Let $a_1,a_2, ... , a_n ∈ \Bbb{N} $ . Prove that there exists $ l ∈ \Bbb{N} $ such that $a_i | l$ for all $i ∈ \{1,2,...,n\}$ and if $x ∈ \Bbb{N} $ is such that each $a_i$ divides $x$, then $l | x$.
Hello, I know this proof includes some method of well ordering principle, perhaps to find $l$. But I am very confused as to how to solve it. Especially with the later half of the question. All answers are appreciated.
On
Let $m=a_1a_2...a_n.$ Then $a_i|m$ for all $i\in \{1,2,...,n\}$. Therefore the set of positive integers $z$ such that $a_i|z$ for all $i\in \{1,2,...,n\}$ is non-empty (it contains $m$). By the well-ordering principle, therefore, this set contains a least element; call it $l$. Because $l$ is in the set, $a_i|l$ for all $i\in \{1,2,...,n\}.$ This proves the existence of $l$.
We are also asked to prove that if $x \in \mathbb N$ and each $a_i$ divides $x$, then $l|x$. Assume to the contrary that $l$ does not divide $x,$ and take the remainder when $x$ is divided by $l$; call it $r$. So $0<r<l$, but each $a_i$ divides $x$ and $l,$ hence each $a_i$ divides $r$, contradicting that $l$ is the least such number.
On
The set $$A=\{\,x\in\Bbb N: a_1\mid x,\, a_2\mid x,\,\ldots ,\, a_n\mid x\,\} $$ is not empty (contains $a_1a_2\cdots a_n$) hence has a minimal element $l$. Let $$ B=\{\,x\in\Bbb N: a_1\mid x,\, a_2\mid x,\,\ldots ,\, a_n\mid x,\,l\nmid x\,\}$$ Assume $B\ne \emptyset$. Then $B$ has a minimal element $m$. From $B\subset A$ and $l\notin B$, we conclude $m>l$. Then $m':=m-l\in \Bbb N$. From $a_i\mid m$ and $a_i\mid l$, we get $a_i\mid m'$, $1\le i\le n$. If $l\mid m'$, then we'd also have $l\mid m'+l=m$. Therefore $l\nmid m'$, so $m'\in B$ contradicting minimality of $m$. We conclude that $B=\emptyset$.
Define $\mathcal{M}=\{m\in\mathbb{N}\::\: a_i|m,\,\forall i\}$ and note that $a_1\cdots a_n$ belongs to $\mathcal{M}$. Now you can use the Well Ordering Principle to conclude the existence of $l$.