Relations and Functions

Trivial Relations

• Trivial relations are of two types:
  • Empty relation
  • Universal relation

• A relation in a set A is called an empty relation if no element of A is related to any element of A, i.e., R = Φ ⊂ A × A.
  • For example: Consider a relation R in set A = {2, 4, 6} defined by R = {(a, b): a + b is odd, where a, b ∈ A}. The relation R is an empty relation since for any pair (a, b) ∈ A × A, a + b is always even.

• A relation R in a set A is called a universal relation if each element of A is related to every element of A, i.e., R = A × A.
  • For example: Let A be the set of all students of class XI. Let R be a relation in set A defined by R = {(a, b): the sum of the ages of a and b is greater than 10 years}

The relation R is a universal relation because it is obvious that the sum of the ages of two students of class XI is always greater than 10 years.

Equivalence Relation

• A relation R in a set A is called reflexive if (a, a) ∈ R for every a ∈ A.
  • For example: A relation R in set A defined by R = {sin a = sin b; a, b ∈ A} is a reflexive relation since.

• A relation R in a set A is called symmetric if (a₁, a₂) ∈ R implies that (a₂, a₁) ∈ R, for all a₁, a₂ ∈A.
  • For example: A relation in the set defined by R = {sin a = sin b; a, b ∈ A} is a symmetric relation. Since for a, b ∈ A, sin a = sin b implies sin b = sin a. So, (a, b) ∈ R ⇒ (b, a) ∈ R.

• A relation R in a set A is called transitive if (a₁, a₂) ∈ R, and (a₂, a₃) ∈ R together imply that (a₁,a₃) ∈ R, for all a₁, a₂, a₃ ∈ A.
  • For example: A relation in the set defined by R = {sin a = sin b, a, b ∈ A} is a transitive relation. Since for a, b, c ∈ A, let (a, b), (b, c) ∈ R.
    
    

                          ⇒ sin a = sin b and sin b = sin c

                          ⇒ sin a = sin c

                          ⇒ (a, c) ∈ R

• A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.
  • For example: Relation R in the setdefined by R = {sin a = sin b; a, b ∈ A} is an equivalence relation.

• To understand the concept of equivalence relation in a better way, let’s have a look at the following video.

Equivalence Classes

• Every arbitrary equivalence relation R in a set X divides X into mutually disjoint subsets (A_i) called partitions or subdivisions of X satisfying the following conditions:
  • All elements of A_i are related to each other for all i
  • No element of A_i is related to any element of A_j whenever i ≠ j
  • A_i ∪ A_j = X and A_i ∩ A_j = Φ, i ≠ j

                     These subsets (A_i) are called equivalence classes.

• For an equivalence relation in a set X, the equivalence class containing a ∈ X, denoted by [a], is the subset of X containing all elements b related to a.

Solved Examples

Example 1

Check whether the relation R in the set of all vowels defined by R = {(u, u), (u, a), (a, u)} is reflexive, symmetric or transitive?

Solution:

The relation R is defined in the set {a, e, i, o, u} as R = {(u, u), (u, a), (a, u)}.

The relation R is not reflexive as (a, a), (e, e) (i, i), (o, o) ∉ R.

Now, (u, a) and (a, u) ∈ R

Hence, R is symmetric.

Now, (u, u) (u, a) ∈ R implies (u, a) ∈ R

Also, (u, a), (a, u) ∈ R implies (u, u) ∈ R

Hence, R is transitive.

Thus, the relation R is symmetric and transitive but not reflexive.

Example 2

Show that the relation R defined in the set of real numbers as R = {(a, b) a = b or a = −b for a, b ∈R} is an equivalence relation. Also, find its equivalence classes.

Solution:

A relation R in R is defined as R = {(a, b): a = b or a = −b, for a, b ∈ R}

Clearly, (a, a) ∈ R for every a ∈ R, since a = a.

