Calculating the square root of 2 Since $\sqrt{2}$ is irrational, is there a way to compute the first 20 digits of it?
 What I have done so far 
I started the first digit decimal of the $\sqrt{2}$ by calculating iteratively so that it would not go to 3 so fast. It looks like this:
\begin{align}
\sqrt 2 & = 1.4^{2} \equiv 1.96\\
\sqrt 2 & = 1.41^{2} \equiv 1.9881\\
\sqrt 2 & = 1.414^{2} \equiv 1.999396\\
& \ldots
\end{align}
First I tell whether it passes such that $1.x^{2}$ would be not greater than 3.
If that passes, I will add a new decimal to it. Let's say $y.$ $1.xy^{2}$
If that y fails, I increment $y$ by 1 and square it again.
The process will keep repeating. Unfortunately, the process takes so much time.
 A: On a similar note to the answer by R. Romero: in the special case of taking the square root of an integer $N$, it is fairly straightforward to calculate the continued fraction representation of $\sqrt{N}$.
In the particular case $N=2$, we have:
$$ \sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \ddots}}}. $$
(This follows from the fact that if $x = \sqrt{2}-1$, then $x = \sqrt{2}-1 = \frac{1}{\sqrt{2}+1} = \frac{1}{2+x}$.)
Now, from this we can calculate subsequent rational approximations to $\sqrt{2}$:
$$
\begin{matrix}
 & & 1 & 2 & 2 & 2 & 2 & 2 & \cdots \\
0 & 1 & 1 & 3 & 7 & 17 & 41 & 99 & \cdots \\
1 & 0 & 1 & 2 & 5 & 12 & 29 & 70 & \cdots
\end{matrix} $$
So, for example $\frac{99}{70} \approx 1.4142857$ whereas $\sqrt{2} \approx 1.4142136$.
(It also happens that this procedure generates solutions to Pell's equation $a^2 - 2 b^2 = \pm 1$; for example, $99^2 - 2 \cdot 70^2 = 1$.  The connection is: if $a^2 - 2 b^2 = \pm 1$ then $a - b \sqrt{2} = \pm \frac{1}{a + b \sqrt{2}}$; so if $a$ and $b$ are large positive integers satisfying Pell's equation, then $a - b\sqrt{2} \approx \pm\frac{1}{2a}$ which implies $\frac{a}{b} - \sqrt{2} \approx \pm\frac{1}{2ab} \approx \pm\frac{1}{a^2\sqrt{2}}$.)
A: The number $\sqrt{2}$ is the solution to the equation $x^2-2=0$, so any method for numerically approximating the roots of an equation (such as the Newton method) will be able to approximate $\sqrt{2}$.
A: Okay, I searched through the answers, but none seems to mention this one: long quadratic root calculation.
From the name it is obvious that it resembles long division, like this:
$$
\begin{align}
\sqrt{2.00\;00\;00\;00\;..}
\end{align}
$$
Notice how they are grouped into tuples. Now estimate the first digit, namely $1$:
$$
\begin{align}
&~~~1.\\
1&\sqrt{2.00\;00\;00\;00\;..}\\
&~~~1\\
&~~~\overline{1\,00}
\end{align}
$$
We calculate $1\times1=1$, write it down, and calculate the "remainder", just like divisions. Notice that we append 2 digits behind instead of 1.
