Rational number that approximates $\sqrt{3}$ 
Questions:
  
  
*
  
*Show that is is theoretically possible to find a rational number that approximates $\sqrt{3}$ with an error less than $0.001$.
  
*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.
  



*

*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 $n = N_\epsilon \implies |q_{N_\epsilon} - \sqrt{3}| < \frac{1}{10^3}$

*Can someone please explain this to me?
 A: 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 too high or too 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.
A: 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 $\sqrt 3$.
A: 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 $3$. 
A: 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.
A: 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 $\surd3$.
A: 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. 
A: 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:
$$\left|\sqrt{3}-\frac{71}{41}\right|\leq\frac{1}{41^2}<\frac{1}{1000}.\tag{3}$$
A: the following series is used for $\sqrt{3}$
$$\sum_{n=0}^{\infty }\frac{(2n-1)!!}{n!3^n}=1+1/3+1/6+5/54+35/684+..$$
