How to create the most compact mapping n → isprime(n) up to a limit N?

Question

Naturally, for bool isprime(number) there would be a data structure I could query.
I define the best algorithm, to be the algorithm that pr

Answer 1

Most of previous answers are correct but here is one more way to test to see a number is prime number. As refresher, prime numbers are whole number greater than 1 whose only factors are 1 and itself.(source)

Solution:

Typically you can build a loop and start testing your number to see if it's divisible by 1,2,3 ...up to the number you are testing ...etc but to reduce the time to check, you can divide your number by half of the value of your number because a number cannot be exactly divisible by anything above half of it's value. Example if you want to see 100 is a prime number you can loop through up to 50.

Actual code:

def find_prime(number):
    if(number ==1):
        return False
    # we are dividiing and rounding and then adding the remainder to increment !
    # to cover not fully divisible value to go up forexample 23 becomes 11
    stop=number//2+number%2
    #loop through up to the half of the values
    for item in range(2,stop):
        if number%item==0:
           return False
        print(number)
    return True


if(find_prime(3)):
    print("it's a prime number !!")
else:
    print("it's not a prime")