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Types of Relation in mathematics help us to understand the connection between two sets. This connection can be referred to as the relation between the two sets of elements. The Types of Relations, in Set Theory, are several. They include:
- Empty Relation
- Universal Relation
- Identity Relation
- Inverse Relation
- Reflexive Relation
- Symmetric Relation
- Transitive Relation
- Equivalence Relation
A Relation can be defined as the relationship between two or more set values. Assume that x and y are the two sets of ordered pairs. Here, consider that set x has a relation with set y. Thus, in that case, the values of set x are known as Domain, while the values of set y are known as range.
For instance,
In case of Ordered Pairs = {(1,2),(-3,4),(5,6),(-7,8),(9,2)}
- Domain = {-7,-3,1,5,9}
- Range = {2,4,6,8}
Read More: Operations on Matrices
| Table of Content |
Key Terms: Relations, Functions, Set-Builder Form, Roster Form, Sets, Set Theory, Cartesian product, Arrow Representation, Trivial Relations
What is a Relation?
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A relation, in mathematics, can be defined as:
| “The relationship between two different information sets.” |
- In case two sets are assumed, the relation that is going to be established between them is the connection between elements of two or more non-empty sets.
- Simply, a set of ordered pairs can be expressed as a Relation.
- The sets, in this case, must be non-empty.
- A subset of the Cartesian product is also known to form a relation R.
- A relation can be classified either by the Roster method or by the Set-builder method.
Consider that A and B be are two sets. Thus, A = {2, 5, 7, 8, 10, 13} and B = {1, 2, 3, 4, 5}. Hence, it can be represented as:
→ R = {(x, y): x = 4y – 3, x ∈ A and y ∈ B} (Set-builder form)
→ R = {(5, 2), (10, 3), (13, 4)} (Roster form)
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Representation of Relations
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Relations can be represented by three methods:
Roster Form
All elements of given sets are arranged in one set and separated by commas within the brackets.
Set-Builder Form
In case of Set-Builder form:
- It is used when a set has infinite numbers.
- It’s used with integers and real numbers., etc.
- It’s a representation of a set that forms an equation.
Arrow-Diagram
The relation between the two sets is shown by the arrows.
Read More: Binary Operations
Types of Relations
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There are several types of Relations, namely:
Empty Relation
An empty relation (also called, void relation) is a type of relation where there is no relation between any elements of a set. For instance, consider set A = {1, 2, 3}. Here, one of the void relations is R = {x, y} where |x – y| = 8. Thus, for empty relation,
→ R = φ ⊂ A × A
Universal Relation
A Universal (also called Full Relation) is a kind of relation wherein every element of a set is related to one another.
- A universal Relation can be expressed by, R = A × A
- It’s also called Full relation because every element of Set 1 is related to Set 2.
- For example: {1, 2,3, 4...} is Set 1, consisting of Natural Numbers. While, {0, 1, 2, 3, 4, 5, …} is Set 2, consisting of Whole numbers. Now, the relation between Set 1 and Set 2 is Universal since the numbers of both sets are directly linked or are the same.
Transitive Relation
A Transitive relation represents a relationship between three sets.
- In case of a transitive relation, if (x, y) ∈ R, (y, z) ∈ R, then (x, z) ∈ R.
- Thus, for a transitive relation, aRb and bRc ⇒ aRc ∀ a, b, c ∈ A.
Symmetric Relation
In case of a Symmetric Relation,if a = b is considered true, then b = a is also equally true as well. Simply, a relation R is symmetric only if (b, a) ∈ R is considered to be true when (a, b) ∈ R. It is a type of Binary Relation.
Thus, for a Symmetric Relation,
aRb ⇒ bRa, ∀ a, b ∈ A
Reflexive Relation
For a Reflexive Relation, every element is known to map itself. For Example, consider A = {1,4}.
Here, R= {(1,1), (4,4), (1,4), (4,1)}
Hence, the Reflexive Relation is (a, a) ∈ R (reverse A) a ∈ A.
Inverse Relation
Inverse relation can be found when a set contains elements which are inverse pairs of some other set.
For example, consider a set A = {(a, b), (c, d)}, and the inverse relation is going to be R-1 = {(b, a), (d, c)}. Thus, for inverse relation, we can say:
→ R-1 = {(b, a): (a, b) ∈ R}
Equivalence Relation
To be an Equivalence Relation, a relation must reflect the properties of Symmetry, Transitive, and Reflexive relations.
