<s>[INST] 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.
 [/INST] 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 $\frac{a}{b} - \sqrt{2} \approx \pm\frac{1}{2ab} \approx \pm\frac{1}{a^2\sqrt{2}}$.)
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