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please help me to do this problem...

Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f(q)=\sin q$ for $q\in\mathbb Q$ (rational numbers). Find the value of f(π/4).

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    to Martin,,I referred your link about cts function between metric spaces are equal....but that says about two functions right???here there's only one and I know Q is dense in R and the thing is I cant find any function that is convergent to π/4.2012-10-17

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I'd do it as follows:

$f$ is continuous which means that you can swap limits and function: $\lim_{n \to \infty} f(x_n) = f( \lim_{n \to \infty} x_n) = f(x)$. So to find $f( \pi / 4)$ you pick a sequence $x_n \to \pi / 4$ and investigate what $f(x_n)$ converges to.

I looked it up (because I didn't know any such sequence off the top of my head), and for example, you could take the formula found by Leibniz:

$ \frac{\pi}{4} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k - 1}$

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    @cccjay I think when you wrote $n$ was correct: you want $y = \lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^{k+1}}{2k - 1}$. If $f$ is the sine function, shouldn't $f(y) = f(\pi/4) \neq 0 $?2012-10-19
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The problem OP seems to have is with finding a sequence of rationals converging to $\pi/4$.

If you know that the rationals are dense in the reals, you can just state that fact as a proof of the existence of such a sequence of rationals.

If you don't want to (or can't) use that, then think about the decimal expansion of $\pi/4$. How can you get rational numbers out of that, rational numbers that converge to $\pi/4$?