Antisymmetric Relation
No pair of distinct elements of Sets should be equal to R if a =b, in the case of an Antisymmetric Relation.
Identity Relation
In the case of an Identity Relation, every element of a set is only related to itself. For Example: {1,4,7}
Thus, Sets of ordered pairs = {(1,1), (4,4), (7,7)}.
Hence, for an Identity Relation,
I = {(a, a), a ∈ A}
Note:
- Empty or Universal Relations are known as Trivial Relations.
- Every Function is a Relation, but not Every Relation is a function.
Relations and Functions Detailed video explanation
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Types of Relation Representation
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The representation of the types of relations are:
| Relation Type | Condition To Be Followed |
|---|---|
| Transitive Relation | aRb and bRc → aRc ∀ a, b, c ∈ A |
| Universal Relation | R = A × A |
| Symmetric Relation | aRb → bRa, ∀ a, b ∈ A |
| Identity Relation | I = {(a, a), a ∈ A} |
| Inverse Relation | R-1 = {(b, a): (a, b) ∈ R} |
| Reflexive Relation | (a, a) ∈ R |
| Empty Relation | R = φ ⊂ A × A |
Things to Remember
- There are many types of Relations, such as Empty Relations, Universal Relations, Identity Relations, Inverse Relations, Reflexive Relations, Symmetric Relations, Transitive Relations and Equivalence Relations.
- A Relation is simply a relationship between two or more set values.
- Relations can be represented in three different ways, Roster Form, Set-Builder Form and Arrow-Diagram.
- In case of Relations and Functions, every function is a relation, but not every relation is a function.
- Empty or Universal Relations are also known by the name of Trivial Relations.
Read Also: Relation and Function
Previous Year Questions
- If a, b and c ∈ N which one of the following is not true… [KCET 2006]
- Consider the following lists… [AP EAPCET]
- If a + π/2 < 2tan−1… [KCET 2019]
- If A={x|x ∈ N, x ≤ 5},B = {x|x ∈ Z, x2 − 5x + 6 = 0}… [KCET 2019]
- f : R → R and g : [0, ∞) → R is defined by… [KCET 2019]
- cos[2sin−13/4 + cos−13/4] =… [KCET 2019]
- On the set of positive rationals, a binary operation… [KCET 2019]
- A class has 175 students… [BITAT 2013]
- If A and B are finite sets and… [KCET 2017]
Sample Questions
Ques: Determine the domain of set of ordered pairs given below:
{(3, 7), (4, -3), (1, 5), (-10, 6)} (1 mark)
Ans: The domain is known as the first component in the ordered pairs.
Thus, the Domain is = { 3, 4, 1, -10}
Ques: Define Reflexive Relation. (2 marks)
Ans: In case of a Reflexive Relation, every element is seen to map itself. For Example, assume A = {1,4}.
Thus,
R = {(1,1), (4,4), (1,4), (4,1)}
Therefore, the Reflexive Relation is (a, a) ∈ R (reverse A) a ∈ A.
Ques: Assume that A = {1, 3, 5, 7} and B = {p, q, r}. Now, consider that R be a relation from A into B expressed by R = {(1, p), (3, r), (5, q), (7, p), (7, q)}. Thus, determine the domain and range of R. (1 mark)
Ans: Here,
Domain = {1, 3, 5, 7}
And, Range = {p, r, q}
Ques: Consider that A = {3, 4, 5, 6} and B = {1, 2, 3, 4, 5, 6}. Now, let R = {(a, b) : a ∈ A, b ∈ B and a < b}. Thus, represent R in the roster form and also determine its domain and range. (2 marks)
Ans: Representing in Roster form:
R ={(3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6)}
And the domain and range are:
- Domain = {3, 4, 5}
- Range = {4, 5, 6}
Ques: What are Binary Operations? (5 marks)
Ans: Binary Operations are the calculations that are performed to get something new or productive out of two sets of numbers. There are four major types of
Binary Operations:
- Addition
- Subtraction
- Multiplication
- Division
Rules of Addition:
- 0 + 0 =0
- 0 + 1 =1
- 1 + 0 =1
- 1 + 1 = 10
Rules of Subtraction:
- 0 - 0 = 0
- 1 - 0 = 1
- 1 - 1 = 0
- 0 - 1 = 1 (Borrow 1)
Rules of Multiplication:
- 0 x 0 =0
- 0 x 1 = 0
- 1 x 0 = 0
- 1 x 1 = 1
Rules of Division:
- 1/1 = 1
- 1/1 = Meaningless
- 0/1 = 0
- 0/0 = Meaningless (No point in dividing zero as we can’t get any value out of it.)
