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

Toolbox:
• 1.R is an equivalance relation if R is
• a) reflexive ie $(a,b) \in A \times A$ $(a,b) R(a,b)$
• b) symmetric ie $(a,b) R (c,d) => (c,d) R (a,b)$ $(a,b)(c,a) \in A \times A$
• c) transitive ie $(a,b) R (c,d) ;(c,d) R (e,f) =>(a,b) R (e,f)$ $(a,b)(c,d),(e,f) \in A \times A$
• 2.Equivalent class of [(2,5)] under R is given by set of ordered pair $(a,b) \in A \times A$ such that $(2,5) R (a,b)$ $2+b=5+a$
$A=\{1,2,3......9\}$

R in $A \times A$

$(a,b) R (c,d)$ if $\qquad (a,b)(c,d) \in A \times A$

$a+b=b+c$

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

$a+b=b+a$

Hence R is reflexive

Consider $(a,b) R (c,d)$ given by $(a,b)(c,d) \in A \times A$

$a+d=b+c=>c+b=d+a$

$=> (c,d) R (a,b)$

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 $c+f=d+e$

$a+b=b+c$

$=>a-c=b-d \qquad (1) \qquad c+f=d+e \qquad(2)$

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 equivalnce relation

we select from set $A=\{1,2,3,....9\}$

a and b such that

$2+b=5+a$

Consider (1,4)

$(2,5) R (1,4) => 2+4 =5+1$

$[(2,5)]=\{(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)\}$ is the equivalent class. under relation R

answered Mar 5, 2013 by
edited Mar 27, 2013 by meena.p