Question

Show that the relation R defined by $(a,b)R(c,d)$ such that $a+d=b+c$ on AXA, where $A=\{1,2,....10\}$ is an equivalence relation. Hence write the equivalence class $[3,4];a,b,c,d\in A.$

Answer 1

R is an equivalance relation if R is reflexive,symmetric and transitive.

a)checking if it is reflexive;

Given R in $A\times Aand(a,b)R(c,d)suchthata+d=b+c$

For reflexive,consider $(a,b)R(a,b)(a,b)\in A$

and applying given condition$\Rightarrow a+b=b+a$;which is true for all A

$\therefore Risreflexive.$

b)checking if it is symmetric;

given$(a,b)R(c,d)suchthata+d=b+c$

consider $(c,d)R(a,b)onA\times A$

applying given condition$\Rightarrow c+b=d+awhichsatisfiesgivencondition$

Hence R is symmetric.

c)checking if it is transitive;

$Let(a,b)R(c,d)and(c,d)R(e,f)$

$and(a,b),(c,d),(e,f)\in A\times A$

applying given condition:$\Rightarrow a+d=b+c\to 1andc+f=d+e\to 2$

equation 1$\Rightarrow a-c=b-d$

$nowaddequation1and2;$

$\Rightarrow a-c+c+f=b-d+d+e$

$\Rightarrow a+f=b+e$

$\therefore (a,b)R(e,f)alsosatisfiesthecondition$

Hence R is transitive.

Therfore by above inspection we can say that R is an equivalence relation.

$\Rightarrow stepsforfindingequivalenceclass[3,4]:$

Let $(3,4)R(a,b)onA\times AwhereA=\{1,2,3,.......10\}$

$\Rightarrow 3+b=4+a$

$leta=1\Rightarrow b=2$

therfore one pair $(a,b)=(1,2)$

similarly we can find the pairs $(a,b)$

Therefore equivalence class of $[3,4]=\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)\}$