Final answer:
To prove that if f: A → A is flippy, then f is bijective, we need to show that f is both onto (surjective) and one-to-one (injective). Specifically, we need to demonstrate that for every element b in A, there exists an element a in A such that f(a) = b, and if f(a) = f(a'), then a = a'. Additionally, we can disprove the statement that if f: A → A and g: A → A are flippy, then f(g(x)) must be flippy.
Step-by-step explanation:
(a) To prove that if f: A → A is flippy, then f is bijective, we need to show that f is both onto (surjective) and one-to-one (injective).
First, let's prove that f is onto. To do this, we need to show that for every element b in A, there exists an element a in A such that f(a) = b.
Since f is flippy, we know that for every a in A, f(f(a)) = a. This means that f is its own inverse. So, if we let b = f(a), then f(f(a)) = a implies f(b) = a. This shows that f is onto (surjective).
Next, let's prove that f is one-to-one. To do this, we need to show that if f(a) = f(a'), then a = a'.
Suppose f(a) = f(a'). Applying f to both sides of this equation, we have f(f(a)) = f(f(a')). By the flippy property, this simplifies to a = a'. Therefore, f is one-to-one (injective).
Since f is both onto and one-to-one, it is bijective.
(b) We need to prove or disprove whether if f: A → A and g: A → A are flippy, then f(g(x)) is flippy.
Consider counterexamples:
Let A = {1, 2, 3}, f(x) = 1, and g(x) = 2 for all x in A.
For f(g(x)), we have:
f(g(1)) = f(2) = 1
As we can see, f(g(x)) = 1 for all x in A, which means it is not flippy since it does not satisfy the property f(f(a)) = a for all a in A.