Two functions discontinuous, but sum continuous I have some exercise that asks me this:
Find $f$ and $g$ discontinuos such that $f+g$ is continuous. This is what I tought:
$$f = \mbox{sign}(x)$$
$$g = -\mbox{sign}(x)$$
Where $\mbox{sign}(x)$ is the function that maps to $1$ if $x\ge0$ and $-1$ if $x<0$. Then, the two are discontinuous in $0$, but the sum, is continuous, because it's equal to $0$ in every point. First of all, is this example correct?
The exercise also asks me to find examples such that:
$f,g$ discontinuous, but $f$ composed with $g$ is continuous. (need help)
$f$ continuous, $g$ discontinuous but the composite $f$ with $g$ is continuous. 
(for the second one, if I take $g=\mbox{sign}(x)$ and $f=|x|$, then $fog = 1$. Is this right?

Could you guys give at least a hint, or a less poor example? 
 A: You can come up with some really interesting examples of compositions that end up being continuous where the components are not. Here's my favorite:
$f(x)=\begin{cases}
0 & \text{if }x \text{ is irrational}\\
1 & \text{if }x \text{ is rational}
\end{cases}$
Then $f\circ f$ is the constant function $1$ which is continuous.
A: Your examples are wonderful, with one correction: your sign function is actually not a function, because $\operatorname{sign}(0)$ maps to both $1$ and $-1$ (in high school algebra parlance, this fails the "vertical line test"). This is easy to correct, however. Just let $\operatorname{sign}(0)=0$. An example of discontinuous functions that compose to a continuous function runs thus: let $f(x)=0$ if $x=0$, and $f(x)=1$ otherwise; let $g(x)=1$ if $x=0$ or $x=1$, and $g(x)=2$ otherwise. $f(g(x))$ is thus continuous, while $f$ and $g$ are discontinuous.
A: If $f$ is discontinuous and has a discontinuous inverse, then let $g = f^{-1}$, 
so that $f \circ g$ is the identity function.
As a candidate for such an $f$, I offer $f(x) = 1 + 2 \lfloor x \rfloor - x$. But there are uncountably many such functions.
A: You can also take $f(x)=sgnx$ and $g(x)=x(1-x^2)$.  Here $f$ is discontinuous at $0$ and $g$ is continuous at $f(0)$ but the composite $gf$ is continuous at $0$.
