Show that the relation R on the set Z of all integers,given by R={(a,b):2 divides(a-b)} is an equivalence relation.

Answers (1)

R Ravindra Pindel

(i) Reflexive:   $\left ( a,a \right )\epsilon R$ given: $R= \left \{ \left ( a,b \right ):2\: divides\: a-b \right \}$
   $\frac{a-a}{2}= \frac{0}{2}= 0$
$\therefore 2\: divides\: a-a\: \: \therefore \left ( a,a \right )\epsilon R\Rightarrow R \: is \: reflexive$
(ii) symnatic:   $\left ( a,b \right )\epsilon R\: \: \Rightarrow \: \left ( b,a \right )\epsilon R$
$\frac{a-b}{2}= p\: \: \: p\epsilon Z$
$a-b= 2p\: \: \: \: \Rightarrow b-a= 2\left ( -p \right )\: \: -p\epsilon Z$
$\frac{b-a}{2}= -p$
2 divides    $b-a \Rightarrow \: \left ( b,a \right )\epsilon R$
$\therefore R \: is\: symnatic$
(iii) Transitive: $\left ( a,b \right )\epsilon R,\: \: \left ( b,c \right )\epsilon R\: \: \Rightarrow \left ( a,c \right )\epsilon R$
$\frac{a-b}{2}= p---(1)$ $\frac{b-c}{2}= q$    $p,q\epsilon Z$
Add (1)& (2)
$\frac{a-b}{2}+\frac{b-c}{2}= p+q$
$\frac{a-b+b-c}{2}= p+q$    $\Rightarrow \frac{a-c}{2}= p+q$
2 divides   $a-c$    $\therefore p+q\epsilon Z$
$\therefore \left ( a,c \right )\epsilon R$
$\Rightarrow$ R is transitive
Hence R is equivalence relations