Final answer:
The statement 'If a | b = 1 and a | c = 1, then a | (b + c) = 1' is true because if b and c are multiples of a, then their sum (b + c) is also a multiple of a, which is confirmed by the commutative property of addition.
Step-by-step explanation:
The question 'If a | b = 1 and a | c = 1, then a | (b + c) = 1 for all integers a, b, and c, where a ≠ 0? 1) True 2) False' relates to the concept of divisibility in integers. Here, 'a | b = 1' means 'a divides b exactly one time' which is equivalent to saying that b is a multiple of a or b = ak for some integer k. Similarly, 'a | c = 1' means that c is a multiple of a, or c = al for some integer l. Thus, if we add these two multiples together, (b + c) = ak + al which can be factored out as a(k + l), showing that a also divides (b + c) exactly one time.
The commutative property of addition, which states that A+B = B+A, supports the idea that the order in which we add numbers does not change the sum. This fact further suggests that if a | b and a | c individually, it does not particularly matter in which order we consider the sum, the divisibility by a is still maintained.