Assume that there exists a function f:R→R that is bijective and satisfies
f(x)+f−1(x)=x
for all x. Here f−1 is the inverse function. Show that f is odd.This was a brain-teaser given to me by a friend. Two other related questions are:
- Show that f is discontinuous
- Give an example of such a function (if indeed one exists).
Edit: As an initial idea, maybe approaching the problem graphically would help? A function and its inverse are reflections of each other about y=x on the x–y plane. Does this lead to anywhere?
Answer
Here is finally a constructive example of a solution, continuous except on a countable set.
Let ϕ=1+√52, and let
f(x)={0if x=0−ϕxif |x|∈[ϕ3k,ϕ3k+1),k∈Zϕxif |x|∈[ϕ3k+1,ϕ3k+2),k∈Zϕ−2xif |x|∈[ϕ3k+2,ϕ3k+3),k∈Z
Then f is a bijection, whose inverse is
f−1(x)={0if x=0ϕ2xif |x|∈[ϕ3k,ϕ3k+1),k∈Z−ϕ−1xif |x|∈[ϕ3k+1,ϕ3k+2),k∈Zϕ−1xif |x|∈[ϕ3k+2,ϕ3k+3),k∈Z
Checking each case shows that f(x)+f−1(x)=x, as required.
Attribution
Source : Link , Question Author : Iconoclast , Answer Author : yoann