CBSE Class 12 Maths –Chapter 1 Relations and Functions- Study Materials

Relation: A relation R from set X to a set Y is defined as a subset of the cartesian product X × Y. We can also write it as R ⊆ {(x, y) ∈ X × Y : xRy}.

Note: If n(A) = p and n(B) = q from set A to set B, then n(A × B) = pq and number of relations = 2pq.

Types of Relation
Empty Relation: A relation R in a set X, is called an empty relation, if no element of X is related to any element of X,
i.e. R = Φ ⊂ X × X

Universal Relation: A relation R in a set X, is called universal relation, if each element of X is related to every element of X,
i.e. R = X × X

Reflexive Relation: A relation R defined on a set A is said to be reflexive, if
(x, x) ∈ R, ∀ x ∈ A or
xRx, ∀ x ∈ R

Solved example of reflexive relation on set:

1. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Examine if R is a reflexive relation on Z.

Answer/Explanation

A

Solution:

Let a ∈ Z. Now 2a + 3a = 5a, which is divisible by 5. Therefore aRa holds for all a in Z i.e. R is reflexive.

2. A relation R is defined on the set Z by “aRb if a – b is divisible by 5” for a, b ∈ Z. Examine if R is a reflexive relation on Z.

Answer/Explanation

Solution:

Let a ∈ Z. Then a – a is divisible by 5. Therefore aRa holds for all a in Z i.e. R is reflexive.

A

3. Consider the set Z in which a relation R is defined by ‘aRb if and only if a + 3b is divisible by 4, for a, b ∈ Z. Show that R is a reflexive relation on on setZ.

Answer/Explanation

Solution:

Let a ∈ Z. Now a + 3a = 4a, which is divisible by 4. Therefore aRa holds for all a in Z i.e. R is reflexive.

4. A relation ρ is defined on the set of all real numbers R by ‘xρy’ if and only if |x – y| ≤ y, for x, y ∈ R. Show that the ρ is not reflexive relation.

Answer/Explanation

Solution:

The relation ρ is not reflexive as x = -2 ∈ R but |x – x| = 0 which is not less than -2(= x).

Symmetric Relation: A relation R defined on a set A is said to be symmetric, if
(x, y) ∈ R ⇒ (y, x) ∈ R, ∀ x, y ∈ A or
xRy ⇒ yRx, ∀ x, y ∈ R.

Transitive Relation: A relation R defined on a set A is said to be transitive, if
(x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R, ∀ x, y, z ∈ A
or xRy, yRz ⇒ xRz, ∀ x, y,z ∈ R.

Equivalence Relation: A relation R defined on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Equivalence Classes: Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets A, called partitions or sub-divisions of X satisfying

  • all elements of Ai are related to each other, for all i.
  • no element of Ai is related to any element of Aj, i ≠ j
  • Ai ∪ Aj = X and Ai ∩ Aj = 0, i ≠ j. The subsets Ai and Aj are called equivalence classes.

Function: Let X and Y be two non-empty sets. A function or mapping f from X into Y written as f : X → Y is a rule by which each element x ∈ X is associated to a unique element y ∈ Y. Then, f is said to be a function from X to Y.
The elements of X are called the domain of f and the elements of Y are called the codomain of f. The image of the element of X is called the range of X which is a subset of Y.
Note: Every function is a relation but every relation is not a function.

Types of Functions
One-one Function or Injective Function: A function f : X → Y is said to be a one-one function, if the images of distinct elements of x under f are distinct, i.e. f(x1) = f(x2 ) ⇔ x1 = x2, ∀ x1, x2 ∈ X
A function which is not one-one, is known as many-one function.

Onto Function or Surjective Function: A function f : X → Y is said to be onto function or a surjective function, if every element of Y is image of some element of set X under f, i.e. for every y ∈ y, there exists an element X in x such that f(x) = y.
In other words, a function is called an onto function, if its range is equal to the codomain.

Bijective or One-one and Onto Function: A function f : X → Y is said to be a bijective function if it is both one-one and onto.

Composition of Functions: Let f : X → Y and g : Y → Z be two functions. Then, composition of functions f and g is a function from X to Z and is denoted by fog and given by (fog) (x) = f[g(x)], ∀ x ∈ X.
Note
(i) In general, fog(x) ≠ gof(x).
(ii) In general, gof is one-one implies that f is one-one and gof is onto implies that g is onto.
(iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof) = (hog)of.

Invertible Function: A function f : X → Y is said to be invertible, if there exists a function g : Y → X such that gof = Ix and fog = Iy. The function g is called inverse of function f and is denoted by f-1.
Note
(i) To prove a function invertible, one should prove that, it is both one-one or onto, i.e. bijective.
(ii) If f : X → V and g : Y → Z are two invertible functions, then gof is also invertible with (gof)-1 = f-1og-1

Domain and Range of Some Useful Functions
Relations and Functions Class 12 Notes Maths Chapter 1
Binary Operation: A binary operation * on set X is a function * : X × X → X. It is denoted by a * b.

Commutative Binary Operation: A binary operation * on set X is said to be commutative, if a * b = b * a, ∀ a, b ∈ X.

