Check the injectivity and surjectivity of the following functions: (i) f:N→Ngiven by f(x) = x2 (ii) f:Z→Zgiven by f(x) = x2 (iii) f:R→Rgiven by f(x) = x2 (iv) f:N→Ngiven by f(x) = x3 (v) f:Z→Zgiven by f(x) = x3

Check the injectivity and surjectivity of the following functions:
(i) f:N→Ngiven by f(x) = x²
(ii) f:Z→Zgiven by f(x) = x²
(iii) f:R→Rgiven by f(x) = x²
(iv) f:N→Ngiven by f(x) = x³
(v) f:Z→Zgiven by f(x) = x³
Answer
(i) f:N→Nis 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.