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\\ & ... \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. The bad is that the process takes so much time.

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.

Since $\sqrt{2}$ is irrational, Is there a way to compute the 20 digits of it?

What I have done so far I started the first digit decimal of the $\sqrt{2}$ by calculating iterately 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\\ & ... \end{align}

What I did is to first tell whether it pass such that $1.x^{2}$ would not greater than to 3.
If that passes, I will add a new decimal to it. Lets say y. $1.xy^{2}$
If that y fails, I increment y by 1 and square it again. The process will keep repeating. The bad is that the process takes so much time.

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\\ & ... \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. The bad is that the process takes so much time.

Since $\sqrt{2}$ is irrational, Is there a way to compute the 20 digits of it?

What I have done so far I started the first digit decimal of the $\sqrt{2}$ by calculating iterately 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\\ & ... \end{align}

Since $\sqrt{2}$ is irrational, Is there a way to compute the 20 digits of it?

What I have done so far I started the first digit decimal of the $\sqrt{2}$ by calculating iterately 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\\ & ... \end{align}

What I did is to first tell whether it pass such that $1.x^{2}$ would not greater than to 3.
If that passes, I will add a new decimal to it. Lets say y. $1.xy^{2}$
If that y fails, I increment y by 1 and square it again. The process will keep repeating. The bad is that the process takes so much time.