real analysis: show that a continuous function is defined for irrationals

Let f be a continuous function defined on (a, b). Supposed f(x)=0 for all rational numbers x in (a, b). Prove that f(x)=0 on (a, b).

i dont even know where to start...any tips just to point me in the right direction?
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 Recognitions: Homework Help There's a couple of ways you can approach this problem. Did you try proof by contradiction? Let f be a continuous function defined on (a, b). Assume f(x)=0 for all rational numbers x in (a, b) and assume that f(x)≠0 on (a, b), so...
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus You'll need two things for this problem: 1) If f is continuous and if $x_n\rightarrow x$, then $f(x_n)\rightarrow f(x)$. 2) For every real number x there exists a sequence of rational numbers that converges to x. This is saying that $\mathbb{Q}$ is dense in $\mathbb{R}$ Try to do something with these things...