We use two methods to find if function has inverse or not

  1. If function is one-one and onto, it is invertible.
  2. We find g, and check fog = I Y and gof = I X

 

We discussed how to check one-one and onto previously.

 

Let’s discuss the second method

 

We find g, and check fog = I Y and gof = I X

 

Steps are

Checking inverse of f : X → Y

Step 1 : Calculate g: Y → X

Step 2 : Prove gof = I X

Step 3 : Prove fog = I Y

 


Example

Let f : N → Y,

f (x) = 2x + 1, where, Y = {y ∈ N : y = 4x + 3 for some x ∈ N }.

Show that f is invertible

 

Checking by One-One and Onto Method

Checking one-one

f(x 1 ) = 2x 1 + 1

f(x 2 ) = 2x 2 + 1

 

Putting f(x 1 ) = f(x 2 )

2x 1 + 1 = 2x 2 + 1

  2x 1 = 2x 2

  x 1 = x 2

 

If f(x 1 ) = f(x 2 ) , then x 1 = x 2

∴  f is one-one

 

Checking onto

f(x) = 2x + 1

 

Let f(x) = y, where y ∈ Y

  y = 2x + 1

  y – 1 = 2x

  2x =  y – 1

  x = (y - 1)/2

 

For every y in Y = {y ∈ N : y = 2x + 1 for some x ∈ N }.

There is a value of x which is a natural number

Thus, f is onto

 

Since f is one-one and onto

f is invertible

 

Checking by fog = I Y and gof = I X  method

Checking inverse of f: X → Y

Step 1 : Calculate g: Y → X

Step 2 : Prove gof = I X

Step 3 : Prove fog = I Y

g is the inverse of f

 

Step 1

f(x) = 2x + 1

Let f(x) = y

  y = 2x + 1

  y – 1 = 2x

  2x =  y – 1

  x = (y - 1)/2

 

Let g(y) = (y - 1)/2

where g: Y → N

 

Step 2 :

gof = g(f(x))

        = g(2x + 1)

        = ((2x + 1) - 1)/2

= (2x + 1 - 1)/2

= 2x/2

= x

= I N

 

Step 3 :

fog = f(g(y))

        = f((y - 1)/2)

= 2 ((y - 1)/2) + 1

= y – 1 + 1

= y

= I Y

 

Since gof   = I N and fog = I Y ,

f is invertible

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.