The problem asks us to prove three statements using the direct proof method: i) If $n$ is an even integer, then $n^2$ is even. ii) The sum of two odd numbers is even. iii) The sum of an even number and an odd number is odd.

1. Problem Description

The problem asks us to prove three statements using the direct proof method:

i) If $n$ is an even integer, then $n^2$ is even.

ii) The sum of two odd numbers is even.

iii) The sum of an even number and an odd number is odd.

2. Solution Steps

i) If $n$ is an even integer, then $n^2$ is even.

A direct proof starts by assuming the hypothesis is true and showing that the conclusion must also be true.

Assume $n$ is an even integer. This means that $n$ can be written as $n = 2k$ for some integer $k$.

Then, $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$.

Since $2k^2$ is an integer, $n^2$ is a multiple of 2, which means $n^2$ is even.

ii) The sum of two odd numbers is even.

Let $a$ and $b$ be two odd integers.

Then $a$ can be written as $a = 2k + 1$ for some integer $k$, and $b$ can be written as $b = 2m + 1$ for some integer $m$.

The sum of $a$ and $b$ is $a + b = (2k + 1) + (2m + 1) = 2k + 2m + 2 = 2(k + m + 1)$.

Since $k + m + 1$ is an integer, $a + b$ is a multiple of 2, which means $a + b$ is even.

iii) The sum of an even number and an odd number is odd.

Let $a$ be an even integer and $b$ be an odd integer.

Then $a$ can be written as $a = 2k$ for some integer $k$, and $b$ can be written as $b = 2m + 1$ for some integer $m$.

The sum of $a$ and $b$ is $a + b = 2k + (2m + 1) = 2k + 2m + 1 = 2(k + m) + 1$.

Since $k + m$ is an integer, $a + b$ can be written in the form $2n + 1$ for some integer $n$, which means $a + b$ is odd.

3. Final Answer

i) If $n$ is an even integer, then $n^2$ is even. Proof: $n = 2k$, $n^2 = 4k^2 = 2(2k^2)$, therefore $n^2$ is even.

ii) The sum of two odd numbers is even. Proof: $a = 2k + 1$, $b = 2m + 1$, $a + b = 2k + 1 + 2m + 1 = 2(k + m + 1)$, therefore $a + b$ is even.

iii) The sum of an even number and an odd number is odd. Proof: $a = 2k$, $b = 2m + 1$, $a + b = 2k + 2m + 1 = 2(k + m) + 1$, therefore $a + b$ is odd.