<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] 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.
</s>