This is from an example in Chapter 4 of Book of Proof by Richard Hammack. The thing is that I do not quite understand why it is been solved in this way, in my understanding it seems redundant or I am misunderstanding somethings. My approach here is to quote the text directly from Hammack followed from comments of how I think it should be approached (I could be wrong). I would greatly appreaciate if you could point me where I got it wrong The text reads as:  If $a,b,c \in \mathbb{N},$ then lcm (ca,cb) = c · lcm(a,b)   Proof. Assume a,b,c ∈ N. Let m = lcm(ca,cb) and n = c·lcm(a,b). We will show m = n. By definition, lcm(a, b) is a positive multiple of both a and b, so lcm(a,b)=ax=by for some x,y ∈ N.  Why he writes that lcm(a,b) is only a positive multiple of a &amp; b instead of the smallest one?  From this we see that n=c·lcm(a,b) = cax = cby is a positive multiple of both ca and cb.  Again, Why he assumes &quot;n&quot; is only positive? The definition of lcm states clearly that it must be the smallest number, not any number.  But m = lcm(ca, cb) is the smallest positive multiple of both ca and cb. Thus m ≤ n.  Why not state here that m=n, since we know from the defintion of lcm(a,b) that we are referring to the same value? I mean m = lcm (ca,cb) is the same as n= c * lcm (a,b) since we know lcm(a,b)=ax=by is the smallest integer multiplied by c  On the other hand, as m = lcm(ca, cb) is a multiple of both ca and cb, we have m = cax = cby for some x, y ∈ Z.  Why not consider &quot;m&quot; as the smallest multiple again?  Then $1(/c)m$ = ax = by is a multiple of both a and b. Therefore lcm(a, b) ≤ $1(/c)m$, so c · lcm(a, b) ≤ m, that is, n ≤ m.  Why not state that they are the same, again?  We’ve shown m≤n and n≤m, so m=n. The proof is complete.  Was it necessary to do it in both hands side? In my perspective it could have done considering only half of Hammack's answer
I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max + mby$ → $d$|$m(ax+by)$. This implies that $d$|$m$ or $d$|$(ax+by)$. This is the same as  $d$|$m$ or $d$|$a$ and $d$|$b$, so $d$|$m$ or $d$|$(a,b)$. This is the same as $d$|$m|$ or $d|(a,b)$, so $d$|$|m|(a,b)$. I don't know what to do.
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