Question

Let $A=\{1,2,3,....9\}$ and $R$ be 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 $\left[\right(2,5\left)\right].$

Answer 1

$A=\{1,2,\mathrm{3...9}\}$
$R$ in $A\times A$
$(a,b)$ $R$ $(c,d)$ if $(a,b)(c,d)$ $\in$ $A\in A$
$a+b=b+c$
Consider $(a,b)$ $R$ $(a,b)$ $(a,b)\in A\times A$
$a+b=b+a$
Hence, $R$ is reflexive.
Consider $(a,b)$ $R$ $(c,d)$ given by $(a,b)$ $(c,d)$ $\in$ $A\times A$

$a+d=b+c=>c+b=d+a$
$\Rightarrow (c,d)R(a,b)$
Hence $R$ is symmetric.
Let $(a,b)$ $R$ $(c,d)$ and $(c,d)$ $R$ $(e,f)$
$(a,b),(c,d),(e,f),\in A\times A$
$a+b=b+c$ and $c+f=d+e$
$a+b=b+c$
$\Rightarrow a-c=b-d$-- (1)
$c+f=d+e$-- (2)
Adding (1) and (2)
$a-c+c+f=b-d+d+e$
$a+f=b+e$
$(a,b)R(e,f)$
$R$ is transitive.
$R$ is an equivalence relation.
We select from set $A=\{1,2,3,....9\}$
$a$ and $b$ such that $2+b=5+a$

so $b=a+3$
Consider $(1,4)$
$(2,5)$ $R$ $(1,4)\Rightarrow 2+4=5+1$
$\left[\right(2,5)=(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)]$ is the equivalent class under relation $R.$