Question

If $A=\{1,2,3,....,9\}$ 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 $\left[(2,5)\right]$.

Answer 1

(i) Reflexive : $(a,b)R(c,d)$ if $(a,b)(a,d)\in A\times A$

$a+b=b+c$

Consider $(a,b)R(a,b)\Rightarrow (a,b)\in A\times A$

$\Rightarrow a+b=b+a$ Hence $R$ is reflexive

(ii) Symmetric : $(a,b)R(c,d)$ given by $(a,b)(c,d)\in A\times A$

$a+d=b+c\Rightarrow c+b=d+a$

$\Rightarrow (c,d)R(a,b)$ Hence $R$ is symmetric

(iii) equivalence relation = reflexive + symmetric + transitive

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

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

$a+b=b+c$ and $4c+f=d+e$

$a+b=b+c$

$\Rightarrow a-c=b-d...\left(1\right)$ & $c+f=d+e...\left(2\right)$

adding (1) & (2) $a+f=b+e\Rightarrow (a,b)R(e,f)$ Transitive

So, $R$ is equivalence relation

from $A=\{1,2,\mathrm{3...9}\}$ are will select $a$ and $b$ such that

$2+b=5+a$

Let $b=4$ and $a=1$ Hence $(2+4)=(5+1)$

and $\left[\right(2,5\left)\right]=\left\{\right(1,4\left)\right(2,5),(3,6\left)\right(4,7\left)\right(5,8\left)\right(6,9\left)\right\}$ is

the equivalent class under relation $R$