Final answer:
To prove that 9n - 2 is even if and only if n is even, we can show that when n is even, 9n - 2 is even, and when 9n - 2 is even, n is even.
Step-by-step explanation:
To prove that 9n - 2 is even if and only if n is even, we need to show that when n is even, 9n - 2 is even, and when 9n - 2 is even, n is even. Even numbers are divisible by 2 without any remainder.
For the first case, if n is even, we can write n as n = 2k, where k is an integer.
Substituting this into 9n - 2, we get 9(2k) - 2 = 18k - 2. Since 18k - 2 can be expressed as 2(9k - 1), we can see that it is divisible by 2, making it an even number.
For the second case, if 9n - 2 is even, we can write 9n - 2 = 2m, where m is an integer.
Expanding this equation, we get 9n - 2 = 2m as 9n = 2m + 2. From this equation, we can see that 9n must be even because it is the sum of two even numbers. This means that n must also be even to satisfy the equation.