∴ R is reflexive

Now, let (a, b) ∈ R for a, b ∈ R

⇒ a = b or a = − b

⇒ b = a or b = − a

⇒ (b, a) ∈ R

∴ R is symmetric

Now, let (a, b), (b, c) ∈ R, for a, b, c ∈ R

∴ a = b or a = − b and b = c or b = − c

Case I

a = b, b = c

⇒ a = c

⇒ (a, c) ∈ R

Case II

a = b, b = − c

⇒ a = − c

⇒ (a, c) ∈ R

Case III

a = − b, b = c

⇒ a = − c

⇒ (a, c) ∈ R

Case IV

a = − b, b = − c

⇒ a = c

⇒ (a, c) ∈ R

Thus, (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

Equivalence class of 0 = [0] = {0}

Equivalence class of 1 = [1] = {1, − 1}

Equivalence class of 2 = [2] = {2, − 2} and so on…

There are an infinite number of equivalence classes.

For every a ∈ R, [a] = {a, − a}.

Example :

Show that the relation ‘congruence module m’ on the set Z₊ of all positive integers is an equivalence relation.

 

Reflexive:

Let a be an arbitrary positive integer.

⇒ a − a is divisible by m

⇒ a ≡ a (mod m) for all a ∈ Z₊.

Symmetric:

Let a, b ∈ Z₊ such that a ≡ b (mod m)

⇒ a − b is divisible by m

⇒ a − b = λ m, for an integer λ

⇒ b − a = −λ m

⇒ b ≡ a (mod m) for all a, b ∈ Z₊.

Transitive:

Let a, b, c ∈ Z₊ such that a ≡ b (mod m) and b ≡ c (mod m)

⇒ a − b = k m and b − c = l m for integers k, l.

⇒ a − b + b − c = (k + l) m

⇒ a − c = (k + l) m

⇒ a ≡ c (mod m), as (k + l) is also an integer.

Thus, the relation congruence module m is an equivalence relation on all positive integers as it is reflexive, symmetric and transitive.

Example :

Show that a relation R on the set of all non-zero complex numbers defined by z₁ R z₂ iff  is real is a transitive relation.

 

Let z₁ = a + ib, z₂ = c + id, z₃ = e + if be any three complex numbers, where e, b, c, d, e, f are real.

z₁ R z₂ iff  is real

⇒ Iff  is real

⇒ Iff  is real

⇒ Iff  is real

⇒ Iff 

Now,  and 

When cd ≠ 0⇒ z₁ R z₃

And when cd = 0 the transitivity is obvious.

Hence, R is transitive relation.

• A function f: X → Y is said to be one-one (or injective) if the images of distinct elements of Xunder f are distinct. In other words, a function f is one-one if for every x₁, x₂ ∈ X, f (x₁) = f (x₂) implies x₁ = x₂.

             An example of a one-one function from X to Y is shown in the following diagram.

A function f: X → Y is said to be many-one if the image of distinct elements of X under f are not distinct i.e., a function that is not one-one is called a many-one function.

An example of a many-one function from X to Y is shown in the following diagram.

In this case, two elements f(d) = f(e) = s.

• A function f: X → Y is defined as onto (or surjective) if every element of Y is the image of some element of x in X under f. In other words, f is onto if &mnForE; y ∈ Y, there exist x ∈ Xsuch that f (x) = y.
  • f: X → Y is onto if and only if the range of f = Y.
  • An example of an onto function from X to Y is shown in the following diagram.

• A function from X to Y that is not onto is shown in the following diagram.

• A function f: X → Y is said to be bijective if it is both one-one and onto.

             A bijective function from X to Y is shown in the following diagram.

• Let us have a look at the following video to understand the different types of functions:

Solved Examples

Example 1

Check whether the function h: R → R defined by h(x) =  is an injective function.

Solution:

The given function i.e., h: R → R isdefined by

h(x) = 

It can be observed that 5, 0 ∈ R (considering domain). Hence, we have

h(5) = 

h(0) = 

∴h(5) = h(0).

Hence, the given function i.e., h(x) is not an injective function.

Example 2

Check whether the function f: R → R defined by f(x) = x⁵ + 4 is a bijective function.

Solution:

We know that a function is bijective if it is both one-one and onto.

Now, let x₁, x₂ ∈ R such that f (x₁) = f (x₂). Accordingly,

f (x₁) = f (x₂)

Therefore, the function f is a one-one function.

