Question
Question: Let S be the set of all real numbers. A relation R has been defined on S by \(aRb \Leftrightarrow \l...
Let S be the set of all real numbers. A relation R has been defined on S by aRb⇔∣a−b∣⩽1 , then R is which of the following?
A) Reflexive and transitive but not symmetric.
B) An equivalence relation.
C) Symmetric and transitive but not reflective.
D) Reflexive
Solution
A relation R is said to be reflexive if every element of A should be related to itself i.e. (a,a)∈R∀a∈A. A relation R is said to be symmetric if (a,b)∈R⇒(b,a)∈R
A relation R is said to be transitive if (a,b)∈R,(b,c)∈R⇒(a,c)∈R.
Complete step by step answer:
Given to us a relation R defined on S such that aRb⇔∣a−b∣⩽1 defined on a set of real numbers.
For a relation R to be reflexive every element of A should be related to itself i.e. (a,a)∈R∀a∈A.
Now, we can write the given relation as −1⩽a−b⩽1.
In given inequality if a∈R then b∈R and this satisfies the above reflexive relation property i.e. every element that satisfies a also satisfies b.
Therefore our given relation is reflexive.
A relation R is said to be symmetric if (a,b)∈R⇒(b,a)∈R , that is aRb⇒bRa∀(a,b)∈R
From the given relation, −1⩽a−b⩽1 which also gives −1+b⩽a⩽1+b or −1+a⩽b⩽1+a
From this inequality, it is clear to us that for every value of a there exist multiple values of b and vice versa. So the given relation is not symmetric.
A relation R is said to be transitive if (a,b)∈R,(b,c)∈R⇒(a,c)∈R
In the given inequality −1⩽a−b⩽1 for every value of a there exists multiple values of b and these values become c.
Hence, for every (a,b)∈R,(b,c)∈R⇒(a,c)∈R
Therefore, the relation R is transitive.
Hence the given relation R is reflexive and transitive but not symmetric i.e. option (A).
Note: Full forms of the notations used in the above solution:
⇔ - If and only if, ∈ - Belongs to, ∀ - for all. It is to be noted that these notations are important and to be used while representing a relation between elements/sets.