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We use two methods to find if function has inverse or not
- If function is one-one and onto, it is invertible.
- 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