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given a relation on R on the set A={1,2,3,4} in the form of matrix representation as ,

$M_R$=$\begin{bmatrix} 0 & 1 & 0 & 0\\ 0& 0& 1 &0 \\ 0& 0& 0 &1 \\ 0& 0& 0& 0 \end{bmatrix}$

Then the cardinality of the smallest equivalence relation on A which contains R is equal to

answer given-16
asked in Set Theory & Algebra by Loyal (9.3k points)
edited by | 50 views
0
16 is the correct answer
0
can you please express your approach.
+1
(1,1) , (2,2) , (3,3) , (4,4) ,

(1,2) , (2,3) , (3,4) --- > matrix representation

(2,1) , (3,2) , (4,3) --> build that relation into  symmetric  relation

(2,4) , ( 1,3) , (4,2) , (3,1) ,  (1,4), (4,1)  ---- > transitive relation
+1

@Prateek Raghuvanshi

you didn't the condition where  (2,3) ,(3,4)  , [ (2,4) --- > As per transitive rule  ]

0
You will also have to add (1,3),(2,4)(3,1)(4,2),(1,4),(4,1).. and also all reflexive relations
+1

 thank you ,i got what i did mistake .

1 Answer

+2 votes
Given adjacency from the matrix:

(1,2) (2,3) (3,4)

to satisfy reflexive: adding (1,1) (2,2) (3,3) (4,4)

to satisfy symmetric: adding (2,1) (3,2) (4,3)

to satisfy transitive: adding (1,3) (2,4) (4,2) (3,1) (1,4) (4,1)

Hence, the cardinality is 16
answered by Junior (953 points)

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