Next, double the number on the top, and write it on the left of $1\,00$:
$$
\begin{align}
&~~~1.\;*\\
1&\sqrt{2.00\;00\;00\;00\;..}\\
&~~~1\\
2*&\,\,|\overline{1\,00}
\end{align}
$$
Now we estimate the next digit, *. It is written both on the top and to the left. Of course, we know that it is 4, so:
$$
\begin{align}
&~~~1.\;4\;\;\;*\\
1&\sqrt{2.00\;00\;00\;00\;..}\\
&~~~1\\
24&\,\,|\overline{1\,00}\\
&\,\,|\,\,\,\,96\\
&2\overline{8{*}|\,4\,00}
\end{align}
$$
We double the numbers on the top again to get $28*$, and repeat the process:
$$
\begin{align}
&~~~1.\;4\;\;\;1\\
1&\sqrt{2.00\;00\;00\;00\;..}\\
&~~~1\\
24&\,\,|\overline{1\,00}\\
&\,\,|\,\,\,\,96\\
&2\overline{8{1}|\,4\,00}\\
&\,\,\,\,\,\,\,\,\,|\,2\,81
\end{align}
$$
I found a picture, but not of $\sqrt{2}$:

This is extremely inefficient for computers, but great for manual calculation. After all, we don't do multiplication through fast Fourier transforms!
Also, this method is developed in ancient China.
A: Calculating the square root of a number is one of the first problems tackled with numerical methods, known I think to the ancient Babylonians. The observation is that if $x,\,y>0$ and $y\ne\sqrt{x}$ then $y,\,x/y$ will be on opposite sides of $\sqrt{x}$, and we could try averaging them. So try $y_0=1,\,y_{n+1}=\frac12\left(y_n+\frac{x}{y_n}\right)$. This is actually the Newton-Raphson method 5xum mentioned. The number of correct decimal places approximately doubles at each stage, i.e. you probably only have to go as far as $y_5$ or so.
A: Suppose you want to find the square root of $p$ and suppose your initial guess is $x/y$:
Let $\mathbf M=\begin{bmatrix} 1 & p \\ 1 & 1 \end{bmatrix}$ and $\mathbf q=\begin{pmatrix} x \\ y \end{pmatrix}$
 Then $\mathbf M\mathbf M\mathbf M...\mathbf q$ gives a numerator and denominator the ratio of which converges to the square root of $p$. This gives an approximation to the square root of $2$ as fast as the other methods but with no floating point arithmetic until the final division.
Performs well for calculation tools optimized for Matrix arithmetic. This also gives you solutions for Pell's equation for $p=2$ as mentioned by Daniel Schepler. 
A: Here's the way I learnt to obtain decimal digit after decimal digit when I began middle school:
\begin{array}{lcl}
2&\big( &\color{red}1.414\,2\dots \\[1ex]
1\,00&& 24\times \color{red}4=96<100\\
-96\,&& 25\times5=125>100\\[1ex]
\phantom{-0}4\,00&&281\times\color{red}1<400\\
\;\:-2\,81&&282\times2>400\\[1ex]
\phantom{-0}119\,00&&2824\times\color{red}4<11900\\
\phantom{0}{-}112\,96&&2825\times5>11900 \\[1ex]
\phantom{00\;}604\,00&&28282\times\color{red}2 < 60400 \\
&&28283\times3> 60400 
\end{array}
&c.
Let me explain the procedure on the first two steps. It relies on a clever use of the identity $(x+y)^2=x^2+2xy+y^2$. Suppose more generally we want to find the square root of a number $a$.


*

*We first find the greatest natural number $n$ such that $n^2\le a$.

*If $a$ is not a perfect square, i.e. if $n^2<a$, let $d$ be the first decimal digit of the square root. This is the greatest digit such that $\;\Bigl(n+\frac d{10}\Bigr)^2\le a$. We'll transform this inequality into a more easy-to-use test:
\begin{align}
\Bigl(n+\frac d{10}\Bigr)^2\le a&\iff \frac{2n}{10}d+\frac{d^2}{100}<a -n^2\\
&\iff (10\times 2n+d)\times d\le (a-n^2)\times 100
\end{align}
In practice, this means, we calculate the difference $a-n^2$ and add two 0s. Then we double $n$, add a digit d (this is the result of calculating $10\times 2n+d$) and multiply what we obtain by this digit. Last, we test whether the result is less than $100(a-n^2)$, and retain the largest possible digit.

