Question

Let A={1,2,3,....,9} and R be the relation in $A\times A$ defined by (a,b) R (c,d) if a+d=b+c for a,b,c,d in $A\times A$.
Prove that R is an equivalence relation.Also obtain the equivalence class [(2,5)].

Answer 1

Reflexivity:Let (a,b) be an arbitrary element of $A\times A$.Then,$(a,b)ϵA\times A\Rightarrow a,bϵA.$

So,a+b=b+a$\Rightarrow (a,b)R(a,b)$

Thus,$(a,b)R(a,b)\forall (a,b)ϵA\times A$.Hence R is reflexive.

Symmetry:Let a,b,c,$dϵA\times A$ be such that (a,b) R (c,d).

Then,a+d=b+c$\Rightarrow c+b=d+a\Rightarrow (c,d)R(a,b).$

Thus, $(a,b)R(c,d)\Rightarrow (c,d)R(a,b)\forall a,b,c,dϵA\times A.$ Hence R is symmetric.

Transitivity:Let a,b,c,d,e,f$ϵA\times A$ be such that (a,b) R (c,d) and (c,d) R (e,f).

Then,a+d=b+c and c+f=d+e$\Rightarrow (a+d)+(c+f)=(b+c)+(d+e)\Rightarrow a+f=b+e$

$\Rightarrow (a,b)R(e,f).\text{That is ,(a,b) R (c,d) and (c,d) R (e,f)}\Rightarrow \text{(a,b) R (e,f)}\forall \text{a,b,c,d,e,f}ϵ\text{A}\times \text{A. Hence R is Transitive.}$
$\text{Since R is reflexive,symmetric and transitive so,R is an equivalence relation as well.}$

$\text{For the equivalence class of [(2,5)], we need to find (a,b) s.t.}$ $(a,b)R(2,5)\Rightarrow a+5=b+2\Rightarrow b-a=3.$
$\text{So,[(2,5)]={(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}}$.