
Let \[A = \left\{ {p,q,r} \right\}\] which of the following is an equivalent relation A?
A. \[{R_1} = \left\{ {\left( {p,q} \right),\left( {q,r} \right),\left( {p,r} \right),\left( {p,q} \right)} \right\}\]
B. \[{R_2} = \left\{ {\left( {r,q} \right),\left( {r,p} \right),\left( {r,r} \right),\left( {q,q} \right)} \right\}\]
C. \[{R_3} = \left\{ {\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right),\left( {p,q} \right)} \right\}\]
D. None of these
Answer
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Hint: A relation is equivalent if it satisfy all relations that are reflexive, symmetric and transitive. We will check whether the relations follow reflexive, symmetric and transitive.
Formula Used: Reflexive: R said to reflexive on set A, if \[\left( {a,a} \right) \in R\] for every \[a \in A\].
Symmetric: R said to symmetricon set A, if \[\left( {a,b} \right) \in R\] implies that \[\left( {b,a} \right) \in R\] where \[a,b \in A\].
Transitive: R said to transitive on set A, if \[\left( {a,b} \right),\left( {b,c} \right) \in R\] implies that \[\left( {a,c} \right) \in R\] where \[a,b,c \in A\].
Complete step by step solution: For option A:Given set is \[A = \left\{ {p,q,r} \right\}\].
Given relation is \[{R_1} = \left\{ {\left( {p,q} \right),\left( {q,r} \right),\left( {p,r} \right),\left( {p,q} \right)} \right\}\]
p, q, and r are the elements of A.
Since \[\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right) \notin {R_1}\], thus \[{R_1}\] is not reflexive.
Therefore \[{R_1}\] is not an equivalent relation.
For option B:
Given relation is \[{R_2} = \left\{ {\left( {r,q} \right),\left( {r,p} \right),\left( {r,r} \right),\left( {q,q} \right)} \right\}\].
Since \[\left( {p,p} \right) \notin {R_2}\], thus it is not reflexive.
Thus \[{R_2}\] is not an equivalent relation.
For option C:
Given relation is \[{R_3} = \left\{ {\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right),\left( {p,q} \right)} \right\}\].
Since \[\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right)\] belongs to \[{R_3}\]. \[{R_3}\] is reflexive.
\[\left( {p,q} \right) \in {R_3}\] but \[\left( {q,p} \right) \notin {R_3}\]. Thus \[{R_3}\] is not symmetric.
Therefore \[{R_3}\] is not an equivalent relation.
Option ‘D’ is correct
Note: Students often confused with order pair with relation. If we make a set using order pairs, then the set is known as relation.
Formula Used: Reflexive: R said to reflexive on set A, if \[\left( {a,a} \right) \in R\] for every \[a \in A\].
Symmetric: R said to symmetricon set A, if \[\left( {a,b} \right) \in R\] implies that \[\left( {b,a} \right) \in R\] where \[a,b \in A\].
Transitive: R said to transitive on set A, if \[\left( {a,b} \right),\left( {b,c} \right) \in R\] implies that \[\left( {a,c} \right) \in R\] where \[a,b,c \in A\].
Complete step by step solution: For option A:Given set is \[A = \left\{ {p,q,r} \right\}\].
Given relation is \[{R_1} = \left\{ {\left( {p,q} \right),\left( {q,r} \right),\left( {p,r} \right),\left( {p,q} \right)} \right\}\]
p, q, and r are the elements of A.
Since \[\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right) \notin {R_1}\], thus \[{R_1}\] is not reflexive.
Therefore \[{R_1}\] is not an equivalent relation.
For option B:
Given relation is \[{R_2} = \left\{ {\left( {r,q} \right),\left( {r,p} \right),\left( {r,r} \right),\left( {q,q} \right)} \right\}\].
Since \[\left( {p,p} \right) \notin {R_2}\], thus it is not reflexive.
Thus \[{R_2}\] is not an equivalent relation.
For option C:
Given relation is \[{R_3} = \left\{ {\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right),\left( {p,q} \right)} \right\}\].
Since \[\left( {p,p} \right),\left( {q,q} \right),\left( {r,r} \right)\] belongs to \[{R_3}\]. \[{R_3}\] is reflexive.
\[\left( {p,q} \right) \in {R_3}\] but \[\left( {q,p} \right) \notin {R_3}\]. Thus \[{R_3}\] is not symmetric.
Therefore \[{R_3}\] is not an equivalent relation.
Option ‘D’ is correct
Note: Students often confused with order pair with relation. If we make a set using order pairs, then the set is known as relation.
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