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    <p>I've added comments on the sides to explain what each line does:</p> <pre><code>from math import * # imports everything from the math module def main(): n = abs(input("Enter a number: ")) # gets input from the user i = 2 # starts off at 2 because all input is divisble by 1 msg = 'is a prime number.' # the message is initially set while i &lt;= sqrt(n): if n % i == 0: # if 'i' divides evenly into n msg = 'is not a prime number.' # only set if it isn't a prime i = i + 1 # increases 'i' by 1 so it can check every value up to the square-root of 'n' (to see if it divides evenly) print n, msg main() </code></pre> <p>The program has to go through every value of <code>i</code> (up to the square-root of <code>n</code>) so that every possible factor is checked.</p> <p>This is sort of a rough-prime checker, and inefficient for large numbers that aren't prime: If the input was a number like <code>1234567890</code>, it will iterate through every number up to the square root of that number, which is <code>35147</code> (rounded up). Using <code>return</code> statements break the loop, so the first number you check, <code>2</code>, it is declared not prime since it is evenly divisible by <code>2</code>. By using <code>return</code>, it will stop the function, and save you 35,146 calculations. That isn't a massive number (for computers, at least) but it's still more memory-efficient and takes less time.</p> <pre><code>def isPrime(n): '''Checks if 'n' is prime.''' from math import sqrt if n == 0 or n == 1: return False else: for check in range(2, int(sqrt(n))+1): if n % check == 0: return False return True </code></pre>
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    1. PODetermining whether a number is prime or not
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    1. COSidenote: why import sqrt if you don't use it? ;)
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    2. COWoops, I forgot to add that part in :P
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