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, a) : a, b ∈ Z and (a – b) is divisible by k}.
Show that R is transitive relation.
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) ∈ R
⇒ (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.
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) ∈ ρ and (n, p ) ∈ ρ
⇒ 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.
● Set Theory
● Sets
● Subset
● Practice Test on Sets and Subsets
● Problems on Operation on Sets
● Practice Test on Operations on Sets
● Venn Diagrams in Different Situations
● Relationship in Sets using Venn Diagram
● Practice Test on Venn Diagrams
8th Grade Math Practice
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