Suppose that f : (a, b) → R is continuous and that f(r) = 0 for ever rational number r є (a, b). Prove that f(x) = 0 for all x є (a, b).
Very nice- and not at all what I was thinking of. I was thinking of a proof by contradiction. Suppose there exist some a such that $\displaystyle f(a)= b\ne 0$. Let $\displaystyle \epsilon= |b|/2$. Then for all $\displaystyle \delta> 0$, there exist rational x in the interval $\displaystyle (a-\delta, a+ delta)$. For that x, $\displaystyle |x- a|< \delta$ but $\displaystyle |f(x)- f(a)|= |0- b|= |b|> |b|/2= \epsilon$, contradicting the fact that f is continuous.