Practicing Success
R is a relation from {11, 12, 13) to {8, 10, 12} defined by y = x - 3. Then $R^{-1}$ is: |
{(8, 11), (10, 13)} {(11, 18), (13, 10)} {(10, 13), (8, 11)} None of these |
{(8, 11), (10, 13)} |
R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x - 3 ⇒ x - y = 3 R = {11, 8}, {13, 10}. Hence $R^{-1}=\{(8,11);(10,13)\}$ |