[size=150]Let's consider the following proposition: [br][br]"For all positive integers n, if n is even, then n[sup]2[/sup] is even."[br][br]To prove this proposition using a direct proof, we can follow these steps:[br][br]1. Clearly state the proposition: "For all positive integers n, if n is even, then n[sup]2[/sup] is even."[br][br]2. Assume the premise: Let's assume that n is a positive even integer.[br][br]3. Use the definition of even numbers: An even number can be expressed as 2k, where k is an integer.[br][br]4. Substitute n with 2k in the proposition: We have n = 2k, and we need to prove that (2k)[sup]2[/sup] is even.[br][br]5. Simplify the expression: (2k)[sup]2[/sup] = 4k[sup]2[/sup], which is 2(2k[sup]2[/sup]).[br][br]6. By definition, 2(2k[sup]2[/sup]) is an even number since it can be expressed as 2m, where m = 2k[sup]2[/sup].[br][br]7. Thus, we have shown that if n is even (represented as n = 2k), then n[sup]2[/sup] (represented as (2k)[sup]2[/sup]) is even.[br][br]8. Conclude the proof: We have successfully proven the proposition that for all positive integers n, if n is even, then n[sup]2[/sup] is even.[br][br]In this example, we used the definition of even numbers and algebraic manipulations to establish the truth of the proposition through a direct proof.[/size]