Suppose thatf: (a,b) →Ris continuous and thatf(r) = 0 for ever rational numberrє (a,b). Prove thatf(x) = 0 for allxє (a,b).

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- May 3rd 2009, 03:16 AMbearej50Continuous functionsSuppose 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*).

- May 3rd 2009, 03:23 AMInfophile
Hello,

For all $\displaystyle x\in [a,b]$ reals, it exists a sequence of rationals $\displaystyle (r_n)$ such as $\displaystyle r_n\longrightarrow x$.

Since $\displaystyle f$ is continuous then $\displaystyle f(r_n)=0\longrightarrow f(x)$

And so...

:) - May 3rd 2009, 08:02 AMHallsofIvy
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.