A: In this answer, there is a method using continued fraction approximations for $\sqrt2$ and the generating function for the central binomial coefficients to get some very quickly convergent series for $\sqrt2$. For example,
$$
\sqrt2=\frac75\sum_{k=0}^\infty\binom{2k}{k}\frac1{200^k}\tag1
$$
and
$$
\sqrt2=\frac{239}{169}\sum_{k=0}^\infty\binom{2k}{k}\frac1{228488^k}\tag2
$$

For example, summing to $k=4$ in $(2)$ gives 
$$
\sqrt2=1.414213562373095048801688
$$
which is accurate to $23$ places.
A: Using the fact that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, then we have to find $2 \sin \frac{\pi}{4}$.
We can approximate $\sin x$ using the Taylor series to three terms:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^6),$$
so we have:
$$\sin \frac{\pi}{4} \approx \frac{\pi}{4} - \frac{(\pi/4)^3}{3!} + \frac{(\pi/4)^5}{5!} .$$
If we approximate $\pi$ as $\frac{22}{7}$, then we have $\frac{\pi}{4} = \frac{11}{14}$, then we have:
$$\sin \frac{\pi}{4} \approx\frac{11}{14} - \frac{(11/14)^3}{3!} + \frac{(11/14)^5}{5!},$$
which when you multiply by $2$ to get $\sqrt{2}$, gives $1.4147$, while the actual value is $1.4142$.
If we expand the Taylor series to more terms, or improve the approximation of $\pi$ (such as $\frac{355}{113}$), then we can get to $20$ correct digits.
A: There's a general method that converges about as quickly as Newton-Raphson but is somewhat more general. It's based off of Continued Fractions:
Suppose you want to find the square root of $N$. Let $a+b = N$ where $b$ has an easy to calculate square root. 
let $y_{n+1} = \sqrt b + \frac{a}{ \sqrt b + y_n}$
$y_{n+1}$ converges to $\sqrt N$. 
A: Start with an initial guess $x$ for the square root of $2$. Then add a correction term $y$. Write down $(x+y)^2 - 2 = 0$. Solve this equation for $y$ by expanding it up to third order in the difference $(2-x^2)$. This is a straightforward calculation. Combining all contributions, the result is elegant:
$$x + y = (x^4+12x^2+4)/(4x^3+8x)$$
For a rational initial guess $x$ the result $(x + y)$ is also rational, but much closer to the desired value. 
For example if we take $x = 3/2$, then $(x +y)=577/408$, which differs from the square root of 2 by a factor 1.0000015. If we start with $x = 7/5$, the result is $19601/13860$, which differs from the square of root of $2$ by a factor   $1.0000000013$   
A: Newton-Rhapson is a good idea because of the convergence rate. However, I am more of a fan of using Taylor's expansions here since it is super easy to derive on the go to give fairly ok estimates in quite a reasonable time. So, the way to go to find $\sqrt{x}$ is to find first the closest integer which approximates $\sqrt{x}$ and call this $a$, then apply Taylor to $a^2$. Then Taylor says
$$ \sqrt{x} \approx a + (x-a^2)\cdot \frac{1}{2 a} - (x-a^2)^2/2 \cdot \frac{1}{4 a^3} + \cdots. $$
The thing that is nice here is that you also get bounds on the error you make. So, denote $f(x) = \sqrt{x}$, then the error of a $n$th order approximation (i.e., going as far as $(x-a^2)^n/n! \cdot f^{(n)}(a^2)$ in the approximation above) is given by $$ (x-a)^{n+1}/(n+1)! \cdot f^{(n+1)}(\xi)$$ for a certain $\xi$ between $a^2$ and $x$. This can be estimated quite easily since this $f^{(n+1)}$ is monotone around $x$. Thus look at the boundaries of the domain of $\xi$ and find the 'best' maximal value which you can calculate without a calculator.
