# Problem 9 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 9 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 1.2 Problem 9 Solution
9. Let f : N → N be defined by \begin{cases}\dfrac{n + 1}{2} & ,\ if\ n\ is\ odd\\\dfrac{n}{2} & ,\ if\ n\ is\ even\end{cases} for all n ∈ N. State whether the function f is bijective. Justify your answer.
To check whether f is one-one
Let k be an odd natural number.
Then {k + 1} will be an even natural number.
Now {f(k) = \dfrac{k + 1}{2}} (∵ k is odd, we’ve to use {f(n) = \dfrac{n + 1}{2}})
Now {f(k + 1) = \dfrac{k + 1}{2}} (∵ {k + 1} is even, we’ve to use {f(n) = \dfrac{n}{2}})
⇒ Two different consecutive natural numbers have the same image.
f is not one-one
To check whether f is onto
To be onto, every natural number yN in the co-domain should be an image of an element in the domain N.
When x is odd {y = f(x) = \dfrac{x + 1}{2}}
{2y - 1 = x}
As we know, for any value of yN, {2y - 1} is always odd and also {2y - 1}N
⇒ Every yN in the co-domain will be an image of an element in the domain.
Now, when x is even {y = f(x) = \dfrac{x}{2}}
{2y = x}
As we know, for any value of yN, 2y is always even and also 2yN
⇒ Every yN in the co-domain will be an image of an element in the domain.
Thus we see that in all the cases, every element yN in the co-domain will be an image of an element in the domain. (to be specific every yN will be an image of two elements in the domain)
In otherwords, all the elements in the co-domain N will have an image in the domain
f is onto(surjective) but not one-one(injective).
f is not bijective.
Note: For short answer/MCQ, as soon as you discover that f is not one-one, you can conclude that f is not bijective. In otherwords, you don’t have to check whether f is onto. This might save some time.