If l = lcm(a,b) and l=pa and l=qb, prove that gcd (p,q) = 1

Question

I am trying to prove the following: Let l = lcm$(a,b)$ and write l=pa and l=qb. Prove that gcd (p,q) = 1.

One of my students wrote his thoughts below.

I know it is probably not that difficult, I just think I am not seeing something.

Thanks for your help.

Source: (from Cambridge's Math HL Discrete Option, page 23)

Answer 1

Let $d=\gcd(p,q)\ge 1$ then $p=dp'$ and $q=dq'$

Now $l=dp'a=dq'b$

so $d$ divides $l\qquad$ i.e. $\quad l'=\frac ld$ is an integer.

But $l'=p'a=q'b$ so it is a common multiple of $a$ and $b$

By definition of the lcm, then $l'\ge l\iff\frac ld\ge l\iff d\le 1$, so $d=1$.

Answer 2

Using a Venn diagram is clever. I don't know if s/he came up with it on her/his own or if this was how s/he was taught.

But given two numbers $a$ and $b$ we can divide those into three factors.

$\frac {a}{\gcd(a,b)} = m$ which is the product of all the prime factors of $a$ that are not shared with $b$.

$\frac {b}{\gcd(a,b)} = n$ which is the product of all the prime factors of $b$ that are not shared with $a$.

And $\gcd(a,b) = o$ which is the product of all the prime factors shared by both.

And fundamental observations:

$\gcd (m,n) = 1$ (as they have no factors in common.) And $a = m*o$ and $b = n*o$.

And the least common multiple of $a$ and $b$ is $m*n*o = a*n = b*m$.

........

And ... everything falls into place. $lcm = l = pa = an$

$lcm = l = qb = b*m$.

So $p=n$ and $q= m$. Ad we know $\gcd(m,n) =1$.

Answer 3

$\text{If $\ell = \operatorname{lcm}(a,b)$, $\ell=pa$ and $\ell=qb$, then prove that $\gcd(p,q) = 1$}$

$\gcd(a,b) = \dfrac{ab}{\operatorname{lcm}(a,b)} = \dfrac{ab}{\ell}$

\begin{align} g &= \gcd(p,q) \\ ga &= \gcd(\ell, qa) \\ ga &= \gcd(qb,qa) \\ ga &= q \ \gcd(a,b) \\ ga &= \dfrac{qab}{\ell} \\ ga &= a \\ g &= 1 \end{align}