The relation R on the set Z of all integer numbers defined by x-y - R- x-y is divisible by n is

The correct option is A Equivalence [ R={( x,y ) :x,y∈ z, x y is divisible by n } #160;; ...

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Standard XII

Mathematics

Equivalence Relation

The relation ...

Question

The relation $R$ on the set $Z$ of all integer numbers defined by $(x, y) ϵ R \Leftrightarrow x - y$ is divisible by $n$ is

Equivalence

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Symmetric only

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Reflexive only

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Transitive only

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Solution

The correct option is A Equivalence
$\begin{matrix} R = {(x, y) : x, y \in z, x - y i s d i v i s i b l e b y n} F o r R e f l e x i v e, x \in z S o \Rightarrow (x - x) i s d i v i s i b l e b y n \Rightarrow (x, x) \in z \end{matrix}$
So, Relation is Reflexive
$\begin{matrix} F o r S y m m e t r i c \Rightarrow L e t (x, y) \in R \Rightarrow (x - y) i s d i v i s i b l e b y n . \Rightarrow \frac{x - y}{n} = c, R e m a i n d e r i s 0. \Rightarrow \frac{y - x}{n} = - c, R e m a i n d e r i s a l s o 0 . \Rightarrow (y - x) i s d i v i s i b l e b y n (y, x) \in R S o, R i s S y m m e t r i c . F o r T r a n s i t i v e \Rightarrow L e t (x, y) \in R & (y, z) \in R \Rightarrow (x - y) i s d i v i s i b l e b y n \Rightarrow (y - q) i s d i v i s i b l e b y n \end{matrix}$
add both these equation (i) & (ii)
$\begin{matrix} (x - y) = x c, c \in z - v, (y - q) = n a, (a \in z) \Rightarrow (x - y + y - q) = n c + n a, c, a \in z \Rightarrow (x - q) = n (a + c), c, a \in z (x - q) i s d i v i s i b l e b y n \Rightarrow (x, q) \in R \end{matrix}$
So, R is Transitive
So, the R is Reflexive, Symmetric and Transitive
Then it is Equivalence Relation.

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The relation R on the set Z of all integer numbers defined by (x,y) ϵ R ⇔x−y is divisible by n is

The relation $R$ on the set $Z$ of all integer numbers defined by $(x, y) ϵ R \Leftrightarrow x - y$ is divisible by $n$ is