Question

Relation R in set z of all integers, defined as $R = \left\{ {\left( {x,y} \right):x - y\,\,is\,\,an\,\,\operatorname{int} eger} \right\}$ check if it is equivalence relation.

Answer 1

Hint: We will apply the condition of reflexive, symmetric and transitive for checking if the relation is equivalence or not. Thereafter, we will solve reflexive, symmetric and transitive.


Complete step by step solution:
Here, \[R = \left\{ {\left( {x,y} \right):x - y\,\,is\,\,an\,\,\operatorname{int} eger} \right\}\]
Equivalence relations: A relation R on a set A is an equivalence relation if and only if R is (i) reflexive (ii) symmetric and (iii) transitive.
Now (i) reflexive
Now, for every $x \in z,\,\,\left( {x,x} \right) \in R$
Then $x - x = 0$
Therefore R is reflexive.
(iii) Symmetric:
 If it symmetric or not, now, for every $x,y \in z,$
If $\left( {x,y} \right) \in R,$ then $x - y$ is an integer
$ \Rightarrow - \left( {x - y} \right)$ is also an integer
$ \Rightarrow - \left[ { - \left( {y - x} \right)} \right]$ is also an integer
$ \Rightarrow \left( {y - x} \right)$ is an integer
Therefore $\left( {x,y} \right) \in R,$ then $\left( {y,x} \right) \in R$
Thus R is symmetric.
Let $\left( {x,y} \right) \in R$ and $\left( {y,z} \right) \in R,$ where $x,y,z \in z$
Then $x - y$ is an integer
$y - z$ is an integer,
Then sum of integers is also an integer
$\left( {x - y} \right) + \left( {y - z} \right)$ is an integer
$\left( {x - y + y - z} \right)$ is an integer
$\left( {x - z} \right)$ is an integer
So, if $\left( {x - y} \right)$ is an integer and $\left( {y - z} \right)$ is an integer then, $x - z$ is an integer
$\therefore $ Now $\left( {x,y} \right) \in R,\left( {y,z} \right) \in R$
then $\left( {x,z} \right) \in R$
Therefore, R is transitive.
Thus R is reflexive, symmetric and transitive.
Hence, it is an equivalence relation.


Note: Students must know that negative of an integer is an integer and before you apply the equivalence relation you must know the condition of equivalence.