Ex 1.1, 3
Last updated at April 16, 2024 by Teachoo
To prove relation reflexive, transitive, symmetric and equivalent
What is reflexive, symmetric, transitive relation?
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Chapter 1 Class 12 Relation and Functions
Concept wise
- Relations - Definition
- Empty and Universal Relation
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To prove relation reflexive, transitive, symmetric and equivalent
- Finding number of relations
- Function - Definition
- To prove one-one & onto (injective, surjective, bijective)
- Composite functions
- Composite functions and one-one onto
- Finding Inverse
- Inverse of function: Proof questions
- Binary Operations - Definition
- Whether binary commutative/associative or not
- Binary operations: Identity element
- Binary operations: Inverse
Transcript
Ex 1.1, 3 Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. R = {(a, b):b = a + 1} where a, b ∈ {1, 2, 3, 4, 5, 6} Check reflexive If the relation is reflexive, then (a, a) ∈ R i.e. a = a + 1 Since a = a + 1 cannot be possible for any value of a, the given relation is not reflexive. Check symmetric To check whether symmetric or not, If (a , b) ∈ R, then (b , a) ∈ R i.e., if b = a + 1, then a = b + 1 Since a = b + 1 is not true for all values of a & b. Hence, the given relation is not symmetric Check transitive To check whether transitive or not, If (a,b) ∈ R & (b,c) ∈ R , then (a,c) ∈ R i.e., if b = a + 1, & c = b + 1 then c = a + 1 Since, b = a + 1 & c = b + 1 Putting value of b in (2) c = (a + 1) + 1 c = a + 2 Hence, c ≠ a + 1 Hence, the given relation it is not transitive
