How do I find the square root of numbers without using a calculator? I need help finding how to determine which number is greater without using a calculator then explaining how I got the answer. I know how to start the first step but I don't know how to do the rest. For example, to find the square root of 20 i know it has to be between 4 and 5 but i do not know how to finish it. Here is an example: which is greater? the square root of 20 or the square root of 10?
 A: $\sqrt{20}>\sqrt{10} $ since $(\sqrt{20})^2=20>(\sqrt{10})^2=10 $.
To calculate by hand a square root, there is an algorithm similar to the division algorithm, that I learnt in grade 6. Here is how it goes for the square root of $20$:
$$\begin{array}[t]{r}
\\\\\\\\ 8*\times *=\\\\88*\times *=\\\\894*\times*=\\\\8944*\times*=
\end{array}
\quad
\begin{array}[t]{*{20}{c@{}}}\color{blue}{\mathbf{4.}}& \color{blue}{\mathbf 4}& \color{blue}{\mathbf 7}& \color{blue}{\mathbf 2}&\color{blue}{\boldsymbol{\dots}}\\
\hline
\mathbf 2&\mathbf 0.&0&0&0&0&0&0\\\color{blue}1& \color{blue}6\\\hline 
& \color{blue}{4}&\color{red}0&\color{red}0\\
& \color{blue}3& \color{blue}3& \color{blue}6\\
\hline
&& \color{blue}6& \color{blue}4&\color{red}0&\color{red}0\\
&& \color{blue}6& \color{blue}2& \color{blue}0& \color{blue}9\\
\hline
&&& \color{blue}1& \color{blue}9& \color{blue}1&\color{red}0&\color{red}0\\
&&&\color{blue}1& \color{blue}7& \color{blue}8& \color{blue}8& \color{blue}4\\
\hline
&&&&\color{blue}1&\color{blue}2&\color{blue}1&\color{blue}6&\color{red}0&\color{red}0 \\
&&&&& \dots&\dots & \dots&\dots&\dots
\end{array}$$
Explanation:
Group the digits of the number in pairs from the right (starting from the decimal dot if it is not an integer. The first digit (from the left) of the square root is the greatest integer $r$ such that $r^2\le $ the leftmost pair. Here, $4^2=16\le 20$ and $5^2>20$, so the first digit is $4$ and there is a remainder of $4$.
Next digit: add the following pair of digits to the remainder (so you get the integer $400$). Double the first digit, and write on its right another digit $k$, so that you get the number $8k$ in decimal notation. In formula, you have the integer $20r+k$. Multiply this integer by $k$. The next digit will be the greatest $k$ so that the result is $\le 400$. Subtract this result from $400$ for the following step, and proceed similarly.
A: Here is the Babylonian method (which also gives the same steps as Newton's method, but the Babylonians get the credit this time)
say you want to find $\sqrt n$  take a guess, call it $x_0$
$\sqrt n$ lies somewhere between $x_0$ and $\frac {n}{x_0}$  so split the difference.
$x_1 = \frac 12 (x_0 + \frac {n}{x_0})$
repeat as necessary, every time you repeat you will approximately double the precision of the result.
example:
$n = 20\\
x_0 = 4$
$x_1 = \frac 12 (4 + 5) = 4.5\\
x_2 = \frac 12 (4.5 + \frac {20}{4.5}) \approx 4.472$
Which is greater $\sqrt 20$ or $3 \sqrt 2$?
You don't need to know how to approximate square roots to do this.
You need to know that $3\sqrt 2 = \sqrt 18$  and if $n>m$ then $\sqrt n > \sqrt m$ 
