Alternative proof of $gcd(a,b)midtext{lcm}(a,b)$











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I proved this problem using the theorem $gcd(a,b)text{lcm}(a,b)=ab$ Let $d=gcd(a,b)$. Then $dmid a$ and $dmid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then
$$d(text{lcm}(a,b))=abRightarrow text{lcm}(a,b)=frac{ab}{d}Rightarrow text{lcm}(a,b)=frac{d^2rs}{d}Rightarrow text{lcm}(a,b)=drsRightarrow dmid text{lcm}(a,b)$$
as required.



Is my alternative proof for this problem correct?






number-theory elementary-number-theory proof-verification







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  • math.stackexchange.com/questions/349858/…
    – 1ENİGMA1
    5 hours ago















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I proved this problem using the theorem $gcd(a,b)text{lcm}(a,b)=ab$ Let $d=gcd(a,b)$. Then $dmid a$ and $dmid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then
$$d(text{lcm}(a,b))=abRightarrow text{lcm}(a,b)=frac{ab}{d}Rightarrow text{lcm}(a,b)=frac{d^2rs}{d}Rightarrow text{lcm}(a,b)=drsRightarrow dmid text{lcm}(a,b)$$
as required.



Is my alternative proof for this problem correct?






number-theory elementary-number-theory proof-verification







share|cite|improve this question
























  • math.stackexchange.com/questions/349858/…
    – 1ENİGMA1
    5 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I proved this problem using the theorem $gcd(a,b)text{lcm}(a,b)=ab$ Let $d=gcd(a,b)$. Then $dmid a$ and $dmid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then
$$d(text{lcm}(a,b))=abRightarrow text{lcm}(a,b)=frac{ab}{d}Rightarrow text{lcm}(a,b)=frac{d^2rs}{d}Rightarrow text{lcm}(a,b)=drsRightarrow dmid text{lcm}(a,b)$$
as required.



Is my alternative proof for this problem correct?






number-theory elementary-number-theory proof-verification







share|cite|improve this question















I proved this problem using the theorem $gcd(a,b)text{lcm}(a,b)=ab$ Let $d=gcd(a,b)$. Then $dmid a$ and $dmid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then
$$d(text{lcm}(a,b))=abRightarrow text{lcm}(a,b)=frac{ab}{d}Rightarrow text{lcm}(a,b)=frac{d^2rs}{d}Rightarrow text{lcm}(a,b)=drsRightarrow dmid text{lcm}(a,b)$$
as required.



Is my alternative proof for this problem correct?






number-theory elementary-number-theory proof-verification




number-theory elementary-number-theory proof-verification






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share|cite|improve this question













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edited 5 hours ago









Tianlalu

2,180631




2,180631










asked 6 hours ago









Ling Min Hao

34418




34418












  • math.stackexchange.com/questions/349858/…
    – 1ENİGMA1
    5 hours ago


















  • math.stackexchange.com/questions/349858/…
    – 1ENİGMA1
    5 hours ago
















math.stackexchange.com/questions/349858/…
– 1ENİGMA1
5 hours ago




math.stackexchange.com/questions/349858/…
– 1ENİGMA1
5 hours ago










2 Answers
2






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1
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Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).



Hope it helps






share|cite|improve this answer





















  • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
    – Ling Min Hao
    6 hours ago










  • he literally says "alternate proof", so the second part of your answer seems pointless
    – mathworker21
    5 hours ago


















up vote
-1
down vote













$gcd(a,b)cdottext{lcm}(a,b)=acdot b=>$
$$
\frac{text{lcm}(a,b)}{gcd(a,b)}=frac{a}{gcd(a,b)}cdotfrac{b}{gcd(a, b)}inmathbb N
$$






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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).



    Hope it helps






    share|cite|improve this answer





















    • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
      – Ling Min Hao
      6 hours ago










    • he literally says "alternate proof", so the second part of your answer seems pointless
      – mathworker21
      5 hours ago















    up vote
    1
    down vote













    Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).



    Hope it helps






    share|cite|improve this answer





















    • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
      – Ling Min Hao
      6 hours ago










    • he literally says "alternate proof", so the second part of your answer seems pointless
      – mathworker21
      5 hours ago













    up vote
    1
    down vote










    up vote
    1
    down vote









    Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).



    Hope it helps






    share|cite|improve this answer












    Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).



    Hope it helps







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 6 hours ago









    Crazy for maths

    4486




    4486












    • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
      – Ling Min Hao
      6 hours ago










    • he literally says "alternate proof", so the second part of your answer seems pointless
      – mathworker21
      5 hours ago


















    • Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
      – Ling Min Hao
      6 hours ago










    • he literally says "alternate proof", so the second part of your answer seems pointless
      – mathworker21
      5 hours ago
















    Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
    – Ling Min Hao
    6 hours ago




    Yes, I do know this proof. I am just worried my proof might reach to circular reasoning so I want to verify at here. Thanks anyway.
    – Ling Min Hao
    6 hours ago












    he literally says "alternate proof", so the second part of your answer seems pointless
    – mathworker21
    5 hours ago




    he literally says "alternate proof", so the second part of your answer seems pointless
    – mathworker21
    5 hours ago










    up vote
    -1
    down vote













    $gcd(a,b)cdottext{lcm}(a,b)=acdot b=>$
    $$
    \frac{text{lcm}(a,b)}{gcd(a,b)}=frac{a}{gcd(a,b)}cdotfrac{b}{gcd(a, b)}inmathbb N
    $$






    share|cite|improve this answer

























      up vote
      -1
      down vote













      $gcd(a,b)cdottext{lcm}(a,b)=acdot b=>$
      $$
      \frac{text{lcm}(a,b)}{gcd(a,b)}=frac{a}{gcd(a,b)}cdotfrac{b}{gcd(a, b)}inmathbb N
      $$






      share|cite|improve this answer























        up vote
        -1
        down vote










        up vote
        -1
        down vote









        $gcd(a,b)cdottext{lcm}(a,b)=acdot b=>$
        $$
        \frac{text{lcm}(a,b)}{gcd(a,b)}=frac{a}{gcd(a,b)}cdotfrac{b}{gcd(a, b)}inmathbb N
        $$






        share|cite|improve this answer












        $gcd(a,b)cdottext{lcm}(a,b)=acdot b=>$
        $$
        \frac{text{lcm}(a,b)}{gcd(a,b)}=frac{a}{gcd(a,b)}cdotfrac{b}{gcd(a, b)}inmathbb N
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Samvel Safaryan

        37710




        37710






























             

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