Question

Let $R_1,R_2$ are relation defined on $Z$ such that $a{R}_{1}b⟺(a-b)$ is divisible by $3$ and $a{R}_{2}b⟺(a-b)$ is divisible by $4$. Then which of the two relation $(R_1\cup R_2),(R_1\cap R_2)$ is an equivalence relation?

• A. $(R_1\cup R_2)$ Only
• B. $(R_1\cap R_2)$ Only
• C. Both $(R_1\cup R_2),(R_1\cap R_2)$
• D. Neither $(R_1\cup R_2)$ nor $(R_1\cap R_2)$

Answer 1

The correct option is B $(R_1\cap R_2)$ Only

Givn relations defined on z are,

$R_1=a{R}_{1}b\leftrightarrow (a-b)$ is divisible by 3.

and $R_2=a{R}_{2}b\leftrightarrow (a-b)$ is divisible by 4.

Consider,

$a{R}_{1}b\leftrightarrow a-b$ is divisible by 3.

Reflexive:

$a{R}_{1}a\leftrightarrow a-a$ is divisible by 3

True
$\therefore$ the given relation $R_1$ is a reflexive relation.

Symmetric:

$a{R}_{1}b\leftrightarrow a-b$ is divisible by 3.

$\Rightarrow a-b=3k$; where $k$ is an integer.

then

$b{R}_{1}a\leftrightarrow b-a$ is divisible by 3.

because $b-a=-(a-b)=-3k$ which is divisible by 3.

$\therefore$ the given relation $R$ is a symmetric relation.

Transitive:

$a{R}_{1}b\leftrightarrow a-b$ is divisible by 3.
$\Rightarrow a-b=3k;k$ is an integer.

$b{R}_{1}c\leftrightarrow b-c$ is divisible by 3.

$\Rightarrow b-c=3p;p$ is integer.

Then $a{R}_{1}c\leftrightarrow a-c$ is divisible by 3.

Because $(a-b)+(b-c)=a-c$

$\Rightarrow 3k+3p=3(k+p)$ which is divisible by 3.

$\therefore$ The given relation $R_1$ is transitive relation.

$\therefore R_1$ satisfies the reflexive, symmetric and transitive relation properties.
$\therefore R_1$ is an equivalence relation.

Similarly $R_2$ is also an equivalence relation.
And as we know that,
"If $R_1$ and $R_2$ are equivalence relations then $R_1\cap R_2$ is also an equivalence relation."
$\therefore R_1\cap R_2$ is an equivalence relation.