Proving that a function is odd

Question 1

Assume that there exists a function $f:\mathbb{R}\to\mathbb{R}$ 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?

Question 2

Here is finally a constructive example of a solution, continuous except on a countable set.

Let $\phi = \frac{1+\sqrt5}{2}$ , and let
$f(x) = \begin{cases} 0 & \text{if }x = 0 \\ -\phi x & \text{if }|x| \in [\phi^{3k}, \phi^{3k+1}), k \in \mathbb Z \\ \phi x & \text{if } |x| \in [\phi^{3k+1}, \phi^{3k+2}), k \in \mathbb Z \\ \phi^{-2} x & \text{if } |x| \in [\phi^{3k+2}, \phi^{3k+3}), k \in \mathbb Z \\ \end{cases}$

Then $f$ is a bijection, whose inverse is
$f^{-1}(x) = \begin{cases} 0 & \text{if }x = 0 \\ \phi^2 x & \text{if }|x| \in [\phi^{3k}, \phi^{3k+1}), k \in \mathbb Z \\ -\phi^{-1} x & \text{if } |x| \in [\phi^{3k+1}, \phi^{3k+2}), k \in \mathbb Z \\ \phi^{-1} x & \text{if } |x| \in [\phi^{3k+2}, \phi^{3k+3}), k \in \mathbb Z \\ \end{cases}$
Checking each case shows that $f(x)+f^{-1}(x) = x$ , as required.

Proving that a function is odd

Answer

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