How can I tell a prime number is a prime number and add it in this code

Question

Example

This number is odd number

This number is a prime number

This the output for the input 7

https://code.sololearn.com/cpVPLWb686K5/?ref=app

Answer 1

def is_prime(x):
if x < 2:
return False
for i in range(2, int(x**0.5) + 1):
if x% i == 0:
return False
return True

x = int(input())

if is_prime(x):
print('This is a prime number')
else:
print('This is not a prime number')

Answer 2

Sreeju ,

your first suggested code to check for prime number is totally incorrect.

> it checks if a number is even or odd. checking for prime number needs an other algorithm.

> find for details here: https://en.m.wikipedia.org/wiki/Prime_number

it would be great if you could rework your code or remove your post since it is confusing for beginners.

Answer 3

Arrchith and Sreeju the code needs one more check for x==2 so that it returns True in that case.

Answer 4

def is_prime(number):
if number <= 1:
return False
if number <= 3:
return True
if number % 2 == 0 or number % 3 == 0:
return False
i = 5
while i * i <= number:
if number % i == 0 or number % (i + 2) == 0:
return False
i += 6
return True

Answer 5

Sure,

if x==2:
return True

Insert these lines somewhere before where it tests if x % 2 == 0.

Answer 6

But this answer is not accurate
It tells 12 is a prime number
Correct me if I am wrong

Answer 7

Thank you sreeju

Answer 8

Sreeju

Nice code from you. Can you explain what this line of code is doing exactly:

for i in range(3, int(x ** 0.5) + 1, 2):

Thanks a lot 🙂

Answer 9

Thanks

Answer 10

You don't need to define functions, you can do it simply like this :
n = int(input())
x = int(n/2)+1
for i in range (2,x):
rem = n%i
if rem == 0:
print("not a prime")
break
else:
print("It is a prime no.")

The indentations are ok. Just another interesting method to use if else.😄

Answer 11

Brian Can you implement it in a code

Answer 12

Hoh
def is_prime(x):
if x <= 1:
return False
if x % 2 == 0:
return False
for i in range(3, int(x ** 0.5) + 1, 2):
if x % i == 0:
return False
return True

x = int(input())

if is_prime(x):
print('This is a prime number')
else:
print('This is not a prime number')

Answer 13

Arrchith
def is_prime(x):
if x <= 1:
return False
if x % 2 == 0:
return False
for i in range(3, int(x ** 0.5) + 1, 2):
if x % i == 0:
return False
return True

x = int(input())

if is_prime(x):
print('This is a prime number')
else:
print('This is not a prime number')

Answer 14

@Lothar

The code correctly checks for prime numbers using the trial division method.

I think you didn't execute the code. ;)

Answer 15

@sandra

The line `for i in range(3, int(x ** 0.5) + 1, 2)` initiates a loop to check odd potential factors of `x` up to its square root. It skips even numbers and factors larger than the square root for efficiency. This is part of a prime-checking algorithm.

;)

Answer 16

Hello

Answer 17

Oh it's simple you run a loop from 2 to just before that no. (You can do also do till half of it ) and check whether any of the numbers divides it .
If any of the numbers divides it then it's not a prime no.
Here's the code for example:
x = int(input())
for i in range(2,x):
if x%i==0:
print("Not a prime no.")
break
else:
print("prime no.")

Note in this case I have used else outside the loop.