Question

Let m be a given fixed positive integer.

Let $R=\left\{\right(a,b):a,bϵZ\}\text{and}(a-b)\text{is divisible by m}$

Show that R is an equivalence relation on Z.

Answer 1

$R=\left\{\right(a,b):a,bϵZ\}\text{and}(a-b)\text{is divisible by m}$

(i) Let $aϵZ$. Then,

$a-a=0$, which is divisible by m.

$\therefore (a,a)ϵR$ for all $aϵZ$.

So, R is reflexive.

(ii) Let $(a,b)ϵR$. Then,

$(a,b)ϵR\Rightarrow (a-b)$ is divisible by m

$\Rightarrow -(a-b)$ is divisible by m

$\Rightarrow (b-a)$ is divisible by m

$\Rightarrow (b,a)ϵR$

Thus, $(a,b)ϵR\Rightarrow (b,a)ϵR$

So, R is symmetric.

(iii) Let $(a,b)ϵR$ and $(b,c)ϵR$. Then,

$(a,b)ϵR$ and $(b,c)ϵR$

$\Rightarrow (a-b)$ is divisible by m and (b - c) is divisible by m

$\Rightarrow \left\{\right(a-b)+(b-c\left)\right\}$ is divisible by m

$\Rightarrow \left\{\right(a-c\left)\right\}$ is divisible by m

$\Rightarrow (a,c)ϵR$

$\therefore (a,b)ϵR$ and $(b,c)ϵR\Rightarrow (a,c)ϵR$.

So, R is transitive.

Thus, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.