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Lcm proof

  1. Feb 11, 2010 #1
    I am aiming to prove that:
    lcm(ab,ad)=a[lcm(b,d)]
    I am not really sure where to start. So I simply want to prove that a divides lcm(ab,ad)? Or that if a divides b and a divides d then it will divide the lcm of (ad,ad)? Nonetheless, I am still not sure as to where to start? Any tips?- Thanks
     
  2. jcsd
  3. Feb 11, 2010 #2
    Hey there! Are you taking introductory number theory?
    If you are, the proof goes like this:

    lcm(ab,ad)gcd(ab,ad)=[tex]a^{2}bd[/tex]
    We know that gcd(ab,ad)=a gcd(b,d)
    [tex]\frac{lcm(ab,ad)}{a}=\frac{bd}{gcd(b,d)}[/tex]
    Now focus on the right hand side , what can you tell?
     
  4. Feb 13, 2010 #3
    Yes, I am taking elementary number theory? So far I am really enjoying it, but I have never taken a math course like this. This is my first course that require writing proofs, so it's been a bit of a challenge.

    Isn't the RHS simply the definition of lcm(b,d)? So does that just imply that a can be factored out of lcm(ab,ad) and when is divided by a cancels?
     
  5. Feb 13, 2010 #4
    Be careful with the details; if you want to prove that for the integers then your equality is false: it should be lcm(ab,ac) = |a|lcm(b,c). Be careful with the signs.
     
  6. Feb 13, 2010 #5
    The problem in the text states that equality... so I guess I should just assume that it means for positive integers only?
     
  7. Feb 13, 2010 #6
    Your text should also state which base set it is considering: the naturals or the integers. Without the modulus, that equality is valid only for the naturals.
     
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