Continuous  function that is zero in at all rationals of the form $m/2^n$ Si $f$ es una función continua y si $f(m/2^n)=0$ para todo entero $m$ y todo natural $n$, ¿cómo demuestro que $f(x)=0$ para todo numero real $x$?

[translation by mixedmath]
If $f$ is a continuous function and if $f\left(\dfrac{m}{2^n}\right) = 0$ for all integers $m$ and all natural $n$, how do I show that $f(x) = 0$ for all real $x$?
 A: Let $x\in\mathbb{R}$, and let $\epsilon\gt 0$. Then there exists $\delta\gt 0$ such that for all $z$, if $|x-z|\lt\delta$ then $|f(x)-f(z)|\lt\epsilon$. Let $n$ be a positive integer such that $\frac{1}{2^{n}}\lt \delta$. Then there is an integer multiple of $\frac{1}{2^{n+1}}$ that lies in $(x-\frac{1}{2^n},x+\frac{1}{2^n})$, since the interval has length $\frac{1}{2^{n-1}}$. Let $z=m/2^{n+1}$ be that integer multiple. Then
$\epsilon\gt |f(x)-f(z)| = |f(x)-0| = |f(x)|$.
Thus, $|f(x)|\lt \epsilon$ for all $\epsilon\gt 0$, hence $f(x)=0$. 

Sea $x\in\mathbb{R}$, y sea $\epsilon\gt 0$. Existe $\delta\gt 0$ tal que para todo $z$, si $|x-z|\lt\delta$ entonces $|f(x)-f(z)|\lt\epsilon$. Sea $n$ un entero positive talk que $\frac{1}{2^n}\lt\delta$. Entonces hay un multiplo entero de $\frac{1}{2^{n+1}}$ en el intervalo $(x-\frac{1}{2^n},x+\frac{1}{2^n}$, pues el intervalo tiene longitud $\frac{1}{2^{n-1}}$. Sea $z=m/2^{n+1}$ ese multiplo entero. Entonces $\epsilon\gt |f(x)-f(z)| = |f(x)-0| = |f(x)|$.
Por lo tanto, $|f(x)|\lt\epsilon$ para toda $\epsilon\gt 0$, de manera que $f(x)=0$.
A: Added: The question changed while the answer below was being typed.  A short note is added at the end to answer the altered question.
The wording of the original question does not accurately convey the intention.  I believe that the OP wants a proof that every real number $x$ is the limit of a sequence
$(a_k)$, where each $a_k$ is of the form $m/2^n$, with $m$ an integer and $n$ a non-negative integer.  (Numbers of the form $m/2^n$ are called dyadic rationals.)
It is enough to show that for any positive integer $k$, we can find a number $a_k$ of the required form such that $|x-a_k|<1/k$.
We prove the result for $x>0$.  A small modification takes care of the case $x<0$.
Let $n$ be the smallest integer such that $2^n>k$.  Consider the numbers $0/2^n$, $1/2^n$, $2/2^n$, $3/2^n$, and so on.  
Let $m$ be the smallest non-negative integer such that $(m+1)/2^n>x$.  Then $m/2^n \le x$. Let $a_k=m/2^n$. Since 
$$\frac{m+1}{2^n} -\frac{m}{2^n}=\frac{1}{2^n} <\frac{1}{k},$$
it follows that
$$0\le x-\frac{m}{2^n}=x-a_k <\frac{1}{k}.$$ 
Added Note:  The question was changed and now asks one to show that if $f$ is a continuous function which is $0$ on the dyadic rationals, then $f$ is $0$ everywhere.
Let $x$ be a real number.  We show that $f(x)=0$.  Let $(a_k)$ be a sequence of dyadic rationals with limit $x$. Such a sequence exists by the calculation above. Then
$$f(x)=\lim_{k\to \infty} f(a_k) =0.$$
A: Let $D = \left\{\frac{m}{2^n}:m\in\mathbb{Z} \land n\in\mathbb{N}\right\}$. (Members of $D$ are called dyadic rationals.) Since $f$ is continuous, you know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$ whenever $\lim\limits_{n\to\infty}x_n=x$. Let $x$ be any real number. If we can find a sequence $\langle x_n:n\in\mathbb{N} \rangle$ of dyadic rationals such that $\lim\limits_{n\to\infty}x_n=x$, we’ll be done, because then $f(x) = \lim\limits_{n\to\infty}f(x_n)=0$.
One way to do this is to use the binary expansion of $x$. Suppose that $\displaystyle x = \sum\limits_{k=m}^{\infty} b_k2^{-k}$ for some integer $m\le 1$, where each $b_k$ is either $0$ or $1$. For each $n\in \mathbb{N}$ let $\displaystyle x_n = \sum\limits_{k=m}^n b_k2^{-k}$. Then $$x_n = \frac{\sum_{k=m}^n b_k 2^{n-k}}{2^n} \in D,$$ and it's clear that $\lim\limits_{n\to\infty}x_n=x$.
(I’m sorry that I can’t translate this into Spanish.)
