1
$\begingroup$

Show that for positive integers if $d_1e_1=d_2e_2\ and \gcd(e_1,e_2)=1\Rightarrow\ lcm(d_1,d_2)=d_1e_1=d_2e_2$
We know that: $$lcm(d_1,d_2)gcd(d_1,d_2)=d_1d_2$$ but this doesn't help much!
What's the trick?

$\endgroup$
1
$\begingroup$

Let $k=\gcd(d_1,d_2)$. Then $d_1=kd_1'$ and $d_2=kd_2'$ where $d_1'$ and $d_2'$ are relatively prime. Thus from the relationship you stated we have $$\text{lcm}(d_1,d_2)k=k^2d_1'd_2',$$ and therefore $$\text{lcm}(d_1,d_2)=kd_1'd_2'.$$ It remains to show that $kd_1'd_2'=d_1e_1=kd_1'e_1$, or equivalently $d_2'=e_1$.

We were told that $d_1e_1=d_2e_2$, or equivalently that $d_1'e_1=d_2'e_2$. Since $e_1$ and $e_2$ are relatively prime, we have $e_1\mid d_2'$. And because $d_1'$ and $d_2'$ are relatively prime, we have $d_1'\mid e_2$.

But because $d_1'e_1=d_2'e_2$, it follows that $e_1=d_2'$ and $d_1'=e_2$.

$\endgroup$
3
  • $\begingroup$ Can you please explain more about the proof of: $e_1=d_2'$ and $d_1'=e_2$? I was a little confused!! $\endgroup$ – Hamid Reza Ebrahimi Apr 8 '16 at 18:03
  • 1
    $\begingroup$ In greater detail, instead of the last paragraph, since $e_1\mid d_2'$ we have $d_2'=se_1$ for some $s$. Similarly, $e_2=t d_1'$. So $d_1'e_1=se_1td_1'$, and therefore $st=1$, so $s=t=1$, and we are finished. $\endgroup$ – André Nicolas Apr 8 '16 at 18:10
  • 1
    $\begingroup$ You are welcome. There are notationally slicker proofs, I was trying for a proof that would be as "unclever" as possible. No tricks, just chasing things down. $\endgroup$ – André Nicolas Apr 8 '16 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.