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Question: The relation R on the set Z of all integer numbers defined by \((x,y) \in R \Leftrightarrow x - y\) ...

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

Explanation

Solution

We are given a relation (x,y)∈R⇔x−y(x,y) \in R \Leftrightarrow x - y is divisible by n . A relation R is said to be reflexive if an element x∈Zx \in Z then (x,x)∈R\left( {x,x} \right) \in R . A relation R is said to be symmetric if for any elements a,b∈Za,b \in Z if (a,b)∈R\left( {a,b} \right) \in R then (b,a)∈R\left( {b,a} \right) \in R. A relation R is said to be transitive if for any elements a,b,c∈Za,b,c \in Z if (a,b)∈R and (b,c)∈R\left( {a,b} \right) \in R{\text{ and }}(b,c) \in R then (a,c)∈R\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)∈R⇔x−y(x,y) \in R \Leftrightarrow x - yis 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∈Zx \in Z then (x,x)∈R\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∈Zx \in Z
    And (x−x)(x - x) is divisible by n
    ⇒(x,x)∈R\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∈Za,b \in Z if (a,b)∈R\left( {a,b} \right) \in R then (b,a)∈R\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∈Zx,y \in Zif (x,y)∈R\left( {x,y} \right) \in R
    ⇒(x−y) \Rightarrow (x - y){\text{ }} is divisible by n
    That is when (a-b) is divided by n the remainder is zero
    ⇒(x−y) n=c\Rightarrow \dfrac{{(x - y){\text{ }}}}{n} = c
    Now ,
    ⇒(y−x)n=−(x−y)n=−c\Rightarrow \dfrac{{(y - x)}}{n} = \dfrac{{ - (x - y)}}{n} = - c
    Which shows that ( y – x ) is divisible by n
    Hence (y,x)∈R\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∈Za,b,c \in Zif (a,b)∈R and (b,c)∈R\left( {a,b} \right) \in R{\text{ and }}(b,c) \in Rthen (a,c)∈R\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∈Zx,y \in Zif (x,y)∈R\left( {x,y} \right) \in R
    ⇒(x−y) \Rightarrow (x - y){\text{ }}is divisible by n
    That is, when (x-y) is divided by n the remainder is zero
    ⇒(x−y) n=c\Rightarrow \dfrac{{(x - y){\text{ }}}}{n} = c
    ⇒(x−y)=nc,  where  c∈Z\Rightarrow \left( x-y \right)= nc,\;\text{where}\; c \in Z
    Same way , for any x,y∈Zx,y \in Zif (y,z)∈R\left( {y,z} \right) \in R
    ⇒(y−z) \Rightarrow (y - z){\text{ }}is divisible by n
    That is, when (y-z) is divided by n the remainder is zero
    ⇒(y−z) n=d\Rightarrow \dfrac{{(y - z){\text{ }}}}{n} = d
    ⇒(y−z)=nd where d∈Z\Rightarrow (y - z) = nd{\text{ where d}} \in {\text{Z}}………..(2)
    Now adding (1) and (2)
    ⇒(x−y)+(y−z)=nc+nd ⇒(x−y+y−z)=n(c+d) ⇒(x−z)=n(c+d)  \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 (x,z)∈R\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.