# Sequence of rational numbers converging to an arbitrary irrational number $\beta$.

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.

• – 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
• @barakmanos: I don't think computability is relevant here. Petite Etincelle's answer is perfectly valid. – TonyK Jan 15 '15 at 12:39

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

• I thought about it but I couldn't. – user194772 Jan 15 '15 at 12:02
• @user194772 I've added details – Petite Etincelle Jan 15 '15 at 12:10
• @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?? – crskhr Mar 8 '15 at 14:26
• @S.C. nothing is special with $10$, other examples that you mentioned work as well – Petite Etincelle Mar 8 '15 at 14:50

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

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

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