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Here is the sequence of irrational number that converges to $\alpha\in \mathbb{Q}$. Take $x_n=\alpha -\frac{\sqrt{2}}{n}$. Clearly $\{x_n\}\rightarrow\alpha .$

But I'm trying to find a sequence of rationals that converges to arbitrary irrational number $\beta$. Can you give me such a sequence.

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  • $\begingroup$ See math.stackexchange.com/questions/670083/… $\endgroup$ – Autolatry Jan 15 '15 at 11:48
  • $\begingroup$ I believe that the "uncountable majority" of irrational numbers cannot be represented with such sequence of rational numbers. Specifically, I am referring to all those that are not computable numbers. See this answer, which gives an example of such number. $\endgroup$ – barak manos Jan 15 '15 at 12:22
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    $\begingroup$ @barakmanos: I don't think computability is relevant here. Petite Etincelle's answer is perfectly valid. $\endgroup$ – TonyK Jan 15 '15 at 12:39
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Take $\dfrac{\lfloor 10^n \beta\rfloor}{10^n} $ for example, then we have

$$\dfrac{10^n \beta -1}{10^n} \leq\dfrac{\lfloor 10^n \beta\rfloor}{10^n} \leq \dfrac{ 10^n \beta}{10^n}=\beta $$

Since $\dfrac{10^n \beta -1}{10^n} = \beta - \dfrac{1}{10^n} \to \beta$, by squeeze theorem, we have $\dfrac{\lfloor 10^n \beta\rfloor}{10^n} \to \beta$

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  • $\begingroup$ I thought about it but I couldn't. $\endgroup$ – user194772 Jan 15 '15 at 12:02
  • $\begingroup$ @user194772 I've added details $\endgroup$ – Petite Etincelle Jan 15 '15 at 12:10
  • $\begingroup$ @PetiteEtincelle Is there anything special about chosing $10^{n}$?? Will $x_{n} = \frac{[a^{n}\beta]}{a^{n}}$ where $a \in N$ and $a>2$ work?? $\endgroup$ – crskhr Mar 8 '15 at 14:26
  • $\begingroup$ @S.C. nothing is special with $10$, other examples that you mentioned work as well $\endgroup$ – Petite Etincelle Mar 8 '15 at 14:50

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