How to Find the Square Root of Numbers That Aren't Square Numbers Instead of trying to find the square root of square numbers, I try to find the square root of numbers that aren't square numbers like 12 and 35, but it seems hard to do it with pencil and paper, so I use a calculator to figure those out.  I wonder how I can find the square root of a not-square number ONLY with paper and pencil.
 A: There is an algorithm that I saw presented in 6th grade, but I never properly learned it.  It was standard at one time, but calculators caused it to become neglected.
However, there is the "Babylonian method":
Let $A$ be an approximation to the square root of $N$.
Then $\dfrac{A+\frac NA}{2}$ is a better approximation.
If the first approximation is not absurd, then usually the third is pretty good.
Suppose $3$ is our first approximation to $\sqrt{10}$.  Then our second approximation is
$$
\frac{3 + \frac{10}{3}}{2} = \frac{19}6 = 3+\frac16=3.16666666\ldots.
$$
If $\left(\dfrac{19}{6}\right)^2=10$ then $19^2=6^2\cdot10$, so $361=360$.  Not too bad an approximation.
Now let's find the third approximation:
$$
\frac{\frac{19}{6}+\frac{10}{19/6}}{2} = \frac{721}{228} = 3.1622807\ldots
$$
If $\left(\dfrac{721}{228}\right)^2=10$ then $721^2=228^2\cdot 10$, so $519841=519840$.  Off by one part in more than half a million.
All of the above is easily done without a calculator.
A calculator tells me that $\sqrt{10}=3.16227766\ldots$.
A: Here's an alternative method that I actually sometimes use when I do not even have pen and paper. It is less efficient than the Babylonian method, but I can store the required intermediate results in my head, which makes it useful if you only need a few decimal places.  Furthermore, it is possible to perform this computation using only addition.
You want the square root of some number $x$.


*

*Pick a number $n_0$ for which you can calculate or perhaps know $y_0 = n_0^2$. Try increasing of decreasing $n_0$ a bit until you are fairly close to $x$, but strictly smaller.

*Compute $y_1 = (n_0+1)^2$. This can be done easily by using $y_1 = y_0 + n_0 + n_0 + 1$.

*If $y_1 > x$, go back to $y_0$. You are done and $y_0$ is your (temporary) answer.

*Otherwise, repeat 2 and 3 until you are done.


This procedure gives you the integer with the closest square that is smaller than $x$.  To find more digits, multiply $x$ by 100 and the last $n$ by 10.  Then, repeat the procedure.  Don't forget to divide by 10 again at the end (simply move the decimal point to the proper place).
An Example
I'll demonstrate this procedure with a simple example.  Suppose we want to compute the square root of 11 to 3 decimal points.
We start with $n = 3$, such that $y = n^2 = 9$. Because $4^2 > 11$, this is the closest integer with a smaller square, so this will be the first decimal.
Now, we add one zero to $n$ and two zeroes to $x$ and $y$ to get $n=30$, $y=900$ and $x = 1100$.  Find a closer integer by repeating step 2 above:


*

*$31^2 = 30^2 + 30 + 31 = 961$

*$32^2 = 31^2 + 31 + 32 = 1024$

*$33^2 = 32^2 + 32 + 33 = 1089$

*$34^2 = 33^2 + 33 + 34 = 1156$


Oops, we've gone too far. So, apparently $33$ is the integer we want. Recall we multiplied by 10, so this means that actually your answer now is $\sqrt{11} \approx 3.3$. You can repeat this process to find more decimals.


*

*$330^2 = 108900$

*$331^2 = 108900 + 330 + 331 = 109561$

*$332^2 = 109561 + 331 + 332 = 110224$

*Back up to $331$.


So, now we have $\sqrt{11} \approx 3.31$.  The actual answer is $3.31662\!\ldots$, so we see that the first 3 decimal places are indeed correct, just like expected.
