even, then we will have n - 1 = m + 1.

⇒ n - m = 2

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

∴ Both n and m must be either odd or even.

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

f (n) = f (m)

⇒ n - 1 = m - 1

⇒ n = m

Again, if both n and m are even, then we have:

f (n) = f (m)

⇒ n + 1 = m + 1

⇒ n = m

∴ f is one-one.

For onto:

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.

Let us define g : W → W as f (m) = {m - 1, if m is odd and f (m) = m + 1, i m is even}

When r is odd

gof (n) = g (f (n)) = g (n - 1) = n - 1 + 1 = n

When r is even

gof (n) = g (f (n)) = g (n + 1) = n + 1 - 1 = n

When m is odd

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

When m is even

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

∴ gof = I_W and fog = I_W.

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

Inverse of f is f itself

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 2

Let f : W → W be defined as f (n) = n - 1, if 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

Summary:

Given that f : W → W be defined as f (n) = n - 1, if is odd and f (n) = n + 1, if n is even. Hence, we have shown that f is invertible and the inverse of f is given by f - 1 = g, which is the same as f