primes

So I was able to solve this problem with a little bit of help from the internet and this is what I got: def isPrime(n): for i in range(2,int(n**0.5)+1): if n%i==0: return False return True But my question really is how to do it, but WHY. I understand that 1 is not considered a "prime" number even though it is, and I understand that if it divides by ANYTHING within the range it is automatically prime thus the return False statement. but my question is what role does the squaring the "n" play here ? Thank you very much for your attention P.s. I am very inexperienced and have just been introduced

I have trouble understanding this chunk of code: let sieve (p:xs) = p : sieve (filter (\ x -> x `mod` p /= 0) xs) in sieve [2 .. ] Can someone break it down for me? I understand there is recursion in it, but thats the problem I can't understand how the recursion in this example works. mipadi It's actually pretty elegant. First, we define a function sieve that takes a list of elements: sieve (p:xs) = In the body of sieve , we take the head of the list (because we're passing the infinite list [2..] , and 2 is defined to be prime) and append it (lazily!) to the result of applying sieve to the

I am trying to find the fastest way to check whether a given number is prime or not (in Java). Below are several primality testing methods I came up with. Is there any better way than the second implementation(isPrime2)? public class Prime { public static boolean isPrime1(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } for (int i = 2; i <= Math.sqrt(n) + 1; i++) { if (n % i == 0) { return false; } } return true; } public static boolean isPrime2(int n) { if (n <= 1) { return false; } if (n == 2) { return true; } if (n % 2 == 0) { return false; } for (int i = 3; i <= Math