How should i prove that $ab|(a,b)[a,b]$ ? Here $(a,b)=gcd(a,b)$ and $[a,b]=lcm(a,b)$. I tried and got an answer as $\frac{ab}{(a,b)}|[a,b]$. then i can also proceed as $[a,b]=(a,b)^2\frac{k}{ab}$. ;here k is an positive integer.Can i reach my answer from here?
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For convenience let it be that $a,b>0$
If $a$ and $b$ can be factored as $a=p_1^{n_1}\cdots p_k^{n_k}$ and $b=p_1^{m_1}\cdots p_k^{m_k}$ where the $p_i$ are primes and the $n_i$ and $m_i$ are nonnegative integers, then $(a,b)=p_1^{\min(n_1,m_1)}\cdots p_k^{\min(n_k,m_k)}$ and $[a,b]=p_1^{\max(n_1,m_1)}\cdots p_k^{\max(n_k,m_k)}$.
Now take there product in the understanding that $\min(n_i,m_i)+\max(n_i,m_i)=n_i+m_i$
This shows you that $(a,b)[a,b]=ab$.
drhab
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