# Let A = {1,2,3,4}. Define Relations ${{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ on $A\times A$ as:

$\begin{align}

& {{R}_{1}}=\left\{ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 1,2 \right),\left( 2,1 \right) \right\} \\

& {{R}_{2}}=\left\{ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 1,2 \right),\left( 2,1 \right),\left( 1,3 \right),\left( 4,1 \right),\left( 1,4 \right) \right\} \\

\end{align}$

and ${{R}_{3}}=\left\{ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right) \right\}$

which of the relations ${{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ define an equivalence relation on $A\times A$.

Last updated date: 22nd Mar 2023

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Answer

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Hint: An equivalence relation is a relation which is symmetric, reflexive, and transitive. Check which of the relations are symmetric, which are reflexive, and which are transitive. The relations falling in all three classes are equivalence relations.

Complete step-by-step answer:

[1] Reflexive: A relation R defined in $A\times A$ is said to be reflexive if $\forall a\in A,(a,a)\in R$.

Since (1,1),(2,2),(3,3) and (4,4)$\in {{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ all of the relations ${{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ are reflexive

[2] Symmetric: A relation R is to be symmetric if $\forall (a,b)\in R\Rightarrow (b,a)\in R$.

We have (1,3)$\in {{R}_{2}}$ but $\left( 3,1 \right)\notin {{R}_{2}}$. Hence ${{R}_{2}}$ is not symmetric.

However, ${{R}_{1}}\text{ and }{{R}_{3}}$ are symmetric.

[3] Transitive: A relation R is said to be transitive if $\forall \left( a,b \right)\in R$ and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$

We have $\left( 2,1 \right)\in {{R}_{2}}$ and $\left( 1,3 \right)\in {{R}_{2}}$ but $\left( 2,3 \right)\notin {{R}_{2}}$. Hence ${{R}_{2}}$ is not transitive.

However, ${{R}_{1}}\text{ and }{{R}_{3}}$ are transitive.

Hence ${{R}_{1}}$ and ${{R}_{3}}$ form equivalence relations on $A\times A$.

Note:

[1] A relation on the set $A\times B$ is a subset of the Cartesian product $A\times B$.

[2] Functions are relations with special properties

[3] if ${{R}_{1}}$ and ${{R}_{2}}$ are equivalence relations on $A\times A$ then ${{R}_{1}}\bigcap {{R}_{2}}$ is also an equivalence relation on $A\times A$.

[4] Restriction of an equivalence relation is also an equivalence relation

[5] If a relation on $A\times B$ relates every element of A to a unique element in B then the relation is known as function and the set A is called domain of the function and set B as the codomain of the function. The set of elements in B to which the function maps elements of A is called Range. It is therefore clear that Range $\subseteq $ Codomain.

Complete step-by-step answer:

[1] Reflexive: A relation R defined in $A\times A$ is said to be reflexive if $\forall a\in A,(a,a)\in R$.

Since (1,1),(2,2),(3,3) and (4,4)$\in {{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ all of the relations ${{R}_{1}},{{R}_{2}}\text{ and }{{R}_{3}}$ are reflexive

[2] Symmetric: A relation R is to be symmetric if $\forall (a,b)\in R\Rightarrow (b,a)\in R$.

We have (1,3)$\in {{R}_{2}}$ but $\left( 3,1 \right)\notin {{R}_{2}}$. Hence ${{R}_{2}}$ is not symmetric.

However, ${{R}_{1}}\text{ and }{{R}_{3}}$ are symmetric.

[3] Transitive: A relation R is said to be transitive if $\forall \left( a,b \right)\in R$ and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$

We have $\left( 2,1 \right)\in {{R}_{2}}$ and $\left( 1,3 \right)\in {{R}_{2}}$ but $\left( 2,3 \right)\notin {{R}_{2}}$. Hence ${{R}_{2}}$ is not transitive.

However, ${{R}_{1}}\text{ and }{{R}_{3}}$ are transitive.

Hence ${{R}_{1}}$ and ${{R}_{3}}$ form equivalence relations on $A\times A$.

Note:

[1] A relation on the set $A\times B$ is a subset of the Cartesian product $A\times B$.

[2] Functions are relations with special properties

[3] if ${{R}_{1}}$ and ${{R}_{2}}$ are equivalence relations on $A\times A$ then ${{R}_{1}}\bigcap {{R}_{2}}$ is also an equivalence relation on $A\times A$.

[4] Restriction of an equivalence relation is also an equivalence relation

[5] If a relation on $A\times B$ relates every element of A to a unique element in B then the relation is known as function and the set A is called domain of the function and set B as the codomain of the function. The set of elements in B to which the function maps elements of A is called Range. It is therefore clear that Range $\subseteq $ Codomain.

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