Show that the two definitions of LCM are equivalent. We can define LCM in two ways:
Defn$1$:   Let $a,b\in \mathbb N$ not both zero and $l\in \mathbb N$ is such that $a\mid l, b\mid l$ and if $a\mid c, b\mid c$ then $l\mid c$. Then $l$ is called the LCM of $a$ and $b$.
Defn$2$: Let $S_{a,b}=\{n\in\mathbb N: a\mid n, b\mid n\}$, $S_{a,b}\neq\varnothing$. By WOP $S_{a,b}$ has a least element $l'$, which is called the LCM of $a$ and $b$.
I want to show that these definitions are equivalent. But I am stuck with the proof. Can someone help?
 A: Okay so I guess here's how it should proceed.
From Defn 2, you can surely see that $l'\mid n \forall n \in S_{a,b}$ and $\min(S_{a,b}) = l$. Also notice that $(a \mid n)\land(b \mid n)\forall n \in S_{a,b}$ with $l'$ being the least. From the first statement, it's obvious that $\exists c : l' \mid c$ and hence $a\mid c, b \mid c$. Also, $a \mid l' , b \mid l'$ which simply satisfies the first definition (I'm not sure if this is all right).
If it's from Defn 1 (as Sayan Dutta doubts in the second part of his comment), I guess here's how it should go: Let $S = \lbrace c : l \mid c, c \in \mathbb{N} \rbrace$. By WOP, $\exists k: \min(S) = k$ and it's obvious that $k = l$ since $l \mid l$. Also, since $a \mid l$ and $b \mid l$,$(a \mid n)\land(b \mid n)\forall n \in S $ with $l$  being the least. Thus we see that every element in $S$ is a common multiple of $a$ and $b$ with $l$ being the least possible number of such a kind, hence we conclude $l = lcm(a,b)$.
I am a grade 11 student and a few chapters that deal with this sort of proof-writing has been deleted from the syllabus temporarily, and hence I'm not so good at proof-writing. It would be worth waiting for a while till a proper answer comes.
