Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x²

(ii) f : Z → Z given by f(x) = x²

(iii) f : R → R given by f(x) = x²

(iv) f : N → N given by f(x) = x³

(v) f : Z → Z given by f(x) = x³ 

(i) f: N → N is given by,

f(x) = x²

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x² = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

 

(ii) f: Z → Z is given by,

f(x) = x²

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x² = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iii) f: R → R is given by,

f(x) = x²

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x² = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iv) f: N → N given by,

f(x) = x³

It is seen that for x, y ∈N, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x³ = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

 

(v) f: Z → Z is given by,

f(x) = x³

It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x³ = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.