Question

Prove that the function f : N → N, defined by f(x) = x² + x + 1 is one – one but not onto.

Answer 1

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Now, f: N → N given by f(x) = x² + x + 1

Check for Injectivity:

Let x,y be elements belongs to N i.e x, y ∈ N such that

So, from definition

⇒ f(x) = f(y)

⇒ x² + x + 1 = y² + y + 1

⇒ x² – y² + x – y = 0

⇒ ( x – y )( x + y + 1) = 0

As x, y ∈ N therefore x + y + 1>0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

y be element belongs to N i.e y ∈ N be arbitrary

Since for y > 1, we do not have any pre image in domain N.

Hence, f is not Onto function.