(i) f : N → N given by f(x) = x2.
(a) f(x1) = f(x2) ⇒ \(x_{1}^{2}\) = \(x_{2}^{2}\) ⇒ x1 = x2.
∴ f is one-one, i.e., it is injective.
(b) There are some members of codomain which have no image in domain N.
e.g., 3 ∈ codomain N. But there is no pre-image in domain of f.
⇒ f is not onto, i.e., not surjective.
(ii) f : Z → Z given by f(x) = x2.
(a) f(-1) = f(1) = 1 ⇒ -1 and 1 have the same image.
∴ f is not one-one, i.e., not injective.
(b) There are many elements belonging to codomain have no pre-image in its domain Z.
e.g. 3 ∈ codomain Z but \(\sqrt{3}\) ∉ domain Z of f
∴ f is not onto, i.e., not surjective.
(iii) f : R → R given by f(x) = x2.
(a) f is not one-one since f(- 1) = f(1) = 1
-1 and 1 have the same image, i.e., f is not injective.
(b) – 2 ∈ codomain R of f, but \(\sqrt{-2}\) does not belong to domain R of f.
⇒ f is not onto, i.e., f is not surjective.
(iv) f: N → N given by f(x) = x3,
(a) f(x1) = f(x2) ⇒ \(x_{1}^{3}\) = \(x_{2}^{3}\) ⇒ x1 = x2
i.e., every x ∈ N has a unique image in its codomain.
∴ f is one-one. It is injective.
(b) There are many members or codomain of f which have no pre-image in its domain.
e.g., 2, 3 etc. These members of codomain have no pre-image in its domain.
∴ f is not onto, i.e., f is not surjective.
(v) f : Z → Z given by f(x) = x3.
(a) Here, also f(x1) = f(x2) ⇒ \(x_{1}^{3}\) = \(x_{2}^{3}\) ⇒ x1 = x2
∴ f is one-one, i.e., it is injective.
(b) Many members of codomain of f do not have any pre-image in its domain.
e.g., 2 belonging to its codomain has no pre-image in its domain of Z. Therefore, f is not surjective.
