Approximating the sqrt of a number with at least 6 decimal places in python with binary search: def sqrt(n): max_err = 0.0000001 lo = 0.0 hi = float(n) while 1: mid = (lo + hi) / 2 err = (n / mid) - mid if err > max_err: lo = mid elif err < -max_err: hi = mid else: return mid print sqrt(2) You can optimize it by giving a better rough estimate for the starting value of "lo" and "hi" instead of 0.0 and n.

Newton's method, we are searching for the sqrt(S): x^2 - S == 0 f(x) = x^2 - S f'(x) = 2x x(n+1) = x - (x^2 - S)/(2x) = (2x^2 - x^2 + S)/(2x) = (x^2 + S)/(2x) = 0.5 * (x + S/x) def sqrt(n): max_err = 0.0000001 guess = 1 while 1: print guess guess = 0.5 * (guess + n / guess) err = (n / guess) - guess if abs(err) < max_err: return guess def isqrt(n): max_err = 1 guess = 1 while 1: print guess guess = (guess + n // guess) // 2 err = (n / guess) - guess if abs(err) < max_err: return guess print sqrt(2) print isqrt(66) The newton method is used for integer square root (isqrt) too.

def sqrt(num): x=num while abs(x*x-num) > 0.000001: x = 1/2*(x+num/x) return x