Question

Check the injectivity and surjectivity of the following function.

f : N → N given by f(x) = x² 

Answer 1

f : N → N given by f(x) = x² 

Let f(x₁) = f(x₂), x₁, x₂ ∈ N

∴ x₁² = x₂²

∴ x₁² – x₂² = 0

∴ `(x_1 - x_2) underbrace((x_1 + x_2))_("for" x_1 * x_2 ∈ "N")` = 0

∴ x₁ = x₂

∴ f is injective.

For every y = x² ∈ N, there does not exist x ∈ N. Example: 7 ∈ N (codomain) for which there is no x in domain N such that x² = 7

∴ f is not surjective.

Answer 2

f: Z → Z given by f(x) = x²

Z = {O, ±1, ±2, ±3, ....}

(a) f : Z → Z

Let -1, 1 ∈ Z, f (-1) = f(1)

⇒ 1 = 1

But -1 ≠ 1       ∴f is not one-one i.e., f is not injective.

(b) There are many such elements belongs to co-domain have no pre-image in its domain z.

e.g., 2 ∈ Z (co-domain). But `2^(1//2) != Z` (domain)

∴  Element 2 has no pre-image in its domain Z.

f is not onto i.e., f is not surjective.