Question

show that the relation R in the set of
-Integers given by R={(a, b)} : 5 divides
(a-b)} is symetric & transitive.​

Answer 1

The relation $\displaystyle\sf{R:\mathbb{Z}\longrightarrow\mathbb{Z}}$ is defined as,

$\displaystyle\longrightarrow\sf{R=\{(a,\ b):5\mid(a-b),\ a,\ b\in\mathbb{Z}\}}$

Let there exist two integers $\displaystyle\sf{a}$ and $\displaystyle\sf{b}$ such that,

$\displaystyle\longrightarrow\sf{a-b=5m\quad\implies\quad5\mid(a-b)}$

Then we see that,

$\displaystyle\longrightarrow\sf{b-a=-5m\quad\implies\quad5\mid(b-a)}$

Hence $\displaystyle\sf{R}$ is symmetric as it contains both $\displaystyle\sf{(a,\ b)\quad\&\quad(b,\ a).}$

Let there exist an integer $\displaystyle\sf{c}$ such that,

$\displaystyle\longrightarrow\sf{b-c=5n\quad\implies\quad5\mid(b-c)}$

Then,

$\displaystyle\longrightarrow\sf{a-c=a-b+b-c}$

$\displaystyle\longrightarrow\sf{a-c=5m+5n}$

$\displaystyle\longrightarrow\sf{a-c=5(m+n)}$

$\displaystyle\Longrightarrow\sf{5\mid(a-c)}$

Hence $\displaystyle\sf{R}$ is transitive as it contains $\displaystyle\sf{(a,\ b),\quad(b,\ c)\quad\&\quad(a,\ c).}$

Also, $\displaystyle\sf{R}$ is reflexive because,

$\displaystyle\longrightarrow\sf{a-a=0\quad\implies\quad5\mid(a-a)}$

so that it contains $\displaystyle\sf{(a,\ a).}$

Hence $\displaystyle\sf{R}$ is an equivalence relation.