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**ynn6871** Let $\displaystyle f:[a,b] -> R $ be continuous on $\displaystyle [a,b]$ and suppose that $\displaystyle f(x) = 0$ for each rational number x in $\displaystyle [a,b] $.Prove that $\displaystyle f(x)=0$ $\displaystyle \forall x \in [a,b]$

Ok, Do I need to do a proof by contradiction to prove that $\displaystyle f(x)=0$ $\displaystyle \forall x \in [a,b]$. What it comes down to is proving that $\displaystyle f(x) = 0$ for each irrational number $\displaystyle x \in [a,b]$, right???

Otherwise I don't really see how to do it, I mean even if I do by contradiction the way I thought about doing it was maybe say something like Assume that $\displaystyle f(x) \neq 0$ $\displaystyle \forall x \in [a,b]$

I am not sure... any suggestion?