Question

Check the injectivity and surjectivity of the following function.

f : N → N given by f(x) = x³

Answer 1

f : N → N given by f(x) = x³ 

Let f(x₁) = f(x₂)

∴ x₁³ = x₂³

∴ x₁³ – x₂³ = 0

∴ `(x_1 - x_2) underbrace((x_1^2 + x_1 x_2 + x_2^2))_(> 0  "for all"  "x"_1,  "x"_2  "as it's discriminant" < 0)` = 0

∴ x₁ = x₂

∴ f is injective.

Numbers from codomain which are not cubes of natural numbers are not images under f.

∴ f is not surjective.