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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 aRbab1aRb \Leftrightarrow \left| {a - b} \right| \leqslant 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

Explanation

Solution

A relation R is said to be reflexive if every element of A should be related to itself i.e. (a,a)RaA\left( {a,a} \right) \in R\forall a \in A. A relation R is said to be symmetric if (a,b)R(b,a)R\left( {a,b} \right) \in R \Rightarrow \left( {b,a} \right) \in R
A relation R is said to be transitive if (a,b)R,(b,c)R(a,c)R\left( {a,b} \right) \in R,\left( {b,c} \right) \in R \Rightarrow \left( {a,c} \right) \in R.

Complete step by step answer:
Given to us a relation R defined on S such that aRbab1aRb \Leftrightarrow \left| {a - b} \right| \leqslant 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)RaA\left( {a,a} \right) \in R\forall a \in A.
Now, we can write the given relation as 1ab1 - 1 \leqslant a - b \leqslant 1.
In given inequality if aRa \in R then bRb \in 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\left( {a,b} \right) \in R \Rightarrow \left( {b,a} \right) \in R , that is aRbbRa  (a,b)RaRb \Rightarrow bRa\;\forall \left( {a,b} \right) \in R
From the given relation, 1ab1 - 1 \leqslant a - b \leqslant 1 which also gives 1+ba1+b - 1 + b \leqslant a \leqslant 1 + b or 1+ab1+a - 1 + a \leqslant b \leqslant 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\left( {a,b} \right) \in R,\left( {b,c} \right) \in R \Rightarrow \left( {a,c} \right) \in R
In the given inequality 1ab1 - 1 \leqslant a - b \leqslant 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\left( {a,b} \right) \in R,\left( {b,c} \right) \in R \Rightarrow \left( {a,c} \right) \in 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:
\Leftrightarrow - If and only if, \in - Belongs to, \forall - for all. It is to be noted that these notations are important and to be used while representing a relation between elements/sets.