📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 1table

Chapter 1 · Part 18

Chapter 1: RELATIONS AND FUNCTIONS · MATHEMATCS PART-1

⇒ uy = vx and hence ( u , v ) R ( x , y ). This shows that R is symmetric. Similarly, ( x , y ) R ( u , v ) and ( u , v ) R ( a , b ) ⇒ xv = yu and ub = va ⇒ a a xv yu u u ⇒ b a xv yu v u ⇒ xb = ya and hence ( x , y ) R ( a , b ). Thus, R is transitive.

Thus, R is an equivalence relation. Example Let X = { , , , , , , , , }. Let R be a relation in X given by R = {( x , y ) : x – y is divisible by } and R be another relation on X given by R = {( x , y ): { x , y } ⊂ { , , }} or { x , y } ⊂ { , , } or { x , y } ⊂ { , , }}. Show that R = R .

Solution Note that the characteristic of sets { , , }, { , , } and { , , } is that difference between any two elements of these sets is a multiple of . Therefore, ( x , y ) ∈ R ⇒ x – y is a multiple of ⇒ { x , y } ⊂ { , , } or { x , y } ⊂ { , , } or { x , y } ⊂ { , , } ⇒ ( x , y ) ∈ R . Hence, R ⊂ R . Similarly, { x , y } ∈ R ⇒ { x , y } ⊂ { , , } or { x , y } ⊂ { , , } or { x , y } ⊂ { , , } ⇒ x – y is divisible by ⇒ { x , y } ∈ R .

This shows that R

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