It is clear that for every y ∈ R, there exists ∈ R such that

Therefore, the function f is an onto function.

Hence, the given function f is a bijective function.

Example 3

Check whether the function f: N → N defined by f(x) = 4^x is an onto function.

Solution:

The given function f: N → N is defined by

f(x) = 4^x

We can clearly observe that 2 ∈ N (codomain). However, there does not exist any element y ∈ N(domain) whose image is 2.

Hence, the given function f is not an onto function.

Example :

If f: A→B is a mapping defined by f(x) = , where A = R − {4} and B = R − {1} then show that f is bijective.

 

Injective:

Let x, y be any two elements of A.

Thus, f is an injective mapping.

Surjective:

Let y be any element of B.

If  then 3 = 4, which is not possible

Thus, every element y in B has its pre-image x in A, where x = 

Thus, f is a surjective mapping.

Hence, f is a bijective mapping.

Example :

Show that the function f : R→R defined by  is onto for every 

 

Let the function f : R→R is an onto function.

∴ Range of f = R

⇒  assumes all real values for real values of x

Let y = 

Thus, f : R→R is onto for a ∈ [2, 14].

• Let f: A → B and g: B → C be two functions. Accordingly, the composition of f and g is denoted by gof and is defined as the function gof: A → C given by gof(x) = g(f(x)), for all x∈A.

• For example: If f: N → N is defined by f(x) = x + 1 for all x∈N and g: N → N is defined by g(x) = x²for all x∈N, then gof : N → N is given by

            gof(x) = g(f(x)) = g(x + 1) = (x + 1)², where x∈N.

            Also, fog(x) = f(g(x)) = f(x²) = x² + 1 for all x∈N

• If f: A → B and g: B → C are one-one, then gof: A → C is also one-one.
• To know the proof of this result, look at the following video.
  
• If f: A → B and g: B → C are onto, then gof: A → C is also onto.
• To know the proof of this result, look at the following video.
  
• If the composite function gof is one-one, then the function f is also one-one. However, the function g may or may not be one-one.
• If the composite function gof is onto, then the function g is also onto. However, the function f may or may not be onto.

Solved Examples

Example 1

Let f: R → R be given by f(x) = 12x² − x − 11 and g: R → R be given by g(x) = x². Find fo(gog).

Solution:

It is given that

f: R → R is defined by f(x) = 12x² − x − 11

g: R → R is defined by g(x) = x²

Now, (gog) (x) = g(g(x))

= g(x²)

= (x²)² = x⁴

(fo(gog))(x) = f((gog)(x))

= f(x⁴)

= 12(x⁴)² = x⁴ − 11

= 12x⁸ − x⁴ − 11

Example 2

Let f: R −  → R −  be defined by and g: R − → R − be defined by g(x) =. Show that fog = I_A and gof = I_B, where I_A and I_B are identity functions on A and B respectively and A = R −  and B = R −.

Solution:

(fog)(x) = f(g(x))

(gof)(x) = g(f(x))

Thus, (fog)(x) = x for all x ∈ A ⇒ fog = I_A and (gof) (x) = x for all x ∈ B ⇒ gof = I_B.

Hence, proved.

Example 3

Let f: R → R be defined as f(x) =; g: R → R be defined as g(x) = x + 2 and h: R → R be defined as h(x) = 4x + 9. Find fo(g + h) and (fog) + (foh).

Solution:

(g + h): R → R is given by:

(g + h)(x) = g(x) + h(x)

= (x + 2) + (4x + 9) = 5x + 11

∴ (fo(g + h)) (x) = f((g + h) (x))

= f(5x + 11)

= 

Now, (fog)(x) = f(g(x))

= f(x + 2) = 

(foh)(x) = f(h(x))

= f(4x + 9)

= 

∴(fog + foh) (x) = (fog)(x) + (foh)(x)

Example :

then find gof(x).

 

Example :

Find the value of gof(x) if.

 

Key Concepts

• A function f: X → Y is said to be invertible if there exists a function g: Y→ X such that gof = I_Xand fog = I_Y.
  • The function g is called the inverse of f and it is denoted by f⁻¹.

