Relation:
The relation is a linear operation that establishes a relationship between the elements of two sets according to some definite rule of relationship.
R: {(a, b)| (a, b) ∈ A × B and a R b}
Function:
Let A and B be two sets, and let a rule or manner or correspondence ‘f’ exist that associates each element of A with a unique element in B. Then, f is called a mapping or function from A to B.
f: A → B or
Which reads f maps A to B or f is a function from A to B.
A Relation R from set A to set B is established as a subset of the Cartesian product A × B. In other words, relation R consists of ordered pairs where the first element belongs to set A and the second element belongs to set B. Therefore, we can say that R is a relation from A to B, represented as .
Total number of relations if n(A) = m and n(B) = n is 2mn.
Let R be a relation from a set A to set B. The domain of R comprises all the first elements of the ordered pairs in R.
The Range of R is the set of all second elements of the ordered pairs.
Thus, Domain = {a: (a, b) ∈ R}
and Range = {b: (a, b) ∈ R}
It is clear from the above definition that the domain of a relation from set A to B is a subset of A, and its Range is a subset of B.
Different types of relations are:
Let us discuss all types of relations one by one.
A relation R on set A is termed a Universal Relation if every element of A is related to every other element within set A.
i.e., R = A × A.
A relation R on set A is called Void/ Empty Relation if no element of A is related to any element of A.
i.e.,
A relation defined on a set A is called an Identity Relation if each element of A is related to only and only itself.
i.e., I = {(a, a): ∀ a ∈ A}.
A relation defined on a set A is called Reflexive Relation if each element of A is related to itself.
I.e., (a, a) ∈ R ∀ a ∈ A.
A relation defined on a set A is said to be Symmetric Relation if
(a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A.
i.e., a R b ⇒ b R a ∀ a, b, ∈ A.
Let A be any set. A relation R on set A is said to be a Transitive Relation if
(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
i.e., a R b, b R c ⇒ a R c ∀ a, b, c ∈ A.
A relation R on a set A is said to be an Equivalence Relation if:
A function f from set A to set B is defined as a relation where each element in A maps to exactly one element in B. This means that for every input from A, B has a unique output. We can represent this function as f: A → B. If (a, b) ∈ f, then f(a) = b, where b represents the image of a under f, while a is referred to as the preimage of b under f.
Let us discuss these functions in detail
Consider the set R of real numbers. Define the real-valued function f: R → R as y = f(x) = x for all x ∈ R. This type of function is known as the identity function, denoted by f(x) = x. Both domain and range of f are R. Its graph is a straight line that passes through the origin.
The function f: R → R can be expressed as y = f(x) = c for all x ∈ R, where c is a constant. In this function, the domain of f comprises all real numbers, denoted as R, and its range is the singleton set {c}.
A polynomial function f: R → R is defined such that for each x in R, y = f(x) = a0 + a1x + a2x2+ …. + anxn, where n represents a non-negative integer, while a0, a1, a2 , … , an are real numbers.
The modulus function f: R → R, defined as f(x) = |x| ∀ x ∈ R, is called the absolute value function. When x is non-negative, f(x) equals x, but when x is negative, f(x) becomes the negative of x, i.e., f(x)=
The function f: R → R is defined by
Is called the signum function.
The function f: R → R defined as f(x) = [x], where x ∈ R, represents the greatest integer function. Here, the notation [x] denotes the greatest integer less than or equal to x.
and so on.
Question 1. In the set N × N consider the relation R defined as (a, b) R (c, d) ⇔ ad = bc; a, b, c, d ∈ N
Show that this is an equivalence relation.
Solution:
For proving the relation R on N × N to be an equivalence relation, we will prove that the relation R is reflexive, symmetric and transitive.
Reflexive:
Let x, y ∈ N, then
⇒ (x, y) ∈ N × N ⇒ xy = yx, by commutativity in N
⇒ (x, y) R (x, y) ∀ (x, y) ∈ N × N
∴ R is reflexive.
Symmetric:
Let (x, y) R (p, q) where x, y, p, q ∈ N, then
(x, y) R (p, q) ⇒ xq = yp ⇒ yp = xq
⇒ py = qx [by commutativity in N]
⇒ (p, q) R (x, y)
∴R is symmetric.
Transitive:
Let (x1, y1) R (x2, y2) and (x2, y2) R (x3, y3) where
⇒ (x1, y1) (x2, y2), (x3, y3) ∈ N × N, then
⇒ (x1, y1) R (x2, y2) ⇒ x1 y2 = y1 x2 ...(1)
∴ (x2, y2) R (x3, y3) ⇒ x2 y3 = y2 x3 ...(2)
putting the value of x2 from (2) into (1)
⇒ x1 y3 = y1 x3 ...(3)
⇒ (x1, y1) R (x3, y3)
Therefore, R is transitive.
Hence the relation defined on N × N is an equivalence relation.
Question 2. Find the domain of function y=\sqrt{5-2 x} .
Solution:
5 – 2x > 0 ⇒ x \leq \frac{5}{2}
∴ Domain is (–∞, 5/2]
Question 3. Find the range of function f(x) = 3 – cos x.
Solution:
f(x) = 3 – cos x
⇒ –1 ≤ cos x ≤ 1
⇒ 2 ≤ 3 – cos x ≤ 4
∴ Range of f(x) = [2, 4]
Question 4. Find the domain where [.] denotes greatest integer function.
Solution:
[|x|-5]|>11
So, [|x|-5]>11 or [|x|-5]<-11
[|x|]<-6
|x|<-6 (Not Possible)
⇒ x ≤ -17 or x ≥ 17
So, x ∈ (–∞, –17] ∪ [17, ∞)
(Session 2025 - 26)