Proof writing: Let $d=\gcd(a,b)$ and $m=\operatorname{lcm}(a,b)$. Prove that $dm=|ab|$.

Question

Let $d=\gcd(a,b)$ and $m=\operatorname{lcm}(a,b)$. Prove that $dm=|ab|$.

Can I just show that since $\operatorname{lcm}(a,b)=|ab|$ iff $\gcd(a,b)=1$, then $dm=(1)|ab|=|ab|$? Or is there a more simple proof to this?

Answer 1

\begin{eqnarray*} a&=& p_1^{ \alpha_1} p_2^{\alpha_2} \cdots p_i^{\alpha_i} \\ b&=& p_1^{ \beta_1} p_2^{\beta_2} \cdots p_i^{\beta_i} \\ hcf(a,b) &=& p_1^{\min(\alpha_1, \beta_1)} p_2^{\min(\alpha_2 ,\beta_2)} \cdots p_i^{\min(\alpha_i ,\beta_i)} \\ lcm(a,b) &=& p_1^{\max(\alpha_1, \beta_1)} p_2^{\max(\alpha_2, \beta_2)} \cdots p_i^{\max(\alpha_i ,\beta_i)} \\ \end{eqnarray*} Now use $\min(\alpha ,\beta)+\max(\alpha ,\beta)=\alpha +\beta$.