If $x$ is any common multiple of $a_1, a_2 \cdots a_n$ all $\neq 0$ then prove that $[a_1, a_2,\ldots,a_n]$ divides $x$. Note, $[a_1, a_2,\ldots,a_n]$ is LCM.
The solution provided in my text:
Let $[a_1, a_2,\ldots,a_n] = a$
We write $x = aq + r$ with $q, r$ being integers and $0 \leq r < a$
Now $a_i$ divides $x$ and $a$ for each $i$. This means that $a_i$ divides $x - aq = r$. But $0 \leq r < a$ implying that $r = 0$. Therefore $a | x$.
My question
I follow the proof everywhere except where it is said that if $a_i | r$ and $r < a$ then $r = 0$. Wouldn't this be true only if $ r < a_i \leq a$? Otherwise we don't have a guarantee that $r$ isn't $> a_i$. Or is this not possible in the first place? Am I missing something?
By definition, $a$ is the least common multiple of the $a_i$. That is, there is no smaller positive integer that is a multiple of every $a_i$. We know that $r$ is smaller than $a$ and divisible by every $a_i$, so it can't be positive.