Practicing Success
If R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e., (a, b) ∈ R ⇔ a < b, then $RoR^{-1}$ is: |
{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} {(3, 3), (3, 5), (5, 3), (5, 5)} {(3, 3), (3, 4), (4, 5)} |
{(3, 3), (3, 5), (5, 3), (5, 5)} |
We have, R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} $R^{-1}=\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}$ Hence $RoR^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$ |