Invertible Functions

A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f–1.

An important note is that, if f is invertible, then f must be one-one and onto and conversely if f is one-one and onto, then f must be invertible.

Example 1:. Find f −1 if it exists:  f: A → B, where 

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Solution:

(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, different elements of the domain have different images in the co-domain.

Clearly, this is one-one.

Range of f = Range of f = B

so, f is a bijection and,

Thus, f -1 exists.

Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, different elements of the domain have different images in the co-domain.

Clearly, f is one-one.

But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

So, f -1does not exist.

Example 2:. Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1

Solution:

Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.

So, f and g are invertible.

Now,

-1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b), (cat , c)}

So, f-1 o g-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)

f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}

So, gof: {1, 2, 3} → {apple, ball, cat}

⇒ (gof) (1) = g (f (1)) = g (a) = apple

(gof) (2) = g (f (2))

= g (b)

= ball,

And (gof) (3) = g (f (3))

= g (c)

= cat

∴ gof = {(1, apple), (2, ball), (3, cat)}

Clearly, gof is a bijection.

So, gof is invertible.

(gof)-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)

Form (1) and (2), we get

(gof)-1 = f-1 o g -1

Example 3:. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f: A → B, g: B → C be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.

Solution:

Given that f (x) = 2x + 1

⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}

= {(1, 3), (2, 5), (3, 7), (4, 9)}

Also given that g(x) = x2−2

⇒ g = {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}

= {(3, 7), (5, 23), (7, 47), (9, 79)}

Clearly f and g are bijections and, hence, f−1: B→ A and g−1: C→ B exist.

So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}

And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f−1 o g−1): C→ A

f−1 o g−1 = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)

Also, f: A→B and g: B → C,

⇒ gof: A → C, (gof) −1 : C→ A

So, f−1 o g−1and (gof)−1 have same domains.

(gof) (x) = g (f (x))

=g (2x + 1)

=(2x +1 )2− 2

⇒ (gof) (x) = 4x+ 4x + 1 − 2

⇒ (gof) (x) = 4x2+ 4x −1

Then, (gof) (1) = g (f (1))

= 4 + 4 − 1

=7,

(gof) (2) = g (f (2))

= 4(2)2 + 4(2) – 1 = 23,

(gof) (3) = g (f (3))

= 4(3)2 + 4(3) – 1 = 47 and

(gof) (4) = g (f (4))

= 4(4)2 + 4(4) − 1 = 79

So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof)– 1 = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)

From (1) and (2), we get:

(gof)−1 = f−1 o g−1

Example 4:. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1

Solution:

Given function f: Q → Q, defined by f(x) = 3x + 5

Now we have to show that the given function is invertible.

Injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ 3x + 5 = 3y + 5

⇒ 3x = 3y

⇒ x = y

so, f is one-one.

Surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ 3x +5 = y

⇒ 3x = y – 5

⇒ x = (y -5)/3 belongs to Q domain

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f-1(x) = y…… (1)

⇒ x = f(y)

⇒ x = 3y + 5

⇒ x −5 = 3y

⇒ y = (x – 5)/3

Now substituting this value in (1) we get

So, f-1(x) = (x – 5)/3

 Example 5: Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution:

Given f: R → R given by f(x) = 4x + 3

Now we have to show that the given function is invertible.

Consider injection of f:

Let x and y be two elements of domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is one-one.

Now surjection of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ 4x + 3 = y

⇒ 4x = y -3

⇒ x = (y-3)/4 in R (domain)

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f -1

Let f-1(x) = y……. (1)

⇒ x = f (y)

⇒ x = 4y + 3

⇒ x − 3 = 4y

⇒ y = (x -3)/4

Now substituting this value in (1) we get

So, f-1(x) = (x-3)/4

Example 6: Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = √ (x-4) where R+ is the set of all non-negative real numbers.

Solution:

Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.

Now we have to show that f is invertible,

Consider injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ x+ 4 = y+ 4

⇒ x= y2

⇒ x = y      (as co-domain as R+)

So, f is one-one

Now surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ x2 + 4 = y

⇒ x2 = y – 4

⇒ x = √ (y-4) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f−1 (x) = y…… (1)

⇒ x = f (y)

⇒ x = y2 + 4

⇒ x − 4 = y2

⇒ y = √ (x-4)

So, f-1(x) = √ (x-4)

Now substituting this value in (1) we get,

So, f-1(x) = √ (x-4)

Example 7:. If f(x) = (4x + 3)/ (6x – 4), x ≠ (2/3) show that fof(x) = x, for all x ≠ (2/3). What is the inverse of f?

