f, g continuous for all rationals follow by continuous on all reals? 
Possible Duplicate:
Can there be two distinct, continuous functions that are equal at all rationals? 

Let $f, g:\Bbb{R}\to\Bbb{R}$ to be continuous functions such that $f(x)=g(x)\text{ for all rational numbers}\,x\in\Bbb{Q}$. Does it follow that $f(x)=g(x)$ for all real numbers $x$?
Here is what I think: 
f continuous when $\lim\limits_{x\to x_0}f(x)=f(x_0)$ and $\lim\limits_{x\to x_0}g(x)=g(x_0)$
So it does not neccesarily mean that $f(x)=g(x)$ when x is irrational. So I can pick a function f so that 
$f(x) =
\begin{cases}
g(x)  & \text{if $x\in\Bbb{Q}$} \\
x & \text{if $x\in\Bbb{R}\setminus \Bbb{Q}$}  \\
\end{cases}  
$

Hint: prove that if $\,h\,$ is a real continuous function s.t. $\,h(q)=0\,\,,\,\,\forall\,q\in\Bbb Q\,$ , then $\,h(x)=0\,\,,\,\,\forall\,x\in\Bbb R\,$
Further hint: For any $\,x\in\Bbb R\,$ , let $\,\{q_n\}\subset\Bbb Q\,$ be s.t. $\,q_n\xrightarrow [n\to\infty]{} x\,$ . What happens with
$$\lim_{n\to\infty}f(q_n)\,\,\,?$$
