Proving: lcm(ab,ad)=a[lcm(b,d)]

[cwatki14]

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

 

[icystrike]

I am aiming to prove that:
lcm(ab,ad)=a[lcm(b,d)]
I am not really sure where to start.

Hey there! Are you taking introductory number theory?
If you are, the proof goes like this:

lcm(ab,ad)gcd(ab,ad)=a^{2}bd
We know that gcd(ab,ad)=a gcd(b,d)
\frac{lcm(ab,ad)}{a}=\frac{bd}{gcd(b,d)}
Now focus on the right hand side , what can you tell?

 

[cwatki14]

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?

 

[JSuarez]

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.

 

[cwatki14]

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.

The problem in the text states that equality... so I guess I should just assume that it means for positive integers only?

 

[JSuarez]

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.