Continuous and discontinuous function problem The following problem can be true?
Problem: For every positive integer $n>1$ there exists a function $ f\left( x \right)$ on $\mathbb{R}$ which satisfies both following conditions:
(i) $f\left( x \right),{\rm{ }}f\left( {f\left( x \right)} \right), \ldots ,{\rm{ }}f\left( { \ldots f\left( x \right) \ldots } \right)$ ( $ (n-1)$ times $ f $) discontinuous at every $x$ belong to $\mathbb{R}$.
(ii) $ f\left( { \ldots \left( {f\left( x \right) \ldots } \right)} \right)$ ( $n$ times $f$ ) continuous in $ \mathbb{R}$.
 A: Here is one that works
for $n=2$:
Let $f(x) = 0$ if $x$ is rational 
and $1$ if $x$ is irrational.
Then $f(x)$ is discontinuous
for all $x$
but $f(f(x))=0$ for all $x$
and is continuous.
At the moment,
I don't see how to delay
the result of $n-1$ iterations
of $f$
to only rational values
while the previous iterations
can be anywhere.
A: Fix $x_1,x_2,\ldots,x_n\in\mathbb R$ which are linearly independent over $\mathbb Q$. For each $\alpha\in\mathbb Q$, let $f(\alpha x_1)=0$ and $f(\alpha x_i)=x_{i-1}$ for $i=2,\ldots,n$; let $f(x)=0$ for all other $x\in\mathbb R$. Clearly $f$ composed with itself $n$ times is the zero function which is continuous. For $k<n$, $f$ composed with itself $k$ times sends the dense subset $\{\alpha x_{k+1}\,:\,\alpha\in\mathbb Q\}$ to $x_1\neq0$ and the dense subset $\{\alpha x_{k}\,:\,\alpha\in\mathbb Q\}$ to $0$. Hence this function is nowhere continuous.
A: Let $A_{n}$ denote the set of real numbers which solve an integer polynomial of degree $n$, but are not roots of any polynomial of degree $< n$. Then $A_{1} = \mathbb{Q}$, and $A_{2}$ is the set of all irrational (real) solutions to quadratic polynomials, etc. Let
\begin{align*}
f(x) & = \begin{cases}
0 & \textrm{if } x \not \in A_{1}, \ldots, A_{n}, \\
2 & \textrm{if } x \in A_{1} , A_{2} , \\
\sqrt{2} & \textrm{if } x \in A_{3} , \\
2^{1 / 3} & \textrm{if } x \in A_{4} , \\
\vdots & \vdots \\
2^{1 / (n - 1)} & \textrm{if } x \in A_{n} .
\end{cases}
\end{align*}
That should do it, I'm pretty sure.
