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 $A\times A$.

Question

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 $A\times A$.

Vedantu Content Team · Accepted Answer

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} and R_{3}

all of the relations

R_{1}, R_{2} 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

(3, 1) \notin R_{2}

. Hence

R_{2}

is not symmetric.
However,

R_{1} and R_{3}

are symmetric.
[3] Transitive: A relation R is said to be transitive if

\forall (a, b) \in R

and

(b, c) \in R \Rightarrow (a, c) \in R

We have

(2, 1) \in R_{2}

and

(1, 3) \in R_{2}

but

(2, 3) \notin R_{2}

. Hence

R_{2}

is not transitive.
However,

R_{1} 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} ⋂ 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.

Let A = {1,2,3,4}. Define Relations R1,R2 and R3 on A×A as:R1={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)}R2={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(4,1),(1,4)}and R3={(1,1),(2,2),(3,3),(4,4)}which of the relations R1,R2 and R3 define an equivalence relation on A×A.