Let A = {0, 3, 4, 6, 7, 8, 9, 10} and R be the relation defined of A such that $\mathrm{R}=\{(\mathrm{x},\mathrm{y})\in \mathrm{A}\times \mathrm{A}:\mathrm{x}-\mathrm{y}\mathrm{is}\mathrm{odd}\mathrm{positive}\mathrm{integer}\mathrm{or}\mathrm{x}-\mathrm{y}=2\}$. The minimum number of elements that must be added to the relation R, so that it is a symmetric relation, is equal to …….. 

A = {10, 9, 8, 7, 6, 4, 3, 0}
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)}
All the elements of R, (a, b) are of type a > b.
Hence, we need to add total of 19 more elements. 
to R to make in symmetric.$\begin{array}{l}\mathrm{R}=\left\{\right(3,0\left)\right(4,3\left)\right(6,3\left)\right(7,0\left)\right(7,4\left)\right(7,6\left)\right(8,3\left)\right(8,7\left)\right(9,0)\\ (9,4)(9,6)(9,8)(10,3)(10,7)(10,9)(6,4)\\ (8,6)(9,7),(10,8)\}\end{array}$