is inumber Primary Number or not with forloop . n/2 . confusion

in example below code tells you if number is Primary or not in a for loop . i understand most of it but one thing confuses me . why , first: when i check if 2 is Prmary number or not progrmamme tells you that it is Prmary number which is correct but i don't understand why when it is checked against i -2 , first iteration it tells you that it is . what in a code decides that it is Primary number ? we know that 2%2 leaves no remainder. code still tell you it is a Primary number.

2nd and even more confusing , when i checking if 9 is Prmary number , code says it is not a Primary number which is correct but in essence code should tell you that it is Primary number immediately after 9 is checked against i-2 . 9%2 leaves a remainder and " break " should stop the loop without checking it with next iteration i=3 which is when it is correctly decided that it is not aPrimary number . 9%3 leaves no remainder and it is not a Primary number . what in this programme decides that it shold be checked against i-3 ?

Thanks.

using namespace std;

int main() {
    
    int i, n;
    bool is_prime = true;
    
    cout << "Enter a positive integer: ";
    cin >> n;
    
  
    if (n == 0 || n == 1) {
        is_prime = false;
    }
    
    
    for (i = 2; i <= n/2; ++i) {
        if (n % i == 0) {
            is_prime = false;
            break;
        }
    }
    
    if (is_prime){
        cout << n << " is a prime number" << " checked against : " << i << endl;
        cout << endl;
    }
    else
        cout << n << " is not a prime number"<< " checked against : " << i << endl;
    cout << endl;
    
    return 0;
}

I think the issue is that you are not understanding the loop.
for (i = 2; i <= n/2; ++i) says that you start with a value of 2, and as long as i is less than half of n, execute the loop.

When n=2, the value of i is never less than n/2 (1) so it does not execute the loop at all.

When n=9, the value if i can range from 2 to 4 (integer division rounds down). So it will check those values, dropping out when it hits 3.

The break statement in the loop is only triggered when there is no remainder, so (9 % 2) will not execute that code.

I spent almost a month figuring it out . it makes sence now . with 2 i missed the fact that 2 is the only even namber that has 2 only factors. that's what i was confused mainly not because the way for loop in given code works . confusion about why 9 and some other numbers are not reported as the Prime number when it is checked against first iteration i= 2 was that number (lets say 9) has to go though all the iterations defined by n/2 untill it hits i(numer)that divides it without remainder and only then loop is stoped " by "brak;" i=3 in case of 9 .

Thanks . i'm very new to coding had no prior experience so it takes little time untill i get my head around basics of coding