Page No 10:
Question 2:
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.
