Let R and S be two equivalence relations on set A. Then
Answer
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Hint: We are given that, set R and S are two equivalence relations on set A. We know that an equivalence relation means it should be reflexive, symmetric and transitive. So, we will prove each of them and thus we will get the final output. In other words, two elements of the given set are equivalent to each other if they belong to the same equivalence class.
Complete step-by-step answer:
We know that,
Equivalence relation means reflexive, symmetric and transitive.
Given R, and S are relations on set A
\[\therefore R \subseteq A \times A\;and\;S \subseteq A \times A\]
\[ \Rightarrow R \cap S \subseteq A \times A\]
\[ \Rightarrow R \cap S\] is also a relation on A
Reflexivity: Let a be an arbitary element of A.
Then \[a \in A\]
\[ \Rightarrow \left( {a,a} \right) \in R\] and \[\left( {a,a} \right) \in S\] .... [∵R and S are reflexive]
\[ \Rightarrow \left( {a,a} \right) \in R \cap S\]
Thus, \[\left( {a,a} \right) \in R \cap S\] for all \[a \in A\]
So, \[R \cap S\] is a reflexive relation on A
\[\therefore R \cap S\] is reflexive ------ (1)
Symmetry: Let \[a,b \in A\] such that \[\left( {a,b} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,b} \right) \in R\] and \[\left( {a,b} \right) \in S\]
\[ \Rightarrow \left( {b,a} \right) \in R\] and \[\left( {b,a} \right) \in S\]
( Since R and S are symmetric),
\[ \Rightarrow \left( {b,a} \right) \in R \cap S\]
Thus,
\[\left( {a,b} \right) \in R \cap S\]
\[ \Rightarrow \left( {b,a} \right) \in R \cap S\] for all \[\left( {a,b} \right) \in R \cap S\].
So, \[R \cap S\] is symmetric on A
\[\therefore R \cap S\] is symmetric ----- (2)
Transitivity: Let \[a,b,c \in A\] such that \[\left( {a,b} \right),\left( {b,c} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,b} \right),\left( {b,c} \right) \in R\]
\[ \Rightarrow \left( {a,c} \right) \in R\]
And,
\[ \Rightarrow \left( {a,b} \right),\left( {b,c} \right) \in S\]
\[ \Rightarrow \left( {a,c} \right) \in S\]
Since R and S are transitive,
\[ \Rightarrow \left( {a,c} \right) \in R \cap S\]
Thus,
\[\left( {a,b} \right) \in R \cap S\] and \[\left( {b,c} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,c} \right) \in R \cap S\]
So \[R \cap S\] is transitive on A
\[\therefore R \cap S\] is transitive --- (3)
From (1), (2) and (3), R∩S is an equivalence relation.
Hence, \[R \cap S\] is an equivalence relation on A.
Note: An equivalence relation is a kind of binary relation that should be reflexive, symmetric and transitive. The well-known example of an equivalence relation is the “equal to (=)” relation.
Sets are represented as a collection of well-defined objects or elements and it does not change from person to person. A set is represented by a capital letter.
Complete step-by-step answer:
We know that,
Equivalence relation means reflexive, symmetric and transitive.
Given R, and S are relations on set A
\[\therefore R \subseteq A \times A\;and\;S \subseteq A \times A\]
\[ \Rightarrow R \cap S \subseteq A \times A\]
\[ \Rightarrow R \cap S\] is also a relation on A
Reflexivity: Let a be an arbitary element of A.
Then \[a \in A\]
\[ \Rightarrow \left( {a,a} \right) \in R\] and \[\left( {a,a} \right) \in S\] .... [∵R and S are reflexive]
\[ \Rightarrow \left( {a,a} \right) \in R \cap S\]
Thus, \[\left( {a,a} \right) \in R \cap S\] for all \[a \in A\]
So, \[R \cap S\] is a reflexive relation on A
\[\therefore R \cap S\] is reflexive ------ (1)
Symmetry: Let \[a,b \in A\] such that \[\left( {a,b} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,b} \right) \in R\] and \[\left( {a,b} \right) \in S\]
\[ \Rightarrow \left( {b,a} \right) \in R\] and \[\left( {b,a} \right) \in S\]
( Since R and S are symmetric),
\[ \Rightarrow \left( {b,a} \right) \in R \cap S\]
Thus,
\[\left( {a,b} \right) \in R \cap S\]
\[ \Rightarrow \left( {b,a} \right) \in R \cap S\] for all \[\left( {a,b} \right) \in R \cap S\].
So, \[R \cap S\] is symmetric on A
\[\therefore R \cap S\] is symmetric ----- (2)
Transitivity: Let \[a,b,c \in A\] such that \[\left( {a,b} \right),\left( {b,c} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,b} \right),\left( {b,c} \right) \in R\]
\[ \Rightarrow \left( {a,c} \right) \in R\]
And,
\[ \Rightarrow \left( {a,b} \right),\left( {b,c} \right) \in S\]
\[ \Rightarrow \left( {a,c} \right) \in S\]
Since R and S are transitive,
\[ \Rightarrow \left( {a,c} \right) \in R \cap S\]
Thus,
\[\left( {a,b} \right) \in R \cap S\] and \[\left( {b,c} \right) \in R \cap S\]
\[ \Rightarrow \left( {a,c} \right) \in R \cap S\]
So \[R \cap S\] is transitive on A
\[\therefore R \cap S\] is transitive --- (3)
From (1), (2) and (3), R∩S is an equivalence relation.
Hence, \[R \cap S\] is an equivalence relation on A.
Note: An equivalence relation is a kind of binary relation that should be reflexive, symmetric and transitive. The well-known example of an equivalence relation is the “equal to (=)” relation.
Sets are represented as a collection of well-defined objects or elements and it does not change from person to person. A set is represented by a capital letter.
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