Let n be an integer. Generalizing Ex 1.11, show that if \( \{a_i\}^k_{i=1} \)is a pairwise relatively prime family of integers, where each \(a_i\) divides \(n\), then their product \(\prod_{i=1}^ka_i\) also divides \(n\).

From Ex 1.11, we know that if \(a,b\) are relatively prime integers, each of which divides \(n\), then \(ab\) divides \(n\). Here, if \(a_i, a_j,\) and \(a_k\) are relatively prime, \(a_i \times a_j, a_k\) will also be relatively prime, since none of these terms share any common prime factor. This means not only does \(a_i \times a_j, a_k\) divides \(n\), so does \(a_i \times a_j \times a_k\), as \(a_i \times a_j\) and \(a_k\) are relatively prime, and each divides n individually. Recursively applying the same logic to the rest of the terms would necessarily mean \(\prod_{i=1}^ka_i\) also divides \(n\).

# posted by rot13(Unafba Pune) @ 12:46 AM