Calculate more digits of square root of 2?

Question

How would I calculate the next digit of the decimal representation of square root of two?

1.4142135...

Say I want to calculate the digit after 5... how would I do that?

I am not allowed to use a calculator.

Answer 1

There is a very nice method, by hand, that is seldom taught nowadays.

You want to take the square root of $2$ to $n$ digits, so we will write $2$ followed by $2n$ zeros. Notice, $2n$. Write them by pairs, like this for $5$ digits:

2 00 00 00 00 00

Now, if your number has several digits, you write them by pairs too (starting from the right), for example for the square root of 20451, to five digits, you would write

2 04 51 00 00 00 00 00

From now on, we continue with the square root of $2$, but the method works for any number. Actually, we are really computing the square root of any integer, truncated to the lower integer: since we add $2n$ zeros, it's like computing the square root of $2\times10^{2n}=10^n\sqrt{2}$, and by truncating you get $n$ decimals of $\sqrt{2}$.

This is the general setting, now let's see the method.

---

So you write your number, with additional zeros

2 00 00 00 00 00

You take mentally the integer (truncated) square root of the last part on the left. Here it's the integer square root of $2$, so it's $1$ (it's the floor of $1.414...$). If you know your table of squares up to $10^2=100$, it's enough to find this.

You will write this one on the right, like for a division

2 00 00 00 00 00    | 1

Now we enter the "general step" (we will repeat this one for each pair of zeros.

You subtract $1^2$ from $2$, and write the rest ($1=2-1^2$) under the preceding, plus one pair of zeros:

2 00 00 00 00 00    | 1
1 00

On the right, there is the "current square root", we will complete it by adding digits.

Now, find the highest possible digit ($0$ to $9$) such that the operation $2d \times d$ is not greater than $100$. $2d$ denotes the number obtained by concatenating twice the current square root, to the chosen digit.

For example, you try $20\times0=0$, $21\times1=21$, $22\times2=44$, $23\times3=46$, $24\times4=96$, $25\times5=125$. Stop! Too large, so the next digit is actually a $4$.

The rest is $100-96=4$.

So we update the current square root on the right (chosen digit $4$), and the current rest (also $4$), and write down two more zeros:

2 00 00 00 00 00    | 14
1 00
   4 00

This was the first step, and we will repeat exactly the same several times:

Step 2

Twice the current root $14\times2=28$, and try $28d\times d$ so that it's not greater than 400. Since $282\times2>400$, the next digit is a $1$, and we subtract $281\times1=281$ from $400$ ($400-281=119$), and update both the rest and the current root (next digit $1$).

2 00 00 00 00 00    | 141
1 00
   4 00
   1 19 00

Step 3

Twice the current root, so $282$, and try $282d\times d$ so that it's not greater than $11900$. Since $2824\times4=11296$ and $2825\times5=14125$, too large, the next digit is a $4$.

The next rest is $11900-11296=604$.

2 00 00 00 00 00    | 1414
1 00
   4 00
   1 19 00
      6 04 00

Step 4

Twice the current root, so $2828$, and we try $2828d\times d$ so that it's not greater than $60400$. The good choice is $28282\times 2=56564$, and $60400-56564=3836$ is the next rest, and next digit is $2$:

2 00 00 00 00 00    | 14142
1 00
   4 00
   1 19 00
      6 04 00
        38 36 00

Step 5

Twice the current root is $28284$, so we try $28284d\times d$ so that it's not greater than $383600$. The good choice is $1$: $282841\times1=282841$ is the largest possible, and $383600-282841=100759$.

2 00 00 00 00 00    | 141421
1 00
   4 00
   1 19 00
      6 04 00
        38 36 00
        10 07 59

Hence, we have the last rest, and the updated square root to $5$ digits, and we have the equation:

$$2 00 00 00 00 00=141421^2+10 07 59$$

We could of course continue, adding as many pairs of zeros as we wish. With two more pairs, you will get the next digit you are after.

---

In the usual setting (at least what I was taught by my father, who learned it in high school in the sixties, in France), you write the successive products under the square root like this:

2 00 00 00 00 00    | 141421
1 00                |------------------
   4 00             | 24×4=96
   1 19 00          | 281×1=281
      6 04 00       | 2824×4=11296
        38 36 00    | 28282×2=56564
        10 07 59    | 282841×1=282841
                    |

That way, it's relatively compact, yet it's easy to do the subtractions.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \root{2}&=\root{288 \over 144}={1 \over 12}\root{289 - 1} ={17 \over 12}\root{1 - {1 \over 289}} ={17 \over 12}\sum_{n = 0}^{\infty}{1/2 \choose n}\, {\pars{-1}^{n} \over 289^{n}} \end{align}

$$ \begin{array}{rcl} {17 \over 12} &\approx& \color{#66f}{1.41}\color{#c00000}{6666666666\ldots} \\ {17 \over 12}\,\pars{1 - {1 \over 578}} &\approx& \color{#66f}{1.41421}\color{#c00000}{5686274\ldots} \\ \underbrace{{17 \over 12}\,\pars{1 - {1 \over 578} - {1 \over 668168}}} _{\ds{=\ \color{#00f}{{222337 \over 157216}}}} &\approx& \color{#66f}{1.414213\ {\Large\underbrace{5}_{\uparrow}}}\ 6\color{#c00000}{6049\ldots} \end{array} $$

$$ \color{#c00000}{\large\color{#66f}{\root{2}}\approx{222337 \over 157216}\approx \color{#66f}{1.414213}\color{#00f}{\LARGE 5}\color{#66f}{6}60492570731\ldots} $$

$$ \mbox{Notice that}\quad \root{2} = {222337 \over 157216} + {\cal R}\quad \mbox{where}\quad {\cal R} \equiv {17 \over 12}\sum_{n = 3}^{\infty}{1/2 \choose n}\, {\pars{-1}^{n} \over 289^{n}} $$

Also, \begin{align} \verts{\cal R}&< {17 \over 12}\sum_{n = 3}^{\infty} \verts{\Gamma\pars{3/2} \over \Gamma\pars{n + 1}\Gamma\pars{3/2 - n}}\, {1 \over 289^{n}} \\[3mm]&={17 \over 12}\Gamma\pars{3/2}\sum_{n = 3}^{\infty} \verts{\Gamma\pars{n - 1/2}\sin\pars{\pi\bracks{n - 1/2}} \over \pi\,\Gamma\pars{n + 1}}\,{1 \over 289^{n}} \\[3mm]&<{17 \over 24\root{\pi}}\sum_{n = 3}^{\infty} {\Gamma\pars{n - 1/2} \over \Gamma\pars{n + 1}}\,{1 \over 289^{n}} <{17 \over 24\root{\pi}}\,{\Gamma\pars{5/2} \over \Gamma\pars{4}} \sum_{n = 3}^{\infty}{1 \over 289^{n}} \\[3mm]&={17 \over 192}\,{1 \over 24054048}\approx 3.6809\times 10^{-9} \end{align}