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?

If $f$ were continuous, then it would have to be strictly monotonic, since it's bijective.  Suppose $f$ is decreasing, $x>0$.  Then, since $f(0)=0$ we get $f(x)<0$ and $f^{-1}(x)<0$, but $x=f(x)+f^{-1}(x)$.  So $f$ is increasing and $f(x)>0$ when $x>0$.  But then, $f^{-1}(x)$ is also positive for $x>0$, so $f(x)<x$.  Since $f$ is increasing, $f^{-1}(x)>x$.  But that's not possible.  
