sqrt

sqrt

Babylonian method for square root

http://www.geeksforgeeks.org/square-root-of-a-perfect-square/

why (x + y)/2 is always bigger than square root of n:

http://en.wikipedia.org/wiki/Square_root#Geometric_construction_of_the_square_root

public double sqrt(double n) {

    double x = n;

    double y = 1;

    double e = 0.0001

    while (x - y > e) {

        x = x + (y - x) / 2;

        y = n / x;

    }

    return x;

}