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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!

Sangchul Lee
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adam dhalla
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    Your idea is valid – J. W. Tanner Aug 04 '21 at 21:49
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    Your construction is equivalent to $$x_n = \frac{\lfloor 10^n \sqrt{2} \rfloor}{10^n}, \qquad n = 1, 2, 3, \ldots, $$ which is a perfectly valid way of constructing a sequence. If your goal is to define $\sqrt{2}$ through a rational approximation, i.e., a rational sequence whose limit squares to $2$ while pretending we haven't established the number $\sqrt{2}$ yet, then this would be an invalid approach and an alternative is required. – Sangchul Lee Aug 04 '21 at 22:03
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    not only is it legal, it is actually these reasoning that allows us to say decimal expansions are a legitimate way to express real numbers. The basic property of real numbers (almost the definiton) is that every real number has (many) sequences of rational numbers that converge to it (and every cauchy sequence or rationals converge to a real) so we can construct such a sequence of decimalls for every real number. That is the only reason we are allowed to say "take the decimal expansion"... how else would we even know a real number has a decimal expansion. – fleablood Aug 04 '21 at 22:33
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    This is completely valid: As $1^2 < 2 < 2^2$ then $a_0=1$. Then test $a_0 + \frac k{10}$ for $k=0, ...9$ until you find $(a_0 + \frac k{10})^2 < 2 < (a_0 + \frac {k+1}{10})^2$. Let $a_1 = a+\frac k{10}$. For each $a_m$ test $k=0...9$ to find $(a_m +\frac k{10^{m+1}})^2 < 2 < (a_m+\frac k{10^{m+1}})^2$. So a sequence is surely, rational, calculatably, and converging to $\sqrt 2$. So that's legit. And it will turn out that it is the decimal expansion. I think the worries you have is how can you define the decimal expansion without knowing it first. And... well, you calculate it. – fleablood Aug 04 '21 at 22:42

3 Answers3

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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 $m$.

Robert Z
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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.

dxiv
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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.

J. W. Tanner
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