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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)].

1 Answer

  • 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 $(2,5) R (a,b)$ $2+b=5+a$
R in $A \times A$
$(a,b) R (c,d)$ if $\qquad (a,b)(c,d) \in A \times A$
Consider $(a,b) R (a,b) \qquad (a,b) \in A \times A$
Hence R is reflexive
Consider $(a,b) R (c,d)$ given by $(a,b)(c,d) \in A \times A$
$=> (c,d) R (a,b)$
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 $c+f=d+e$
$=>a-c=b-d \qquad (1) \qquad c+f=d+e \qquad(2)$
adding (1) and (2)
$=> (a,b) R(e,f)$
R is transitive
R is an equivalnce relation
we select from set $A=\{1,2,3,....9\}$
a and b such that
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 meena.p
edited Mar 27, 2013 by meena.p

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