Htykuut

lower = int(input("from:"))

upper = int(input("to:"))

for num in range(lower,upper + 1):

   if num > 1:

       for i in range(2,num):

           if (num % i) == 0:

               break

       else:

           print(num)

Why does this code print "2" as a prime number? (it is but it should not print it)

2%2==0 so it should skip it...

When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.

Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:

lower = int(input("from:"))

upper = int(input("to:"))

for num in range(lower,upper + 1):

   if num > 1:

       for i in range(2,max(num,3)):

           if (num % i) == 0:

               break

       else:

           print(num)

This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.

For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).

#### INPUTS    

lower = int(input("from:"))

upper = int(input("to:"))



### Code

n = upper

prime_booleans = [True for i in range(n+1)] 

p = 2

while (p * p <= n): 



    # Is current number a prime, eliminate the numbers that are multiples of it 

    if (prime_booleans[p] == True): 

        for i in range(p * 2, n+1, p): 

            prime_booleans[i] = False

    p += 1



# Print all prime numbers 

for p in range(lower, n): 

    if prime_booleans[p]: 

        print p,