f ( ) = . But f is onto, as given any y ∈ N , y ≠ , we can choose x as y + such that f ( y + ) = y + – = y . Also for ∈ N , we have f ( ) = . Example Show that the function f : R → R , defined as f ( x ) = x , is neither one-one nor onto.
Solution Since f (– ) = = f ( ), f is not one- one. Also, the element – in the co-domain R is not image of any element x in the domain R (Why?). Therefore f is not onto. Example Show that f : N → N , given by ,if is odd, ( ) ,if is even f x + is both one-one and onto.
Fig . Solution Suppose f ( x ) = f ( x ). Note that if x is odd and x is even, then we will have x + = x – , i.e., x – x = which is impossible. Similarly, the possibility of x being even and x being odd can also be ruled out, using the similar argument.
Therefore, both x and x must be either odd or even. Suppose both x and x are odd. Then f ( x ) = f ( x ) ⇒ x + = x + ⇒ x = x . Similarly, if both x and x are even, then also f ( x ) = f ( x ) ⇒ x – = x – ⇒ x = x .
Thus, f is one-one. Also, any odd number r + in the co-domain N is the image of r + in the domain N and any even number r in