Solution:

It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3

Now we have to show fof(x) = x

(fof)(x) = f (f(x))

= f ((4x+ 3)/ (6x – 4))

= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)

= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)

= (34x)/ (34)

= x

Therefore, fof(x) = x for all x ≠ 2/3

=> fof = 1

Hence, the given function f is invertible and the inverse of f is f itself.

Example 8:. Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with

f-1(x) = (√(x +6)-1)/3 

Solution:

Given f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5

We have to show that f is invertible.

Injectivity of f:

Let x and y be two elements of domain (R+),

Such that f(x) = f(y)

⇒ 9x+ 6x – 5 = 9y+ 6y − 5

⇒ 9x+ 6x = 9y+ 6y

⇒ x = y (As, x, y ∈ R+)

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (Q)

Such that f(x) = y

⇒ 9x2 + 6x – 5 = y

⇒ 9x2 + 6x = y + 5

⇒ 9x2 + 6x +1 = y + 6 (By adding 1 on both sides)

⇒ (3x + 1)2 = y + 6

⇒ 3x + 1 = √(y + 6)

⇒ 3x = √ (y + 6) – 1

⇒ x = (√ (y + 6)-1)/3 in R+ (domain)

f is onto.

So, f is a bijection and hence, it is invertible.

Now we have to find f-1

Let f−1(x) = y….. (1)

⇒ x = f (y)

⇒ x = 9y+ 6y − 5

⇒ x + 5 = 9y+ 6y

⇒ x + 6 = 9y2+ 6y + 1         (adding 1 on both sides)

⇒ x + 6 = (3y + 1)2

⇒ 3y + 1 = √ (x + 6)

⇒ 3y =√(x +6) -1

⇒ y = (√ (x+6)-1)/3

Now substituting this value in (1) we get,

So, f-1(x) = (√ (x+6)-1)/3

Example 9:. If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).

Solution:

Given f: R → R be defined by f(x) = x3 −3

Now we have to prove that f−1 exists

Injectivity of f:

Let x and y be two elements in domain (R),

Such that, x3 − 3 = y3 − 3

⇒ x3 = y3

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x3 – 3 = y

⇒ x3 = y + 3

⇒ x = ∛(y+3) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1(x) = y…….. (1)

⇒ x= f(y)

⇒ x = y− 3

⇒ x + 3 = y3

⇒ y = ∛(x + 3) = f-1(x)         [from (1)]

So, f-1(x) = ∛(x + 3)

Now, f-1(24) = ∛ (24 + 3)

= ∛27

= ∛33

= 3

And f-1(5) =∛ (5 + 3)

= ∛8

= ∛23

= 2

Binary Operations

A binary operation ∗ on a set A is a function ∗ : A × A → A.

We denote * (a, b) by a * b.

Example Problems

Example 1: Show that subtraction and division are not binary operations on R.

Solution: N × N → N, given by (a, b) → a – b, is not binary operation, as the image of (2, 5) under ‘–’ is 2 – 5 = – 3 ∉ N.

Similarly, ÷: N × N → N, given by (a, b) → a ÷ b is not a binary operation, as the image of (2, 5) under ÷ is 2 ÷ 5 = 2/5 ∉ N.

Example 2: Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Find gof.

Solution: From the given, we have:
gof(2) = g (f(2)) = g (3) = 7
gof (3) = g (f(3)) = g (4) = 7
gof(4) = g (f(4)) = g (5) = 11
gof(5) = g (5) = 11

Example 3: Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation.

Solution: R is reflexive, as 2 divides (a – a) for all a ∈ Z.
Further, if (a, b) ∈ R, then 2 divides a – b.
Therefore, 2 divides b – a.
Hence, (b, a) ∈ R, which shows that R is symmetric.

Similarly, if (a, b) ∈ R and (b, c) ∈ R, then (a – b) and (b – c) are divisible by 2.
Now, a – c = (a – b) + (b – c) is even. (from the above statements)
From this,
(a – c) is divisible by 2.
This shows that R is transitive.
Thus, R is an equivalence relation in Z.