Show that the function f: R → R: f(x) = x5 is one-one and onto.

1.	Show that the function f: R → R: f(x) = x5 is one-one and onto.
Answer» Injectivity: Consider x, y ∈ R where f(x) = f(y) We get x⁵ = y⁵ It can be written as x⁵ – y⁵ = 0 By multiplying and dividing by 2 on both sides (x^5/2)² – (y^5/2)² = 0 We know that (x^5/2 + y^5/2) (x^5/2 – y^5/2) = 0 So we get x^5/2 – y^5/2 = 0 where x = y Hence, f(x) = f(y) is x = y for all x, y ∈ R f is injective. Surjectivity: Consider y as an arbitrary element of R We know that f(x) = y It can be written as x⁵= y So we get x⁵ – y = 0 Odd degree equation has one real root Thus, for every real value of y x⁵ – y = 0 has real root α where α⁵ – y = 0 So α⁵ = y Thus, f(α) = y For every y ∈ R there exists α ∈ R where f(α) = y f is surjective Thus, f: R → R is bijective.

Answer»

Injectivity:

Consider x, y ∈ R where f(x) = f(y)

We get

x⁵ = y⁵

It can be written as

x⁵ – y⁵ = 0

By multiplying and dividing by 2 on both sides

(x^5/2)² – (y^5/2)² = 0

We know that

(x^5/2 + y^5/2) (x^5/2 – y^5/2) = 0

So we get

x^5/2 – y^5/2 = 0

where x = y

Hence, f(x) = f(y) is x = y for all x, y ∈ R

f is injective.

Surjectivity:

Consider y as an arbitrary element of R

We know that

f(x) = y

It can be written as

x⁵= y

So we get

x⁵ – y = 0

Odd degree equation has one real root

Thus, for every real value of y

x⁵ – y = 0 has real root α where

α⁵ – y = 0

So α⁵ = y

Thus, f(α) = y

For every y ∈ R there exists α ∈ R where f(α) = y

f is surjective

Thus, f: R → R is bijective.

Discussion