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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?

Tianlalu
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Nothing
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  • https://math.stackexchange.com/questions/349858/easiest-and-most-complex-proof-of-gcd-a-b-times-operatornamelcm-a-b-a – 1ENİGMA1 Nov 14 '18 at 07:52

2 Answers2

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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

Martund
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  • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway. – Nothing Nov 14 '18 at 07:11
  • he literally says "alternate proof", so the second part of your answer seems pointless – mathworker21 Nov 14 '18 at 07:20
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$\gcd(a,b)\cdot\text{lcm}(a,b)=a\cdot b=>$ $$ \\\frac{\text{lcm}(a,b)}{\gcd(a,b)}=\frac{a}{\gcd(a,b)}\cdot\frac{b}{\gcd(a, b)}\in\mathbb N $$

Samvel Safaryan
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