Question

Let $A=\{x$ $\in Z:0\le x\le 12\},$ $R=\left\{\right(a,b):a,b\in A,|a-b|\text{is divisible by}4\}$, then which among the following options are correct

• A. equivalence class of $1$ is $\{1,5,9\}$
• B. $R$ is reflexive and symmetric but not transitive relation.
• C. $R$ is a Transitive relation
• D. equivalence class of $1$ is $A$

Answer 1

Answer: (A), (C)

$R$ is a Transitive relation

We have $R=\left\{\right(a,b):a,b\in A,|a-b|\text{is divisible by}4\}$, where $a,b\in \{0,1,2,3,....,12\}$.

For any $a\in$ A, we have
$|a-a|=0$, which is divisible by $4$
$\Rightarrow (a,a)\in R$.
So, $R$ is a reflexive relation.

Let $(a,b)\in R$
$\Rightarrow |a-b|=4\lambda \text{for some}\lambda \in Z$
$\Rightarrow |b-a|=4{\lambda }_1\text{for some}{\lambda }_1\in Z[\because |a-b|=|b-a|]$
$\Rightarrow (b,a)\in R$
So, $R$ is a symmetric relation.

Let $(a,b)\text{and}(b,c)\in R$
$\Rightarrow a-b=4\alpha ,b-c=4\beta ,\alpha ,\beta \in Z$
Now $a-c=4(\alpha +\beta )$
$\Rightarrow |a-c|$ is divisible by $4$
$\Rightarrow (a,c)\in R$
So, $R$ is a transitive relation.
Hence, $R$ is an equivalence relation.

Equivalance class of $1$ is all possible values of $x$ such that $(x,1)\in R,x\in A$ $\Rightarrow |x-1|$ is divisible by $4$
$\Rightarrow |x-1|=0,4,8$
$\Rightarrow x-1=0,4,8$
$\Rightarrow x=1,5,9$
Thus, elements related to $1$ are $\{1,5,9\}$.