Question

If ${{R}_{1}}\ and\ {{R}_{2}}$ be two equivalence relations on set A, prove that ${{R}_{1}}\cap {{R}_{2}}$ is also an equivalence relation on A.

Answer 1

As ${{R}_{1}}\ and\ {{R}_{2}}$ are equivalence relation on S. ${{R}_{1}}\ and\ {{R}_{2}}$ will be symmetric, reflexive and transitive on S. Use these conditions to prove that $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ is also symmetric, reflexive and transitive and thus an equivalence relation on A.

Complete step-by-step answer:
Given, ${{R}_{1}}\ and\ {{R}_{2}}$are equivalence relations on set A.
A relation is called the equivalence relation, if it is reflexive, symmetric and transitive.
Since, ${{R}_{1}}$ is a transitive relation.
(1) ${{R}_{1}}$ is reflexive.
i.e. $\left( a,a \right)\in {{R}_{1}}\ for\ all\ a\in A$
(2) ${{R}_{1}}$ is symmetric.
i.e. if $\left( a,b \right)\in {{R}_{1}},\ then\ \left( b,a \right)\in {{R}_{1}};\ \ \ a,b\in A$
(3) ${{R}_{1}}$ is transitive.
i.e. if $\left( a,b \right)\in {{R}_{1}},\ and\ \left( b,c \right)\in {{R}_{1}}\ then\ \left( a,c \right)\in {{R}_{1}};\ \ \ a,b,c\in A$
Similarly, ${{R}_{2}}$ is also an equivalence relation. So,
(1) ${{R}_{2}}$is symmetric.
i.e. if $\left( a,b \right)\in R\ then\ \left( b,a \right)\in R;\ \ a,b\in A$
(2) ${{R}_{2}}$is reflexive.
i.e. $\left( a,a \right)\in R\ for\ all\ a\in A$
(3) ${{R}_{2}}$is transitive.
i.e. if $\left( a,b \right)\in {{R}_{2}},\ and\ \left( b,c \right)\in {{R}_{2}}\ then\ \left( a,c \right)\in {{R}_{2}};\ \ \ a,b,c\in A$
We have to prove that ${{R}_{1}}\cap {{R}_{2}}$ is an equivalence relation.
Check reflexive:
For all $a\in A,\left( a,a \right)\in {{R}_{1}}\ and\ \left( a,a \right)\in {{R}_{2}}\ \ \ \ \left[ As\ both\ {{R}_{1}}\ and\ {{R}_{2}}\ are\ reflexive\ on\ A \right]$
We know that if an element belongs to set A and also to set B, then the element will also belong to $\left( A\cap B \right)$.
So, as $a\in A,\left( a,a \right)\in {{R}_{1}}\ and\ \left( a,a \right)\in {{R}_{2}}\ ,\ \left( a,a \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)$
$\therefore$ $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ is reflexive on A.
Check symmetric:
If $\left( a,b \right)\in {{R}_{1}}\ then\ \left( b,a \right)\in {{R}_{1}}............\left( 1 \right)$ [As ${{R}_{1}}$ is symmetric on A]
If $\left( a,b \right)\in {{R}_{2}}\ then\ \left( b,a \right)\in {{R}_{2}}............\left( 2 \right)$ [As ${{R}_{2}}$ is symmetric on A]
From (1) and (2),
If $\left( a,b \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)\ then\ \left( b,a \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)$
So, $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ is symmetric on A.
Check transitivity:
If $\left( a,b \right)\in {{R}_{1}}\ and\ \left( b,c \right)\in {{R}_{1}}\ then\ \left( a,c \right)\in {{R}_{1}}............\left( 3 \right)$ [As ${{R}_{1}}$ is transitive on A]
If $\left( a,b \right)\in {{R}_{2}}\ and\ \left( b,c \right)\in {{R}_{2}}\ then\ \left( a,c \right)\in {{R}_{2}}............\left( 4 \right)$ [As ${{R}_{2}}$ is transitive on A]
From (3) and (4),
If $\left( a,b \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)\ and\ \left( b,c \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)\ then\ \left( a,c \right)\in \left( {{R}_{1}}\cap {{R}_{2}} \right)$
So, $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ is transitive on A.
Since, $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ is symmetric, reflexive and transitive on A, $\left( {{R}_{1}}\cap {{R}_{2}} \right)$ will be an equivalence relation on set A.

Note: Let’s look at an example of equivalence relation: so, the relation ‘is equal to’ denoted by $=$ , is an equivalence relation on the set of real numbers since for any $x,y,z\in \mathbb{R}$ :
A. Reflexivity: $x=x$
B. Symmetry: if $x=y$ then $y=x$
C. Transitivity: if $x=y$ and $y=z$ then $x=z$
. Since all of these are true, therefore it is an equivalence relation. Note that to represent an equivalence relation we use: $\sim$ .