Question

Show that each of the relation $R$ in the set $A=\{x\in Z:0\le x\le 12\}$, given by

(i) $R=\left\{\right(a,b):|a-b|\text{is a multiple of 4}\}$

(ii) $R=\left\{\right(a,b):a=b\}$

is an equivalence relation. Find the set of all elements related to $1$ in each case.

Answer 1

(i)

$A=\{x\in Z:0\le x\le 12\}=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$
$R=\left\{\right(a,b):|a-b|\text{is a multiple of 4}\}$
For any element $a\in A$, we have $(a,a)\in R$ as $|a-a|=0$ is a multiple of $4$.
$\therefore R$ is reflexive.
Now, let $(a,b)\in R\Rightarrow |a-b|$ is a multiple of $4$.
$\Rightarrow |-(a-b)|=|b-a|$ is a multiple of $4$.
$\Rightarrow (b,a)\in R$
$\therefore$ is symmetric.
Now, let $(a,b),(b,c)\in R$
$\Rightarrow |a-b|$ is a multiple of $4$ and $|b-c|$ is a multiple of $4$.
$\Rightarrow (a-b)$ is a multiple of $4$ and $(b-c)$ is a multiple of $4$.
$\Rightarrow (a-c)=(a-b)+(b-c)$ is a multiple of $4$.
$\Rightarrow |a-c|$ is a multiple of $4$.
$\Rightarrow (a,c)\in R$.
$\therefore R$ is transitive.
Hence, $R$ is an equivalence relation.
The set of elements related to $1$ is $\{1,5,9\}$ since
$|1-1|=0$ is a multiple of $4$,
$|5-1|=4$ is a multiple of $4$, and
$|9-1|=8$ is a multiple of $4$

Hence, $R$ is an equivalence relation.

(ii)

$A=\{x\in Z:0\le x\le 12\}=\{0,1,2,3,4,......,11,12\}$
$R=\left\{\right(a,b):a=b\}=\left\{\right(0,),(1,1),(2,2),........,(11,11),(12,12\left)\right\}$
For any element $a\in A$, we have $(a,a)\in R$, since $a=a$.
$\therefore R$ is reflexive.
Now, let $(a,b)\in R$.
$\Rightarrow a=b$
$\Rightarrow b=a$
$\Rightarrow (b,a)\in R$
$\therefore R$ is symmetric.
Now, let $(a,b)\in R$ and $(b,c)\in R$.
$\Rightarrow a=b$ and $b=c$
$\Rightarrow a=c$
$\Rightarrow (a,c)\in R$
$\therefore R$ is transitive.
Hence, $R$ is an equivalence relation.
The elements in $R$ that are related to $1$ will be those elements from set $A$ which are equal to $1$.
Hence, the set of elements related to $1$ is $\left\{1\right\}$.