Prove that for any integers $m$ and $n$ we have to: $mn=(m,\,n)[m,\,n]$.

Question

Prove that for any integers $m$ and $n$ we have to: $mn=(m,\,n)[m,\,n]$ ($(m,\,n)$ being the gcd & $[m,\,n]$ the lcm). How can i prove it?

Hint: use the prime factorizations for $a$ and $b$, and express $\text{LCM}(a,b)$ and $\text{GCD}(a,b)$ in terms of these prime factorizations. — Anthony, Jul 30 '20 at 14:09

score 3 · Answer 1 · answered Jul 30 '20 at 14:09

3

Write prime factorizations $m=\prod_ip_i^{a_i},\,n=\prod_ip_i^{b_i}$ over all primes that divide at least one of $m,\,n$. The problem reduces to $a_i+b_i=\min\{a_i,\,b_i\}+\max\{a_i,\,b_i\}$.

answered Jul 30 '20 at 14:09

J.G.

115,835

score 1 · Answer 2 · answered Jul 30 '20 at 14:36

1

Hint:

Let $d=\gcd(m,n)$ so that $m=m'd$, $\:n=n'd\:$ and $\:m',n'$ are coprime. Then $\operatorname{lcm}(m,n)=m'n=mn'$, so that $$mn=m'd\,n'd=\dots$$

answered Jul 30 '20 at 14:36

Bernard

175,478

Prove that for any integers $m$ and $n$ we have to: $mn=(m,\,n)[m,\,n]$.

2 Answers2