 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/1/c/i/solution

The function is injective and surjective, with inverse function $f^{- 1} (y) = 1 - y$ . Equality

1 - x = 1 - x^{'}

x = x^{'}

. For any

y \in (0, 1)

, the input

x = 1 - y

also lies in

(0, 1)

and satisfies

f (x) = y

. Applying the function twice returns the original input.

 New to topics? Read the docs here!