Let f(x) = (x² - x)

⇒ f(x) = x(x - 1)

Since “x” is a positive integer “x” and “x - 1” are consecutive positive integers. Now between any two consecutive integers definitely one of them will be divisible by 2. So let us say here that x is divisible by 2 ⇒ x = 2n 

⇒ f(x) = (2n)(2n - 1)

⇒ f(x) = 2(2n² - n)

Clearly f(x) is of the form 2k so f(x) is divisible by 2.

Proof that x² - x is divisible by 2 where x is a positive integer

To prove that x² - x is divisible by 2, we need to show that x² - x is an even number.

1. Express x² - x as x(x - 1)
We can factor x² - x as x(x - 1). This is because when we multiply x by (x - 1), we get x² - x.

2. Use the principle of parity
The principle of parity states that every integer is either even or odd. An even number is any integer that can be expressed as 2n, where n is an integer. An odd number is any integer that can be expressed as 2n + 1, where n is an integer.

3. Consider two cases for x and x-1
Now, we can consider two cases for x and (x - 1):

Case 1: x is even
If x is even, then x can be expressed as 2n for some integer n. Therefore, x - 1 is odd and can be expressed as 2k + 1 for some integer k. Substituting these values into x(x - 1), we get:

x(x - 1) = 2n(2n - 1) = 4n² - 2n = 2(2n² - n)

Since 2n² - n is an integer, we can see that x(x - 1) is divisible by 2. Therefore, x² - x is divisible by 2.

Case 2: x is odd
If x is odd, then x can be expressed as 2n + 1 for some integer n. Therefore, x - 1 is even and can be expressed as 2k for some integer k. Substituting these values into x(x - 1), we get:

x(x - 1) = (2n + 1)(2n) = 4n² + 2n = 2(2n² + n)

Since 2n² + n is an integer, we can see that x(x - 1) is divisible by 2. Therefore, x² - x is divisible by 2.

Conclusion
In both cases, we have shown that x² - x is divisible by 2. Therefore, we can conclude that x² - x is always even and is divisible by 2 for any positive integer value of x.