Question

The relation R on the set Z of all integer numbers defined by $(x,y) \in R \Leftrightarrow x - y$ is divisible by n is
A) Equivalence
B) Symmetric only
C) Reflexive only
D) Transitive only

Answer 1

Hint:
We are given a relation $(x,y) \in R \Leftrightarrow x - y$ is divisible by n . A relation R is said to be reflexive if an element $x \in Z$ then $\left( {x,x} \right) \in R$ . A relation R is said to be symmetric if for any elements $a,b \in Z$ if $\left( {a,b} \right) \in R$ then $\left( {b,a} \right) \in R$. A relation R is said to be transitive if for any elements $a,b,c \in Z$ if $\left( {a,b} \right) \in R{\text{ and }}(b,c) \in R$ then $\left( {a,c} \right) \in R$. Whenever a relation is reflexive , symmetric and transitive then the relation is said to be equivalence relation .

Complete step by step solution:
We are given a relation $(x,y) \in R \Leftrightarrow x - y$is divisible by n
Now we need to check the properties satisfied by the above relation.
1) Reflexive
A relation R is said to be reflexive if an element $x \in Z$ then $\left( {x,x} \right) \in R$
The given relation , $R = \left\{ {(x,y):x,y \in Z,x - y{\text{ is divisible by n}}} \right\}$
We know that $x \in Z$
And $(x - x)$ is divisible by n
$ \Rightarrow (x,x) \in R$
Therefore the relation R is reflexive.

2) Symmetric
A relation R is said to be symmetric if for any elements $a,b \in Z$ if $\left( {a,b} \right) \in R$ then $\left( {b,a} \right) \in R$
The given relation , $R = \left\{ {(x,y):x,y \in Z,x - y{\text{ is divisible by n}}} \right\}$
For any $x,y \in Z$if $\left( {x,y} \right) \in R$
$ \Rightarrow (x - y){\text{ }}$ is divisible by n
That is when (a-b) is divided by n the remainder is zero
 $ \Rightarrow \dfrac{{(x - y){\text{ }}}}{n} = c$
Now ,
$ \Rightarrow \dfrac{{(y - x)}}{n} = \dfrac{{ - (x - y)}}{n} = - c$
Which shows that ( y – x ) is divisible by n
Hence $\left( {y,x} \right) \in R$
Therefore the relation R is symmetric.

3) Transitive
A relation R is said to be transitive if for any elements $a,b,c \in Z$if $\left( {a,b} \right) \in R{\text{ and }}(b,c) \in R$then $\left( {a,c} \right) \in R$
The given relation , $R = \left\{ {(x,y):x,y \in Z,x - y{\text{ is divisible by n}}} \right\}$
For any $x,y \in Z$if $\left( {x,y} \right) \in R$
$ \Rightarrow (x - y){\text{ }}$is divisible by n
That is, when (x-y) is divided by n the remainder is zero
 $ \Rightarrow \dfrac{{(x - y){\text{ }}}}{n} = c$
$\Rightarrow \left( x-y \right)= nc,\;\text{where}\; c \in Z$
Same way , for any $x,y \in Z$if $\left( {y,z} \right) \in R$
$ \Rightarrow (y - z){\text{ }}$is divisible by n
That is, when (y-z) is divided by n the remainder is zero
 $ \Rightarrow \dfrac{{(y - z){\text{ }}}}{n} = d$
\[ \Rightarrow (y - z) = nd{\text{ where d}} \in {\text{Z}}\]………..(2)
Now adding (1) and (2)
$
   \Rightarrow (x - y) + (y - z) = nc + nd \\
   \Rightarrow (x - y + y - z) = n(c + d) \\
   \Rightarrow (x - z) = n(c + d) \\
 $
This shows that ( x – z ) is divisible by n which show that $\left( {x,z} \right) \in R$
Therefore the relation R is transitive
Whenever a relation is reflexive, symmetric and transitive then the relation is said to be an equivalence relation.

Hence the given relation is an equivalence relation.
The correct option is a.

Note:
The relation is a subset of a cartesian product of two and more sets. The connection between the elements of two or more sets is Relation. The sets must be non-empty. A subset of the Cartesian product also forms a relation R. A relation may be represented either by Roster method or by Set-builder method.