divisibility application along with gcd and lcm

Question

show that (a,b)[a,b]|ab .I used fundamental theorem of arithmetic and got a relationship as ab=[a,b]*(a,b). but i'm not sure is it a perfect way to get the answer

Answer 1

If you use FTA then it reduces to $\, \min(e,f)+\max(e,f) = e+f$ where $\,e,f\,$ are the exponents of a prime factor. Alternatively it can be proved using gcd laws as follows

$\quad \rm\,\ m,n\mid x\!\iff\! mn\mid mx,nx\!\iff\! mn\mid(mx,nx) = (m,n)x\!\iff\! mn/(m,n)\mid x\!\iff\!\ell\mid x$

for $\rm\:\ell = mn/(m,n).\ $ $\rm\:x = \ell\:$ in $\,(\Leftarrow)\,$ shows $\rm\:m,n\mid \ell,\:$ i.e. $\rm\:\ell\:$ is a common multiple of $\rm\:m,n,\:$ necessarily the least common multiple, since $(\Rightarrow)$ shows $\rm\:m,n\mid x\:\Rightarrow\:\ell\mid x\:\Rightarrow\:\ell\le x.$

Remark $\ $ From above we deduce $\rm\: m,n\mid x\iff lcm(m,n)\mid x.\:$ This is the definition of the lcm in more general rings. See this answer for this efficient universal approach to LCMs and GCDs.