Answer
Let $m$ be an even integer and $n$ an odd integer. By the definition of even, $m=2j$ for some integer $j$, and by the definition of odd, $n=2k+1$ for some integer $k$. Therefore, $m-n=(2j)-(2k+1)=2j-2k-1=2j-2k-2+2-1=2(j-k-1)+2-1=2(j-k-1)+1$. But $j-k-1$ is an integer, because $j$, $k$, and $1$ are all integers, and the integers are closed under subtraction. Therefore, $m+n$ is of the form $2x+1$ for the integer $x=j-k-1$, so by the definition of odd, $m+n$ is odd. We conclude that an even integer minus an odd integer is always an odd integer.
Work Step by Step
This is an example of proof by generalization from the generic particular, as described on pages 151 and 152. The idea is to write the theorem requiring proof as a universal implication ("for all [ _ ], if ..., then..."), name (using a variable) a specific but arbitrary element of the domain with the properties specified in the hypothesis, and show that it must have the properties specified in the conclusion. This proof method works because, even though we go through the logic for a specific element of the domain, we never reference any properties of the element that might not apply to other elements of the domain.
