Question 3

We will use proof by contradiction.

Step 1 — Assume the opposite

Let's assume the opposite of what we want to prove. We want to prove that $a$ is even. So, let's assume $a$ is not even. This means $a$ must be an odd integer. We can write an odd integer as $2n + 1$. So, let $a = 2n + 1$ for some integer $n$.

Step 2 — Calculate $a^2$

Now, let's find $a^2$ using our assumption. We will substitute $a = 2n + 1$. $a^2 = (2n + 1)^2$ $= (2n)^2 + 2(2n)(1) + 1^2$ $= 4n^2 + 4n + 1$ $= 2(2n^2 + 2n) + 1$ Let $k = 2n^2 + 2n$. Since $n$ is an integer, $2n^2$ is an integer. Also, $2n$ is an integer. So, $k$ is an integer. $a^2 = 2k + 1$

$\boxed{a^2 \text{ is odd}}$

Step 3 — Find the contradiction

We found that $a^2$ is odd. But the problem states that $a^2$ is even. An integer cannot be both odd and even. This is a contradiction. Our initial assumption must be wrong. Therefore, $a$ cannot be odd. If $a$ is not odd, then $a$ must be even.

Answer

(i) We assumed $a$ is odd. (ii) We showed that $a^2$ is odd. (iii) This contradicts the given information that $a^2$ is even.