# Increasing Sequence of Rationals

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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Try to read about Continued Fractions. – Rijul Saini Aug 1 '12 at 4:10
In your first sentence, there is no requirement that the sequence be monotone increasing. Is that a requirement, and if so, does it have to be strictly increasing? Did you mean the square brackets to be the floor function? – Ross Millikan Aug 1 '12 at 4:12
Yes,it is a requirement and I ment the greatest integer function – Ester Aug 1 '12 at 10:16

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

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

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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 $x_n = \dfrac{[10^n x] - 1}{10^n}$.

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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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Thanks everyone for your kind help.Wasn't there another answer that does not use my example? – Ester Aug 1 '12 at 10:46
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). – Dan Brumleve Aug 4 '12 at 6:35