According to Wikipedia:

Based on this definition, if we consider the first 10 natural numbers, we can see that 2, 3, 5, and 7 are primes, while 1, 4, 6, 8, 9, and 10 are not. In order to have a computer tell you whether a number, N, is prime, you can divide that number by all natural numbers in the range [2, N). If any of those divisions yields zero as a remainder, then the number is not a prime. Enough chatter, let's get down to business. I'll write two versions of this, the second of which will exploit the for...else syntax:

# primes.py
primes = []  # this will contain the primes in the end
upto = 100  # the limit, inclusive
for n in range(2, upto + 1):
    is_prime = True  # flag, new at each iteration of outer for
    for divisor in range(2, n):
        if n % divisor == 0:
            is_prime = False
            break

    if is_prime:  # check on flag
        primes.append(n)
print(primes)

There are a lot of things to notice in the preceding code. First of all, we set up an empty primes list, which will contain the primes at the end. The limit is 100, and you can see it's inclusive in the way we call range() in the outer loop. If we wrote range(2, upto) that would be [2, upto), right? Therefore range(2, upto + 1) gives us [2, upto + 1) == [2, upto].

So, there are two for loops. In the outer one, we loop over the candidate primes, that is, all natural numbers from 2 to upto. Inside each iteration of this outer loop, we set up a flag (which is set to True at each iteration), and then start dividing the current n by all numbers from 2 to n - 1. If we find a proper divisor for n, it means n is composite, and therefore we set the flag to False and break the loop. Notice that when we break the inner one, the outer one keeps on going normally. The reason why we break after having found a proper divisor for n is that we don't need any further information to be able to tell that n is not a prime.

When we check on the is_prime flag, if it is still True, it means we couldn't find any number in [2, n) that is a proper divisor for n, therefore n is a prime. We append n to the primes list, and hop! Another iteration proceeds, until n equals 100.

Running this code yields:

$ python primes.py
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

Before we proceed, one question: of all the iterations of the outer loop, one of them is different from all the others. Could you tell which one, and why? Think about it for a second, go back to the code, try to figure it out for yourself, and then keep reading on.

Did you figure it out? If not, don't feel bad, it's perfectly normal. I asked you to do it as a small exercise because it's what coders do all the time. The skill to understand what the code does by simply looking at it is something you build over time. It's very important, so try to exercise it whenever you can. I'll tell you the answer now: the iteration that behaves differently from all others is the first one. The reason is because in the first iteration, n is 2. Therefore the innermost for loop won't even run, because it's a for loop that iterates over range(2, 2), and what is that if not [2, 2)? Try it out for yourself, write a simple for loop with that iterable, put a print in the body suite, and see whether anything happens (it won't...).

Now, from an algorithmic point of view, this code is inefficient, so let's at least make it more beautiful:

# primes.else.py
primes = []
upto = 100
for n in range(2, upto + 1):
    for divisor in range(2, n):
        if n % divisor == 0:
            break
    else:
        primes.append(n)
print(primes)

Much nicer, right? The is_prime flag is gone, and we append n to the primes list when we know the inner for loop hasn't encountered any break statements. See how the code looks cleaner and reads better?