## Discrete Mathematics with Applications 4th Edition

Let $m$ be an odd integer and $n$ an even integer. Then by the definition of odd, $m=2j+1$ for some integer $j$, and by the definition of even, $n=2k$ for some integer $k$. Hence, $m-n=(2j+1)-2k=(1+2j)-2k$$=1+(2j-2k)=(2j-2k)+1=2(j-k)+1$. Each of these steps is justified either by the associative property of addition and subtraction, or the commutative property of addition and subtraction. By the closure of the integers under subtraction ("an integer minus an integer is also an integer"), we know that $j-k$ is an integer, so $2(j-k)+1$ is of the form twice an integer plus one. But that is precisely the definition of an odd number. Therefore, since $m-n=2(j-k)+1$, it must be that $m-n$ is odd. But since $m$ and $n$ were arbitrarily chosen, we can conclude that the difference between any odd integer and any even integer is odd.
Note that it does not matter that we do not know the values of $j$ and $k$, or whether they themselves are even and odd. It is sufficient to know that they are integers to prove the theorem.