Alternative proof of $\gcd(a,b)\mid\text{lcm}(a,b)$

Question

I proved this problem using the theorem $\gcd(a,b)\text{lcm}(a,b)=ab$ Let $d=\gcd(a,b)$. Then $d\mid a$ and $d\mid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then $$d(\text{lcm}(a,b))=ab\Rightarrow \text{lcm}(a,b)=\frac{ab}{d}\Rightarrow \text{lcm}(a,b)=\frac{d^2rs}{d}\Rightarrow \text{lcm}(a,b)=drs\Rightarrow d\mid \text{lcm}(a,b)$$ as required.

Is my alternative proof for this problem correct?

Answer 1

Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).

Hope it helps

Answer 2

$\gcd(a,b)\cdot\text{lcm}(a,b)=a\cdot b=>$ $$ \\\frac{\text{lcm}(a,b)}{\gcd(a,b)}=\frac{a}{\gcd(a,b)}\cdot\frac{b}{\gcd(a, b)}\in\mathbb N $$