Question

Let $n$ be a fixed positive integer. Define a relation $R$ on the set $Z$ of integers by, $aRb\iff n|\left(a–b\right)$. Then $R$ is:

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Equivalence
• E. All the above

Answer 1

The correct option is E

All the above

Explanation For The Correct Option:

The defined relation: $aRb\iff n|\left(a–b\right)$

Checking reflexivity of relation:

Let $a\in N,$for all $a\in Z$

$\Rightarrow a-a=0=0\times n$

$\therefore -a$ is divisible by $n$

Thus, $(a,a)\in R$ for all $a\in Z$

Hence, $R$ is reflexive.

Checking Symmetricity of relation:

Let $(a,b)\in R$

$\therefore$ $(a-b)$ is divisible by $n$.

⇒ $(a-b)=np$ for some $p\in Z$

$\Rightarrow b-a=n(-p)$

$\Rightarrow b-a$ is divisible by $n$ .

$\therefore (a,b)\in R\Rightarrow (b,a)\in R$ for all $(a,b)\in Z$

Thus, $R$ is symmetric.

Checking Transitivity of relation:

Let $(a,b,c)\in Z$ such that $(a,b)\in R\&(b,c)\in R.$

Then $(a,b)\in R\Rightarrow (a-b)$ is divisible by $n$.

$\Rightarrow a-b=np$ for some $p\in Z$

$(b,c)\in R\Rightarrow (b-c)$ is divisible by $n$.

$\Rightarrow b-c=nq$ for some $q\in Z$

$(a,b)\in R\&b-c\in R$

$$\begin{gathered} \Rightarrow (a-b)+(b-c)=np+nq \\ \Rightarrow a-c=n(p+q) \end{gathered}$$

⇒ $a-c$ is divisible by $n$.

$$\begin{gathered} \Rightarrow (a-c)\in R \\ \therefore (a,b)\in R\&(b,c)\in R \end{gathered}$$

$\Rightarrow (a,c)\in R$for all $a,b,c\in Z$

$R$ is a transitive relation on $Z$.

As $R$ reflexive, symmetric and transitive, we can conclude that it is an equivalence relation also.

Hence, the correct answer is option (E).