Given a positive integer N, The task is to write a Python program to check if the number is prime or not.

Definition: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are {2, 3, 5, 7, 11, ….}.

Examples : 

Input:  n = 11
Output: true

Input:  n = 15
Output: false

Input:  n = 1
Output: false

The idea to solve this problem is to iterate through all the numbers starting from 2 to (N/2) using a for loop and for every number check if it divides N. If we find any number that divides, we return false. If we did not find any number between 2 and N/2 which divides N then it means that N is prime and we will return True.

Below is the Python program to check if a number is prime: 

# Python program to check if
# given number is prime or not

num = 11

# If given number is greater than 1
if num > 1:

	# Iterate from 2 to n / 2
	for i in range(2, int(num/2)+1):

		# If num is divisible by any number between
		# 2 and n / 2, it is not prime
		if (num % i) == 0:
			print(num, "is not a prime number")
			break
	else:
		print(num, "is a prime number")

else:
	print(num, "is not a prime number")

Output

11 is a prime number

Optimized Method 

We can do the following optimizations: 

1. Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of a smaller factor that has been already checked.
2. The algorithm can be improved further by observing that all primes are of the form 6k ± 1, with the exception of 2 and 3. This is because all integers can be expressed as (6k + i) for some integer k and for i = ?1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides (6k + 3). So a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of form 6k ± 1. (Source: wikipedia)

Now lets see the code for the first optimization method ( i.e. checking till √n )

from math import sqrt
# n is the number to be check whether it is prime or not
n = 1

# no lets check from 2 to sqrt(n)
# if we found any facto then we can print as not a prime number

# this flag maintains status whether the n is prime or not
prime_flag = 0

if(n > 1):
	for i in range(2, int(sqrt(n)) + 1):
		if (n % i == 0):
			prime_flag = 1
			break
	if (prime_flag == 0):
		print("true")
	else:
		print("false")
else:
	print("false")

Output

false