    Question

# For real numbers x and y, we define xRy iff x−y+√5 is an irrational number. The relation R is

A
reflexive
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B
symmetric
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C
transitive
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D
None of these
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Solution

## The correct option is A reflexivex,y∈R f(x,y)=x−y+√5 is an irrational number Case1: Reflexive f(x,x)=x−x+√5=√5 is an irrational number f(y,y)=y−y+√5=√5 is an irrational number ∴f(x,y) is a reflexive relation Case2: Symmetric if (x,y) is an irrational number holds f(y,x) an irrational number Let x=√5,y=1 f(x,y)=√5−1+√5=2√5−1 is an irrational number f(y,x)=1−√5+√5=1 is not an irrational number ∴f(x,y) is not a symmetric relation Case3: Transitive relation if if (x,y) is an irrational number and f(y,z) is an irrational number holds f(x,z) an irrational number Let x=1,y=2√5 f(x,y)=1−2√5+√5=1−√5 is an irrational number f(y,z)=y−z+√5 can be irrational number if z=√5 ∴f(x,z)=x−z+√5⇒f(1,√5)=1 is not an irrational number ∴f(x,y) is not a transitive relation  Suggest Corrections  0      Similar questions  Explore more