# Solution

An "even" number is a number of the form $2n$, where $n$ is an integer. An "odd" number is one of the form $2n+1$, where $n$ is an integer. Show that the sum of two even or two odd numbers is even, the sum of an odd and an even integer is odd, and the product of two odd numbers is odd.

Part I.
Let $a$ and $b$ be even numbers. As such, $a=2n_1$ and $b=2n_2$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= 2n_1 + 2n_2 \\&= 2 (n_1 + n_2)\end{align} We know that $(n_1 + n_2)$ must be an integer due to the closure of the integers under addition. So, defining $n_3 = n_1 + n_2$, we see that the sum $a+b$ can be written as $2n_3$ for some integer $n_3$, making it an even number. Hence, the sum of two even numbers is even.

Part II.
Now let $a$ and $b$ be odd numbers. As such, $a=n_1+1$ and $b=2n_2+1$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= (2n_1 + 1) + (2n_2 + 1) \\ &= 2n_1 + 2n_2 + 2\\ &= 2(n_1 + n_2 + 1)\end{align} We know that $(n_1 + n_2 + 1)$ must be an integer due to the closure of the integers under addition. So defining $n_3 = n_1 + n_2 + 1$, we see that the sum $a+b$ can be written as $2n_3$ for some integer $n_3$, making it an even number. Hence, the sum of two odd numbers is even.

Part III.
Now let $a$ be odd and $b$ be even. As such, $a=2n_1+1$ and $b=2n_2$ for some integers $n_1$ and $n_2$. Now consider their sum: \begin{align}a+b &= (2n_1 + 1)+ 2n_2 \\ &= 2n_1 + 2n_2 + 1\\ &= 2(n_1 + n_2) + 1\end{align} We know that $(n_1 + n_2)$ must be an integer due to the closure of the integers under addition. So defining $n_3 = n_1 + n_2$, we see that the sum $a+b$ can be written as $2n_3+1$ for some integer $n_3$, making it an odd number. Hence, the sum of an even and an odd integer is odd.

Part IV.
Now let $a$ and $b$ be odd numbers. As such, $a=2n_1 + 1$ and $b=2n_2+1$ for some integers $n_1$ and $n_2$. Now consider their product: \begin{align} ab &= (2n_1 + 1)(2n_2 + 1) \\ &= 4n_1 n_2 + 2n_1 + 2n_2 + 1 \\ &= 2(2n_1 n_2 + n_1 + n_2) + 1\end{align} We know that $(2n_1 n_2 + n_1 + n_2)$ must be an integer due to the closure of the integers under addition and multiplication. So defining $n_3 = (2n_1 + n_1 + n_2)$, we see that the product $ab$ can be written as $2n_3+1$ for some integer $n_3$, making it an odd number. Hence, the product of two odd numbers is odd.