Q:

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:
$$
f(x) =
\begin{cases}
1 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\cup \{0\}\\     
0 &\text{otherwise}
\end{cases}
$$

A:

You can modify the fractional part function: $ \{x\} = \lceil x \rceil - x $, which is discontinuous at integers; to
$$
f(x) = \begin{cases}
0 & ; x = 0 \\
\left\{\dfrac{1}{x}\right\} & ; otherwise
\end{cases}
$$

