Example 2 - Chapter 1 Class 12 Relation and Functions
Last updated at April 16, 2024 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
Example 6 Important
Example 2 You are here
Ex 1.1, 12 Important
Ex 1.1, 13
Ex 1.1, 11
Example 3
Ex 1.1, 14
Misc 3 Important
Example 19 Important
Example 18
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 2 Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation. R = {(T1, T2) : T1 is congruent to T2}. Check reflexive Since every triangle is congruent to itself Triangle T is congruent to triangle T ∴ (T , T) ∈ R So, R is reflexive, Check symmetric If T1 is congruent to T2 , then T2 is congruent to T1 So, if (T1, T2) ∈ R , then (T2, T1) ∈ R. Hence , R is symmetric. Check transitive If T1 is congruent to T2 , & T2 is congruent to T3 then T1 is congruent to T3 So, if (T1, T2), (T2, T3) ∈ R , then (T1, T3) ∈ R. So, R is transitive Since R is reflexive, symmetric & transitive. Therefore, R is an equivalence relation.
