Proving that a function is odd 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?

Let's call $(E)$ the equation $f(y) + f^{-1} (y) = y$
Let $a \in \mathbb{R}$. Since $f$ is bijective, there exists $x \in \mathbb{R}$ such that $f(x) = a$. Now, since $f^{-1}(a) = x$, we have (using $(E)$ with $y = a$) $f(a) = a - x$. Now, using $(E)$ with $y = x$:
$$f^{-1}(x) = x - a$$
Applying $f$ on both side:
$$f(x-a) = x $$
Using $(E)$ with $y = x-a$:
$$f^{-1}(x-a) = x - a - f(x -a) = x -a -x = a$$
Applying $f$ on both side:
$$f (-a ) = x -a $$.
Thus $f(a) = a - x = -f(-a)$. This is true for all $a \in \mathbb{R}$: $f$ is therefore odd.