Example for $x=2$. Apparently $1$ is the closest integer to $\sqrt{2}$ and thus we will take $a=1$. Then, let's take a second order approximation
$$\sqrt{2} \approx 1 + (2-1)\cdot \frac{1}{2} - (2-1)^2/2\cdot \frac{1}{4} = 1 + 0.5 - 0.125 = 1.375 $$
and the absolute error is given by
$$ E=\left|(2-1)^3/3!\cdot \frac{3}{8 \cdot \xi^2\sqrt{\xi}}\right| = \frac{1}{16} \cdot \frac{1}{|\xi^2\sqrt{\xi}|}$$
for a certain $\xi$ between $1$ and $2$. Since this is a decreasing function on $(1,2)$. The maximum is attained at $1$ and hence the error is bounded by
$$ E \leq \frac{1}{16} $$
which seems to be a good estimate since $E = 0.039\dots$ and $1/16 = 0.0625$.
Edit As some of you noted this method 'looks' more difficult than Newton-Rhapson and the convergence is slower. The last part is obviously true and I would answer this question with: How quick do you need it to be and do you want to calculate it in your head or do you have a computer? Do you need to have a quick guess which is approximately equal to the value of $\sqrt{2}$ or do you need a precise estimate. If you don't have a computer but pen and paper, the best method is Newton-Rhapson.
I would argue that my method is better if you don't have pen and paper or a computer and you are asked to give an estimation of $\sqrt{10}$ on the go (especially for $\sqrt{x}$ with $x$ big, the Taylor approximation is better since the $\sqrt{\bullet}$ function becomes more linear as $x$ grows).
I agree that my method looks way more difficult but it isn't if you get more familiar with it. Also, this method is super quick in terms of calculation time in your head and if you practice a little with it, it becomes way easier. Also, this method works particularly nice for $\sqrt{x}$ where $x$ differs one from a perfect square because then the $(x-a^2)^n$ term will always be one.
Let's look at an example here. Suppose you need to calculate $\sqrt{122}$, then first order approximation of my method gives
$$ \sqrt{122} \approx 11 + \frac{1}{2\cdot 11}. $$
It took me less than one second to find this approximation and the second order approximation works almost as quick here. You just need to add $\frac{-1}{8\cdot 11^3}$. Please note that the error of the first order approximation here is approximately equal to $10^{-4}$.
If you apply Newton-Rhapson here you get the same approximation after one step if you choose $x_0=11$. The only thing is that I always forget what the exact form is of Newton-Rhapson. So when I want to apply it, I have to think about it where I could have immediately applied Taylor but I would say that is just my particular preference.
A: I came up with an interesting, but terribly inefficient method. 
Consider the sequence {$x_n$}: $1,\frac{1}{2},\frac{1}{2},\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}$, ...
Suppose you want k digits of the square root of 2. Then add up the first $100^k$ terms and then divide the sum by $10^k$.
A: I know an easy way to calculate the binary digits of $\sqrt{2}$. Take the ordered pair (1, 2) $1^2$ is less than 2 and $2^2$ is more than 2. Calculate the square of the average $1.5^2$ in base 2. The square of the average is just the average of the squares minus $\frac{1}{4}$. The result expressed in binary is 10.01 so the first binary digit after the decimal is 0. Take the next ordered pair to be (1, 1.5) and calculate the square of its average which is the average of its squares minus $\frac{1}{16}$. The result expressed in binary is 1.1001 so the next binary digit is 1.
A: Towers' bisection method above is similar to your own approach, but more efficient.   Another method that is not as good as binary search, but is better than your own method, is to increment the last digit in bigger steps.  I would try incrementing by 3.  The worst case is that you reach the correct digit in 5 steps instead of 9.   
My favorite method for mental approximation is to find the next lowest square, determine the error, and add to its square root the error divided by double the guess.   For sqrt(200), the lowest square is 196.  The error is 4, so my mental estimate is 14 + 4/14 = 14.142857...   
I apologize for off-topic, but note that square roots can be used to calculate logarithms by a process similar to bisection.  I suspect that is how it was done in the late 16th century, as they did not yet have calculus.  In our times, there are extremely accurate formulas for logarithm that still require square roots.   This exercise should make you appreciate the power of a square root button on a calculator, even if you have no "scientific" functions.
