Suppose 𝑎 ∈ℤ. Prove that 14|𝑎 if and only if 7|𝑎 and 2|𝑎. Recall this proposition from class: For all integers 𝑚 and 𝑛, if 𝑚𝑛 is an even integer, then 𝑚 is even or 𝑛 is even.

Question

please write step by step

Answer 1

Let's start with the forward case: by the fundamental theorem of arithmetic (which is that every integer n has a unique prime factorization), if a=14k (where k is some integer), since 14 can be uniquely written as the prime factorization 2•7, we have a = (2•7)k = 2(7k) = 7(2k). Since a can be written as 2 times a constant and 7 times a constant, 2|a and 7|a. Because the prime factorization is unique, this should cover the backwards case as well. If it's not divisible by either 2 or 7, then part of the prime factorization of 14 is missing, so for example, if b = 2•p and 7 does not divide p, then it does not divide 14 because 7 is part of the prime factorization; the same could be said about c=7•q where 2 does not divide q.

Using the proposition from your class follows very similar reasoning, at least for the forwards case. 14 can be expressed as mn. Without loss of generality, since multiplication is commutative, we can let m be the even number. Since m is even, it is divisible by 2. Then, we can simply choose m=2, since it is an integer. Then n=7 is the other integer. Then, we can write a=14k as a=(2•7)k, and the same reasoning follows as before.

I hope this is helpful.