Prove that the relation R in the set $A= \left \{ 1,2,3,4,5,6,7 \right \}$ given by $R= \left \{ \left ( a,b \right ):\left | a-b \right |is\; given \right \}$ is an equivalence relation.

Answers (1)

Reflexive: $\because \left | a-a \right |= 0\: which\; is \; even \; \therefore \left ( a,a \right )\epsilon R$ so ,R is reflexive.
Symmetric: Let $\left ( a,b \right )\epsilon R$ implies $\left | a-b \right |$ is even which means, $\left | b-a \right |$ is also even.That means, $\left ( b,a \right )\epsilon R\cdot \; \therefore R\: is\: symmetric$
Transitive: Let $\left ( a,b \right )\epsilon R$ $\left ( b,c \right )\epsilon R$
It means $\left | a-b \right |$ and $\left | b-c \right |$ are even
That is , $\left | a-b \right |= 2m\; \; \left | b-c \right |= 2n$ where $m,n\epsilon\: Z$
$\Rightarrow a-b= \pm 2m,\: \: b-c=\pm 2n$
$\Rightarrow a-b+b-c= a-c\pm 2\left ( m+n \right )$
$\Rightarrow \left | a-c \right |= 2\left ( m+n \right ), ie \left | a-c \right |is \: even\cdot$
That is $\left ( a,c \right )\epsilon R$ so R is transitive
Hence R is equvivalence relation.