Function that is discontinuous only for integer fractions I have this question:

Find a function $f :\mathbb R \to\mathbb R$ which is discontinuous at the points of the
  set $\{\frac1n : n \text{ a positive integer}\} \cup \{0\}$ but is continuous everywhere
  else.

I really don't know what to do. I was thinking maybe:
$$
f(x) =
\begin{cases}
1 \quad&\text{if }x=0  \\
0 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\\     
x &\text{otherwise}
\end{cases}
$$
But that kind of seems like 'cheating'. Is there a better example?
EDIT: Would it be better to have:
$$
f(x) =
\begin{cases}
1 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\cup \{0\}\\     
0 &\text{otherwise}
\end{cases}
$$

I'm going to assume your true question is finding an answer that you do not consider "cheating."
Question/Problem
Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$
General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function.
Let $\epsilon>0$ be an arbitrary positive real number.  
Your edited answer has $g$ be the zero function and $\epsilon$ be $1$ 
Define $f:\mathbb{R}\to\mathbb{R}$ by
$$f(x)=\begin{cases}
g(x)+\epsilon&\text{if }x\in K\\
g(x)&\text{if }x\in(\mathbb{R}\setminus K)
\end{cases}$$
for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.
The reason why I introduce "$K$" is because this method can be used for any given nowhere dense set $K$ where you want discontinuities.
 The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem.
