# Python program to check whether a number is Prime or not

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 = 11Output:true

Input:n = 15Output:false

Input:n = 1Output: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:

## Python3

`# 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:

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.

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

## Python3

`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

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)