Prove $f(x) =0$ for all $x \in [a, b]$ If $f:[a,b] \rightarrow R$ be continuous , let $f(x) =0$ when $x$ is rational. Prove $f(x) =0$ for all $x$ that is an element of $[a,b]$. 
Thanks for the help I can't seem to find the f[a,b] notation in my notes and it is throwing me off. 

Go back to the definition of continuity.  If such a function is continuous, then for any given $\varepsilon > 0$ and $x \in [a, b]$, there exists a $\delta > 0$ such that $|x-y| < \delta \implies |f(x) - f(y)|< \varepsilon$.  
So what problems arise if there is an $x \in [a, b] \setminus \mathbb{Q}$ such that $f(x) \neq 0$?  Well, we can always find a $y \in \mathbb{Q}$ that is as close to $x$ as we'd like (in math jargon, $\mathbb{Q}$ is dense in $\mathbb{R}$).  As a hint, let $\varepsilon$ be the value of $f(x)$.  
Can you fill in the details?
