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Is $$S = \left\{ \frac{m}{10^n}\,\Bigm|\, m,n\in\mathbb{Z}, n\geq 0\right\}$$ dense in the set of real numbers ?

I guess that $S$ is dense in $\mathbb{R}$ and so I was tring to produce a sequence in $S$ that converges to any given real number . Thanks for any help .

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Truncating a real $x\in\Bbb R$ to the first $n$ decimal digits results in an element of $S$, so naturally we can use this to find a converging sequence. In particular, given $x$ define $$x_n:= \frac{\lfloor 10^nx \rfloor}{10^n}=\frac{10^nx-\{10^n x\}}{10^n}=x-\frac{\epsilon_n}{10^n}$$ with $\lfloor\cdot\rfloor$ and $\{\cdot\}$ the floor and fractional part functions respectively. Given $\epsilon_n\in[0,1)$ and $10^{-n}\to0$ we find that $x_n\to x$. –  anon May 21 '12 at 2:34
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3 Answers

up vote 2 down vote accepted

Or, more explicitly:

Choose $x \in \mathbb{R}$, and let $x_n = \frac{\lfloor 10^n x \rfloor}{10^n}$. Then $x_n \in S$, $\forall n$, and $|x-x_n| < \frac{1}{10^n}$. Hence $x_n \rightarrow x$, so $S$ is dense.

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Hint. Decimal expansion. Also: for every $\epsilon\gt 0$ there exists $b\in\mathbb{Z}$, $b\gt 0$ such that $\frac{1}{10^b}\lt \epsilon$.

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Thanks to everyone for helping me out. –  Ester May 21 '12 at 2:47
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Hint: Any real number $r$ has a decimal expansion $d_n\cdots d_0.d_{-1}\cdots$. What happens if you consider the sequence $d_n\cdots d_0.d_{-1},d_n\cdots d_0.d_{-1}d_{-2},\ldots$?

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