since using a for loop can only check the integer roots Finding integer roots seems like all you need if you are looking for the floor of the square root. If you start at zero, and check every integer for i*i >= x, the first i for which that would be true would be the ceiling of the square root. Then you just need to decide if x is a perfect square (return i), or not (return i - 1). For example, if you are looking for sqrt(10), i*i >= 10 becomes true when i is 4. Because i*i is not actually equal to 10 (because 10 is not a perfect square), you know that the floor of the square root is 3. If you want to get more efficient and mathy, you could use Newton's method. You are looking for x such that x2 = a. Another way of saying this is that you are looking for x such that f(x) = x2 - a is zero. The derivative of this is f'(x) = 2*x. Now Newton's method tells us to just repeatedly execute: x = x - f(x) / f'(x) This should converge on the square root. More on reddit.com

Casting a float as an integer will always round down, you can make use of that to create your own 'roundup' function by just casting your number as an int, then adding 1. The exception to that is if your number is already an integer - in that case we don't want to do anything to it, we can handle this case separately beforehand. def round_up(x): if int(x)==x: return int(x) else: return int(x) + 1 More on reddit.com

ceil always rounds up. More on reddit.com

Specifically, because it uses the "ties to even" rounding rule (there is more than one rule allowed by the IEEE standard) as it's more fair statistically

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