Is the set of Rational Approximations for $\sqrt{2}$ with Base Ten Exponent Denominators Valid? Can it be the basis for a sequence?

Question

I am looking to create a sequence that contains only rational numbers but converges to an irrational number, e.g $\sqrt{2}$. In this effort, I thought a possible idea could be to define some countable set S that contains all fractional approximations of $\sqrt2$ with some $10^n$ base and then let a sequence $x_n$ return the $n^{th}$ element in $S$.

So, $$S = \biggl\{\frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, \frac{14142}{10000}, \ldots \biggr\},$$

$$x_n = S_n,$$

$$\lim(x_n) = \sqrt{2}$$

I don't really know why, but I feel like this is illegal in some way. Something to do with the fact that the way you construct the set $S$ is by looking at the decimal approximation $\sqrt{2}$ is a bit fishy to me.

I'm pretty confident this solution is not valid somewhere, but could someone help me out in finding where/why exactly it is invalid?

Thanks!

Answer 1

Your sequence is indeed valid. Another sequence of rational numbers which tends to $\sqrt{2}$ is $$\begin{cases} x_{n+1}=\dfrac{x_n^2+2}{2x_n}\quad \text{for $n\geq 0$}\\ x_0=1 \end{cases}$$ The terms of the recurrence are clearly all positive rational numbers. Can you show that $\lim_{n\to\infty}x_n=\sqrt{2}$?

You can easily modify the sequence in such a way that it tends to $\sqrt{m}$: replace the number $2$ at the numerator with $m$.

Answer 2

Your idea is valid. Another sequence of rational numbers which tends to $\sqrt{2}$ is $$\begin{cases} x_{n+1}=\dfrac{2x_n+2}{x_n+2}\quad \text{for $n\geq 0$}\\\\ \\ x_0=1. \end{cases}$$

Cf. this question.

Answer 3

you construct the set S is by looking at the decimal approximation

No, you construct the sequence by using what you know about $\sqrt{2}$, namely that $\left(\sqrt{2}\right)^2=2\,$.

• The first term is $14/10$ because $14/10 \lt \sqrt{2} \lt 15/10 \iff 14^2 \lt 2 \cdot 100 \lt 15^2\,$.

• Next one is $141/100$ because $141/100 \lt \sqrt{2} \lt 142/100 \iff 141^2 \lt 2 \cdot 10000 \lt 142^2\,$.

• And so on $\dots$

That's building the decimal representation of $\sqrt{2}\,$, and there is nothing wrong or circular about it.