Show that the relation R on the set Z of all integers defined by <img alt="\left ( x,y \right )\, \epsilon \, R\Leftrightarrow \left ( x-y \right )" src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%28%20x%2Cy%20%5Cright%20%29%5C%2C%20%5Cep

Show that the relation R on the set Z of all integers defined by $\left ( x,y \right )\, \epsilon \, R\Leftrightarrow \left ( x-y \right )$ is divisible by 3 is an equivalence relation.

Answers (1)

we've relation R defined on x as $\left ( x,y \right )\epsilon R\Rightarrow \left ( x-y \right )$ is divisible by 3.
- R is reflexive , as 3 divides (x-x) ie o for all $x\, \epsilon z$ ie. x R x for all $x\, \epsilon z$
-Further if $\left ( x,y \right )\varepsilon R$ for all $x,y \epsilon \, z$ then 3 divides $\left ( x-y \right )\therefore 3 \: divides\: \left ( y-x \right )$ as well
Hence $\left ( y-x \right )\epsilon R$ which follows that R is symmetric
- similarly , if $\left ( x,y \right )\varepsilon R$ and $\left ( y-z \right )\epsilon R$ then $\left ( x,y \right )$ and $\left ( y-z \right )$ are both divisible by 3.
That is , $x-y= 3p,y-z= 3m$ where $p,m\epsilon z$
Note that:
$x-z= \left ( x-y \right )+\left ( y-z \right )= 3p+3m= 3\left ( p+m \right )$
so $\left ( x-z \right )$ is divisible by 3
This shows that R is transitive as $\left ( x,z \right )\epsilon R$
Thus , R is an equivalance relation is z