# There exists a rational sequence that converges to $\sqrt3$

I got a proof of this but I am quite sure that it is not what was expected on the exam. Also, this proof seems really kludgy and non-kosher.

Because of the density of the raitonals in the reals, there exists a $q\in\mathbb{Q}$ such that $\sqrt3-\frac1 n < q < \sqrt3$

For each n, let $a_n = q$. ($\sqrt3-\frac1 n < a_n < \sqrt3$)

So for n=1. There exists a $q$ such that $\sqrt3-1 < q < \sqrt3$. Choose this q for $a_1$

For n=2, choose $q$ such that $\sqrt3-\frac1 2 < q < \sqrt3$.

Since $\sqrt3-\frac1 n$ converges to $\sqrt3$, $a_n$ must also converge to $\sqrt3$.

So we are constructing a sequence out of things that we are only know the existence of. Also does this require the axiom of choice?

Anyways, on the exam there was a hint "consider $S = \{r\in\mathbb{Q}|r>0 \,\mathrm{and}\,r^2<3\}$" And I am not exctly sure what to make of it other than the fact that $\sup S = \sqrt 3$

Edit: To be specific about my question, is this proof ok? And what is the standard proof (the proof that my prof was hinting at)?

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– MJD Oct 12 '12 at 13:14

1. Well, the above proof relies on the fact that 'Because of the density of the raitonals in the reals', and this in other words, in all metric spaces, mean that for any point there is a sequence from the dense set (now $\Bbb Q$) that converges there... Well, ok, proving the density of $\Bbb Q$ is not that hard, but I would say, it is required beforehand..
2. A classical proof, working for any real number $s$, is that we write up $s$ in decimal fraction format, and in the $n$-th step we keep only the first $n$ digits, i.e. $s_n:= \displaystyle\frac{[10^n\cdot s]}{10^n}$
3. And the solution what was hinted.. if you already know that $\sup S=\sqrt 3$, then you're done, because it also means that, for every $\varepsilon$ (in particular for $\varepsilon=1/n$, there is an element $r\in S$ such that $r < \sqrt 3 < r+\varepsilon$
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A simpler sequence to calculate than that using the continued fraction is to let $r_1=1/1$ and $r_2=4/2$, and generally if $r_n=p/q$ let $r_{n+1}=(p+3q)/(p+q)$. The first several terms:

$1/1,4/2,10/6,28/16,76/44,208/120,...$

The terms will be alternately below and above $\sqrt{3}$, and approach it. It's fairly fast, for example for the sixth term, $208/120-\sqrt{3}=.001285...$ It doesn't converge as quickly as does the sequence arising from the continued fraction.

But this approach has the advantage that one can do any squareroot of a positive (nonsqure) integer $m$ this way. Start with $r_1=1/1$ and if $r_n=p/q$ let $r_{n+1}=(p+mq)/(p+q)$ Then the sequence $r_n$ will approach $\sqrt{m}$, the terms alternately below and above it.

As with the continued fraction method, there is a recursive formula going from one fraction in the sequence to the next; the recusion here looks a bit simpler than does the continued fraction one, in that one only need keep track of the current numerator and denominator.

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But the fractions after cancelling the common factors are exactly the convergents of the continued fraction! At least the first few terms I checked. – PAD Oct 13 '12 at 14:31

Uses the Newton methold $x_{n+1}= x_n -\frac{f(x_n)}{f^\prime(x_n)}$ wthi start point $x_0\in\mathbb{Q}$ to calculate the root of $f(x)=x^2-3$.

You can use a basic scientific calculator to make these interactions. I suggest using as a starting point $x_0=\frac{3}{2}$. This ensures the convergence of the sequence to $\sqrt{3}$.

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Can we choose the initial point so that the sequence is close to the one defined by the continued fraction? – PAD Oct 13 '12 at 14:32

This one is based on the same idea as couple answers above

$$a_n=\frac{ \lfloor n \sqrt{3} \rfloor}{n}$$

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.... or try Newton's iteration to solve $f(x) := x^2-3 = 0$. Start with, say, $x=x_0=2$. Then $x_1=x_0-f(x_0)/f'(x_0)=2-1/4=7/4$; $x_2=7/4-(1/16)/(7/2)=97/56$; and so on.

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Let $a_n=\dfrac{b_n}{10^n}$, where $b_n$ is the greatest integer such that $b_n^2\le 3\times 10^{2n}$.

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Write down the continued fraction $[a_0, a_1,a_2 \cdots ]$ of $\sqrt{3}$. It turns out to be periodic $[1; 1,2, 1,2, \cdots ]$. Then the sequence $\frac{p_k}{q_k}$ converges to $\sqrt{3}$, where $q_0=a_0=1$, $q_0=1$,

$p_1=a_0 a_1+1=2$, $q_1=a_1=1$ and then recursively

$$p_k=a_k p_{k-1}+p_{k-2}$$

$$q_k=a_k q_{k-1} +q_{k-2}$$

e.g. $$\frac{p_2}{q_2}=\frac{5}{3}$$

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I would rather construct such sequence: use Taylor's formula for $\sqrt{1+3}$.

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I guess you mean $\sqrt{x+3}$? – Max Oct 12 '12 at 13:34
I don't really see what you are trying to say. However, the partial sums of the binomial series expansion for $\frac{7}{4}\sqrt (1-x)$, with $x=1/49$, would do the job. – John Bentin Oct 12 '12 at 14:07

The existence of such sequence does not require the axiom of choice.

First note that you can always enumerate the rational and just pick the least rational (in the enumeration) in the interval $(\sqrt3-\frac1n,\sqrt3)$.

But also note that if you write $\sqrt3=1.d_1d_2d_3\ldots$, then truncating the digits at the $n$-th place yields a rational number, and clearly this is a sequence approaching $\sqrt3$.

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