Question

Which one of the following relations defined on $N$ is an equivalance relation?

• A. $a{R}_{1}b⟺|a|=|b|$
• B. $a{R}_{2}b⟺a\ge b$
• C. $a{R}_{3}b⟺a$ divides $b$
• D. $a{R}_{4}b⟺3$ divides $(a-b)$

Answer 1

Answer: (A), (D)

$a{R}_{4}b⟺3$ divides $(a-b)$

For $a{R}_{1}b⟺|a|=|b|$
$|a|=|a|\forall a\in N$
If $a{R}_{1}b$, then $b{R}_{1}a$, as $|a|=|b|\Rightarrow |b|=|a|$
If $a{R}_{1}b,b{R}_{1}c\Rightarrow |a|=|b|=|c|$
$\therefore a{R}_{1}c$
Hence, $R_1$ is an equivalance relation.

For $a{R}_{2}b⟺a\ge b$
$R_2$ is reflexive as $a\ge a\forall a\in N$
$R_2$ is not symmetric as $a\ge b$ but $b\ngeqq a\forall a,b\in N$

For $a{R}_{3}b⟺a$ divides $b$
$a$ divides $b$, but $b$ may not divide $a$, for example $2$ divides $6$, $6$ doesnot divide $2$.
Hence, $R_3$ is not symmetric.

For $a{R}_{4}b⟺3$ divides $(a-b)$
$R_4$ is reflexive as $3$ divides $(a-a)=0\forall a\in N$
$R_4$ is symmetric as $3$ divides $(a-b)$, then $3$ divides $(b-a)\forall a,b\in N$
$R_4$ is transitive as $3$ divides $(a-b)$ and $3$ divides $(b-c)\Rightarrow 3$ divides $a-c$, as $a-c=(a-b)-(b-c)$