Let f : X → Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1)-1 = f.
Let f : X → Y be an invertible function.
Then, there exists a function g: Y → X such that gof = IX and fog = IY
Here, f-1 = g.
Now, gof = IX and fog = IY
⇒ f-1 of = IX and fof -1= IY
Hence, f-1 : Y → X is invertible and f is the inverse of f-1
i.e., (f-1)-1 = f.
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
During the festival season, a mela was organized by the Resident Welfare Association at a park near the society. The main attraction of the mela was a huge swing, which traced the path of a parabola given by the equation:\[ x^2 = y \quad \text{or} \quad f(x) = x^2 \]