Prove that the function f : N → N , defined by f ( x ) = x 2 + x + 1 , is one-one but not onto.

f : N → N , defined by f ( x ) = x 2 + x + 1 Injectivity: Let x and y be any two elements in the domain N , such that f ( x ) = f ( y ) . Then , ⇒ x 2 + x + 1 = y 2 + y + 1 ⇒ x 2 - y 2 + ( x - y ) = 0 ⇒ ( x + y ) ( x - y ) + ( x - y ) = 0 ⇒ ( x - y ) ( x + y + 1 ) = 0 ⇒ x - y = 0 [ x + y + 1 cannot be zero because x and y are natural numbers] ⇒ x = y So, f is one-one. Surjectivity: We claim that f will not assume the value 1. f x = 1 ⇒ x 2 + x + 1 = 1 ⇒ x 2 + x = 0 ⇒ x ( x + 1 ) = 0 ⇒ x = 0 or x = - 1 But 0 and - 1 are not natural numbers. So, f is not onto.

Let A = − 1 , 0 , 1 and f = x , x 2 : x ∈ A . Show that f : A → A is neither one-one nor onto.

We have, A = − 1 , 0 , 1 and f : A → A defined by f = x , x 2 : x ∈ A Clearly, f ( 1 ) = 1 2 = 1 and f ( - 1 ) = ( - 1 ) 2 = 1 ⇒ 1 and - 1 have the same images. So, f is not one-one. Again, y = - 1 ∈ A in the co-domain does not have any pre-image in domain A . Hence, f is not onto