Let
S = {a, b, c.......}
be any set then the relation R is a subset of the product set S x S.
If R contains all ordered pairs of the form (a, a) in S x S, then R is called reflexive. In a reflective relation a is related to itself.
For example, is equal to is a reflective relation for a = a is true.
If (a, b) ∈ R ==> (b, a) ∈ R for every a, b ∈ S then R is called symmetric.
For example, a = b ==> b = a. Hence the relation is equal to a symmetric relation.
If (a, b) ∈ R and (b, c) ∈ R ==> (a, c) ==> R for every a, b , c ∈ S then R is called transitive.
For example, a = b, b = c ==> a = c
Hence the relation is equal to transitive relation.
Note :
A relation which is reflexive, symmetric and transitive is called an equivalence relation or simply an equivalence is equal to is an equivalence relation.
The relation I = {(a, a) : a ∈ A is called the identity relation on A.
Let A = {1, 2, 3} then I = {(1, 1) (2, 2) (3, 3)}
If R be a relation on A, then the relation R^{-1} on A, defined by
R^{-1} = {(b, a) : (a, b) ∈ R is called an inverse relation on A.
Clearly, domain of R^{-1 }= Range of R
Range of R^{-1 }= Domain of R
Problem 1 :
Let
𝑆 = {(𝑎, 𝑎), (𝑎, 𝑐), (𝑏, 𝑏), (𝑏,𝑒), (𝑐, 𝑎), (𝑐, 𝑐), (𝑑, 𝑑), (𝑒, 𝑏), (𝑒,𝑒)}
be a relation defined on the set 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑,𝑒}
Check if the relation 𝑆 is an equivalence relation
Solution :
Reflexive :
Symmetric :
𝑆 = {(𝑎, 𝑎), (𝑎, 𝑐), (𝑏, 𝑏), (𝑏,𝑒), (𝑐, 𝑎), (𝑐, 𝑐), (𝑑, 𝑑), (𝑒, 𝑏), (𝑒,𝑒)}
S is a symmetric relation.
Transitive :
S is transitive relation. So, the given relation S is equivalence relation.
Problem 2 :
Check the relation is whether, reflexive, symmetric or transitive.
Let A = {1, 2, 3}
R_{1} = { (1, 1) (2, 2) (3, 3) (1, 2) }
Solution :
Reflexive :
Set A contains 3 elements 1, 2 and 3.
(1, 1)
(2, 2) (3, 3) ∈ A
The relation is reflexive.
Symmetric :
(1, 2) ∈ A, then (2, 1) does not exists. So, the relation is not symmetric.
Transitive :
Then the relation is reflexive and transitive but not symmetric.
Problem 3 :
Check the relation is whether, reflexive, symmetric or transitive.
Let A = {1, 2, 3}
R_{2} = { (1, 1) (2, 2) (1, 2) (2, 1) }
Solution :
Reflexive :
Set A contains 3 elements 1, 2 and 3.
(1, 1), (2, 2) ∈ A but (3, 3) does not exists in A. So, it is not reflexive.
Symmetric :
Then the relation A is not symmetric.
Transitive :
Problem 4 :
Check the relation is whether, reflexive, symmetric or transitive.
Let A = {1, 2, 3}
R_{3} = { (1, 1) (2, 2) (3, 3) (1, 2) (2, 1)(2, 3) (3, 2) }
Solution :
Reflexive :
Set A contains 3 elements 1, 2 and 3.
(1, 1), (2, 2), (3, 3) does not exists in A. So, it is reflexive.
Symmetric :
Then the relation A is symmetric.
Transitive :
The function is symmetric and reflexive but not transitive.
Problem 5 :
Let A = {1, 2, 3} and R = {(1, 2) (2, 2)(3, 1)(3, 2)}
From the given relation, find domain and range of function and inverse function.
Solution :
Set of first values (inputs) from the ordered pairs is domain.
Set of second values (outputs) from the ordered pairs is range.
R = {(1, 2) (2, 2)(3, 1)(3, 2)}
Domain of R = {1, 2, 3}
Range of R = {1, 2}
Domain of R = Range of R^{-1}
Range of R = Domain of R^{-1}
Domain of R^{-1} = {1, 2}
Range of R^{-1} = {1, 2, 3}
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