Final answer:
To prove that the sum of two odd integers is even, express them as 2n + 1 and 2m + 1, where n and m are integers. The sum is 2n + 2m + 2, which equals 2 times the integer (n + m + 1), therefore showing the sum is even.
Step-by-step explanation:
To prove that the sum of two odd integers is even, we can use a direct proof. An odd integer can be expressed in the form of 2n + 1, where n is an integer. Let's consider two odd integers, a and b. We can write a as 2n + 1 and b as 2m + 1, where n and m are integers.
The sum of a and b is (2n + 1) + (2m + 1), which simplifies to 2n + 2m + 2. Factoring out a 2 from the sum, we get 2(n + m + 1). This is an even integer since it is two times another integer. Therefore, the sum of two odd integers is even.