Define $lcm(a, b) :={(ab)}/ {(gcd(a, b))}$ .Prove that $lcm(a, b)$ is the smallest positive integer that is evenly divisible by both $a$ and $b$.

Question

Let $a,b \in Z^+$. Define $lcm(a, b) :={(ab)}/ {(gcd(a, b))}$. Prove that $lcm(a, b)$ is the smallest positive integer that is evenly divisible by both $a$ and $b$.

In order to do this I am trying to prove three different parts:

1. $lcm(a, b) \in \mathbb{Z}^+$,
2. $a\,|\,lcm(a, b)$ and $b\,|\,lcm(a, b)$, and
3. If $m \in \mathbb{Z}^+$, $a\,|\,m$, and $b\,|\,m$, then $lcm(a,b) ≤ m$.

Answer 1

Let $l:=\mathrm{lcm}(a,b)$ and $g:=\gcd(a,b).$

For $(1)$ note that $a,b,g>0.$ Thus $ab/g=l>0.$

For $(2)$ write $a=ga'$ and $b=gb'.$ Then $l=ab'=ba'.$

Now, for $(3)$ let $m$ be a positive integer such that $a,b\mid m.$ Then $m=ak_0=bk_1$ for some $k_0,k_1\in\mathbb Z$ and so $a'k_0=b'k_1$ and since $\gcd(a',b')=1$ then $b'\mid k_0.$ Hence $m=ga'b'q$ for some $q\in\mathbb Z$ and since $ga'b'=l,$ the result follows.