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Question

Let ρ1 and ρ2 be two equivalence relations defined on a non-void set S. Then

A
both ρ1ρ2 and ρ1ρ2 are equivalence relations.
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B
ρ1ρ2 is equivalence relation but ρ1ρ2 is not so.
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C
ρ1ρ2 is equivalence relation but ρ1ρ2 is not so.
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D
neither ρ1ρ2 nor ρ1ρ2 is equivalence relation
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Solution

The correct option is B ρ1ρ2 is equivalence relation but ρ1ρ2 is not so.
Given ρ1,ρ2 are equivalence relations on S.
ρ1,ρ2 are reflexive, symmetric and transitive.

Reflexive:
Let xS
(x,x)ρ1 and (x,x)ρ2
(x,x)ρ1ρ2
ρ1ρ2 is reflexive.

Symmetric:
Let (x,y)ρ1ρ2
We have to show (y,x)ρ1ρ2
(x,y)ρ1ρ2
(x,y)ρ1 and (x,y)ρ2
(y,x)ρ1 and (y,x)ρ2
(y,x)ρ1ρ2
ρ1ρ2 is symmetric.

Transitive:
Let (x,y),(y,z)ρ1ρ2
(x,y),(y,z)ρ1 and (x,y),(y,z)ρ2
(x,z)ρ1 and (x,z)ρ2
(x,z)ρ1ρ2
ρ1ρ2 is transitive.

Therefore, ρ1ρ2 is equivalence relation.

ρ1ρ2 is always reflexive and symmetric but not transitive.
e.g. Let S={1,2,3}
ρ1={(1,1),(2,2),(3,3),(1,2),(2,1)}
ρ2={(1,1),(2,2),(3,3),(2,3),(3,2)}

ρ1,ρ2 is equivalence relation.
But ρ1ρ2 is not transitive as (1,2),(2,3)ρ1ρ2 but (1,3)ρ1ρ2

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