Associative Binary Operation: A binary operation * on set X is said to be associative, if a * (b * c) = (a * b) * c, ∀ a, b, c ∈ X.
Note: For a binary operation, we can neglect the bracket in an associative property. But in the absence of associative property, we cannot neglect the bracket.

Identity Element: An element e ∈ X is said to be the identity element of a binary operation * on set X, if a * e = e * a = a, ∀ a ∈ X. Identity element is unique.
Note: Zero is an identity for the addition operation on R and one is an identity for the multiplication operation on R.

Invertible Element or Inverse: Let * : X × X → X be a binary operation and let e ∈ X be its identity element. An element a ∈ X is said to be invertible with respect to the operation *, if there exists an element b ∈ X such that a * b = b * a = e, ∀ b ∈ X. Element b is called inverse of element a and is denoted by a-1.
Note: Inverse of an element, if it exists, is unique.

Operation Table: When the number of elements in a set is small, then we can express a binary operation on the set through a table, called the operation table.

Solved example on symmetric relation on set:

1. A relation R is defined on the set Z by “a R b if a – b is divisible by 5” for a, b ∈ Z. Examine if R is a symmetric relation on Z.

Answer/Explanation

Solution:

Let a, b ∈ Z and aRb hold. Then a – b is divisible by 5 and therefore b – a is divisible by 5.

Thus, aRb ⇒ bRa and therefore R is symmetric.

2. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Examine if R is a symmetric relation on Z.

Answer/Explanation

Solution:

Let a, b ∈ Z and aRb holds i.e., 2a + 3b  is divisible by 5.

Now, 2b + 3a = 5a – 2a + 5b – 3b = 5(a + b) – (2a + 3b) is also divisible by 5 since 5(a + b) is divisible by 5 and (2a + 3b) is divisible by 5 ( given)

Therefore aRa holds for all a in Z i.e. R is reflexive.

3. Let R be a relation on Q, defined by R = {(a, b) : a, b ∈ Q and a – b ∈ Z}. Show that R is Symmetric relation.

Answer/Explanation

Solution:

Given R = {(a, b) : a, b ∈ Q, and a – b ∈ Z}.

Let ab ∈ R ⇒ (a – b) ∈ Z, i.e. (a – b) is an integer.

               ⇒ -(a – b) is an integer

               ⇒ (b – a) is an integer

               ⇒ (b, a) ∈ R

Thus, (a, b) ∈ R ⇒ (b, a) ∈ R

Therefore, R is symmetric.

4. Let m be given fixed positive integer.

Let R = {(a, a) : a, b  ∈ Z and (a – b) is divisible by m}.

Show that R is symmetric relation.

Answer/Explanation

Solution:

Given R = {(a, b) : a, b ∈ Z, and (a – b) is divisible by m}.

Let ab ∈ R . Then,

     ab ∈ R ⇒ (a – b) is divisible by m

               ⇒ -(a – b) is divisible by m

               ⇒ (b – a) is divisible by m

               ⇒ (b, a) ∈ R

Thus, (a, b) ∈ R ⇒ (b, a) ∈ R

Therefore, R is symmetric relation on set Z.

What is transitive relation on set?

Let A be a set in which the relation R defined.

R is said to be transitive, if

(a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R,

That is aRb and bRc ⇒ aRc where a, b, c ∈ A.

The relation is said to be non-transitive, if

(a, b) ∈ R and (b, c) ∈ R do not imply (a, c ) ∈ R.

For example, in the set A of natural numbers if the relation R be defined by ‘x less than y’ then

a < b and b < c imply a < c, that is, aRb and bRc ⇒ aRc.

Hence this relation is transitive.

Solved example of transitive relation on set:

1. Let k be given fixed positive integer.

Let R = {(a, b) : a, b  ∈ Z and (a – b) is divisible by k}.

Show that R is transitive relation.

Answer/Explanation

Solution:

Given R = {(a, b) : a, b ∈ Z, and (a – b) is divisible by k}.

Let (a, b) ∈ R and (b, c) ∈ R. Then

      (a, b) ∈ R and (b, c) ∈ Z

   ⇒ (a – b) is divisible by k and (b – c) is divisible by k.

   ⇒ {(a – b) + (b – c)} is divisible by k.

   ⇒ (a – c) is divisible by k.

   ⇒ (a, c) ∈ R.

Therefore, (a, b) ∈ R and (b, c) ∈ R   ⇒ (a, c) ∈ R.

So, R is transitive relation.

2. A relation ρ on the set N is given by “ρ = {(a, b) ∈ N × N : a is divisor of b}”. Examine whether ρ is transitive or not transitive relation on set N.

Answer/Explanation

Solution:

Given ρ = {(a, b) ∈ N × N : a is divisor of b}.

Let m, n, p ∈ N and (m, n) ∈ ρ and  (n, p ) ∈ ρ. Then

                                                 (m, n)  ∈ N × N and  (n, p ) ∈ N × N

                                              ⇒ m is divisor of n and n is divisor of p

                                              ⇒ m is divisor of p

                                              ⇒ (m, p) ∈ ρ

Therefore, (m, n) ∈ ρ and (n, p) ∈ ρ ⇒ (m, p) ∈ ρ.

