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