质因数，帮助理解平方根的使用

Question

根据 geeks for geeks 网站上给出的寻找质因数的解决方案，我不太明白为什么他们在第 16 行使用 n 的平方根（for i in range(3,int(math.sqrt(n))+1,2): )

https://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/

# Python program to print prime factors 

import math 

# A function to print all prime factors of 
# a given number n 
def primeFactors(n): 
    
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        print 2, 
        n = n / 2
        
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
    for i in range(3,int(math.sqrt(n))+1,2): 
        
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            print i, 
            n = n / i 
            
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        print n

如果有人能理解下面给出的反例，这就是我无法遵循的逻辑。

Every composite number has at least one prime factor less than or equal to square root of itself. This property can be proved using counter statement. Let a and b be two factors of n such that a*b = n. If both are greater than √n, then a.b > √n, * √n, which contradicts the expression “a * b = n”.

Answer 1

这是简单的算术运算。如果 r 是 n 的平方根，则 r * r == n。如果将 r 乘以任何大于它的值，结果将大于 n。所以不可能有任何其他大于 r.

的质因数

所以继续循环超过平方根是没有意义的，因为它永远找不到任何东西。

质因数，帮助理解平方根的使用

Prime factors, help understand the use of square root

python

python-3.x

math

logic

prime-factoring