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Example 5 - Chapter 1 Class 12 Relation and Functions (Term 1)
Last updated at May 29, 2018 by Teachoo
To prove relation reflexive, transitive, symmetric and equivalent
What is reflexive, symmetric, transitive relation?
Example 4 Important
Ex 1.1, 6
Ex 1.1, 15 (MCQ) Important
Ex 1.1, 7
Ex 1.1, 1 (i)
Ex 1.1, 2
Ex 1.1, 3
Ex 1.1, 4
Ex 1.1, 5 Important
Ex 1.1, 10 (i)
Ex 1.1, 8
Ex 1.1, 9 (i)
Example 5 You are here
Example 6 Important
Example 2
Ex 1.1, 12 Important
Ex 1.1, 13
Ex 1.1, 11
Example 3
Ex 1.1, 14
Misc. 8 Important
Example 42 Important
Example 41
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
Example 5, Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides a b} is an equivalence relation. R = {(a, b) : 2 divides a b} Check reflexive Since a a = 0 & 2 divides 0 , eg: 0 2 = 0 2 divides a a (a, a) R, R is reflexive. Check symmetric If 2 divides a b , then 2 divides (a b) i.e. b a Hence, If (a, b) R, then (b, a) R R is symmetric Check transitive If 2 divides (a b) , & 2 divides (b c) , So, 2 divides (a b) + (b c) also So, 2 divides (a c) If (a, b) R and (b, c) R, then (a, c) R Therefore, R is transitive. Thus, R is an equivalence relation in Z.
