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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.

Saikai Prime
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user194772
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  • See http://math.stackexchange.com/questions/670083/proof-that-rational-sequence-converges-to-irrational-number – Autolatry Jan 15 '15 at 11:48
  • 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. – barak manos Jan 15 '15 at 12:22
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    @barakmanos: I don't think computability is relevant here. Petite Etincelle's answer is perfectly valid. – TonyK Jan 15 '15 at 12:39

2 Answers2

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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 $\dfrac{\lfloor 10^n \beta\rfloor}{10^n} \to \beta$

Petite Etincelle
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Consider $\{x_n\}$ where $x_n={[n\beta]+1 \over n}$ $\forall n \in \mathbb N$.

$[n\beta]<n\beta<[n\beta]+1$ $\forall n \in \mathbb N \Rightarrow \frac{[n\beta]}{n}<\beta<x_n$.

Again $\frac{[n\beta]}{n}<\beta \Rightarrow x_n=\frac{[n\beta]+1}{n}<\beta+{1 \over n}$. Thus $\beta<x_n<\beta+\frac 1n$ $\forall n \in \mathbb N$. By sandwich theorem $\lim_{n \to \infty} x_n =\beta$.

Saikai Prime
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