Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). Suppose \(a \in \mathbb{Z}\). Prove that \(14 \mid a\) if and only if \(7 \mid a\) and \(2 \mid a\).

Short answer

Based on two directions proved, we can conclude that for any integer \(a\), \(14\) divides \(a\) if and only if \(7\) divides \(a\) and \(2\) divides \(a\).

Step-by-step solution

Proving the forward direction

Suppose \(14 \mid a\). Then, by the definition of divisibility, \(a = 14k\) for some integer \(k\). Expressing \(14\) as \(7 * 2\), we obtain \(a = 7 * 2 * k\), which implies \(a\) can be written as an integer multiple of both \(7\) and \(2\). Therefore, \(7 \mid a\) and \(2 \mid a\). This proves the forward direction.

Proving the backward direction

Now, suppose \(7 \mid a\) and \(2 \mid a\). This implies \(a = 7n\) and \(a = 2m\) for some integers \(n\) and \(m\). Since \(7\) and \(2\) are prime numbers and hence have no common divisors other than \(1\), \(a = 2 * 7 * r = 14r\) for some integer \(r\). Therefore, if \(7 \mid a\) and \(2 \mid a\), then \(14 \mid a\). This proves the backward direction.