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*