KCET 2014 Math Solved Paper

Given that aRb⇔|a−b|≥1
Now, aRa⇔|a−a|=0≤1
Therefore, R is reflexive
Again,aRb⇔|a−b|≤1
Then,bRa⇔|b−a|≤1
⇒|a−b|≤1, which is true.
Therefore, R is symmetric
Take a b = = 1 2 , . Then,|a−b|=|1−2|=1=1
Take b c = = 2 3 and . Then,|b−c|=|2−3|=1
But aRc |a−c|=|1−3|=2>1
Therefore, R not transitive
Hence, R is reflexive and symmetric but not transitive.