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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 $x$?

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    Lei sin cuidado la pregunta. Como esta escrita, el unico numero que es limite de una sucesion de la forma que usted pone es el $0$. Apuesto a que no escribio lo que queria preguntar, y sugiero que trate con cuidado de escribir correctamente lo que de verdad quiere preguntar. Yo ya me voy a dormir... (Translation: I read the question carelessly. As written, the only number that is the limit of such a sequeence is $0$. I'll bet you did not write what you meant to ask, and I suggest you try carefully to write down what you actually want to ask. Me, I'm going to bed...)2011-09-17
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    The quantifier is surely not in the intended place. The number $m$ is not intended to be fixed.2011-09-17
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    What definition of real number do you have?2011-09-17
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    Is this Andres from the "Topology Exercise" question? You asked the same question as he did in the comment. Andres also wrote his question incorrectly multiple times. Anyway, do you mean that $x = m \sum_{n \in \mathbb{Z}} a_n \frac{1}{2^m}$ where $a_n$ is $0$ or $1$? If this is the case, then as Magidin noted, this can only hold if $x$ and $m$ have the same sign (or both zero). If $m \neq 0$, then just use the binary for $\frac{x}{m}$ to get the necessary $a_n$'s.2011-09-17
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    New version: If $f$ is a continuous function and $f(m/2^n)=0$ for all integers $m$ and all natural numbers $n$, how do I prove that $f(x)=0$ for all real numbers $x$?2011-09-17
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    Did Jerson change this question? Although Jerson wrote his question incorrectly, I don't think this question is in same spirit as the original. I think he wanted something like an expansion of a real number.2011-09-17
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    @William: Yes, he changed it completely. I think that he may have been trying to answer the present question, realized that getting each real as the limit of a sequence of dyadic rationals would do the trick, asked the subsidiary question badly, and changed it to the main question when his wording was challenged.2011-09-17
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    @Arturo Magidin: What is mixedmath? I hoped it was some clever translator but could not find any sense when I googled it.2011-09-17
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    @AD.: http://math.stackexchange.com/users/9754/mixedmath2011-09-17
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    @William: Yes, he changed it; if the original question was meant to be what most people think it was (with $m$ not fixed), or something similar, then answering the original question implies this one (since you would be proving the dyadic rationals are dense).2011-09-17

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