Question

Prove:
If p|a and q|a then lcm(p,q)|a.
Is the converse true? I think so.

Answer 1

Let \(\text{lcm}(p,q)=l\).
By definition, \(l\) is the smallest positive integer that is divisible by both \(p\) and \(q\).

By division algorithm, there exists integers \(t,r\) with \(0\le r\lt l\) such that
\[a=tl+r\]

Rearranging we get
\[r=a-tl\]
Since both \(a\) and \(l\) are divisible by \(p,q\), it must be the case that their linear combination, \(r\), is also divisible by \(p,q\). And \(0\le r\lt l \implies r=0.\)

( \(r\gt 0\) is a contradiction because \(l\) is the "least positive integer" that is divisible by both \(p,q\) by definition of \(\text{lcm}\). Therefore \(r=0\).)

Answer 2

Counter example :_
15,20, lcm=60
a=60 or 6 or whatever it doesn't sound true , however if u mean by converse if a does not divide l.c.m then a does not does not divide q or p this is true cause of logic statement I wanna you to make your question more clear

Answer 3

Counter example :_
15,20, lcm=60
a=60 or 6 or whatever it doesn't sound true , however if u mean by converse if a does not divide l.c.m then a does not does not divide q or p this is true cause of logic statement I wanna you to make your question more clear

Answer 4

I think the converse of given statement is :
If \(\text{lcm}(p,q)\) divide \(a\), then \(p\) and \(q\) also divide \(a\).

Answer 5

That is easy to prove :
Let \(l=\text{lcm}(p,q)\).

\[l\mid a\implies a=l'l\]
for some integer \(l'\)

By definition of \(\text{lcm}\), we have \(p\mid l\) and \(q\mid l\). These give \(l=p'p\) and \(l=q'q\). Plug them above and get
\[a=l'p'p\]
and
\[a=l'q'q\]
together imply that \(p\mid a\) and \(q\mid a\)

Answer 6

In general :
if \(m\mid n\), then all the divisors of \(m\) also divide \(n\).