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Let $x$ be any real number. Construct a sequence $x_n$ of rational numbers such that $x = \sup\{x_n : n \in \mathbb{N} \}.$ I was trying $x_n = [ 10^n x ]/10^n$, but is it actually monotone increasing? If so, how to prove it analytically?

Thanks for any help.

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    Yes,it is a requirement and I ment the greatest integer function2012-08-01

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Yes, your sequence $x_n = \dfrac{\lfloor 10^n x \rfloor}{10^n}$ will do the job. Your sequence is also monotone increasing.

HINT:

To prove monotone increasing, try to prove the following first: $a \lfloor x \rfloor \leq \lfloor ax \rfloor$ where $a \in \mathbb{Z}^+$ and $x \geq 0$.

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Your sequence is easy justify

$\left|\frac{\lfloor 10^nx\rfloor}{10^n}-x\right|=\left|\frac{\lfloor 10^nx\rfloor-10^nx}{10^n}\right|=\frac{\text{fractional part of }10^nx}{10^n}\le\frac1{10^n}$

You see from here that the convergence is exponential, which very rapid!

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Yes. First show that $\lfloor x \rfloor \le \frac{ \lfloor a \cdot x \rfloor }{a}$. Then it should be clear that $x_{n+1} \ge x_n$ when $a = 10$. (This is essentially the same as @Marvis' answer.)

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    To be a smart-ass, one might say that the sequence $x_n$ is provided by the _definition_ of $x$ as a Dedekind cut or Cauchy sequence. QED (almost).2012-08-04
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As others mentioned, your sequence is monotone, but not necessarily strictly increasing. For example, if $x = 0$ all $x_n = 0$. If you want it strictly increasing, you might try $x_n = \dfrac{[10^n x] - 1}{10^n}$.