Ques: Assume that A = {3, 4, 5, 6} and B = {x, y, z}. Now let R be a relation from A into B represented by R = {(3, x), (4, y), (5, z), (6, x), (6, y)}. Thus, determine the domain and range of R. (2 marks)
Ans: As the given relation, R = {(3, x), (4, y), (5, z), (6, x), (6, y)}
The domain can be defined as the first component of the ordered pairs in the Relation R. While, Range can be expressed as the second component of the ordered pairs.
Thus,
- Domain of the Relation = {3, 4, 5, 6}
- Range of the Relation = {x, y, z}
Ques. Is there any difference between Relation and Functions?(2 marks)
Ans. The difference between Relations and Functions are as follows:
| Relations | Functions |
|---|---|
| It relates to the elements of the given Sets and makes the Output out of it. | It relates to the Input and Output of Sets. |
| It’s denoted by Capital R. | It’s denoted by Capital F or Small f. |
| All relations are not functions. | All functions are Relations. |
Ques: Illustrate that relation R on R defined as R = [(a, b): a ≤ b], is reflexive, and transitive but not symmetric. (2019) (3 marks)
Ans: As per the question,
(a, a) : a ≤ a is true. Thus, ∀a∈R.
Therefore, it is a reflexive relation.
(a,b) ⇒ a ≤ b, b
(b,a) ∈ R
Thus, it can e seen that R is not symmetric
i.e. (1,3) ⇒ 1≤3. However, (3,1) 3 ≤ 1
(a, b) ⇒ a ≤ b, (b, c) ⇒ b ≤ c ⇒ a ≤ c = (a, c) ∈ R
Hence R is transitive.
Ques: Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. (2019 outside Delhi) (4 marks)
Ans: Assume that A = {1, 2, 3, 4, 5, 6}.
A relation R can be expressed on set A as:
R = {(a, b): b = a + 1}
Thus, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
It can be determined that (a, a) ∉ R, where a ∈ A.
An example include,
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
Thus, R is not reflexive.
It can be seen that (1, 2) ∈ R, but (2, 1) ∉ R.
Hence, R is not symmetric.
Now, (1, 2), (2, 3) ∈ R
However,(1, 3) ∉ R
Thus, R is not transitive
Therefore, R is not reflexive, symmetric, or transitive.
Thus, the relation is not transitive as well.
Ques: Let A = {1,2,3,...,9} and R be the relation in A*A defined by (a,b) R (c,d) if a + d = b + c for (a,b), (c,d) in A*A. Prove that R is an equivalence relation. Also obtain the equivalence class [(2,5)]. (2014) (5 marks)
Solution:
A = {1,2,3,...9}
R in A*A
(a,b) R (c,d) if (a,b)(c,d) A A
A+b = b+c
Considering (a,b) R (a,b) (a,b) A*A
a+b = b+a
Therefore we can say that R is reflexive.
Considering (a,b) R (c,d) given by (a,b) (c,d) A*A
a+d = b+c
(c,d) R (a,b)
Therefore we can say that R is symmetric.
Let (a,b) R (c,d) and (c,d) R (e,f)
(a,b), (c,d), (e,f), A*A
a+b = b+c and c+f = d+e
a+b = b+c
a- c = b-d …(i)
c+f = d+e …(ii)
When we add (i) and (ii)
a - c + c + f = b - d + d + e
a+ f = b + e
(a,b) R (e,f)
R is transitive
R is an equivalence relation.
We can select from set A = {1,2,3,...9}
a and b such as 2+b = 5+a
Therefore b = a+3
Considering (1,4)
(2,5) R (1,4) 2+4 = 5+1
[(2,5) = (1,4),(2,5),(3,6),(4,7),(5,8),(6,9)] can be considered as the equivalent class under the relation R.
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