For loop in python is ignoring the if statement

Question

I'm new to python, and I can't see why my for loop is ignoring the if statement and saying every number is a prime number. My code is as such.


def prime_number(number):
    sqrt_prime = math.sqrt(number)
    for x in (2,sqrt_prime-1):
        if isinstance((number/x), int) and x <= sqrt_prime-1:
            con_num = str(number)
            print(con_num,end=' ');print('is not a prime number.')
            break
    else:
        con_num = str(number)
        print(con_num,end=' ');print('is a prime number.')

It could probably be a lot more streamlined, just want to know why it isn't either outputting true for the isinstance or that x is smaller or equal to the sqrt of the number - 1. Thanks in advance.

check type in isinstance it may be float is you're using python3 — neilharia7
– neilharia7, Commented Mar 8, 2021 at 4:15
number/x will always be a float in python3, so your condition will not be satisfied(The same opinion as @neilharia7) — Alex.Kh
– Alex.Kh, Commented Mar 8, 2021 at 4:16
I wonder if we can reduce the range of (2, sqrt_prime-1) to half. — Razzle Shazl
– Razzle Shazl, Commented Mar 8, 2021 at 5:24
You might consider checking a potential factor like this: if number % x == 0 instead of using / though personally I would be cautious about 0 vs 0.0 and getting incorrect answers from rounding. You might also try if number / x == number // x:. — Razzle Shazl
– Razzle Shazl, Commented Mar 8, 2021 at 5:26

astrochun · Accepted Answer · 2021-03-08 04:21:33Z

1

If this is python 3.x then it makes sense why it ignores it.

number/x will always return a float and not an integer. Thus, isinstance((number/x), int) is always False.

edited Mar 8, 2021 at 4:21

answered Mar 8, 2021 at 4:15

astrochun

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1 Comment

DBrowne Over a year ago

The else statement is attached to the for loop, to be executed if the loop does not break. So it is properly indented.

Frank · Accepted Answer · 2021-03-08 04:22:23Z

0

The problem is isinstance((number/x), int).

The type is always float if you check them by

print(type(number/x))

Correction: Change it from

        if isinstance((number/x), int) and x <= sqrt_prime-1:

to

        if number % x == 0 and x <= sqrt_prime-1:

Another problem is the coding:

def prime_number(number):
    sqrt_prime = math.sqrt(number)
    flg = 0
    for x in (2,sqrt_prime-1):
        if number % x == 0 and x <= sqrt_prime-1:
            con_num = str(number)
            print(con_num,end=' ');print('is not a prime number.')
            flg = 1
            break
    if flg==0:
            con_num = str(number)
            print(con_num,end=' ');print('is a prime number.')

I am not sure if you have finished the code, seems like this is algorithm is not running rightly.

edited Mar 8, 2021 at 4:22

answered Mar 8, 2021 at 4:15

Frank

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2 Comments

hackerman Over a year ago

Thanks, fixed the float issue. I also realised I forgot to make it range(2,int(sqrt_prime)-1), so it was only dividing by 2 and then moving on, rather than going up all the values from 2 to the sqrt -1

Olvin Roght Over a year ago

if not number % x and x < sqrt_prime, print(number, 'is not a prime number.') Check this

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