Okay, I know this is very basic question. I learned 2 methods in school. But now, I forget one.
Here is a simple method that I know.
- Find the prime divisors of the number
- Omit the half of numbers that have been appeared even times
- multiply the rest
For example you want to find square root of 36. You find the divisors. They are 2x2x3x3. In step 2 they appeared as 2×3. That is 6. Problem is this method works when the square root is an integer number. It doesn't work for numbers that doesn't have integer square root. Like 38.
So my question is how can I find the square root of any arbitrary number using pen and paper?
Best Answer
You can use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to arrive at a "long-division" like method. Let me show you how it is done for 3838 before giving the algorithm.
So you've arrived at, at this point, $3838 = (61.9)^2 + 6.39$. And you can continue the process indefinitely.
How does this work? Given a number $A$, you want to find its square root in base-10 representation. Suppose your square root looks like $$ a_{100}a_{10}a_1.a_{0.1}a_{0.0.1}\ldots $$ when expanded as a string of digits. Then you find the biggest $a_{100}$ such that $$ (a_{100}\times 100)^2 \leq A $$ (similar to how you do long division). Then to make solve for the next digit, you use that $$ (a_{100} \times 100 + a_{10}\times 10)^2 \leq A $$ (since you are truncating the decimal expansion, which makes the number smaller). So you solve for the best $a_{10}$ such that $$ (a_{100})^2\times 100^2 + \left[ 20 \times (a_{100}\times a_{10}) + (a_{10})^2\right] \times 100 + \leq A $$ The above expression shows why in the first step you want to group the digits in twos: in some sense we are thinking of $A$ as in base-100, the "square" of base-10.
At every step you use the identity $(x+c)^2 = x^2 + 2xc + c^2$ to compute the next digit correction to the square root.
Another method, as the Babylonians did it, was recently detailed by John Baez at his blog, which uses the "equality case" of the arithmetic-mean-geometric-mean inequality to power the iteration.