• A function f is invertible if and only if f is one-one and onto.
• If f: X → Y and g: Y → Z are invertible functions, then gof is also invertible and (gof)⁻¹ = f⁻¹ og⁻¹
• To understand the proof of the above mentioned result, related to the inverse of a composite function, let’s look at the following video.

Solved Examples

Example 1

Determine whether the following functions have inverse or not. Find the inverse, if it exists.

(i) f : {10, 12, 15} → {3, 7, 9, 10, 14}is defined as f = {(12, 9), (15, 7), (10, 10)}.

(ii) g : {2, 4, 6, 8} → {1, 3, 5} is defined as g : {(4, 3), (8, 3), (2, 1), (6, 5)}.

(iii) h : {11, 16}→ {7, 14} is defined as h :{(11, 7), (16, 14)}.

Solution:

(i) The given function f is one-one. However, f is not onto since the elements 3, 14 ∈ {3, 7, 9, 10, 14} are not the image of any element in {10, 12, 15} under f.

Hence, function f is not invertible.

(ii) The given function g is onto. However, g is not one-one since, g(4) = g(8) = 3.

Hence, the function g is not invertible.

(iii) Clearly, the given function h is both one-one and onto. Hence, h is invertible.

The inverse of h is given by h⁻¹ = {(7, 11), (14, 16)}.

Example 2

Determine whether the functions f and g, defined below, are inverses of each other or not.

f : R − {4} → R − {−3} is given as , and

g: R − {−3} → R − {4} is given as 

Solution:

We have 

Thus,  where B = R −{−3} and A = R −{4}.

∴gof = I_A and fog = I_B.

Thus, functions f and g are the inverses of each other.

Example 3

Let f: R₊ → [−3, ∞) be defined as f (x) = 4x² − 5x − 3 where R₊ is the set of all positive real numbers. Show that f is invertible and find the inverse of f.

Solution:

f: R₊ → [−3, ∞) is defined as f (x) = 4x² − 5x − 3.

Let y be an arbitrary element of [−3, ∞).

Let y = 4x² − 5x − 3

∴f is onto.

Hence, Range f = [−3, ∞).

Let us define g: [−3, ∞) → R₊ as

Now, we have

Thus, f is invertible and its inverse is given by

Example :

Show that f is invertible function if f : R→R is defined by f(x) = (x + 2)² − 4, x ≥ −2.

 

Injectivity:

For any x, y ∈ R such that x ≥ −2, y ≥ −2

So, f(x) is injective.

Surjectivity:

For all y ≥ −2 there exists x such that f(x) = y.

So, f(x) is surjective.

Hence, f is a bijective function and thus, f is invertible.

Example :

Find the inverse of the function.

 

f : x → y is invertible if it is both one-one and onto.

One-one :

For f(x) to be onto, Y = [0, 4]

∴ f⁻¹(x) will exist and domain of f⁻¹ = y and range of f⁻¹ = X.

Definition and Properties

• A binary operation * on a set A is a function * from A × A → A. We denote *(a, b) by a * b.

For example, the operation * defined on N as a * b = a²b is a binary operation since * carries each pair (a, b) to a unique    element a²b in N.

• A binary operation * on a set A is called commutative, if a * b = b * a, for every a, b ∈ A.

 For example, *: R × R → R defined by a * b = 11 (a + b + ab) is commutative since a * b = 11(a + b + ab) and b * a = 11(b + a +  ba) and therefore, a * b = b * a

• A binary operation * on a set A is called associative, if (a * b) * c = a * (b * c), &mnForE a, b, c ∈ A.

For example, *: N × N → N defined by a * b = 5 + a + b is associative since

a * (b * c) = 10 + a + b + c = 5 + (5+ a + b) + c = ((a * b) * c

• For a binary operation *: A × A → A, an element e ∈ A, if it exists, is called its identity element, if a *e = a = e * a &mnForE a ∈ A.

For example: 1 is the identity for multiplication on R.

• Given a binary operation *: A × A → A with the identity element e in A, an element a ∈ A is said to be invertible with respect to the operation *, if there exists an element b∈A, such that a * b = e = b *a, and b is called the inverse of a and is denoted by a ⁻¹.