So, R is transitive relation.

Equivalence Relation on Set

Equivalence relation on set is a relation which is reflexive, symmetric and transitive.

A relation R, defined in a set A, is said to be an equivalence relation if and only if

(i) R is reflexive, that is, aRa for all a ∈ A.

(ii) R is symmetric, that is, aRb ⇒ bRa for all a, b ∈ A.

(iii) R is transitive, that is aRb and bRc ⇒ aRc for all a, b, c ∈ A.

The relation defined by “x is equal to y” in the set A of real numbers is an equivalence relation.

Let A be a set of triangles in a plane. The relation R is defined as “x is similar to y, x, y ∈ A”.

We see that R is;

(i) Reflexive, for, every triangle is similar to itself.

(ii) Symmetric, for, if x be similar to y, then y is also similar to x.

(iii) Transitive, for, if x be similar to y and y be similar to z, then x is also similar to z.

Hence R is an equivalence relation.

A relation R in a set S is called a partial order relation if it satisfies the following conditions:

(i) aRa for all a∈ A, [Reflexivity]

(ii) aRb and bRa ⇒ a = b, [Anti-symmetry]

(iii) aRb and bRc ⇒ aRc, [Transitivity]

In the set of natural numbers, the relation R defined by “aRb if a divides b” is a partial order relation, since here R is reflexive, anti-symmetric and transitive.

A set, in which a partial order relation is defined, is called a partially ordered set or a poset.

Solved example on equivalence relation on set:

1. A relation R is defined on the set Z by “a R b if a – b is divisible by 5” for a, b ∈ Z. Examine if R is an equivalence relation on Z.

Answer/Explanation

Solution:

(i) Let a ∈ Z. Then a – a is divisible by 5. Therefore aRa holds for all a in Z and R is reflexive.

(ii) Let a, b ∈ Z and aRb hold. Then a – b is divisible by 5 and therefore b – a is divisible by 5.

Thus, aRb ⇒ bRa and therefore R is symmetric.

(iii) Let a, b, c ∈ Z and aRb, bRc both hold. Then a – b and b – c are both divisible by 5.

Therefore a – c = (a – b) + (b – c) is divisible by 5.

Thus, aRb and bRc  ⇒ aRc and therefore R is transitive.

Since R is reflexive, symmetric and transitive so, R is an equivalence relation on Z.

2. Let m be a positive integer. A relation R is defined on the set Z by “aRb if and only if a – b is divisible by m” for a, b ∈ Z. Show that R is an equivalence relation on set Z.

Answer/Explanation

Solution:

(i) Let a ∈ Z. Then a – a = 0, which is divisible by m

Therefore, aRa holds for all a ∈ Z.

Hence, R is reflexive.

(ii) Let a, b ∈ Z and aRb holds. Then a – b is divisible by m and therefore, b – a is also divisible by m.

Thus, aRb ⇒ bRa.

Hence, R is symmetric.

(iii) Let a, b, c ∈ Z and aRb, bRc both hold. Then a – b is divisible by m and b – c is also divisible by m. Therefore, a – c = (a – b) + (b – c) is divisible by m.

Thus,  aRb and bRc ⇒ aRc

Therefore, R is transitive.

Since, R is reflexive, symmetric and transitive so, R is an equivalence relation on set Z

3. Let S be the set of all lines in 3 dimensional space. A relation ρ is defined on S by “lρm if and only if l lies on the plane of m” for l, m ∈ S.

Examine if ρ is (i) reflexive, (ii) symmetric, (iii) transitive

Answer/Explanation

Solution:

(i) Reflexive: Let l ∈ S. Then l is coplanar with itself.

Therefore, lρl holds for all l in S.

Hence, ρ is reflexive

(ii)  Symmetric: Let l, m ∈ S and lρm holds. Then l lies on the plane of m.

Therefore, m lies on the plane of l. Thus, lρm ⇒  mρl and therefore ρ is symmetric.

(iii) Transitive: Let l, m, p ∈ S and lρm, mρp both hold. Then l lies on the plane of m and m lies on the plane of p. This does not always implies that l lies on the plane of p.

That is, lρm and mρp do not necessarily imply lρp.

Therefore, ρ is not transitive.

Since, R is reflexive and symmetric but not transitive so, R is not an equivalence relation on set Z

4. Give an example of a relation which is reflexive, but neither symmetric nor transitive.

Answer/Explanation

Solution:

Consider A = {l, 2, 3, 4, 5},

R= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (3, 4), (4, 5)} 

\(Here, (x,x)\in R,\forall x\in A\)

\(\therefore R \; is \; reflective\)

\(Now,\; (3,4)\in R,\;but (4,3)\notin R\)
\(therefore R \;is\; not \; symmetric\)
\(Also \; (3,4)\in R \; and\;(4,5)\in R, \;but (3,5)\notin R\)
\(therefore \;R\;is\;not\;transitive\)

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