Let A = 0 , 3 , 4 , 6 , 7 , 8 , 9 , 10 and R be the relation defined on A such that R x , y ∈ A × A : x - y is odd positive integer or x - y = 2 . The minimum number of elements that must be added to the relation R , so that it is a symmetric relation, is equal to _ _ _ _ _ _ _ _ _

Step-by-step solution

Given, Set A = 10 , 9 , 8 , 7 , 6 , 4 , 3 , 0 Now relation x - y  is odd or   x - y = 2 can be given by, R = ( 10 , 9 ) , ( 10 , 8 ) , ( 10 , 7 ) , ( 10 , 3 ) , ( 9 , 8 ) , ( 9 , 7 ) , ( 9 , 6 ) , ( 9 , 4 ) , ( 9 , 0 ) , ( 8 , 7 ) , ( 8 , 6 ) , ( 8 , 3 ) , ( 7 , 6 ) , ( 7 , 4 ) , ( 7 , 0 ) , ( 6 , 4 ) , ( 6 , 3 ) , ( 4 , 3 ) , ( 3 , 0 ) So, total there are 19 elements and all the elements of R ,   ( a ,   b ) are of type a   >   b . Hence, we need to add total of 19 more