For example: −a is the inverse of a for the addition operation on R,where 0 is the identity element.

Binary Operation Table

• When the number of elements in set A is small, we can express a binary operation * on A through a table called operation table.
• For an operation *: A × A → A, if A = {a₁, a₂… a_n}, then the operation table will have n rows and ncolumns with (i, j)^th entry being a_i * a_j.
• Given any operation with n rows and n columns with each entry being an element of A = {a₁, a₂ …a_n}, we can define a binary operation * on A given by a_i * a_j = entry in i^th row and j^th column of the operation table

Example: We can define a binary operation * on A = {a, b, c} as follows:

*	a	b	c
a	a	b	c
b	b	a	c
c	c	c	c

Here, a * b = b = b * a

a * c = c = c * a

b * c = c = c * b

∴ The operation * is commutative.

To understand this concept better, let us look at the following video.

 

Solved Examples

Example 1:

A binary operation A × A → A, where A = {a, b, c}, is defined as follows:

*	a	b	c
a	a	a	a
b	a	b	c
c	a	c	b

Determine whether the operation * is commutative and associative. Also, find the identity for the operation *, if it exists.

Solution:

From the table, it can be observed that

a *b = a = b * a

a * c = a = c * a 

b * c = c = c * b

The given binary operation * is commutative since for all x, y, ∈ A = {a, b, c},

x * y = y * x

Now, consider a * (b * c) = a * c = a

(a * b) * c = a * c = a

Thus, a * (b * c) = (a * b) * c

Similarly, we can prove that (x * y) * z = x *(y * z) for all x, y, z ∈ A.

Thus, the given binary operation * is associative.

Also, we can observe that for any element x∈A, we have x * a = x = a * x

Thus, a is the identity element for the given binary operation *.

Example 2:

Determine whether the binary operation on the set R, defined by a * b = , a, b ∈ R, is commutative or not.

Solution:

We have *: R × R → R defined by , a, b ∈ R.

We know that a binary operation * defined on set A is commutative, if a * b = b * a &mnForE a, b ∈A.

Now, a * b =  and b * a = 

∴ a * b ≠ b * a

Hence, the given binary operation * is not commutative.

Example 3:

A binary operation * on the set {5, 6, 9} is defined by the following table:

*	5	6	9
5	5	6	9
6	6	9	5
9	9	5	6

Compute (5 * 9) * 6 and 5 * (9 * 6). Are they equal?

Solution:

From the given binary operation table, we have (5 * 9) = 9

∴(5 * 9) * 6 = 9 * 6 = 5

Then, (9 * 6) = 5

∴5 * (9 * 6) = 5 * 5 = 5

Thus, (5 * 9) * 6 = 5 * (9 * 6)

Example :

What is the total number of non-commutative binary operations on a finite set S having 10 elements?

 

A binary operation on set S with n elements is a function from S × S to S.

∴ Total number of binary operations on S = 

Total number of commutative binary operations on S = 

∴ Total number of non-commutative binary operations on S = 

Example :

If * is a binary operation on the set Q⁺ of all positive rational numbers such that a * b =  for all a,b ∈ Q⁺, find the identity and inverse.

 

Identity element:

Let e be the identity element in Q⁺ with respect to *.

∴ 4 is the identity element for * in Q⁺.

Inverse element:

Let b be the inverse of a ∈ Q⁺.

∴ a * b = e = b * a

Thus, for every element a ∈ Q⁺ has  is its inverse.

Example :

Show that * is commutative and associative on the set N of all natural numbers defined by m * n = l.c.m. (m, n) for all m, n ∈ N.

 

Commutative:

For all m, n ∈ N

m * n = l.c.m. (m, n)

= l.c.m. (n, m)

= n * m.

So, * is commutative on N.

Associative:

For all m, n, p ∈ N

(m * n) * p = {l.c.m. (m, n)}* p

= l.c.m. {l.c.m. (m, n), p}

= l.c.m. {m, l.c.m. (n, p)}

= l.c.m. (m, n * p)

= m * (n * p)

So, * is associative on N.