Question

Let A = {–1, 0, 1} and f = {(x, x²) : x ∈ A}. Show that f : A → A is neither one – one nor 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, We have, A = {–1, 0, 1} and f = {(x, x²) : x ∈ A}.

To Prove: – f : A → A is neither One – One nor onto function

Check for Injectivity:

We can clearly see that

f(1) = 1

and f( – 1) = 1

Therefore

f(1) = f( – 1)

⇒ Every element of A does not have different image from A

Hence f is not One – One function

Check for Surjectivity:

Since, y = – 1 be element belongs to A

i.e -1 ∈ A in co – domain does not have any pre image in domain A.

Hence, f is not Onto function.