Question

Let $Z$ be the set of integers and $aRb$, where $a,bϵZ$ if an only if $(a-b)$ is divisible by $5$.
Consider the following statements:
$1.$ The relation $R$ partitions $Z$ into five equivalent classes.
$2.$ Any two equivalent classes are either equal or disjoint.
Which of the above statements is/are correct?

• A. $1$ only
• B. $2$ only
• C. Both $1$ and $2$
• D. Neither $1$ nor $2$

Answer 1

The correct option is C Both $1$ and $2$

A relation is defined on $Z$ such that $aRb\Rightarrow (a-b)$ is divisible by $5$,

For Reflexive: $(a,a)\in R$.

Since, $(a-a)=0$ is divisible by $5$. Therefore, the realtion is reflexive.

For symmetric: If $(a,b)\in R\Rightarrow (b,a)\in R$.

$(a,b)\in R\Rightarrow (a-b)$ is divisible by $5$.

Now, $(b-a)=-(a-b)$ is also divisible by $5$. Therefore, $(b,a)\in R$

Hence, the relation is symmetric.

For Transitive: If $(a,b)\in R$ and $(b,a)\in R\Rightarrow (a,c)\in R$.

$(a,b)\in R\Rightarrow (a-b)$ is divisible by $5$.

$(b,c)\in R\Rightarrow (b-c)$ is divisible by $5$.

Then $(a-c)=(a-b+b-c)=(a-b)+(b-c)$ is also divisible by $5$. Therefore, $(a,c)\in R$.

Hence, the relation is transitive.

There, the relation is equivalent.

Now, depending upon the remainder obtained when dividing $(a-b)$ by $5$ we can divide the set $Z$ iinto $5$ equivalent classes and they are disjoint i.e., there are no common elements between any two classes.

Therefore, the correct option is $\left(C\right)$.