Let $a_1….a_n$ be pairwise coprime. That is $gcd(a_i, a_k) = 1$ for distinct $i,k$, I would like to show that if each $a_i$ divides $b$ then so does the product.
I can understand intuitively why it's true – just not sure how to formulate the proof exactly.
I want to say if we consider the prime factorizations of each $a_i$, then no two prime factorizations share any prime numbers. So the product of $a_1…a_n$ must appear in the prime factorization of $b$. Is this correct? Or at least if if the idea is correct, any way to formulate it more clearly?
Best Answer
One way of arguing that avoids looking at prime factorizations is to note that $gcd(a,b)=1$ iff $a\alpha+b\beta=1$ for some integers $\alpha,\beta$.
But then, if $a,b$ both divide $c$, say $at=c$ and $br=c$ where $t,r$ are integers, we have $$ ab(\alpha r+\beta t)=br(\alpha a)+at(\beta b)= c(\alpha a+\beta b)=c. $$
This explicitly shows that $ab$ also divides $c$. Now use induction.