a*b=gcd(a,b)lcm(a,b)个人证明

设: a=\({p_1}^e1{p_2}^e2...{p_r}^er\) b=\({p_1}^f1{p_2}^f2...{p_r}^fr\) 所以 ab=\({p_1}^e1{p_2}^e2...{p_r}^er{p_1}^f1{p_2}^f2...{p_r}^fr\) 因为gcd(a,b)=\({p_1}^{m\in(e1,f1)}{p_2}^{m\in(e2,f2)}...{p_r}^{m\in(er,fr)}\) lcm(a,b)=\({p_1}^{max(e1,f1)}{p_2}^{max(e2,f2)}...{p_r}^{max(er,fr)}\) 又因为\({p_k}^ek{p_k}^fk={p_k}^{max(ek,fk)}{p_k}^{m\in(ek,fk)}\) 所以a*b=gcd(a,b)lcm(a,b)