Given R={(x,y):4y=5x−3, x,y∈M} where M={1,2,…,16}.
From 4y=5x−3, we get y=45x−3.
For y to be a positive integer, 5x−3≡0(mod4), i.e., x≡3(mod4).
So x=3,7,11,15.
x=3⇒y=3,
x=7⇒y=8,
x=11⇒y=13,
x=15⇒y=18∈/M.
R={(3,3), (7,8), (11,13)}.
For symmetry, (7,8)∈R requires (8,7)∈R and (11,13)∈R requires (13,11)∈R.
(3,3) is already symmetric.
Minimum elements to be added =2.