Final answer:
By representing odd integers with algebraic expressions (2m + 1 and 2n + 1 for any integers m and n), adding them together yields an expression (2m + 2n + 2) that factors out to 2 times another integer, confirming that the sum of two odd integers is indeed an even integer.
Step-by-step explanation:
To prove that the sum of two odd positive integers is an even positive integer, we can use algebraic representation. An odd integer can be represented by 2n + 1, where n is any integer. If we take two odd integers, the first can be written as 2m + 1 and the second as 2n + 1. When we add them together, we get:
(2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1).
Since m, n, and 1 are all integers, the sum (m + n + 1) is also an integer. The product of 2 and any integer is, by definition, an even integer. Therefore, the sum of two odd integers is an even integer. This algebraic proof confirms the commutative property of addition, which states that A + B = B + A, and illustrates that the result is the same regardless of the order in which numbers are added.