Question
Question: Let R be the relation on the set R of all real numbers defined by \(aRb\ if\ \left| a-b \right|\le 1...
Let R be the relation on the set R of all real numbers defined by aRb if ∣a−b∣≤1. Then R is,
A. Reflexive and symmetric
B. Symmetric only
C. Transitive only
D. Anti – symmetric only
Solution
Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify if each relation holds or not and hence deduce the answer.
Complete step-by-step solution -
Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. (a,a)∈R while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we know different types of relations and we will check the given relation for these.
Now, we have a relation R as aRb if ∣a−b∣≤1.
Now, we will check for reflexive as,
aRa⇒∣a−a∣≤10≤1
Which is true for a∈R set of real numbers. Hence, R is reflexive.
Now, for symmetric we have if,
aRb⇒∣a−b∣≤1
Now, we have ∣a−b∣≤1
Or we can write,
∣−(b−a)∣≤1=∣b−a∣≤1
So, we have aRb⇒bRa. Hence, the given relation is symmetric also.
Now, for transitive we have if,
aRb⇒∣a−b∣≤1bRc⇒∣b−c∣≤1
Now, if we take a = 1.5, b = 2, c = 2.8.
So,
aRb⇒∣1.5−2∣=0.5≤1bRc⇒∣2−2.8∣=0.8≤1aRc⇒∣1.5−2.8∣=1.3≰1
Hence, the given relation is not transitive.
Hence, the correct option is (A) the given relation is reflexive and symmetric.
Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. Also these types of questions are solved easily by giving examples and counter examples. Also, we have to check the relation for reflexive, symmetric and transitive relation to check it for equivalence relation.