Question

Let $A$ = $\{x$ $\in Z:0\le x\le 12$}. Show that
$R=\left\{\right(a,b):a,b\in A,|a-b|\text{is divisible by}4\}$ is an equivalence relation. Find the set of all elements related to $1$. Also write the equivalence class $\left[2\right]$.

Answer 1

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 reflexive.

For any $(a,b)\in R$

$|a-b|$ is divisible by $4$

$\Rightarrow |a-b|=4\lambda$ for some $\lambda \in N$
$\Rightarrow |b-a|=4\lambda$ for some $\lambda \in N[\because |a-b|=|b-a|]$
$\Rightarrow (b,a)\in R$

So, $R$ is symmetric.

Let $(a,b)\in Rand(b,c)\in R$, then

$|a-b|$ is divisible by $4$ and $|b-c|$ is divisible by $4$

$\Rightarrow |a-b|=4\lambda$ and $|b-c|=4\mu$

$\because a-b$ and $b-c$ are both multiples of $4$
$\therefore a-b+b-c=a-c\text{is a multiple of 4}$
$\Rightarrow |a-c|$ is divisible by $4$

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

So, $R$ is transitive.

Hence, $R$ is an equivalence relation.

Let $x$ be an element of A such that $(x,1)\in R,$ then $|x-1|$ is divisible by $4$

$\Rightarrow |x-1|=0,4,8,12$

$\Rightarrow x-1=0,4,8,12$

$\Rightarrow x=1,5,9$

Thus, elements related to 1 are $\{1,5,9\}$.

Also, $\left|\begin{array}{c}x-2\end{array}\right|$ is divisible by $4$

$\Rightarrow \left|\begin{array}{c}x-2\end{array}\right|=0,4,8,12$

$\Rightarrow x-2=0,4,8,12$

$\Rightarrow x=2,6,10$

Hence, the equivalence class $\left[2\right]=\{2,6,10\}$