1. Show that is is theoretically possible to find a rational number that approximates $\sqrt{3}$ with an error less than $0.001$.
  2. Explain how you would go about determining a rational that approximates $\sqrt{3}$ correct to 2 decimal places. Explain the procedure only, do not find the rational number.
  1. Let $q \in \mathbb{Q}$. We require $| \sqrt{3} - q | < \frac{1}{10^3}$. Now because $\bar{\mathbb{Q}}= \mathbb{R}, \exists (q_n) \in \mathbb{Q}$ such that $q_n \to \sqrt{3}$. Let $\epsilon = \frac{1}{10^3}$. For the given $\epsilon >0$ there exists a $N_\epsilon \in \mathbb{N}$ such that $| q_n - \sqrt{3} | < \frac{1}{10^3}$ for all $n \geq N_\epsilon$. Choose $n = N_\epsilon \implies |q_{N_\epsilon} - \sqrt{3}| < \frac{1}{10^3}$

  2. Can someone please explain this to me?

up vote 5 down vote accepted

One way is: use decimal search to find the first 2 decimal places, then multiply by 100, truncate to an integer, and use this value over 100.

Decimal search is: Find each digit by seeing if a certain value is two high or two low. Eg: calculate $1.5^2$. If this is < 3, try $1.6^2$, $1.7^2$ until we find a value which squares to more than 3. Then do the same with the next digit: try 1.75 and see whether to go higher or lower. This technique can be used to find any square root to an arbitrary number of decimal places.

  • What is decimal search? – Christoph Apr 9 '15 at 13:16
  • @Christoph good point, will add an explanation. – IanF1 Apr 9 '15 at 13:16
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    If you multiply a number with $2$ decimal places by $100$, it will always be an integer, no need to truncate. – Christoph Apr 9 '15 at 13:35

It is an overkill for sure, but since the continued fraction of $\sqrt{3}$ is given by: $$ \sqrt{3}=[1;\overline{1,2}],\tag{1}$$ we have that: $$ [1,1,2,1,2,1,2] = \frac{71}{41} \tag{2} $$ is an accurate approximation, and: $$\left|\sqrt{3}-\frac{71}{41}\right|\leq\frac{1}{41^2}<\frac{1}{1000}.\tag{3}$$

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    Definite overkill.. LOL! But also definite AWESOMENESS! :D . Thanks! :) – DJS Apr 9 '15 at 13:39
  • Much more elegant than my answer, +1 – IanF1 Apr 9 '15 at 13:49
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    Why overkill? Continued fractions is the first thing that came to my mind and I don't think it is that advanced a subject, is it? Very nice +1 – Timbuc Apr 9 '15 at 14:10

May be, we could consider solving for $x$ $$f(x)=x^2-3=0$$ and use Newton method which will give $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{x_n^2+3}{2 x_n}$$ Starting with $x_0=1$, we should get as iterates $2$, $\frac{7}{4}$, $\frac{97}{56}$, $\frac{18817}{10864}$, $\frac{708158977}{408855776}$.

Using Halley method instead, which will give $$x_{n+1}=\frac{3 \left(x_n^4+6 x_n^2-3\right)}{8 x_n^3}$$ the iterates would be $\frac{3}{2}$, $\frac{83}{48}$, $\frac{126766609}{73188736}$.

In other words, starting with a rational value of $x_0$ will generate an infinite number of iterates which will be rational too, closer and closer to $\sqrt 3$.

  • Nice numerical approxiamtion! :). I did not even think to bring a Numerical Analysis approach to real analysis :) – DJS Apr 9 '15 at 13:41
  • This is the only area in which I know $\epsilon$ ! Cheers :-) – Claude Leibovici Apr 9 '15 at 13:42

There are $100$ distinct rational numbers between $1$ and $2$ of the form $1.XY5$. Square them and find where the square changes from below $3$ to above $3$.

Take the average of the largest number whose square is below $3$ and the smallest number whose square is above $3$.

To test if $x>\sqrt3$, it suffices to check if $x^2>3$, since these are equivalent when $x$ is positive.

With that in mind, I then see if $1.00>\sqrt3$, if $1.01>\sqrt3$, if $1.02>\sqrt3$, if $1.03>\sqrt3$, in order, until I find the smallest one that's bigger than $\sqrt3$. I then take the one immediately before that.

Hey, nobody said anything about efficiency.

Brahmagupta's equation gives such approximations. An integer solution for $X^2-dY^2=1$ leads to the rational number $X/Y$ an approximation for $\surd d$. $$49-3\times16=1$$ $$7^2-(4\surd3 )^2=1$$ $$(7-4\surd3)(7+4\surd3)=1$$ Now raising to an arbitrary power $$ (7-4\surd3)^n(7+4\surd3)^n=1$$ Expressing $(7+4\surd3)^n$ in the form $a_n+b_n\surd3$ you can see that $a_n/b_n$ is a sequence of improving approximations for $\surd3$.

I outlined a really easy to use method originally created by the Babylonians to approximate the square root of a number $N$ in this answer. Here's the idea: make an initial guess as to the square root, and let this initial guess be given by $r_1$. Now, continue improving your guesses by using $$ r_{n+1}=\frac{1}{2}\left(r_n+\frac{N}{r_n}\right),\tag{1} $$ where $N$ is the number you are trying to find the square root of. This method works surprisingly well.

For your problem specifically, we know $2$ is somewhat close to the square root of $3$. Thus, let $r_1=2$. Using $(1)$ just one time tells us that $r_2=7/4=1.75$, whereas the actual square root of $3$ is $\sqrt{3}=1.73\ldots$; thus, use $(1)$ just once more to get $$ r_3=97/56\approx \color{red}{1.732}\color{blue}{14285714}, $$ whereas $$ \sqrt{3}\approx \color{red}{1.73205080757}, $$ and, just making one last guess (though several more could be made to further improve accuracy): $$ r_4=18817/10864\approx \color{red}{1.7320508}\color{purple}{1001}. $$ I know this doesn't answer your question exactly--your quest sounds more theoretically oriented, but this is something practical that could be of use to you perhaps.

the following series is used for $\sqrt{3}$ $$\sum_{n=0}^{\infty }\frac{(2n-1)!!}{n!3^n}=1+1/3+1/6+5/54+35/684+..$$

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