If $m=\operatorname{lcm}(a,b)$ then $\gcd(\frac{m}{a},\frac{m}{b})=1$

Question

Let $a,b\in \mathbb{N}$ and $m=\operatorname{lcm}(a,b)$. Prove that $\gcd(\frac{m}{a},\frac{m}{b})=1$.

Any suggestion how to prove I would appreciate. I have try to it directly but I don't get anything. Probably I'm doing some dumb mistake.

Answer 1

Let $m = \operatorname{lcm}(a,b)$, then $\gcd(a,b)= \frac{ab}m$.

We know that there exist integers $x,y$ such that $\frac{ab}{m} = ax + by$. Now, multiplying by $m$, we get $ab = max+mby$. Dividing by $ab$, we get $1 = x\left(\frac{m}{b}\right) + y\left(\frac{m}{a}\right)$. Hence, there exist intgers $x,y$ such that the above happens. By Bezout's theorem, it follows that $\frac mb, \frac ma$ are co-prime.

Answer 2

Let's set $d=\gcd(\frac{m}{a}, \frac{m}{b})$, then $d\mid \frac{m}{a}$ and $d\mid \frac{m}{b}$, which lead us to $a\mid \frac{m}{d}$ and $b\mid \frac{m}{d}$. Then by the definition of $\text{lcm}$ we deduce that $m\mid \frac{m}{d}$, so $d\mid \frac{m}{m}=1$ and hence $d=1$.

Answer 3

The result is the special case $\, m = {\rm lcm}(a,b),\,$ in the following Theorem, whose proof is a completely mechanical application of $\,\rm\color{#c00}{cofactor\ duality}\,$ and $\,\rm\color{#0a0}{universal}\,$ properties of gcd, lcm

Theorem $\ \displaystyle \gcd\left(\frac{m}a,\frac{m}b\right) = \frac{m}{{\rm lcm}(a,b)}\,\ $ for $\,m\,$ any common multiple of $a,b,\,$ since

$\displaystyle d\mid \gcd\!\left (\frac{m}a,\,\frac{m}b\right) \!\!\color{#0a0}\iff\! d\,\ \Bigl\vert\ \frac{m}a,\,\frac{m}b \!\color{#c00}\iff\! a,b\ \Bigl\vert\ \frac{m}d \!\color{#0a0}\iff\! {\rm lcm}(a,b)\ \Bigl\vert\ \frac{m}d \!\color{#c00}\iff\!d\ \Bigl\vert\ {\frac{m}{{\rm lcm}(a,b)}}$

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i.e. $\ d\mid\gcd(a',b')\!\iff d\,\mid a',\ b'\color{#c00}\iff\, a,b\mid d' \iff {\rm lcm}(a,b)\mid\, d'\!\color{#c00}\iff d\mid {\rm lcm}(a,b)' $

i.e. $\,\ \bbox[6px,border:1px solid red]{\gcd(a',b') = {\rm lcm}(a,b)'}\ $ where $\ n':= m/n\,$ denotes the cofactor of any divisor $n$ of $m$

The following cofactor duality is the source of the prime-flipped divisibilities in the $\rm\color{#c00}{red}$ arrows

$$ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\ \ \ {\rm by}\ \ \ \dfrac{y}x = \dfrac{x'}{y'} \ \ \ {\rm by}\ \ \ yy' = m = xx' $$

This duality between gcd and lcm is a divisibility analog of DeMorgan's Laws (this will be made more precise if one studies lattice theory, e.g. see Birkhoff's books).

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Corollary $\ \displaystyle \ \gcd(b,\,a)\, =\, \frac{ab}{{\rm lcm}(a,b)}\,\ $ by $\ m = ab\ $ in the Theorem.

If the prior is already known (e.g. here) then it can be applied to derive the sought result:

$\quad ab\, =\!\!\!\!\! \overbrace{m}^{\large {\rm lcm}(a,b)}\!\!\!\!\!\!\gcd(a,b)\, =\, \gcd(ma,mb)\, =\, ab\gcd(\frac{m}b,\frac{m}a)\ $ by the GCD Distributive Law

Thus canceling $\,ab\,$ yields $\, 1 = \gcd\left(\frac{m}b,\frac{m}a\right).$

Answer 4

Let $\operatorname{hcf}(a,b)=d$. Then there exists $p,q$ such that $a=pd$, $b=qd$, such that $\operatorname{hcf}(p,q)=1$. Therefore, $\operatorname{lcm}(a,b)=pqd=m$. Also, $\frac{m}{a} =q$ and $\frac{m}{b} =p$, and the result follows.